LOCALLY CONVEX SPACES MANV 250067-1, 5 ECTS, SUMMER …nigsch/teaching/lcs.pdf · G. Kothe. Topological vector spaces II. Grundlehren der Mathematischen Wis-senschaften 237. New York:

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LOCALLY CONVEX SPACESMANV 250067-1, 5 ECTS, SUMMER TERM 2017

SR 10, FR. 13:15–15:30

EDUARD A. NIGSCH

These lecture notes were developed for the topics course locally convex spaces held at theUniversity of Vienna in summer term 2017. Prerequisites consist of general topology andlinear algebra. Some background in functional analysis will be helpful but not strictlymandatory.

This course aims at an early and thorough development of duality theory for locallyconvex spaces, which allows for the systematic treatment of the most important classesof locally convex spaces. Further topics will be treated according to available time as wellas the interests of the students.

Thanks for corrections of some typos go out to Benedict Schinnerl.

1

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LOCALLY CONVEX SPACES 2

Contents

1. Introduction 3

2. Topological vector spaces 4

3. Locally convex spaces 7

4. Completeness 11

5. Bounded sets, normability, metrizability 16

6. Products, subspaces, direct sums and quotients 18

7. Projective and inductive limits 24

8. Finite-dimensional and locally compact TVS 28

9. The theorem of Hahn-Banach 29

10. Dual Pairings 34

11. Polarity 36

12. S-topologies 38

13. The Mackey Topology 41

14. Barrelled spaces 45

15. Bornological Spaces 47

16. Reflexivity 48

17. Montel spaces 50

18. The transpose of a linear map 52

19. Topological tensor products 53

References 66

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1. Introduction

These lecture notes are roughly based on the following texts that contain the standardmaterial on locally convex spaces as well as more advanced topics.

• J. Horvath. Topological vector spaces and distributions. Vol. 1. Reading, Mass.:Addison-Wesley, 1966. isbn: 978-0-486-48850-9• H. H. Schaefer. Topological Vector Spaces. New York: Springer-Verlag, 1971.isbn: 978-0-387-05380-6• H. Jarchow. Locally Convex Spaces. Stuttgart: B. G. Teubner, 1981. isbn:

978-3-519-02224-4• A. Grothendieck. Topological vector spaces. 3rd ed. Philadelphia, PA: Gordon

and Breach Science Publishers, 1992. isbn: 0-677-30020-4/hbk• F. Treves. Topological Vector Spaces, Distributions and Kernels. New York: Aca-

demic Press, 1976• L. Narici and E. Beckenstein. Topological Vector Spaces. 2nd ed. Pure and applied

mathematics. Boca Raton: Taylor and Francis Group, 2011. isbn: 978-1-58488-866-6• A. P. Robertson and W. Robertson. Topological vector spaces. 2nd ed. Cam-

bridge Tracts in Mathematics and Mathematical Physics 53. Cambridge Univer-sity Press, 1973. isbn: 978-0-521-20124-7• G. Kothe. Topological vector spaces I. Grundlehren der mathematischen Wis-

senschaften 159. New York: Springer-Verlag, 1969. isbn: 978-0-387-04509-2• G. Kothe. Topological vector spaces II. Grundlehren der Mathematischen Wis-

senschaften 237. New York: Springer-Verlag, 1979. isbn: 978-0-387-90400-9

“The principal motivation behind the general theory is the same as thatof Banach himself: namely, a search for general tools which might be ap-plied successfully to functional analysis. [...] These efforts culminated inL. Schwartz’ theory of distributions (1945), which could be expressed onlyin the language of locally convex vector spaces.”

“Analysts are more interested in the properties of a space than in theway it is defined; hence the idea, first clearly formulated by L. Schwartz,of classifying topological vector spaces according to their behaviour withregard to the validity of the main theorems of functional analysis.” [D]

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2. Topological vector spaces

We set K to be either R or C and

D := λ ∈ K | |λ| ≤ 1, N = 1, 2, 3, . . . , N0 = 0, 1, 2, . . . .

Definition 2.1. Let E be a vector space over K. A topology T on E is called a lineartopology if the mappings

(x, y) 7→ x+ y, E × E → E

(λ, x) 7→ λx, K× E → E

are continuous. The pair (E,T ) is called a topological vector space (TVS).

We write E instead of (E,T ) if it is clear from the context which topology is used.

Examples. Normed spaces are topological vector spaces:

1. c = (an)n ∈ KN : limn→∞ an exists with norm ‖(an)n‖ = supn |an|,2. c0 = (an)n ∈ c : an → 0 with the induced norm,3. ϕ = (an)n ∈ c : an 6= 0 only for finitely many n with the induced norm,4. C(K) = f : K → K continuous where K is a compact topological space, with

norm ‖f‖ = supx∈K |f(x)|,5. C0(X) = f : X → K continuous | ∀ε > 0 ∃K ⊆ X compact : supx∈X\K |f(x)| ≤ε where X is a topological space, with norm ‖f‖ = supx∈X |f(x)|,

6. lp = (an)n ∈ KN :∑

n |an|p < ∞ with norm ‖(an)n‖p = (

∑n |an|

p)1/p (for1 ≤ p <∞),

7. l∞ = (an)n ∈ KN : supn |an| <∞ with norm ‖(an)n‖∞ = supn |an|,8. Lp(X,F , µ) = f : X → K measurable |

∫X|f(x)|p dµ(x) <∞/∼ where (X,F , µ)

is a σ-finite measure space and f ∼ g if f = g a.e., with norm ‖[f ]‖p =(∫X|f(p)|p dµ(x))1/p (for 1 ≤ p <∞),

9. L∞(X,F , µ) = f : X → K measurable | ∃M > 0 such that |f(x)| < M a.e./∼with norm ‖[f ]‖∞ = infM > 0 : |f(x)| < M a.e..

But there are also non-normable topological vector spaces:

10. C(X) = f : X → K continuous with the coarsest topology such that all restric-tion mappings C(X)→ C(K), f 7→ f |K for K ⊆ X compact are continuous,

11. lp and Lp for 0 < p < 1,12. RR with the product topology.

Because for z ∈ E and λ 6= 0 the map x 7→ z + λx is a homeomorphism with inversex 7→ λ−1(x− z) we have:

Proposition 2.2. If U is (a basis of) the filter of neighborhoods of 0 then for all z ∈ Eand λ 6= 0, z + λU | U ∈ U is (a basis of) the filter of neighborhoods of z.

We will refer to the filter of neighborhoods of 0 as the 0-filter and any basis of this filterwill be called a 0-basis. Similarly, a neighborhood of 0 will be called a 0-neighborhood.

Corollary 2.3. Let E and F be TVS. A linear map f : E → F is continuous if and onlyif it is continuous at 0.

Corollary 2.4. If U is a 0-basis of a TVS E then for every nonempty subset A ⊆ E wehave

A =⋂A+ U | U ∈ U .

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Proof. Because x − U is a basis of the neighborhood filter of x ∈ E we have x ∈ A ⇔∀U ∈ U : (x− U) ∩ A 6= ∅ ⇔ ∀U ∈ U : x ∈ A+ U .

Corollary 2.5. If U is a 0-basis of a TVS E then so is U | U ∈ U .

Proof. Given U ∈ U , by continuity of addition there is V ∈ U such that V ⊆ V + V ⊆U .

Some things become simpler when we can work with a 0-basis consisting of special sets.

Definition 2.6. Let E be a vector space. A subset A ⊆ E is called

• balanced (or circled) if λx ∈ A for all λ ∈ D and x ∈ A;• absorbent (or radial) if ∀x ∈ E ∃λ0 > 0: x ∈ λA for all λ ∈ K with |λ| ≥ λ0.

Because arbitrary intersections of balanced sets are balanced, for every A ⊆ E thereexists a smallest balanced set A containing A, called its balanced hull ; clearly A = D ·A.

If A is a subset of a TVS E we call the closure¯A of A the closed balanced hull of A. It

is the smallest closed and balanced set containing A because if A ⊆ M and M is closed

and balanced we have A ⊆ M = M and hence¯A ⊆ M = M by the following result.

Proposition 2.7. Let E be a TVS and A ⊆ E balanced. Then A is balanced and if0 ∈ int(A) then int(A) is balanced.

We denote by int(A) denotes the interior of a set A.

Proof. D · A ⊆ A implies D · A ⊆ A by continuity of multiplication. For ρ ∈ D \ 0 wehave ρ int(A) = int(ρA) ⊆ int(A), which gives the second claim.

Corollary 2.8. If U is a 0-basis in E then int(U) | U ∈ U , U | U ∈ U and

¯U : U ∈ U are 0-bases in E.

Proof. Let U ∈ U and choose (Corollary 2.5) U0 ∈ U such that U0 ⊆ U . By continuityof multiplication there are λ0 > 0 and V ∈ U such that λV ⊆ U0 for |λ| ≤ λ0. Set

V0 := λ0V ; then V0 ⊆ U0 and for W ∈ U such that W ⊆ λ0V we have

int(W ) ⊆ int(W ) ⊆ W ⊆ ¯W ⊆ U0 ⊆ U.

Proposition 2.9. In a TVS every 0-neighborhood is absorbent.

Proof. Let U be a 0-neighborhood in a TVS E and fix x ∈ E. Because λ 7→ λx iscontinuous at 0 there is λ0 > 0 such that λx ∈ U for |λ| ≤ λ0.

Next, we identify the minimal requirements on a filter basis such that it generates a lineartopology.

Theorem 2.10. In a TVS E there exists a 0-basis U such that

(i) every U ∈ U is balanced and absorbent,(ii) for every U ∈ U there exists V ∈ U such that V + V ⊆ U .

Conversely, if U is a filter basis on a vector space E satisfying these properties then thereexists a unique linear topology T on E which has U as 0-basis.

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Proof. The first claim is already shown. For the second part, uniqueness follows becausea linear topology is uniquely determined by its 0-neighborhood filter. Hence, we have toshow that the family T given by all subsets A ⊆ E such that ∀x ∈ A ∃U ∈ U : x+U ⊆ Ais a linear topology having U as 0-basis.

T clearly is a topology.

For any (x, y) ∈ E × E and U ∈ U choose V ∈ U such that V + V ⊆ U . Then(x+ V ) + (y + V ) ⊆ (x+ y) + U , hence addition is continuous at (x, y).

For (ρ, x) ∈ K×E and U ∈ U we want that for λ−ρ and y−x in suitable 0-neighborhoodsin K and E, respectively, we have

λy − ρx = (λ− ρ)(y − x) + (λ− ρ)x+ ρ(y − x) ∈ U.

Set V0 := U and choose a sequence (Vk)k∈N in U such that Vk + Vk ⊆ Vk−1 for all k ∈ N.Let n ∈ N be such that |ρ| ≤ 2n − 2 and 0 < σ ≤ 1 such that [0, σ] · x ⊆ Vn. Then forλ− ρ ∈ σD and y − x ∈ Vn we have

(λ− ρ)(y − x) ∈ Vn,(λ− ρ)x ∈ Vn,

ρ(y − x) ∈ (2n − 2)Vn ⊆ Vn + . . .+ Vn (2n − 2 summands)

and hence

λy − ρx ∈ Vn + . . .+ Vn︸︷︷︸2n−2

+Vn + Vn ⊆ . . . ⊆ V0 = U,

so multiplication is continuous at (ρ, x).

Every 0-neighborhood in T contains an element of U by definition. In order to showthat each U ∈ U is a 0-neighborhood in T , set

M := x ∈ E | x+ V ⊆ U for some V ∈ U .

Given x ∈ M choose V,W ∈ U such that x + V ⊆ U and W + W ⊆ V . Thenx + W + W ⊆ U and thus x + W ⊆ M , hence M ∈ T . Because 0 ∈ M ⊆ U , U is a0-neighborhood in T .

Proposition 2.11. A TVS E is Hausdorff if and only if for every x 6= 0 there exists aneighborhood U of 0 which does not contain x, i.e., if

0 =⋂U | U is a 0-neighborhood.

Proof. Suppose E is Hausdorff and x 6= 0. Then there are 0-neighborhoods U, V suchthat U ∩ (x+ V ) = ∅, hence x 6∈ U .

Conversely, let x 6= y. Then there is a 0-neighborhood U such that x− y 6∈ U . Choose a0-neighborhood V such that V − V ⊆ U and suppose that (x + V ) ∩ (y + V ) 6= ∅; thenx + x′ = y + y′ for some x′, y′ ∈ V and x − y = y′ − x′ ∈ V − V ⊆ U , which gives acontradiction.

By Corollary 2.4, we see:

Proposition 2.12. A TVS E is Hausdorff if and only if the set 0 is closed.

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3. Locally convex spaces

Let E be a vector space. A subset A ⊆ E is called convex if ∀α, β ≥ 0 with α + β = 1,αA + βA ⊆ A. If A is convex also a + λA (x ∈ E, λ ∈ K) is convex. If f : E → F isa linear map (where F is a vector space) then f(A) and f−1(B) are convex if A and Bare so. The intersection of an arbitrary family of convex sets is convex. For any B ⊆ Ethere is a smallest convex set A containing B, called the convex hull of B.

Proposition 3.1. Let A be a convex subset of a vector space, x1, . . . , xn ∈ A andλ1, . . . , λn ≥ 0 such that

∑i λi = 1. Then

∑ni=1 λixi ∈ A.

Proof. Clear for n = 1. Assuming that the claim holds for n−1 summands, suppose thatλi > 0 for all i. With α =

∑n−1i=1 λi we have

n∑i=1

λixi = α(n−1∑i=1

λiαxi) + λnxn ∈ A

because∑n−1

i=1λiα

= 1 and∑n−1

i=1λiαxi ∈ A by assumption.

Proposition 3.2. Let (Aι)ι∈I be a family of convex subsets of a vector space. Then theconvex hull of

⋃ι∈I Aι is given by

C :=

∑λιxι

∣∣∣ xι ∈ Aι, λι ≥ 0,∑ι

λι = 1, only finitely many λi 6= 0

.

Proof. Any convex set containing the Aι contains C, so C is contained in the convex hull.Conversely, Aι ⊆ C by taking λι = 1, so we have to show that C is convex.

Let x =∑λιxι, y =

∑µιyι ∈ C, α, β > 0, α + β = 1. Set νι = αλι + βµι and

J = ι|νι > 0. Then

zι :=αλιxι + βµιyιαλι + βµι

∈ Aι ∀ι ∈ J,

and αx+ βy =∑

ι∈J νιzι ∈ C because∑νι = α

∑λι + β

∑µι = 1.

Corollary 3.3. The convex hull of A is the set of all finite linear combinations∑λixi,

xi ∈ A, λi ≥ 0,∑λi = 1.

Proof. Apply Proposition 3.2 to A =⋃x∈Ax.

Proposition 3.4. Let E be a TVS. If A ⊆ E is convex then A is convex.

Proof. Let x, y ∈ A, α, β > 0, α + β = 1 and W a neighborhood of αx + βy. Because(u, v) 7→ (αu, βv) is continuous there are neighborhoods U of x and V of y such thatαU + βV ⊆ W . Let z ∈ U ∩ A 6= ∅, w ∈ V ∩ A 6= ∅, then αz + βw ∈ W ∩ A, soαx+ βy ∈ A.

Definition 3.5. A TVS is called locally convex if it has a 0-basis consisting of convexsets.

We will simply call a locally convex topological vector space a locally convex space, orLCS in short.

Proposition 3.6. A locally convex space (LCS) has a 0-basis consisting of balancedconvex closed sets.

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Proof. Let W be a 0-neighborhood; by Corollary 2.5 it contains a closed 0-neighborhoodV , and V contains a convex 0-neighborhood U ; we have U ⊆ V . Set B :=

⋂|λ|≥1 λU ,

which is closed.

B is a 0-neighborhood: by continuity of multiplication there are α > 0 and a 0-neighborhoodX such that µx ∈ U for |µ| < α, x ∈ X. αX is a 0-neighborhood and αX ⊆ B becausefor |λ| ≥ 1 and x ∈ X, α

λx ∈ U and hence αx ∈ λU .

B is balanced: let |µ| ≤ 1, x ∈ B. Then for |λ| ≥ 1, x ∈ λµU because

∣∣∣λµ ∣∣∣ ≥ 1, so

µx ∈ λU .

Conversely, we have

Proposition 3.7. Let E be a vector space and B a filter basis on E consisting of ab-sorbent, balanced, convex sets. Let N := λV | λ > 0, V ∈ B. Then there exists aunique topology on E for which E is a locally convex space and which has N as a 0-basis.

Proof. We apply Theorem 2.10; note that for V ∈ N , 12V ∈ N and 1

2V + 1

2V ⊆ V .

Corollary 3.8. Let E be a vector space and S a family of absorbent, balanced, convexsubsets of E. Let B be the collection of finite intersection of sets of the form λV , λ > 0,V ∈ S. Then there exists a unique topology on E for which E is a locally convex spaceand B is a 0-basis.

Definition 3.9. A seminorm on a vector space E is a mapping q : E → R+ = [0,∞)such that

q(λx) = |λ| q(x) (λ ∈ K, x ∈ E)

q(x+ y) ≤ q(x) + q(y) (x, y ∈ E).

Note that if q(x) = 0 implies x = 0 then q is a norm. [3.3.]

Remark. A balanced set A is absorbent if and only if ∀x ∃λ0 > 0: x ∈ λ0A (becausefor |λ| ≥ λ0, λ0A ⊆ λDA = λA).

We denote the open unit ball and closed unit ball of a seminorm p by

p<1 := x | p(x) < 1, p≤1 := x | p(x) ≤ 1.Note that ε · p<1 = x | p(x) < ε and ε · p≤1 = x | p(x) ≤ ε. These are balanced(λ ∈ D, x ∈ E ⇒ p(λx) ≤ |λ| p(x) ≤ p(x)), absorbent (x ∈ (ε + p(x))p<1 for any ε > 0)and convex (p(αx+ βy) ≤ αp(x) + βp(y) for α, β > 0).

Definition 3.10. Let E be a vector space. For A ⊆ E, the gauge or Minkowski functionalof A is the map gA : E → R+ ∪ ∞ defined by gA(x) := infρ > 0 | x ∈ ρA.

We note:

(i) If A is absorbent then gA is finite.(ii) If 0 ∈ A then gA(0) = 0.

(iii) If A is convex then gA(x + y) ≤ gA(x) + gA(y). This is clear for gA(x) or gA(y)equal to∞. Otherwise, note that λA+µA = (λ+µ)A for λ, µ ≥ 0. Then ∀ε > 0∃ρ, σ such that

gA(x) ≤ ρ < gA(x) + ε, x ∈ ρA,gA(y) ≤ σ < gA(y) + ε, y ∈ σA,

and hence x+ y ∈ (ρ+ σ)A, which implies gA(x+ y) < gA(x) + gA(y) + 2ε.

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(iv) If A is balanced then gA(λx) = |λ| g(x) for λ ∈ K (λ = 0 clear; λ 6= 0: ρA(λx) =infρ > 0 | |λ|x ∈ ρA = |λ| · infρ |λ|−1 | ρ > 0, x ∈ ρ |λ|−1A = |λ| ρA(x)) andx ∈ αA for α > gA(x), because then ∃η: gA(x) ≤ η < α and x ∈ ηA = α η

αA ⊆

αDA ⊆ αA.

Proposition 3.11. In a vector space E the gauge p of an absorbent, balanced and convexset A is a seminorm with

(1) p<1 ⊆ A ⊆ p≤1.

Note that if (1) holds for a seminorm p on E and a subset A ⊆ E, then gA = p:

α > p(x)⇒ p(x

α) < 1⇒ x

α∈ p<1 ⇒ x ∈ αA,

0 > α < p(x)⇒ p(x

α) > 1⇒ x

α6∈ p≤1 ⇒ x 6∈ αA.

We will now relate seminorms to topologies.

Lemma 3.12. Let p be a seminorm on a TVS E. Then

int(p≤1) ⊆ p<1 ⊆ p≤1 ⊆ p<1.

Proof. p<1 ⊆ p≤1 is clear. If x ∈ p≤1 then x/(1 + ε) ∈ p<1 for all ε > 0, x/(1 + ε)→ x for

ε→ 0, so x ∈ p<1. Similarly, one sees E \p<1 = p≥1 ⊆ p>1 ⊆ E \ p≤1, which is equivalentto int(p≤1) ⊆ p<1.

Proposition 3.13. Let p be a continuous seminorm on a TVS. Then

(i) p<1 is open,(ii) p≤1 is closed,

(iii) p<1 = p≤1,(iv) int(p≤1) = p<1.

Proof. We have p<1 = p−1([0, 1)) and p≤1 = p−1([0, 1]), which gives (i) and (ii). Moreover,p<1 ⊆ p≤1 implies p<1 ⊆ p≤1 = p≤1 and p<1 = int(p<1) ⊆ int(p≤1), and the converseinclusions are given in Lemma 3.12.

Proposition 3.14. Let p be a seminorm on a TVS E. Then the following assertions areequivalent:

(i) p<1 is open,(ii) p≤1 is a 0-neighborhood,

(iii) p is continuous at 0,(iv) p is continuous.

Proof. (i) ⇒ (ii) is clear. (ii) ⇒ (iii): p−1([0, ε]) = ε · p≤1 is a 0-neighborhood for allε > 0. (iii) ⇒ (iv): We have |p(x)− p(y)| ≤ p(x− y), and for x ∈ E and ε > 0 there is a0-neighborhood U such that p(U) ⊆ εD and hence |p(x+ U)− p(x)| ⊆ εD. (iv) ⇒ (i) isProposition 3.13 (i).

Definition 3.15. An absorbent balanced closed convex subset of a TVS is called a barrel.

Every LCS has a 0-basis consisting of barrels (Proposition 3.6).

Proposition 3.16. Let A be a barrel in a TVS E. Then there is a unique seminorm pon E whose closed unit ball equals A.

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Proof. Let p be the gauge of A. Then we have p<1 ⊆ A ⊆ p≤1 ⊆ p<1 (Proposition 3.11and Lemma 3.12). Taking the closures shows that p≤1 = A, and uniqueness follows fromthe remark after Proposition 3.11.

We will now show how locally convex topologies are defined by seminorms.

Let P be a family of seminorms on a vector space E. By Corollary 3.8, the family of allfinite intersections of the sets ε · p≤1 (ε > 0, p ∈P) defines a locally convex topology Ton E, having as 0-basis the sets

x | pk(x) ≤ εk for 1 ≤ k ≤ n (p1, . . . , pn ∈P, ε1, . . . , εn > 0)

or, equivalently, the sets

x | pk(x) ≤ ε for 1 ≤ k ≤ n (p1, . . . , pn ∈P, ε > 0).

We say that T is defined or generated by the family P.

Theorem 3.17. A locally convex topology can always be defined by a family of seminorms.

Proof. By Proposition 3.6 the balanced, closed, convex 0-neighborhoods V form a 0-basis.Each gauge gV is a continuous seminorm on E, and the locally convex topology definedby (gV )V is the same as the original topology because V = (ρV )≤1.

Proposition 3.18. A locally convex topology defined by a family P of seminorms isHausdorff if and only if ∀x 6= 0 ∃p ∈P: p(x) 6= 0.

Proof. If a LCS E is Hausdorff then x 6= 0 implies that there is a 0-neighborhood Usuch that x 6∈ U , and U contains some set x | pk(x) ≤ ε for k = 1 . . . n for somep1, . . . , pn ∈P and ε > 0, so pk(x) > ε for some k. Conversely, if p(x) = α > 0 for somep ∈ P then y | p(y) ≤ α/2 is a 0-neighborhood not containing x, so the topology isHausdorff by Proposition 2.11.

Proposition 3.19. Let E,F be LCS and f : E → F linear. Then f is continuous if andonly if for each continuous seminorm q on F there is a continuous seminorm p on Esuch that q(f(x)) ≤ p(x) for all x ∈ E.

Proof. Suppose f is continous and let q be a continuous seminorm on F . Then q≤1is a 0-neighborhood in F , so there is a balanced convex 0-neighborhood U in E withf(U) ⊆ q≤1. Let p the gauge of U . Then for all x ∈ E and ε > 0,

q(f(x)) = (p(x) + ε) · q(f(x

p(x) + ε)) ≤ p(x) + ε

because xp(x)+ε

∈ p<1 ⊆ U , so q(f(x)) ≤ p(x).

Conversely, let a 0-neighborhood V in F be given, which can be assumed to be balancedand convex. Its gauge q is a continuous seminorm, so there exists a continuous seminormp on E such that q(f(x)) ≤ p(x) for all x ∈ E. p<1 is a 0-neighborhood in E such thatf(p<1) ⊆ q<1 ⊆ V , so f is continuous.

In practice, it is convenient if we don’t have to consider all seminorms but just enoughto describe the topology.

Definition 3.20. Let P be a family of continuous seminorms on a LCS E. We saythat P is a basis of continuous seminorms (or a fundamental system of continuousseminorms) on E if for every continuous seminorm p on E there exists p′ ∈ P andC > 0 such that p(x) ≤ Cp′(x) for all x ∈ E.

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Proposition 3.21. Let E,F be LCS with bases of continuous seminorms P and Q,respectively, and f : E → F a linear mapping. Then f is continuous if and only if forevery q ∈ Q there is a seminorm p ∈P and C > 0 such that q(f(x)) ≤ Cp(x) for x ∈ E.

Proof. “⇒” is clear; “⇐”: given a continuous seminorm q on F , q ≤ Cq′ for someC > 0 and q′ ∈ Q, q′(f(x)) ≤ C ′p(x) for some p ∈ P and C ′ > 0, which givesq(f(x)) ≤ CC ′p(x), and x 7→ CC ′p(x) is a continuous seminorm on E.

Remark. Balanced convex sets are also called absolutely convex because a set A isbalanced and convex and if and only if ∀x1, . . . , xn ∈ A ∀λ1, . . . , λn ∈ K with

∑|λi| ≤ 1:∑

λixi ∈ A. The balanced convex hull (also: absolutely convex hull) of⋃ιAι (where

each Aι is absolutely convex) is∑λιxι

∣∣∣ ∑ |λι| ≤ 1 (finite sum), λι ∈ K, xι ∈ Aι.

We denote the absolutely convex hull of a set A by acx(A).

Examples. Let Ω ⊆ Rn be open and m ∈ N0 ∪ ∞. The following spaces are LCS:

1. C(Ω) with seminorms qK(f) = maxx∈K |f(x)| for all K ⊆ Ω compact.2. Em(Ω) := f : Ω → K | ∂αf exists and is continuous for all α ∈ Nn

0 with |α| ≤m with seminorms qK,α(f) = maxx∈K |∂αf(x)| for K ⊆ Ω compact and |α| ≤ m.

3. E (Ω) =⋂∞m=0 Em(Ω).

4. For K ⊆ Ω compact: Dm(K) = f ∈ Em(Ω) | supp f ⊆ K with seminormsqK,α.

5. D(K) = D∞(K).

4. Completeness

Definition 4.1. Let E be a TVS and A ⊆ E. A filter (filter basis) F on A is called aCauchy filter (Cauchy filter basis) if for every 0-neighborhood U in E there exists X ∈ Fsuch that X − X ⊆ U . A sequence (xn)n in A is called a Cauchy sequence if for every0-neighborhood U in E there exists N ∈ N such that xn − xm ∈ U for n,m ≥ N .

A filter basis is a Cauchy filter basis if and only if the filter generated by it is a Cauchyfilter; moreover, any convergent filter (filter basis) is a Cauchy filter (Cauchy filter basis).

Definition 4.2. Let E be a TVS. A subset A ⊆ E is called complete if every Cauchyfilter on A has a limit in A and sequentially complete if every Cauchy sequence in A hasa limit in A.

Proposition 4.3. Let E be a Hausdorff TVS. If A ⊆ E is complete then it is closed.

Proof. For x ∈ A let F be the neighborhood filter of x. Then its trace FA on A is a filteron A which is even a Cauchy filter (U 0-neighborhood in E ⇒ ∃x ∈ F : X − X ⊆ U⇒ X ∩ A ∈ FA, (X ∩ A) − (X ∩ A) ⊆ U). Hence, the filter on E generated by FA isa Cauchy filter and converges to an element of A; because this filter is finer than F andF → x, x ∈ A.

Proposition 4.4. Let E be a TVS and A ⊆ E complete. Then every closed subset of Ais complete.

Proof. Let B ⊆ A be closed and F a Cauchy filter on B. F generates a Cauchy filterF ′ on A, and F ′ and hence F converges to some x ∈ A, but x (as a limit point of F )has to be in B = B.

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Definition 4.5. Let E and F be TVS and A ⊆ E. A mapping f : A → F is calleduniformly continuous if for every 0-neighborhood U in F there exists a 0-neighborhood Vin E such that for all x, y ∈ A, x− y ∈ V implies f(x)− f(y) ∈ U .

If a linear map f : E → F between TVS is continuous at 0 it is also uniformly continuous.

Lemma 4.6. Let E,F be TVS, A ⊆ E and f : A→ F uniformly continuous. If F is aCauchy filter on A then f(F ) is a Cauchy filter basis in F .

Proof. If U is a 0-neighborhood in F there is a 0-neighborhood V in E such that x−y ∈ Vimplies f(x)−f(y) ∈ U for x, y ∈ A. By assumption there is F ∈ F such that F−F ⊆ V ,which implies f(F )− f(F ) ⊆ U , and f(F ) ∈ f(F ).

Proposition 4.7. If M is a linear subspace of a TVS then its closure M is so.

Proof. Let a, b ∈ M and U an arbitrary neighborhood of a+ b. There are neighborhoodsV of a and W of b such that V +W ⊆ U , and V ∩M 6= ∅, W ∩M 6= ∅ imply U ∩M 6= ∅,so a+ b ∈ M .

Let λ ∈ K, a ∈ M and U a neighborhood of λa. There is a neighborhood V of a withλV ⊆ U . V ∩M 6= ∅ implies λV ∩M 6= ∅, hence U ∩M 6= ∅ and λa ∈ M .

[10.3.]

Lemma 4.8. Let E,F be TVS and A ⊆ E. If f : A → F is uniformly continuous it iscontinuous.

Proof. If U is a 0-neighborhood in F there is a 0-neighborhood V in E such that x−y ∈ Vimplies f(x) − f(y) ∈ U . Given x ∈ A, (x + V ) ∩ A is a neighborhood of a in A andx+ V − x ∈ V implies f(x+ V )− f(x) ⊆ U , hence f(x+ V ) ⊆ f(x) + U .

Proposition 4.9. Let E be a TVS, F a complete Hausdorff TVS, and A ⊆ E. Iff : A→ F is uniformly continuous then there exists a unique uniformly continuous map-ping f : A→ F such that f(x) = f(x) ∀x ∈ A.

If A is a linear subspace of E and f is linear, f also is linear.

Proof. For x ∈ A, let F be its neighborhood filter in E. Then its trace FA is a Cauchyfilter and by Lemma 4.6, f(FA) is a Cauchy filter basis in F and hence converges to

some unique y ∈ F . We define f(x) := y. For x ∈ A we have FA → x and hence

f(FA)→ f(x), so f(x) = f(x).

Let U and V be 0-bases in E and F , resp., consisting of balanced closed 0-neighborhoods(Corollary 2.8). Let V0 ∈ V and choose V ∈ V such that V + V + V ⊆ V0, U ∈ Usuch that x1 − x2 ∈ U implies f(x1) − f(x2) ∈ V for x1, x2 ∈ A, and U1 ∈ U withU1 + U1 + U1 ⊆ U .

For x, y ∈ A with x − y ∈ U1 there are x1, x2 ∈ A such that x1 − x, y1 − x ∈ U1. Letus prove f(x) ∈ f(x1) + V : let X be a neighborhood of f(x), then f(Y ∩ A) ⊆ Xfor some neighborhood Y of x. (x + U1) ∩ Y is a neighborhood of x, so there is someu ∈ (x+U1)∩Y ∩A. Then, u−x1 = (u−x)+(x−x1) ∈ U1+U1 ⊆ U so f(u)−f(x1) ∈ V ,i.e., f(u) ∈ f(x1) + V . Moreover, f(u) ∈ X so f(u) ∈ (f(x1) + V ) ∩ X. Hence,

f(x) ∈ f(x1) + V = f(x1) + V = f(x1) + V . Similarly, f(y) ∈ f(y1) + V . Thenx1− y1 = (x1−x) + (x− y) + (y− y1) ∈ U1 +U1 +U1 ⊆ U , so f(x1)− f(y1) ∈ V . Finally,

f(x)− f(y) = (f(x)− f(x1)) + (f(x1)− f(y1)) + (f(y1)− f(y)) ∈ V + V + V ⊆ V0, so fis uniformly continuous.

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Uniqueness of f follows from the fact that F is Hausdorff, as for two mappings A → Fthe set where they are equal (which contains A) is closed, hence equal to A.

For the second claim let V ∈ V be arbitrary and U ∈ U such that for x, y ∈ A,x− y ∈ U + U implies f(x)− f(y) ∈ V . Choose x1, y1 ∈ A such that x− x1, y − y1 ∈ Uand hence x+ y − (x1 + y1) ∈ U + U . Then

f(x) + f(y)− f(x+ y) = f(x)− f(x1) + f(y)− f(y1)− f(x+ y) + f(x1 + y1)

∈ V + V + V.

As V is arbitrary and F is Hausdorff, f(x) + f(y) = f(x+ y) (Proposition 2.11).

For λ ∈ K \ 0 (the case λ = 0 is trivial) and V ∈ V choose V1 ∈ V with V1 + λV1 ⊆ V

and U1 ∈ U such that x− y ∈ U1 implies f(x)− f(y) ∈ V1, as well as U ∈ U such thatU∩λU ⊆ U1. Given x ∈ A, choose x1 ∈ A such that x−x1 ∈ U and hence λx−λx1 ∈ λU .Then

f(λx)− λf(x) = f(λx)− f(λx1) + λ(f(x1)− f(x)) ∈ V1 + λV1 ⊆ V,

which proves that f(λx) = λf(x).

For constructing the completion we need some terminology.

Definition 4.10. Let E be a TVS. A Cauchy filter F on E is called minimal if for anyCauchy filter G on E, G ⊆ F implies G = F .

Proposition 4.11. Let E be a TVS and F a Cauchy filter on E. There exists a uniqueminimal Cauchy filter F0 on E such that F0 ⊆ F . If B is a basis of F and U a 0-basisin E, F + V | F ∈ B, V ∈ U is a basis of F0.

Proof. The family F +V | F ∈ F , V 0-neighborhood in E is a filter basis (for F1, F2 ∈F , V1, V2 0-neighborhoods we have F1 ∩ F2 ∈ F and V1 ∩ V2 is a 0-neighborhood, and(F1 ∩ F2) + (V1 ∩ V2) ⊆ (F1 + V1) ∩ (F2 + V2)) and generates a filter F0 on E.

F0 is Cauchy: given a 0-neighborhood U in E choose a 0-neighborhood V in E suchthat V + V + V ⊆ U . There is F ∈ F with F − F ⊆ V . We have F + V ∈ F0 and(F + V )− (F + V ) = F − F + V + V ⊆ V + V + V ⊆ U .

F0 ⊆ F : given F0 ∈ F0 we have F + V ⊆ F0 for some F, V , hence F ⊆ F0 which meansF0 ∈ F .

Finally, if G ⊆ F is a Cauchy filter on E, let F +V be a given element of the filter basisgenerating F0. There is G ∈ G ⊆ F with G − G ⊆ V , and for any a ∈ G ∩ F 6= ∅ wehave G ⊆ a+ V ⊆ F + V , hence F + V ∈ G which implies F0 ⊆ G . This shows that F0

is minimal, and unique.

The neighborhood filter F of x ∈ E is minimal: if G ⊆ F is a Cauchy filter and U ∈ F ,then ∃V ∈ G with V − V ⊆ U − x ⇒ V − x ⊆ U − x ⇒ V ⊆ U ⇒ U ∈ G .

Lemma 4.12. Let E be a TVS and F a minimal Cauchy filter on E. Then for anyB ∈ F , B ∈ F .

Proof. By Proposition 4.11, F +V | F ∈ F , V 0-neighborhood is a basis of F becauseF is minimal. Given B ∈ F , F +V ⊆ B for some F, V . Taking an open 0-neighborhoodU ⊆ V we have F +U ⊆ B and hence F +U ⊆ B because F +U is open, which impliesthat B ∈ F .

Lemma 4.13. If E is a TVS and A ⊆ E a dense subset such that every Cauchy filteron A converges in E, then E is complete.

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Proof. A Cauchy filter F on E contains a minimal Cauchy filter F0. Because F ∈ F0 forF ∈ F0, F ∩A 6= ∅ and (F0)A is a Cauchy filter on A. The filter G ⊇ F0 it generates onE converges to some x ∈ E, so x is a cluster point of F0. Each Cauchy filter converges toits cluster points: if U is a closed 0-neighborhood in E then F −F ⊆ U for some F ∈ F0,so F − F ⊆ F − F ⊆ U = U and because x ∈ F , F ⊆ F ⊆ x + U , hence F0 → x andF → x.

Theorem 4.14. Let E be a Hausdorff TVS. There exists a complete Hausdorff TVS E,

called the completion of E, and a mapping ι : E → E such that

(i) ι is a linear homeomorphism onto its image,

(ii) ι(E) is dense in E.

For any other pair (E1, ι1) such that (i) and (ii) hold there is a linear isomorphism j : E →E1 such that ι1 = j ι. In other words, the completion is unique up to isomorphism.

Proof. We define E to be the set of all minimal Cauchy filters on E. For F ,G ∈ Ewe define F + G to be the minimal Cauchy filter contained in the filter generated bythe Cauchy filter basis A + B | A ∈ F , B ∈ G ; similarly, define λF (λ ∈ K) as theminimal Cauchy filter contained in the filter generated by λA | A ∈ F. Let 0 be the

neighborhood filter of 0 in E. E then is a vector space over K (exercise!).

For any 0-neighborhood in E we set

U := F ∈ E | ∃A ∈ F ∃V 0-neighborhood in E : A+ V ⊆ U.One easily verifies:

U1 ⊆ U2 ⇒ U1 ⊆ U2,

U1 ∩ U2 ⊆ U1 ∩ U2,

λU ⊆ λU (λ 6= 0),

U1 + U1 ⊆ U2 ⇒ U1 + U1 ⊆ U2.

In fact, λU ⊆ λU for λ 6= 0: let F ∈ U . Then F + V ⊆ U ⇒ λF + λV ⊆ λU , and

λF ∈ λF , λV is 0-nbhd. U1 +U1 ⊆ U2: given F ,G ∈ U1, F + V1 ⊆ U1 and G+ V2 ⊆ U1

imply F +G+ V1 + V2 ⊆ U2, and F +G+ V1 ∈ F + G , so F + G ∈ U2.

Let U be a 0-basis in E consisting of balanced sets. By Theorem 2.10 there is a unique

linear topology T on E having U | U ∈ U as 0-basis. In fact, U 6= ∅ because 0 ∈ U . U

is balanced because λU ⊆ λU ⊆ U for λ 6= 0. U is absorbent: for F ∈ E choose V withV + V + V ⊆ U . Then ∃F ∈ F : F − F ⊆ V . Take x ∈ F and λ ≥ 1 such that x ∈ λV .Then F ⊆ x+V and F +V ⊆ λV +V +V ⊆ λ(V +V +V ) ⊆ λU , so λ−1F +λ−1V ⊆ U ,

and λ−1F ∈ U . Given U1, U2 ∃U3 with U3 ⊆ U1 ∩ U2, so U3 ⊆ U1 ∩ U2 ⊆ U1 ∩ U2.

T is Hausdorff: suppose x ∈ U for all U ∈ U . We know that ∀U ∈ 0 ∃A ∈ x: A ⊆ U ,

so x is finer than 0. Because x is minimal, 0 = x and T is Hausdorff.

We define ι : E → E by mapping x to its neighborhood filter. Linearity of ι is clear. Infact, x + V1 + λy + V2 + V3 ∈ ι(x + λy) implies ι(x + λy) ⊆ ι(x) + λι(y), hence they areequal. x+ V1 forms a basis of ι(x), λy + V2 a basis of λι(y), and x+ V1 + λy + V2 + V3 abasis of ι(x) + λι(y). ι(x) = 0 implies x = 0, so ι is injective.

Let U ∈ U . For x ∈ U there is a 0-neighborhood V such that x + V + V ⊆ U ; because

x+V ∈ ι(x), ι(x) ∈ U . On the other hand, if ι(y) ∈ U for some y ∈ E then y+V1+V2 ⊆ U

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for some 0-neighborhoods V1 and V2, hence y ∈ U . Consequently, ι(U) ⊆ U∩ι(E) ⊆ ι(U),which shows that ι is a homeomorphism.

Finally, we show that (E, T ) is complete: let F0 be a Cauchy filter on ι(E). ThenF ′ := ι−1(F0) is a Cauchy filter on E containing a minimal Cauchy filter F . ι(F ) is a

Cauchy filter on ι(E) with F0 = ι(F ′) ⊇ ι(F ). Because ι(F )→ F in E, F0 → F . ByLemma 4.13, completeness follows.

For uniqueness, the isomorphism f : ι(E)→ ι1(E) can be extended to a continuous linear

map f : E → E1, and the isomorphism g : ι1(E) → ι(E) extends to a continuous linear

map g : E1 → E (Proposition 4.9). Because f g = id and g f = id, f g = id andg f = id.

Proposition 4.15. Let E be a dense subspace of a TVS F . If U is a 0-basis in E thenU | U ∈ U (where the closures are taken in F ) is a 0-basis in F .

Proof. Let V be an open 0-neighborhood in F . Then U := V ∩ E is an open 0-neighborhood in E. For z ∈ V and any neighborhood W of z in F , V ∩W is a neighbor-hood of z, so ∅ 6= V ∩W ∩ E = W ∩ U , and z ∈ U . Hence, U ⊆ V ⊆ U ⊆ V .

Given U ′ ∈ U choose an open 0-neighborhood U in E with U ⊆ U ′; then U = V ∩E forsome open 0-neighborhood V in F and V ⊆ U ⊆ U ′, so U ′ is a 0-neighborhood.

Given any 0-neighborhood V ′ in F , choose an open 0-neighborhood V with V ⊆ V ′.Then U = V ∩ E contains some U ′ ∈ U and U ′ ⊆ U ⊆ V ⊆ V ′, so the closures ofelements of U form a 0-basis in F .

By Proposition 3.4 the completion of a LCS is a LCS again.

Proposition 4.16. If a family P defines the topology of a LCS E, the family p | p ∈Pdefines the topology of E.

Note for this that a continuous seminorm p is uniformly continuous because |p(x)− p(y)| ≤p(x− y).

Proof. Let p ∈P and x, y ∈ E. For ε > 0 let U be a 0-neighborhood in E such that for

x′, y′ ∈ E we have

(2) x′ − y′ ∈ U ⇒ |p(x′)− p(y′)| ≤ ε.

Choose a 0-neighborhood V in E such that V +V ⊆ U and elements x′, y′ ∈ E such thatx− x′ ∈ V , y − y′ ∈ V and hence x+ y − (x′ + y′) ∈ U . Then

p(x+ y)− p(x)− p(y) = p(x+ y)− p(x′ + y′) + p(x′ + y′)− p(x)− p(y)

≤ ε+ p(x′) + p(y′)− p(x)− p(y) ≤ 3ε.

This shows that p(x+ y) ≤ p(x) + p(y).

Given x ∈ E and λ 6= 0 as well as ε > 0, choose a 0-neighborhood U such that (2)holds. Choose a 0-neighborhood V such that λV ∩ V ⊆ U . Choose any x′ ∈ E such thatx− x′ ∈ V . Then

|p(λx)− |λ| p(x)| = |p(λx)− p(λx′) + |λ| (p(x′)− p(x))| ≤ (1 + |λ|)εshows that p(λx) = |λ| p(x).

It is clear that for p ∈P, p<ε is open. Conversely, let a 0-neighborhood in E be given; wecan assume that it is of the form U , where U = (p1)<ε∩. . .∩(pn)<ε for some p1, . . . , pn ∈P

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and ε > 0. If every x ∈ (p1)<ε∩. . .∩(pn)<ε is an element of U , the family of all p generates

the topology of E as claimed. For each i we can choose a 0-neighborhood Vi in E suchthat x+ Vi ⊆ (pi)<ε. With V := V1 ∩ . . . ∩ Vn we have

x+ V ⊆ (p1)<ε ∩ . . . ∩ (pn)<ε.

From this it follows that x ∈ U .

5. Bounded sets, normability, metrizability

Definition 5.1. Let E be a TVS. A subset A ⊆ E is called bounded if for each 0-neighborhood U there is λ0 > 0 such that A ⊆ λU for |λ| ≥ λ0.

The family of bounded sets is closed under formation of subsets, finite unions, closures,and in LCS also absolutely convex hulls. Moreover, if E,F are TVS and f : E → F islinear and continuous then f(B) is bounded if B is bounded: if U is a 0-neighborhood inF and V a 0-neighborhood in E with f(V ) ⊆ U , then B ⊆ λV implies f(B) ⊆ λU .

Proposition 5.2. A subset A of a TVS E is bounded if and only if for every sequence(λn)n in K with λn → 0 and every sequence (xn)n in A, λnxn → 0 in E.

Proof. Let A be bounded, U a balanced 0-neighborhood in E and (λn)n with λn → 0.There is λ > 0 with A ⊆ λU . Because |λn| ≤ λ−1 for n large enough we have λnxn ∈λ−1λU = U for n large enough.

Conversely, suppose A satisfies the condition but is not bounded. Then there are a 0-neighborhood U and sequences (λn)n, (xn)n with |λn| ≥ n and xn ∈ A \ λnU . Becauseλ−1n → 0 and λ−1n xn 6∈ U for all n, this gives a contradiction.

[17.3.]

Proposition 5.3. If a LCS E is locally bounded (i.e., has a bounded 0-neighborhood)its topology can be defined by a single seminorm.

Proof. Let V be a bounded 0-neighborhood. By Proposition 3.6 there is an absolutelyconvex closed 0-neighborhood W contained in V . The gauge q of W is a continuousseminorm and the sets εq≤1 form a 0-basis: if U is any 0-neighborhood there is ε > 0such that εV ⊆ U and hence εq≤1 = εW ⊆ U .

In this case, q is a norm if and only if E is Hausdorff (Proposition 3.18). Hence, a TVS isnormable if and only if it is locally convex, Hausdorff and has a bounded 0-neighborhood.

A TVS E is called metrizable if there is a metric on E which induces its original topology.A metrizable topological space is Hausdorff and each point has a countable neighborhoodbasis. In particular, the topology of a metrizable locally convex space can be defined bya countable family of seminorms, given by the gauges of a countable 0-basis consisting ofabsolutely convex closed sets (Proposition 3.6). The converse also holds:

Proposition 5.4. Let E be a Hausdorff LCS whose topology T can be defined by asequence (qn)n∈N of seminorms such that qn(x) ≤ qn+1(x) for all x ∈ E. Then the map

|x| :=∞∑n=1

1

2nqn(x)

1 + qn(x)

has the following properties:

(i) |x| = 0⇔ x = 0,(ii) |x| = |−x|,

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(iii) |x+ y| ≤ |x|+ |y|,(iv) |λ| ≤ 1⇒ |λx| ≤ |x|,(v) λ→ 0⇒ |λx| → 0 ∀x ∈ E.

The metric d(x, y) := |x− y| defines the topology T and is translation invariant, i.e.,d(x+ a, y + a) = d(x, y) ∀x, y, a ∈ E.

Proof. The series converges. We have |x| = 0 ⇔ qn(x) = 0 ∀n ⇔ x = 0 by Proposi-tion 3.18, which gives (i). (ii) is clear from qn(x) = qn(−x).

The function ξ 7→ ξ1+ξ

is increasing for ξ > −1, hence for |λ| ≤ 1, qn(λx) ≤ qn(x) implies

qn(λx)

1 + qn(λx)≤ qn(x)

1 + qn(x)

which gives (iv).

Since qn(x+ y) ≤ qn(x) + qn(y), we see that

qn(x+ y)

1 + qn(x+ y)≤ qn(x) + qn(y)

1 + qn(x) + qn(y)≤ qn(x)

1 + qn(x)+

qn(y)

1 + qn(y)

which gives (iii).

Because of (i), (ii) and (iii), d(x, y) = |x− y| defines a translation-invariant metric.Denote the topology it induces on E by T ′. For any k ∈ N set

V =

x | qk+1(x) ≤ 1

2k+1

, U =

x | |x| ≤ 1

2k

.

Let x ∈ V ; by splitting the sum |x| into two parts,

k+1∑n=1

1

2nqn(x)

1 + qn(x)≤ 1

2k+1

k+1∑n=1

1

2n≤ 1

2k+1,

∞∑n=k+2

1

2nqn(x)

1 + qn(x)≤

∞∑n=k+2

1

2n=

1

2k+1,

we see that x ∈ U , so T is finer than T ′. Conversely, for any m ∈ N set

X = x | |x| ≤ 1

2m+k+1, W = x | qm(x) ≤ 1

2k

For x ∈ X we have

1

2mqm(x)

1 + qm(x)≤ 1

2m+k+1

=⇒ qm(x)

1 + qm(x)≤ 1

2k+1

=⇒ qm(x)(1− 1

2k+1) ≤ 1

2k+1

=⇒ qm(x) ≤ 1

2k+1 − 1≤ 1

2k.

and hence X ⊆ W , so T ′ is finer than T .

Remark. The condition qn(x) ≤ qn+1(x) is no restriction because for any sequence ofseminorms pn, the sequence of seminorms defined by qn(x) := maxk≤n pk(x) satisfies thiscondition and defines the same topology.

Combining Proposition 4.16 and Proposition 5.4, we see:

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Corollary 5.5. The completion of a metrizable LCS is metrizable.

Definition 5.6. A complete metrizable locally convex space is called a Frechet space.

Moreover, we can define the following notion:

Definition 5.7. A LCS E is called quasicomplete if every bounded closed subset of E iscomplete.

Note. If we call a filter F bounded if it contains a bounded set, then a LCS E isquasicomplete if and only if every bounded Cauchy filter converges.

Proposition 5.8. A metrizable LCS E is complete if and only if it is sequentially com-plete.

Proof. If E is complete and a Cauchy sequence (xn)n is given, the filter basis consistingof all sets Sm := xn | n ≥ m is a Cauchy filter basis and converges, which means thatthe sequence converges.

Conversely, if E is sequentially complete and E denotes its completion, let x ∈ E be

arbitrary. Since E is dense in E and E is metrizable there exists a sequence (xn)n in Econverging to x. Since (xn)n is a Cauchy sequence in E, x ∈ E.

6. Products, subspaces, direct sums and quotients

We recall:

Proposition 6.1. Let (Xi)i∈I be a family of topological spaces, X a set and fi : X → Xi

a mapping for each i.

(i) There is a coarsest topology T on X such that all fi are continuous.(ii) If Ui is a subbasis of Xi for each i, a subbasis of T is given by f−1i (Ui) | i ∈

I, Ui ∈ Ui.(iii) If Y is a topological space, a mapping g : Y → (X,T ) is continuous if and only if

all fi g are continuous, and T is the unique topology on X with this property.(iv) If F is a filter on X then F → x ∈ X if and only if fi(F )→ fi(x) in Xi for all

i ∈ I.

Proof. (i), (ii): The family B := f−1i (Ui) | i ∈ I, fi ∈ Ui is the subbasis of a topologyT for which all fi are continuous. Any other such topology T ′ has to contain B andhence is finer than T .

(iii): If all fi g are continuous then g−1(f−1i (Ui)) = (fi g)−1(Ui) is open and g iscontinuous because the f−1i (Ui) form a subbasis. If T ′ is another topology with thisproperty, continuity of id : (X,T ′)→ (X,T ′) gives that fi : (X,T ′)→ Xi is continuousfor all i, hence T is coarser than T , and vice versa.

(iv): Given F and x ∈ X with fi(F ) → fi(x) for all i, let U be a neighborhood ofx. Then there is a finite set J ⊆ I and neighborhoods Vi ∈ Ui of fi(x) (i ∈ J) suchthat x ∈

⋂i∈J f

−1i (Vi) ⊆ U . For each i ∈ J there is Fi ∈ F with fi(Fi) ⊆ Vi, and

U ⊇⋂i∈J Fi ∈ F .

The topology T of Proposition 6.1 is called the projective topology on X with respect tothe family (fi)i.

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Proposition 6.2. Let (Ei)i be a family of TVS (LCS), E a vector space, fi : E → Ei alinear mapping for each i, and T the projective topology on E with respect to (fi)i.

(i) T is a linear (locally convex) topology.(ii) If Ui is a 0-basis in Ei then a 0-basis of T is given by the family of finite inter-

sections of sets of the form f−1i (Ui) with Ui ∈ Ui.(iii) A subset B ⊆ E is bounded if and only if fi(B) is bounded for all i ∈ I.

Proof. The mappings

(x, y) 7→ fi(x+ y) = fi(x) + fi(y)

(λ, x) 7→ fi(λ · x) = λ · fi(x)

are continuous, which gives (i) in the linear case. (ii) is clear from the definition of theprojective topology, which gives (i) in the locally convex case. For (iii), fi(B) is bounded(Remark after Definition 5.1); conversely, for B to be bounded it suffices to show thatB ⊆ λf−1i (Ui) for each i, U ∈ Ui and λ large enough, which is implied by fi(B) ⊆ λUi.

Corollary 6.3. If (pi)i is a family of seminorms defining the topology of a LCS E thenthe topology of E is the projective topology with respect to (pi)i.

Proof. Finite intersections of sets of the form ε(pi)≤1 form a 0-basis of both topologies.

Corollary 6.4. Let (Ei)i∈I be a family of LCS, E a vector space and fi : E → Ei alinear map for each i. If (pi,λ)λ∈Li

is a family of seminorms defining the topology of Eithen (pi,λ fi)i∈I,λ∈Li

is a family of seminorms on E defining the projective topology withrespect to (fi)i.

Proof. This follows from a general property of projective topologies: the projective topol-ogy with respect to (fi)i is the projective topology with respect to (pi,λ fi)i,λ (Proposi-tion 6.1 (iii)).

Examples.

(i) The cartesian product∏

i∈I Ei of a family (Ei)i of TVS with the projective topol-ogy with respect to the projections πj :

∏iEi → Ej. Note that each πj is open.

(ii) A linear subspace E of a TVS F with the projective topology with respect to theinjection ι : E → F .

Proposition 6.5. A countable product of metrizable LCS is metrizable.

Proof. A 0-basis in∏

iEi is given by all sets∏

iXi such that each Xi is taken from a(countable) 0-basis of Ei for finitely many i and Xi = Ei for all other i. The collectionof all finite subsets of a countable set is countable.

Proposition 6.6. Let (Ei)i be a family of TVS. The cartesian product∏

iEi is Hausdorffif and only if all factors Ei are Hausdorff. If E is a Hausdorff TVS, every subspace of Eis Hausdorff.

We skip the proof, as this is known from topology.

Concerning completeness, we have:

Proposition 6.7. Let (Ei)i be a family of TVS and E =∏

iEi their product. Then foreach family (Ai)i of complete subsets Ai ⊆ Ei, the product A =

∏iAi ⊆ E is complete.

In particular, if all the Ei are (quasi)complete then E is (quasi)complete.

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Proof. Given a (bounded) Cauchy filter F on A, πi(F ) is a (bounded) Cauchy filter onAi which converges to some xi ∈ Ai, hence F → (xi)i ∈ A by Proposition 6.1 (iv).

We will now consider the dual situation, namely that of direct sums and quotients.

Recall that given a family (Xi)i of topological spaces, a set X and mappings fi : Xi → Xfor each i, there is a finest topology T on X such that all fi are continuous. A subsetU ⊆ X is open for T if and only if all f−1i (U) are open. A mapping f from (X,T ) intoany topological space is continuous if and only if f fi is continuous for all i, and Tis the unique topology on X having this property. This topology is called the inductivetopology on X with respect to (fi)i.

If the Ei are TVS or LCS and the fi are linear, the inductive topology is not linear orlocally convex anymore, in general. However, we have the following:

Proposition 6.8. Let (Ei)i∈I be a family of a) TVS b) LCS, E a set and fi : Ei → E alinear mapping for each i.

(i) There is a finest a) linear topology, b) locally convex topology T on X such thatall fi are continuous.

(ii) A 0-basis for T is given by the family of all a) balanced, absorbent b) absolutelyconvex, absorbent subsets U ⊆ E such that f−1i (U) is a 0-neighborhood in Ei foreach i ∈ I.

(iii) If F is an a) TVS, b) LCS and f : E → F is linear, then f : (E,T ) → F iscontinuous if and only if all compositions f fi are continuous. This propertyuniquely characterizes T .

This topology is called a) the inductive linear topology or b) the inductive locally convextopology on E with respect to the family (fi)i.

Proof. (i) Let T be the class of all a) linear topologies, b) locally convex topologies on Efor which all fi are continuous. The trivial topology (having only ∅ and E as open sets) isin T . Put on E the projective topology T with respect to all mappings id : E → (E,T ′)for T ′ ∈ T . Then the fi are continuous into E by Proposition 6.1 (iii). Moreover, givenany a) linear topology b) locally convex topology T0 on E such that all fi are continuousinto T0, id : (E,T )→ (E,T0) is continuous because T0 ∈ T , so T is finer than T0.

(ii): The given family is the 0-basis of a) a linear topology, b) a locally convex topologyT ′. All fi are continuous into (E,T ′), so T is finer than T ′.

Conversely, if U is a 0-neighborhood in T there are finitely many T ′1 , . . . ,T

′n ∈ T and a)

balanced or b) absolutely convex 0-neighborhoods Vk in T ′k such that x ∈ V :=

⋂nk=1 Vk ⊆

U . Because f−1i (Vk) is a 0-neighborhood for each i, V is a 0-neighborhood in T ′ and T ′

is finer than T .

(iii): Suppose that all f fi are continuous and let U be a) a balanced b) an absolutelyconvex neighborhood of 0 in F . Then f−1i (f−1(U)) is a 0-neighborhood in Ei, hence theabsorbent and a) balanced, b) absolutely convex set f−1(U) is a 0-neighborhood in E forT by (ii) and f is continuous. Uniqueness is seen as in Proposition 6.1 (iii).

Proposition 6.9. In the case of LCS, if the linear hull of⋃i fi(Ei) equals E then the

absolutely convex hulls acx(⋃i fi(Ui)) with Ui ∈ Ui (where Ui is a 0-basis in Ei) form a

0-basis of the inductive locally convex topology.

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Proof. Each such hull U is absolutely convex, f−1i (U) ⊇ Ui is a 0-neighborhood, andU is absorbent (x ∈ E ⇒ x =

∑λifi(xi) = α

∑λiα

−1fi(xi) ∈ α · acx(⋃i fi(Ui)) with

α =∑|λi|), so U is an element of the basis B given in Proposition 6.8 (ii). Conversely,

if V ∈ B then each f−1i (V ) contains some Ui ∈ Ui, hence fi(Ui) ⊆ V ,⋃i∈I fi(Ui) ⊆ V ,

and because V is absolutely convex the absolutely convex hull of⋃i fi(Ui) is contained

in V .

Examples for the inductive locally convex topology are quotients and direct sums. Wewill treat those in more detail now.

Proposition 6.10. Let E be a TVS and M ⊆ E a linear subspace. Then M/E with thequotient topology is a TVS. It is Hausdorff if and only if M is closed.

Proof. Let ϕ : E → E/M denote the canonical surjection; we first show that ϕ is open.If U ⊆ E is open then ϕ−1(ϕ(U)) = U + M =

⋃x∈M(U + x) is open, so ϕ(U) is open in

E/M by definition of the quotient topology. Hence, a subset V ⊆ E/M is a neighborhoodof x = ϕ(x) if and only if ϕ−1(V ) is a neighborhood of x.

If V is a neighborhood of x+ y with x = ϕ(x), y = ϕ(y), then ϕ−1(V ) is a neighborhoodof x + y and there are neighborhoods U1, U2 of x and y such that U1 + U2 ⊆ ϕ−1(V ).Then ϕ(U1) and ϕ(U2) are neighborhoods of x and y and we have ϕ(U1) + ϕ(U2) ⊆ V . [24.3.]

If x = ϕ(x) and V is a neighborhood of λx in E/M then ϕ−1(V ) is a neighborhood of λxin E, so there are neighborhoods U1 of λ and U2 of x such that U1U2 ⊆ ϕ−1(V ), whichimplies that ϕ(U2) is a neighborhood of x such that U1ϕ(U2) ⊆ V .

If E/M is Hausdorff then 0 is closed, hence M = ϕ−1(0) is closed. Conversely,suppose that M is closed and x ∈ E/M is different from 0. Then x = ϕ(x) for somex 6= M , so there is a neighborhood U of x with U ∩ M = ∅. Hence, x − ϕ(U) is aneighborhood of 0 which does not contain x because 0 6∈ ϕ(U), so E/M is a Hausdorffspace.

Definition 6.11. A set P of seminorms is called

(i) directed if ∀n ∈ N ∀p1, . . . , pn ∈P ∃q ∈P: maxi pi ≤ q, and(ii) saturated if ∀n ∈ N ∀p1, . . . , pn ∈P: maxi pi ∈P.

Lemma 6.12. Let P be a set of seminorms on a vector space.

(i) The set maxi pi | p1, . . . , pn ∈ P, n ∈ N is saturated and generates the samelocally convex topology as P.

(ii) If P is saturated it is directed.(iii) If P is directed then the locally convex topology it generates has as 0-basis the

family ε · q≤1 | q ∈P (or ε · q<1 | q ∈P).

Proof. All claims follow immediately from (maxi pi)≤1 = (p1)≤1 ∩ . . . ∩ (pn)≤1 and q≤1 ⊆p≤1 for p ≤ q.

Proposition 6.13. If E is a LCS and M ⊆ E a linear subspace then E/M is a LCS. IfP is a directed family of seminorms defining the topology of E then the family p | p ∈P defines the topology of E/M , where for a seminorm p on E we define the quotientseminorm p on M/E by p(x) = infx∈x p(x).

Proof. If U is a 0-neighborhood in E/M , ϕ−1(U) (which is a 0-neighborhood in E) con-tains a convex 0-neighborhood V , and ϕ(V ) is a convex 0-neighborhood contained inU .

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Next, we show that p is a seminorm. We have

p(λx) = infx∈x

p(λx) = |λ| infx∈x

p(x) = |λ| p(x).

Moreover, for x, y ∈ E/M and any ε > 0 there are x ∈ x and y ∈ y such that p(x) ≤p(x) + ε and p(y) ≤ p(y) + ε and hence p(x+ y) ≤ p(x) + p(y) ≤ p(x) + p(y) + 2ε. Thisimplies that p(x+ y) ≤ p(x) + p(y).

If p is a seminorm on E then ϕ(p<1) = p<1. Hence, if p is continuous then p is continuousby Proposition 3.14.

Given a 0-neighborhood U in E/M we have p<ε ⊆ ϕ−1(U) for some p ∈ P, henceϕ(p<ε) = p<ε ⊆ ϕ(ϕ−1(U)) = U .

Proposition 6.14. If E is a metrizable LCS and M a closed subspace of E, the quotientspace M/E is metrizable. If in addition E is complete then M/E is complete.

Proof. Given a sequence of seminorms defining the topology of E, the quotient semi-norms define the topology of E/M and the first claim follows because E/M is Hausdorff(Proposition 5.4).

Now let P = pn | n ∈ N be a countable family of seminorms defining the topology ofE such that pn ≤ pn+1. By Proposition 5.8 it suffices to show sequential completeness.Let (xn)n be a Cauchy sequence in M/E. It suffices to show that a subsequence of(xn)n converges, so we can assume that

∑∞n=1 pn(xn − xn+1) < ∞. In fact, ∀i ∃Ni:

pi(xn − xm) ≤ 2−i for n,m ≥ Ni, and we can have Ni+1 ≥ Ni. Choose ni ≥ Ni for eachi. Then pi(xni

− xni+1) ≤ 2−i for all i, and we replace (xn)n by (xni

)i.

For given xn ∈ xn there is xn+1 ∈ xn+1 such that pn(xn − xn+1) ≤ 2pn(xn − xn+1).In fact, there is zn = x′n − x′n+1 ∈ xn − xn+1 for some x′n ∈ xn, x′n+1 ∈ x′n+1, suchthat pn(zn) ≤ 2pn(xn − xn+1). For xn+1 := x′n+1 + xn − x′n ∈ xn+1 we have pn(xn −xn+1) = pn(zn) ≤ 2pn(xn − xn+1). Consequently, we find a sequence xn ∈ xn such that∑pn(xn− xn+1) <∞. This implies that

∑p(xn− xn+1) <∞ for every p ∈P, so (xn)n

is a Cauchy sequence in E because

p(xn − xm) = p(m−1∑k=n

(xk − xk+1)) ≤m−1∑k=n

p(xk − xk+1).

Hence, xn has a limit x0 and xn = ϕ(xn)→ ϕ(x0).

Next, we consider the direct sum⊕

iEi of a family (Ei)i of locally convex spaces with theinductive locally convex topology with respect to the canonical injections ιi : Ei →

⊕iEi.

Proposition 6.15. If all Ei are Hausdorff then⊕

iEi is Hausdorff. In this case a subsetB ⊆

⊕iEi is bounded if and only if πi(B) = 0 except for a finite subset J ⊆ I and

πi(B) is bounded for i ∈ J .

Proof. The product topology is Hausdorff (Proposition 6.6) and the topology of⊕

iEiis finer than the topology induced by

∏iEi (πk ιj = 0 if k 6= j, = id if k = j, and is

continuous in both cases).

For the second claim, let B ⊆⊕

iEi be bounded. Because the projections πi are contin-uous, each πi(B) is bounded in Ei. Suppose J ⊆ I is infinite and πj(B) 6= 0 for j ∈ J ;take a sequence (ik)k of distinct indices in J . Choose a sequence (xk)k ⊆ B such thatπik(xk) 6= 0 for all k, and even (as Eik is Hausdorff) πik(xk) 6∈ kVk, where each Vk is someabsolutely convex 0-neighborhood in Eik . Let U = acx(

⋃i ιi(Ui)) be a 0-neighborhood in

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LOCALLY CONVEX SPACES 23⊕iEi (Proposition 6.9) with Uik = Vk for all k. Then πik(U) = Vk but πik(k−1xk) 6∈ Vk,

so k−1xk 6∈ U for all k, contradicting the boundedness of B by Proposition 5.2.

Conversely, given a 0-neighborhood U = acx(⋃i ιi(Ui)) there is λ > 0 such that πi(B) ⊆

λUi ⊆ πi(λU) for i ∈ J and even all i ∈ I, which implies B ⊆ λU .

Proposition 6.16. If all the Ei are complete then so is⊕

Ei.

Proof. We consider on E two topologies: T0, the trace topology of∏

iEi, and Ti, theinductive locally convex topology with respect to the injections ιj : Ej →

⊕iEi (Ti is

finer than T0).

Let F be a Cauchy filter on E with respect to Ti. Then πi(F ) converges to some xi ∈ Ei.Set ai := xi if xi 6∈ 0, and ai := 0 if xi ∈ 0. Set M := i ∈ I : ai 6= 0. Note thatalso for i ∈ I \M , πi(F ) → ai = 0: let U be an arbitrary 0-neighborhood in Ei, and Vone with V − V ⊆ U . Then ∃F ∈ F : πi(F ) ⊆ xi + V , and 0 ∈ xi + V , i.e., xi ∈ −V , sowe have πi(F ) ⊆ V − V ⊆ U , which means that πi(F )→ 0.

For each i ∈M choose an absolutely convex 0-neighborhood Wi in Ei such that ai 6∈ Wi,and for i ∈ I \ M let Wi be an arbitrary 0-neighborhood. Let A ∈ F be such thatA−A ⊆ acx(

⋃i ιi(Wi)), then for i ∈M , πi(A)− πi(A) ⊆ Wi, which implies ai− πi(A) ⊆

Wi. For (yi)i ∈ A and i ∈ M , yi ∈ ai −Wi 6= 0, so M is finite and (ai)i ∈ E. Becauseπi(F )→ ai we have F → a := (ai)i in T0.

Ti has a 0-basis U of T0-closed sets: let U = acx(⋃i ιi(Vi)) be a 0-neighborhood in Ti

and x = (xi)i ∈ E in the T0-closure of U . As H := i ∈ I : xi 6= 0 is finite, T0 andTi induce the same topology on

⊕i∈H Ei: we had T0 ≤ Ti above, and Ti ≤ T0 follows

from n−1(U1 × . . .× Un) ⊆ acx(ι1(U1) ∪ . . . ∪ ιn(Un)). Define P :⊕

i∈I Ei →⊕

i∈H Ei asthe canonical projection, and Q :

⊕i∈H Ei →

⊕i∈I Ei by Q(ιi(yi)) = ιi(yi). Both P and

Q are continuous with respect to Ti, and Q(P (U)) ⊆ U by the form U has. Then

x ∈ UT0 ⇒ P (x) ∈ P (U)⇒ x = Q(P (x)) ∈ Q(P (U))Ti ⊆ U

Ti.

Hence, the Ti and T0-closure of U are the same. Apply Corollary 2.5 to obtain the 0-basisU .

Now given U ∈ U , let A ∈ F be such that A− A ⊆ U , and fix y0 ∈ A. y0 −F is a Ti-Cauchy filter and, by the above, T0-convergent to y0− a; it contains the set y0−A ⊆ U ,and because U is T0-closed, y0 − a ∈ U . Then A ⊆ a + U , which gives F → a withrespect to Ti.

We summarize the permanence properties of being complete, Hausdorff or metrizable inthe following table.

complete Hausdorff metrizableproduct + (Proposition 6.7) + (Proposition 6.6) + if countable

(Proposition 6.5)

subspace + if closed(Proposition 4.4)

+ (Proposition 6.6) + (Corollary 6.3)

direct sum + (Proposition 6.16) + (Proposition 6.15) − ([S, Ex. 11, p. 70])

quotient + if E metrizableand M closed(Proposition 6.14)

+ if M closed(Proposition 6.10)

+ if M closed(Proposition 6.14)

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7. Projective and inductive limits

Recall: A set I is said to be directed by an order relation ≤ if ∀i, j ∈ I ∃k ∈ I:i ≤ k, j ≤ k.

Definition 7.1. Let (I,≤) be a directed set, (Ei)i∈I a family of TVS (LCS), and gij : Ej →Ei a family of continuous linear mappings for each pair (i, j) with i ≤ j such that gii = idfor all i and gij gjk = gik whenever i ≤ j ≤ k. Then (Ei, gij)(I,≤) is called a projectivesystem of TVS (LCS).

The projective limit lim←− gij(Ej) := lim←−Ej of such a system is defined to be the subspace

(xi)i ∈∏i

Ei | xi = gij(xj) whenever i ≤ j

endowed with the projective topology with respect to the restrictions Tk of the projectionsπk :

∏iEi → Ek.

Proposition 7.2. For any family of continuous linear mappings fi : F → Ei (where Fis a TVS (LCS)) (i ∈ I) such that fi = gij fj there is a unique continuous linear mapf : F → lim←−iEi such that fi = Ti f .

lim←−iEi together with the Ti is, up to linear homeomorphisms, unique with this property.

Proof. There is a unique linear continuous map f : F →∏

j Ej such that πi f = fi.

Then, f is linear and continuous into lim←−j Ej because πi(f(x)) = fi(x) = gij(fj(x)) =

gij(πj(f(x))).

For uniqueness: we have

∃|f : E → E : Ti f = Ti ∀i

∃|g : E → E : Ti g = Ti ∀i

⇒

Ti g f = Ti f = Ti, Ti id = Ti ⇒ g f = id

Ti f g = Ti g = Ti, Ti id = Ti ⇒ f g = id .

Remarks.

1. The topology of a projective limit equals the subspace topology induced by theproduct topology on

∏iEi.

2. If all Ei are Hausdorff, lim←−Ej =⋂i≤j(πi − gij πj)−1(0) is closed in

∏iEi.

3. If Ui is a 0-basis in Ei, the sets T−1i (Ui) with Ui ∈ Uj already form a 0-basisin lim←−Ej: let U =

⋂i∈J T

−1i (Ui) be a 0-neighborhood, J finite, Uj ⊆ Ej a 0-

neighborhood. There is k ∈ I with k ≥ i for all i ∈ J . Let Vk ⊆ Ek be a 0-neighborhood such that gik(Vk) ⊆ Ui for all i ∈ J . Then for i ∈ J , Ti(T

−1k (Vk)) =

gik(πj(T−1k (Vk))) ⊆ Ui, so T−1k (Vk) ⊆ U .

Corollary 7.3. The projective limit of a system of complete Hausdorff TVS is complete.

Proposition 7.4. Every Hausdorff LCS is linearly homeomorphic to a dense subspaceof the projective limit of Banach spaces.

Proof. Let U be a 0-basis of E consisting of absolutely convex sets. Then the corre-sponding family of gauges P = gU | U ∈ U is directed by ≤ because (U,⊇) is directedand U ′ ⊇ U if and only if gU ′ ≤ gU .

For p ∈P set Np := p−1(0), which is a linear subspace of E. Let

Φp : E → Ep := E/Np

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be the quotient map. For the quotient seminorm p we have

p(Φp(x)) = infy∈Np

p(x+ y) = p(x)

because z − z′ ∈ Np implies |p(z)− p(z′)| ≤ p(z − z′) = 0 and p(z) = p(z′). Clearly p isa norm on Ep and Φp is continuous. For p ≤ q, Nq ⊆ Np so the mapping

Φpq : Eq → Ep, Φq(x) 7→ Φp(x)

is well-defined, linear and continuous (p(Φpq(Φq(x))) = q(Φq(x))) and extends to a linearcontinuous map

Φpq : Eq → Ep

on the completions. Then (Ep, Φpq) is a projective system of LCS and we can form the

projective limit F := lim←−p Ep.

We define f : E → F by f(x) := (Φp(x))p, noting that for p ≤ q we have Φp(x) =Φpq(Φq(x)). Because each πp f = Φp is continuous, f is continuous. Conversely, let [31.3.]

U = p≤1 be a 0-neighborhood in E and set V = f(E) ∩∏

q∈P Vq with Vq = Eq for

q 6= p and Vp = p≤1. Then f(x) ∈ V implies p(x) ≤ 1, so f−1(V ) ⊆ U and f is ahomeomorphism.

Moreover, f is injective because Φp(x) = 0 for all p implies that p(x) = 0 for all p, hencex = 0 as E is Hausdorff.

Finally, to see that f(E) is dense in F , let (zp)p ∈ F and a 0-neighborhood U in F be

given. U contains a set V =∏

q Vq with Vq a 0-neighborhood in Eq, and Vq 6= Eq only for

q equal to some p. With any x ∈ E such that Φp(x) − zp ∈ Vp we have Φq(x) − zq ∈ Vqfor all q and hence f(x) ∈ (zp)p + U .

Corollary 7.5. A LCS is complete if and only if it is linearly homeomorphic to theprojective limit of Banach spaces.

Every Frechet space is linearly homeomorphic to the projective limit of a sequence ofBanach spaces.

Examples.

1. (N,≤) is directed, for n ≤ m the inclusion Cm(Ω) → Cn(Ω) is continuous. Wehave

lim←−m

Cm(Ω) =⋂m

Cm(Ω) = C∞(Ω).

2. Let K := K ⊆ X | K compact for X a locally compact topological space.Then (K ,⊆) is directed, as K1 ∪ K2 is compact. For K ⊆ L the restrictionρKL : C(L)→ C(K) is continuous and

lim←−K∈K

C(K) ∼= C(X).

The concept which is dual to projective limits is the following:

Definition 7.6. Let (I,≤) be a directed set, (Ei)i a family of LCS, and hji : Ei → Ej acontinuous linear mapping for each pair (i, j) with i ≤ j, such that hii = id for all i andhki = hkj hji whenever i ≤ j ≤ k. Then (Ei, hji)(I,≤) is called an inductive system ofLCS. The inductive limit lim−→Ei of such a system is defined to be the quotient space of⊕

i∈I Ei by the subspace

H := spanιi(xi)− ιj(hji(xi)) | i ≤ j, xi ∈ Ei,endowed with the quotient topology induced by the locally convex direct sum topology.

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Remark. By Proposition 6.8 (iii) the topology of the inductive limit equals the inductivelocally convex topology with respect to the maps hj := ϕ ιj, where ϕ is the quotientmapping and ιj : Ej →

⊕iEi the canonical inclusion.

The inductive limit has the following universal property:

Proposition 7.7. For any family of continuous linear mappings fi : Ei → F into a LCSF (i ∈ I) such that fi = fj hji there is a unique continuous linear map f : lim−→i

Ei → Fsuch that fi = f hi.lim−→i

Ei together with the hi is unique up to linear homeomorphisms in the following sense:

given any LCS E and mappings hi : Ei → E with the same property, there is a linear

homeomorphism f : lim−→iEi → E such that hi f = hi.

Proof. There is a unique linear continuous map f :⊕

iEi → F such that f ιi = fi.

Because H ⊆ ker f there is a unique linear continuous map f : lim−→iEi → F such that

f ϕ = f , so f hi = f ϕ ιi = f ιi = fi. Now if g hi = fi for some g,g ϕ ιi = fi ⇒ g ϕ = f ⇒ g = f .

Uniqueness follows as in Proposition 7.2.

For general inductive limits only few good results hold. We introduce the following:

Definition 7.8. An inductive system is called reduced if all hj : Ej → lim−→iEi are injec-

tive.

In this case, hi = hj hji implies that also all hji are injective.

Given a reduced inductive system, consider the space Fi := hi(Ei) with the direct imagetopology (i.e., hi is a homeomorphism). For i ≤ j and hi(xi) ∈ Fi, hi(xi) = hj(hji(xi)) ∈Fj, so Fi ⊆ Fj and this inclusion ιji := hj hij h−1i is continuous. The family (Fi, ιji)iwith these inclusion mappings forms an inductive system.

By Proposition 7.7, (lim−→iEi, hi) ∼= (lim−→i

Fi, hi hi) (where hi : Fi → lim−→jFj) and as hi hi

is injective, (Fi, ιji)i is reduced. However, because of its simpler structure lim−→iFi can be

described in a nice way as follows:

Proposition 7.9. Let E be a vector space and (Ei)i a family of LCS which are subspacesof E such that

⋃iEi = E. Suppose that for i ≤ j, Ei ⊆ Ej and the inclusion mapping

hji : Ei → Ej is continuous. Then lim−→Ei ∼= (E,Ti), where Ti is the inductive locallyconvex topology w.r.t. the inclusions Ei → E, and the inductive system is reduced.

Proof. We show that E has the universal property of Proposition 7.7. Given mappingsfi : Ei → F compatible with the inclusions, define f : E → F by f |Ei

:= fi. Then f iswell-defined, continuous and the unique linear continuous map whose restriction to Ei isfi.

In practice one starts with a space (⋃iEi,Ti) and establishes its properties (completeness,

...) via results about inductive limits.

If the index set of a reduced inductive system is N with its natural order we call it aninductive sequence. An inductive sequence and its inductive limit are called strict if eachEn is a topological subspace of Ek whenever n ≤ k.

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Definition 7.10. A LCS is called (LB)-space or (LF)-space, respectively, if it can begiven by a strict inductive limit of a strictly increasing sequence of Banach spaces orFrechet spaces, respectively.

Lemma 7.11. Let E be a LCS, M a subspace of E and U an absolutely convex 0-neighborhood in M . Then there is an absolutely convex 0-neighborhood V in E withU = V ∩M . For x0 ∈ E \ M , V can be chosen such that x0 6∈ V .

Proof. Let W be an absolutely convex 0-neighborhood in E such that W ∩M ⊆ U andset V := acx(W ∪U). Then V ∩M = U : U ⊆ V ∩M is clear; for V ∩M ⊆ U , take somez ∈ V ∩M , which is of the form z = αw + βu with |α| + |β| ≤ 1, w ∈ W and u ∈ U .Then αw = z − βu ∈M means that either α = 0 or w ∈M , which both implies z ∈ U .

For x0 6∈ M , W can be chosen such that (x0 +W ) ∩M = ∅. Then x0 6∈ V because x0 =αw+βu ∈ V would imply x0−αw = βu ∈M and x0−αw ∈ x0 +W , contradiction.

For the remainder of this section let (E,T ) = lim−→En be the strict inductive limit of aninductive sequence (En,Tn)n of LCS.

Theorem 7.12. T induces Tn on each En.

Proof. The inclusion En → E is continuous, so Tn is finer than the topology T ′n induced

by T on En. Let Vn ⊆ En now be an absolutely convex 0-neighborhood in Tn. ByLemma 7.11 we can construct a sequence of absolutely convex 0-neighborhoods Vn+kin En+k such that Vn+k+1 ∩ En+k = Vn+k (k = 0, 1, 2, . . . ,). Then V :=

⋃k≥0 Vn+k is

absolutely convex and absorbent, V ∩ Ei is a 0-neighborhood in Ei for all i, so V is a0-neighborhood in lim−→Ei (Proposition 6.8 (ii)). V ∩En = Vn is a 0-neighborhood in T ′

n ,so T ′

n is finer than Tn.

Corollary 7.13. If each En is Hausdorff, lim−→En is Hausdorff.

Proof. For x 6= 0 there is some n with x ∈ En and hence there is an absolutely convex0-neighborhood Vn in En with x 6∈ Vn. By Theorem 7.12 there is a 0-neighborhood V inE such that V ∩ En = Vn, x 6∈ V , so E is Hausdorff.

Definition 7.14. An inductive limit is called regular if every bounded set in lim−→Ei iscontained and bounded in some Ei.

Theorem 7.15. If En is closed in En+1, the inductive limit is regular.

Proof. Suppose a set B is not contained in any En and choose a sequence (xn)n in Bwith xn 6∈ En. Choose a subsequence (yk)k of xk and a strictly increasing sequence (nk)kin N such that yk ∈ Enk+1

\ Enk. Take an increasing sequence (Vk)k of absolutely convex

sets such that Vk is a 0-neighborhood in Enk, Vk+1 ∩ Enk

= Vk and k−1yk 6∈ Vk+1. ThenV =

⋃k Vk is a 0-neighborhood in lim−→Ei and k−1yk 6∈ V for all k, so k−1yk does not

converge to 0 and B cannot be bounded by Proposition 5.2.

Hence, a bounded set B is countained in some En. If Vn is an absolutely convex 0-neighborhood in En, by Theorem 7.12 and Lemma 7.11 there is an absolutely convex0-neighborhood V in E with V ∩En = Vn, and B ⊆ λV for some λ implies B ⊆ λVn.

Theorem 7.16. If all En are complete then so is E = lim−→nEn.

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Proof. If F is a Cauchy filter in E and U the neighborhood filter of 0, F + U is aCauchy filter basis of F (cf. the proof of Proposition 4.11). We claim that the trace ofF + U on some En0 is a filter basis. Otherwise, we would have a decreasing sequenceFn ∈ F and a decreasing sequence of absolutely convex 0-neighborhoods Wn in E suchthat (Fn +Wn) ∩ En = ∅ for all n. Then U = acx(

⋃n(Wn ∩ En)) is a 0-neighborhood in

E (Proposition 6.9 and Theorem 7.12), and (Fn +U)∩En = ∅ for all n: supposing thereis some y ∈ (Fn + U) ∩En, y = zn +

∑λixi, with xi ∈ Wi ∩Ei (i = 1 . . . p) and zn ∈ Fn,

which gives

y −∑i≤n

λixi = zn +∑i>n

λixi.

Because Wi ⊆ Wn for i > n and Wn is absolutely convex, zn +∑

i≥n λixi ∈ Fn +Wn, buty −

∑i≤n λixi ∈ En, which contradicts (Fn +Wn) ∩En = ∅, so (Fn + U) ∩En = ∅ for all

n. Because F is a Cauchy filter, F −F ⊆ U for some F ∈ F . Take w ∈ F , then w ∈ Ekfor some k. For v ∈ Fk ∩ F , w = v + (w − v) ∈ v + (F − F ) ⊆ Fk + U , which gives acontradiction.

Hence, the trace of F +U on En0 is a Cauchy filter basis and converges, and F convergesin E.

Corollary 7.17. Every (LF)-space is complete.

Example. For Ω ⊆ Rn open, D(K) := f ∈ C∞(Ω) | supp f ⊆ K. If (Kn)n is asequence of compact sets with

⋃nKn = Ω and Kn ⊆ Kn+1, we set D(Ω) := lim−→D(Kn),

which is an (LF)-space.

8. Finite-dimensional and locally compact TVS

Lemma 8.1. If a TVS E is the algebraic direct sum of subspaces M and N , then E =M⊕N topologically if and only if the canonical mapping v : E/M → N is an isomorphismof TVS.

Proof. Note that M ⊕N → E is always continuous; its inverse is continuous if and onlyif the projections πM and πN are continuous.

With ϕ : E → E/M and πN : E → N the canonical mappings, πN = v ϕ.

“⇒”: πN continuous, ϕ open⇒ v continuous (U ⊆ N open⇒ π−1N (U) open⇒ ϕ(π−1N (U))open and v(ϕ(π−1N (U))) = πN(π−1N (U)) = U). πN open, ϕ continuous ⇒ v open (U ⊆E/M open ⇒ v(U) = v(ϕ(ϕ−1(U))) = πN(ϕ−1(U)) open).

“⇐”: v and ϕ continuous ⇒ πN and πM = (id−πN) continuous.

Theorem 8.2. If E is a Hausdorff TVS with dimE = n then E ∼= Kn.

For each basis x1, . . . , xn of E, the map (λ1, . . . , λn) 7→ λ1x1 + . . . + λnxn is a linearhomeomorphism of Kn onto E, and all linear homeomorphisms are of this form.

Proof. We perform induction over n.

n = 1: λ 7→ λx1 is continuous and an algebraic isomorphism of K onto E. For continuityof λx1 7→ λ at 0, let 0 < ε < 1 and choose a balanced 0-neighborhood V such thatεx1 6∈ V . Then λx1 ∈ V implies |λ| < ε.

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Now suppose the claim holds for dimension n − 1. If x1, . . . , xn is any basis of E, letM = spanx1, . . . , xn−1 and N = spanxn. By induction hypothesis,

(λ1, . . . , λn−1) 7→n−1∑i=1

λixi

is a linear homeomorphism Kn−1 → M . Because Kn−1 is complete, M is complete andhence closed in E, so E/M is Hausdorff and has dimension 1, and it is isomorphic to K.The canonical map E/M → N is the composition

E/M → K→ N

and hence an isomorphism, so by Lemma 8.1, Kn ∼= Kn−1 ⊕K ∼= M ⊕N = E.

It is clear that every linear homeomorphism is of this form. [7.4.]

Definition 8.3. A Hausdorff TVS E is called locally compact if it has a compact 0-neighborhood

The class of locally compact TVS is rather restrictive, as the following result shows:

Theorem 8.4. A Hausdorff TVS E is locally compact if and only if it is finite-dimensional.

Proof. If dimE = n ∈ N, E ∼= Kn. A 0-basis of Kn is given by products of closed boundedintervals, which are compact.

Conversely, assume that E is locally compact and let V be a balanced compact neighbor-hood of 0. We show that 2−nV : n ∈ N is a 0-basis. Let B be any 0-neighborhood andchoose a balanced 0-neighborhood U such that U + U ⊆ B. Since V is compact, thereare xi ∈ V (i = 1 . . . k) such that V ⊆

⋃ki=1(xi +U), and there is λ ∈ K with 0 < |λ| < 1

such that λxi ∈ U for all i. There exists n ∈ N such that 2−n ≤ |λ|, and

2−nV ⊆ λV ⊆k⋃i=1

(λxi + λU) ⊆ U + U ⊆ B.

Since V is compact there is a finite subset D = x1, . . . , xn ⊆ E such that V ⊆ D+ 12V .

Let M := spanD, which is closed because it is finite dimensional and hence complete.Since V ⊆M + 1

2V and tM = M for all t ∈ K \ 0, we have

1

2V ⊆M +

1

4V,

V ⊆M +1

2V ⊆M + (M +

1

4V ) = M +

1

4V,

V ⊆⋂m∈N

(M + 2−mV ) = M = M

by induction, and M = E because V is absorbent.

9. The theorem of Hahn-Banach

We recall some facts about hyperplanes from linear algebra. If E is a vector space, a linearsubspace H0 ⊆ E with codimH0 = dimE/H0 = 1 is called a hyperplane. If f : E → K islinear and f 6= 0 then f−1(0) is a hyperplane: dimE/ ker f = dim im f = 1. Conversely, ifH0 is a hyperplane it is given by f−1(0) for some nontrivial linear f : E → K: let v ∈ E/H0

be nonzero and ` : E/H0 → K, λv 7→ λ the coefficient map. Then for f := ` ϕ, whereϕ : E → E/H0 is the quotient map, we have ker f = kerϕ = H0.

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If x0 ∈ E and H0 is a hyperplane in E then H = x0 + H0 is called an affine hyperplane.For a nontrivial linear f : E → K and α ∈ K, f−1(α) is an affine hyperplane: givenany x0 ∈ f−1(α) we have f−1(α) = x0 + ker f . Conversely, if H = x0 + H0 is an affinehyperplane then H0 = f−1(0) for some f and H = f−1(0) + x0 = f−1(f(x0)).

Lemma 9.1. Let E be a TVS and f a nontrivial linear form on E. Then the followingare equivalent:

(i) f is continuous.(ii) ker f is closed in E.

(iii) ker f is not dense in E.

Proof. (i) ⇒ (ii) ⇒ (iii) is clear. For (iii) ⇒ (i), let x ∈ int(E \ ker f) and choose abalanced 0-neighborhood U in E such that (x + U) ∩ ker f = ∅. Then f(x) 6∈ f(U),

which implies f(U) ⊆ |f(x)| · D; in fact, |λ| ≥ |f(x)| 6= 0 for some λ ∈ f(U) would imply

f(x) = f(x)λ· λ ∈ D · f(U) ⊆ f(U) because f(U) is balanced. This implies continuity of

f .

Let E be a vector space, A ⊆ E convex and nonempty, and H = f−1(α) a hyperplanesuch that A ∩ H = ∅. Then either A ⊆ x : f(x) < α or A ⊆ x : f(x) > α becausef(A) is convex and does not contain α. The geometric form of the Hahn-Banach theoremstates the following.

Theorem 9.2. Let E be a TVS, A ⊆ E open, convex and nonempty, and M an affinesubspace of E with M ∩ A = ∅. Then there is a closed affine hyperplane H in E suchthat M ⊆ H and H ∩ A = ∅.

Proof. By applying a translation we can assume that 0 ∈M . It suffices to show the caseM = 0. In fact, supposing the theorem is known for this case, let ϕ : E → E/M be thequotient map. Then ϕ(A) is open, convex, and 0 6∈ ϕ(A). By assumption there exists aclosed hyperplane H1 ⊆ E/M such that H1 ∩ ϕ(A) = ∅. Then H := ϕ−1(H1) is a closedhyperplane in E (H1 = ker f ⇒ H = ker(f ϕ), and f ϕ is nontrivial because ϕ issurjective and f is nontrivial). Because ϕ(H ∩ A) ⊆ ϕ(H) ∩ ϕ(A) = H1 ∩ ϕ(A) = ∅ wehave H ∩ A = ∅.For the case M = 0 we first assume that K = R and introduce some terminology. Asubset A ⊆ E is called a cone with vertex at x0 if ∀x ∈ C : x0 + λ(x− x0) | λ > 0 ⊆ C.It is called pointed if x0 ∈ C and blunt if x0 6∈ C. A subset C ⊆ E is a convex conewith vertex at 0 if and only if λC ⊆ C for all λ > 0 and C + C ⊆ C. In fact, the firstcondition is the definition of a cone with vertex at 0. For the second, if C is convex thenC+C = 2(1

2C+ 1

2C) = 2C ⊆ C. Conversely, if C+C ⊆ C holds then αx+βy ∈ C+C ⊆ C

for a convex combination of x, y ∈ C.

Fix any interior point a of A. Let C be the set of all convex pointed cones C with vertexat 0 such that A ⊆ C and −a 6∈ C. Then C contains B :=

⋃λ≥0 λA, so it is not empty.

Order C by inclusion. Then every nonempty totally ordered subset C0 of C has an upperbound, obtained as C0 :=

⋃C : C ∈ C0. In fact, λC0 ⊆ C0 is clear, and for x1 ∈ C1

and x2 ∈ C2 we have either C1 ⊆ C2 or C2 ⊆ C1, which in both cases gives C0 +C0 ⊆ C0.A ⊆ C0 and −a 6∈ C0 is clear. By Zorn’s Lemma there is a maximal element C of C .

We will show that H := C ∩ (−C) is a closed hyperplane such that H ∩ A = ∅.First, we show that C ∪ (−C) = E. If x ∈ E were such that x 6∈ C and x 6∈ −C, thenλx 6∈ C for all λ 6= 0. Set C1 := λx+y | λ ≥ 0, y ∈ C. Then we have λC1 ⊆ C1 ∀λ > 0,C1+C1 ⊆ C1, A ⊆ C and −a 6∈ C1 because λx+y = −a would imply C 3 y = −a 6∈ C for

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λ = 0 and C 63 −λx = y+a ∈ C for λ > 0, both a contradiction. Hence, C1 ∈ C . BecauseC ⊆ C1 and x ∈ C1 \ C this contradicts maximality of C, so necessarily C ∪ (−C) = Eholds.

We now call a point x ∈ C an internal point of C if ∀y ∈ E there is some τ0 > 0 suchthat x+ τy ∈ C for all τ ∈ R with |τ | < τ0.

We claim that if x ∈ C is not internal then x ∈ −C. Assume to the contrary thatx 6∈ −C, or −x 6∈ C. If x is not internal there is y ∈ E such that x + τy 6∈ C for a0-sequence of τ 6= 0, and (by convexity of C) for all τ > 0. We now set

C1 := w − λx | w ∈ C, λ ≥ 0.λC1 ⊆ C1, C1 + C1 ⊆ C1 and A ⊆ C are clear. We have y 6∈ C because y ∈ C wouldimply x + τy ∈ C for all τ > 0. Moreover, y ∈ C1 would imply y = w − λx withλ > 0 and x + λ−1y = λ−1w ∈ C, which contradicts the choice of y, so necessarily wehave y 6∈ C1 and C1 6= E. Now in order to see that −a 6∈ C1, take any b ∈ E. Thena + λ(b − a) ∈ A ⊆ C1 for |λ| ≤ λ0 (A is open), and the assumption −a ∈ C1 wouldimply λ(b − a) ∈ C1 and hence b ∈ C1 for all b ∈ E, which contradicts C1 6= E, so wehave −a 6∈ C1. Hence, C1 ∈ C .

Now C ⊂ C1 ∈ C and −x ∈ C1 \ C contradicts maximality of C, so x ∈ −C has to holdfor every x ∈ C which is not internal.

Let now y ∈ C, z ∈ −C and set

x(τ) := (1− τ)y + τz (0 ≤ τ ≤ 1).

We then claim that x(σ) ∈ H for some σ. Set σ := supτ : x(τ) ∈ C. Then x(τ) ∈ Cfor τ < σ and x(τ) 6∈ C for τ > σ. If now x(σ) ∈ C then x is not an internal point of Cand hence x ∈ −C by what was shown above. If, on the other hand, we have x(σ) ∈ −Cthen we have either x(τ) ∈ −C for some τ < σ or x(τ) 6∈ −C for all τ < σ. In thefirst case we have x(τ) ∈ H; in the second case, x(σ) is not an internal point of −C, i.e.,−x(σ) is not an internal point of C, which means that x(σ) ∈ H.

In order to see that H is a hyperplane, we will show that H ⊕ Ra = E. H is a linearsubspace because it is closed under addition, reflection x 7→ −x and multiplication bynonnegative reals. H ∩ Ra = 0 is clear from −a 6∈ C. Finally, if x ∈ C then the linesegment

x(τ) = (1− τ)x+ τ(−a)

contains a point y = x(σ) ∈ H with σ 6= 1 because −a 6∈ C, which gives

x =y

1− σ+

σ

1− σa.

Similarly, for x ∈ −C the line segment

x(τ) = (1− τ)a+ τx

contains a point y = x(σ) ∈ H with σ 6= 0, which gives

x =y

σ+σ − 1

σa.

The complement of H contains the open set A, so H cannot be dense and hence is closedby Lemma 9.1. To see that A ∩H = ∅, we first note that every point of A is an internalpoint of C. It hence suffices to prove that H contains no internal points of C. For anyx ∈ H we have x− λa 6∈ C for all λ > 0 because x− λa = y ∈ C would give

−a = λ−1(y − x) ∈ C.

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This complete the proof for the case of real scalars.

For complex scalars, let E0 be the real vector space underlying the complex vector spaceE, with the trace topology. Then E0 is a TVS and M a linear subspace of E0. Let H0

be a closed hyperplane in E0 such that M ⊆ H0 and H0 ∩A = ∅. We set H = H0 ∩ iH0.Then by Lemma 9.3, H is a closed hyperplane in E such that we have M = M ∩ iM ⊆ Hand H ∩ A = ∅.

Lemma 9.3. Let E be a complex vector space and E0 the underlying real vector space.If H0 is a hyperplane of E0 with equation f0(x) = 0 then H = H0 ∩ iH0 is a hyperplaneof E with equation f(x) = f0(x)− if0(ix) = 0.

Proof. We show that H = ker f . For x ∈ H0 ∩ iH0, we have x = iy for some y ∈ H0 andhence f(x) = f0(x) − if0(−y) = 0. Conversely, if f0(x) − if0(ix) = 0 then f0(x) = 0,which implies x ∈ H0, and −f0(ix) = f0(−ix) = 0, which implies −ix ∈ H0 and hencei(−ix) = x ∈ iH0.

The analytic form of the Hahn-Banach theorem is the following.

Theorem 9.4. Let E be a vector space, q a seminorm on E and M a linear subspace ofE. If f is a linear form on M satisfying |f(x)| ≤ q(x) for all x ∈ M then there exists alinear form g on E such that g(x) = f(x) for all x ∈M and |g(x)| ≤ q(x) for all x ∈ E.

Proof. First suppose that K = R. Then q defines a locally convex topology on E suchthat A := q<1 is open, convex and nonempty. The case f = 0 is trivial, so we assumef 6= 0. The set N := f−1(1) is an affine hyperplane in M (and an affine subspace of E)such that N ∩ A = ∅. By Theorem 9.2 there is an affine hyperplane H in E containingN such that H ∩ A = ∅, and we have H = g−1(1) for some linear form g : E → R.

For y ∈ N , f(y) = g(y) = 1; for z ∈M and any x0 ∈ N we have the implication

f(z) = 0⇒ z + x0 ∈ N ⊆ H ⇒ g(z) = 0.

For any x ∈M and y ∈ N we have x = f(x)y+(x−f(x)y) with x−f(x)y ∈ ker f ⊆ ker g,which implies g(x) = f(x).

If x ∈ A then g(x) < 1 because g(x) = 1 would mean x ∈ H, but H ∩ A = ∅. Hence,g(x) = 1 implies q(x) ≥ 1. For g(x) = 0, |g(x)| ≤ q(x). For g(x) 6= 0, x = g(x) · x

g(x)and

g(x) = g(x)g( xg(x)

) ≤ |g(x)| q( xg(x)

) = q(x). Moreover, −g(x) = g(−x) ≤ q(−x) = q(x), so

|g(x)| ≤ q(x) for all x ∈ E.

In the complex case, let E0 be the real vector space underlying E and M0 the set M , asa subspace of E0. Set f1(x) = <f(x), where < denotes the real part; then f1 : M0 → Ris a linear form and for x ∈ E we have f(x) = f1(x) − if1(ix). For x ∈ M0, |f1(x)| ≤|f(x)| ≤ q(x). By the real case there is a linear form g1 : E0 → R with g1(x) = f1(x) onM0 and |g1(x)| ≤ q(x) for x ∈ E0. With g(x) := g1(x)− ig1(ix) for x ∈ E we then haveg(x) = f(x) on M and g(ix) = ig(x), so g is a linear form on E. For each x ∈ E we thenhave g(x) = ρeiθ for some ρ ≥ 0, which gives

|g(x)| = ρ = e−iθg(x) = g(e−iθx) = g1(e−iθx) ≤ q(e−iθx) = q(x).

We list some consequences.

Proposition 9.5. Let M be a linear subspace of a LCS E and f a continuous linearform on M . Then there is a continuous linear form g on E such that g(x) = f(x) forx ∈M .

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Proof. There is a continuous seminorm q on E such that |f(x)| ≤ q(x) for all x ∈M . ByTheorem 9.4 there is a linear form g on E with g(x) = f(x) for x ∈M and |g(x)| ≤ q(x)for x ∈ E, i.e., g is continuous.

Proposition 9.6. Let M be a closed linear subspace of a LCS E.

(i) For z ∈ E \M there is a continuous linear form f on E such that f(z) = 1 andf(x) = 0 for x ∈M .

(ii) M is the intersection of all closed hyperplanes containing it.

Proof. (i): Let ϕ : E → E/M be the canonical surjection. Then z := ϕ(z) 6= 0 andthere is a continuous seminorm q on E/M such that q(z) 6= ∅. Define g : Kz → K by

g(λz) = λ. Then |g(λz)| = |λ| = q(λz)q(z)

, so by Theorem 9.4 there is a linear form h on

E/M such that |h(x)| ≤ 1q(z)

q(x) and h(z) = 1; then f := hϕ is a linear form on E such

that f(z) = h(z) = 1 and f(x) = h(0) for x ∈M . [5.5.]

(ii): Let H be the collection of all closed hyperplanes in E containing M . Then M ⊆⋂H∈HH and for z ∈ E \M there is, by (i), a linear form f ∈ E ′ with f(z) = 1 and

M ⊆ ker f . Hence, ker f ∈ H but z 6∈ ker f .

Definition 9.7. If E is a TVS then

E ′ := f : E → K | f linear and continuousis called the (topological) dual of E.

Corollary 9.8. A LCS E is Hausdorff if and only if ∀x ∈ E \ 0 there is u ∈ E ′ suchthat u(x) 6= 0.

Proof. If E is Hausdorff then 0 = 0 =⋂keru | u ∈ E ′ by Proposition 2.12 and

Proposition 9.6 (ii), so for x 6= 0 there is u ∈ E ′ such that u(x) 6= 0.

Conversely, if u(x) 6= 0 there is a balanced 0-neighborhood V such that 0 6∈ u(x + V ),i.e., x 6∈ V , so E is Hausdorff by Proposition 2.11.

Lemma 9.9. If p is a continuous seminorm on a TVS E and x ∈ E, then ∃u ∈ E ′ suchthat |u(y)| ≤ p(y) for all y ∈ E and u(x) = p(x).

Proof. If p(x) = 0 set u = 0. If p(x) 6= 0 set G := ker p + spanx and choose v ∈ G∗such that ker v = ker p and v(x) = p(x). Then for y ∈ G, |v(y)| ≤ |p(y)| and Theorem 9.4gives the claim.

Proposition 9.10. Let E be a TVS. Then E ′ 6= 0 if and only if there is a convex0-neighborhood U 6= E.

Proof. If U ∈ E ′ \ 0 then U := u−1(D) = x ∈ E : |u(x)| ≤ 1 is a convex 0-neighborhood and U 6= E.

Conversely, if a convex 0-neighborhood U is given, U contains an absolutely convex 0-neighborhood V . The gauge qV of V is a continuous seminorm (Proposition 3.11). ByLemma 9.9, for x ∈ E \ V there is u ∈ E ′ with u(x) = qV (x). Because qV (x) = 0 wouldimply x ∈ V we have qV (x) > 0 and hence u 6= 0.

Finally, we have the following separation theorem:

Proposition 9.11. Let E be a LCS over R, A,B ⊂ E disjoint, convex and nonempty.If A is open there is u ∈ E ′ and α ∈ R such that u(x) < α ≤ u(y) for (x, y) ∈ A×B. IfB is open as well, one can have u(x) < α < u(y) for (x, y) ∈ A×B.

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Proof. A − B is open, convex, nonempty and does not contain 0, so by Theorem 9.2there is v ∈ E ′ \ 0 such that ker v ∩ (A − B) = ∅. Because v(A) and v(B) are non-empty, disjoint and convex subsets of R (i.e., intervals) we can find α ∈ R such thatu(x) ≤ α ≤ u(y) for (x, y) ∈ A × B, where u is v or −v. u is surjective and open (uinduces a map E/ keru→ R which is a linear homeomorphism by Theorem 8.2). Becauseu(A) is open we have u(x) < α for x ∈ A. Similarly, if B is open then α < u(y) fory ∈ B.

Corollary 9.12. In a real locally convex space, every closed convex subset A is the in-tersection of the closed half-spaces f(x) ≤ α containing it (f ∈ E ′, α ∈ R).

Proof. For x 6∈ A, there is a closed affine hyperplane H such that x and A lie on differentsides of H.

10. Dual Pairings

Definition 10.1. If E,F are vector spaces and B : E × F → K a bilinear form we saythat E and F are paired or form a pairing with respect to B. B is called nondegenerateif it separates points of E and F , i.e.,

B(x, y) = 0 ∀y ∈ F ⇒ x = 0,

B(x, y) = 0 ∀x ∈ E ⇒ y = 0.

In this case we call the pairing separated and (E,F ) a dual system with respect to B.

Examples.

1. The canonical bilinear form E × E∗ → K, (x, x′) 7→ 〈x, x′〉 = 〈x′, x〉 = x′(x) isnondegenerate: x′(x) = 0 for all x means that x′ = 0; if x′(x) = 0 for all x′, let(bi)i be a basis of E and write x =

∑xibi; defining x′j(bi) := δij we have that

x′j(x) = xj = 0 for all j and hence x = 0.2. If E is a TVS, E ′ := x′ : E → K | x′ is linear and continuous is the topological

dual of E. The restriction of 〈., .〉 : E×E∗ → K to E×E ′ separates points of E ′.If E is locally convex and Hausdorff, 〈., .〉 separates points of E (Proposition 9.6(i) with M = 0).

Suppose that E and F are paired with respect to B and consider the mappings

Ψ: F → E∗, y 7→ [x 7→ B(x, y)],

Φ: E → F ∗, x 7→ [y 7→ B(x, y)].

If B separates points of F , Ψ is injective and we can identify F with the linear subspaceim(F ) of E∗. In this case, B is the restriction of the canonical bilinear form (x, y) 7→ 〈x, y〉on E × E∗ and we write 〈x, y〉 instead of B(x, y).

Similarly, if B separates points of E then Φ is injective and E ⊆ F ∗.

We have seen that if E is a Hausdorff LCS then (E,E ′) is a dual system. We will see thatfor every dual system (E,F ) one can find a topology T on E such that (E,T )′ = F .

Definition 10.2. Let E,F be paired with respect to a bilinear form B. The weak topologyσ(E,F ) on E is the locally convex topology defined by the family of seminorms

x 7→ |B(x, y)| : y ∈ F.

We see that σ(E,F ) is Hausdorff if and only if ∀x ∈ E \ 0 ∃y ∈ F : |B(x, y)| 6= 0(Proposition 3.18), i.e., if and only if B separates points of E.

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Lemma 10.3. Let E,F be paired w.r.t. B such that B separates points of E. ConsideringE as a subspace of F ∗ and the latter as a subspace of KF , σ(E,F ) is the topology inheritedfrom the product topology on KF .

Proof. A subbasis of the product topology is given by the collection of all sets of theform f : F → K | f(y) ∈ εD for y ∈ F , ε > 0, which in E correspond to the setsx ∈ E : |B(x, y)| < ε.

Lemma 10.4. Let E be a vector space and u, u1, . . . , un ∈ E∗. Then u ∈ spanu1, . . . , un ⇔keru ⊇

⋂ni=1 kerui.

Proof. “⇒” is clear. Conversely, define T : E → Kn by T (x) := (〈ui, x〉)i=1,...,n andv : imT → K by v(T (x)) := u(x), which is well-defined by assumption. Extend v to alinear form v ∈ (Kn)∗ such that vT = u and choose (νi)i ∈ Kn such that v(α1, . . . , αn) =∑νiαi. Then u(x) = v(T (x)) =

∑νi〈ui, x〉 = 〈

∑νiui, x〉, so u =

∑νiui.

Proposition 10.5. Let E,F be paired with respect to B. Then for all y ∈ F , Ψ(y) =B(., y) ∈ (E, σ(E,F ))′; conversely, for each u ∈ (E, σ(E,F ))′ there is y ∈ F such thatu(x) = B(x, y) ∀x ∈ E.

In particular, Ψ: F → (E, σ(E,F ))′ is surjective and F/ ker Ψ ∼= (E, σ(E,F ))′. If Bseparates points of F then F = (E, σ(E,F ))′.

Proof. Continuity of B(., y) is clear. For u ∈ (E, σ(E,F ))′ there are y1, . . . , yn ∈ Fsuch that |u(x)| ≤ maxi |B(x, yi)|. With ui := Ψ(yi) ∈ E∗ we have that ui(x) = 0 forall i implies u(x) = 0, so by Lemma 10.4 u =

∑i νiui for some νi ∈ K, which gives

u(x) = B(x,∑

i νiyi).

Proposition 10.6. Let E,F form a pairing which separates points of E (i.e., E ⊆ F ∗).Then the pairing separates points of F if and only if E is dense in (F ∗, σ(F ∗, F )).

Proof. If E is dense and y ∈ F satisfies 〈x, y〉 = 0 for all x ∈ E ⊆ F ∗ then this also is thecase for all x ∈ F ∗ because 〈., y〉 is σ(F ∗, F )-continuous. Conversely, if E is not densetake x∗ ∈ F ∗ \ E. Then by Proposition 9.6 (i) there is an element f ∈ (F ∗, σ(F ∗, F ))′

given by f(y∗) = 〈y∗, y〉 for some y ∈ F and such that 〈x, y〉 = f(x) = 0 for x ∈ E but〈x∗, y〉 = f(x∗) 6= 0, i.e., y 6= 0, which implies that the pairing does not separate pointsof F .

Proposition 10.7. Let E,F form a pairing with respect to B which separates points ofF . If N is a proper subspace of F then σ(E,F ) is strictly finer than σ(E,N).

Proof. For y ∈ F \ N , B(., y) is σ(E,F )-continuous. If B(., y) was σ(E,N)-continuous,Proposition 10.5 would give some z ∈ N such that B(., y) = B(., z), i.e., B(x, y − z) = 0for all x ∈ E, which would imply y = z.

Proposition 10.8. Let (E,F ) be a dual system. Then (E, σ(E,F )) is complete if andonly if E = F ∗.

Proof. If we consider F ∗ as a subspace of KF then σ(F ∗, F ) is the topology inheritedfrom the product topology on KF ; the latter is complete.

Take f in the closure of F ∗ in KF . For x, y ∈ F , ρ ∈ K and ε > 0 there is x∗ ∈ F ∗ suchthat

|f(z)− x∗(z)| ≤ minε3,

ε

3 |ρ|+ 1

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for z ∈ x, y, ρx+ y. Then we have

|f(ρx+ y)− ρf(x)− f(y)| ≤ |f(ρx+ y)− 〈x∗, ρx+ y〉|+ |ρ〈x∗, x〉 − ρf(x)|+ |〈x∗, y〉 − f(y)| ≤ ε

so f ∈ F ∗, and F ∗ is closed and hence complete.

If (E, σ(E,F )) is complete then E = F ∗ follows from Proposition 10.6.

11. Polarity

Definition 11.1. Let E,F be vector spaces paired with respect to B. The polar of asubset A ⊆ E is the set

A = y ∈ F : |B(x, y)| ≤ 1 ∀x ∈ A.

Proposition 11.2. (i) A1 ⊆ A2 ⇒ A1 ⊇ A2,

(ii) A = A,(iii) A ⊆ A := (A),(iv) A = A,(v) A is absolutely convex and σ(F,E)-closed,

(vi) (λA) = 1λA for λ ∈ K \ 0,

(vii) A is absorbent if and only if A is σ(E,F )-bounded, and(viii)

(⋃i∈I Ai

)=⋂i∈I A

i .

Proof. (i): y ∈ A2 ⇒ |B(x, y)| ≤ 1 ∀x ∈ A2 ⇒ |B(x, y)| ≤ 1 ∀x ∈ A1 ⇒ y ∈ A1.(ii): A ⊆ D · A ⇒ (D · A) ⊆ A. Let y ∈ A; for all λ ∈ D and x ∈ A, |B(λx, y)| ≤|λ| · |B(x, y)| ≤ 1, so y ∈ (D · A).

(iii): x ∈ A ⇒ |B(x, y)| ≤ 1 ∀y ∈ A ⇒ x ∈ A.(iv): A ⊆ A ⇒ A ⊆ A by (i) and A ⊆ A by (iii).

(v): Let y, z ∈ A and λ, µ ∈ K with |λ|+|µ| ≤ 1. Then for all x ∈ A: |B(x, λy + µz)| ≤ 1,so λy + µz ∈ A. For all x ∈ E, B(x, .) is σ(F,E)-continuous (Proposition 10.5), soA =

⋂x∈AB(x, .)−1(D) is σ(F,E)-closed.

(vi): y ∈ (λA) ⇔ ∀x ∈ λA : |B(x, y)| ≤ 1 ⇔ ∀x ∈ A : |B(λx, y)| = |B(x, λy)| ≤ 1 ⇔λy ∈ A ⇔ y ∈ 1

λA.

(vii): A is absorbent if and only if ∀y ∈ F ∃µ > 0 such that µy ∈ A, which is equivalentto ∀x ∈ A : |B(x, y)| ≤ 1

µ; using Proposition 6.2 (iii) and Corollary 6.3, this is equivalent

to A being σ(E,F )-bounded.

(viii): y ∈(⋃

i∈I Ai) ⇔ ∀x ∈ ⋃iAi : |B(x, y)| ≤ 1 ⇔ ∀i ∀x ∈ Ai: |B(x, y)| ≤ 1 ⇔

∀i : y ∈ Ai ⇔ y ∈⋂iAi .

Theorem 11.3 (Bipolar theorem). For A 6= ∅, A is the absolutely convex σ(E,F )-closed hull of A.

Proof. By Proposition 11.2, A is absolutely convex, σ(E,F )-closed and contains A. Weneed to show that A ⊆ D for each set D of this kind.

Take a 6∈ D and let E0 be the real vector space underlying E. If U is a convex openneighborhood of a such that (a + U) ∩ D = ∅ then by Proposition 9.11 there is α ∈ Rand a σ(E0, F )-continuous linear form u : E0 → R such that u(x) ≤ α for x ∈ D andu(a) > α. Replacing α by (α + u(a))/2, we also have that u(x) < α for x ∈ D. Because0 ∈ D, 0 < α and by replacing u with u/α we can assume that α = 1.

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In the real case, set v = u; in the complex case, set v(x) = u(x) − iu(ix). Then v ∈ E ′,such that v(x) = B(x, y) for some y ∈ F and u(x) = <v(x) = <B(x, y) for all x ∈ E.Now for each x ∈ D there is some λ ∈ K with |λ| = 1 such that

|B(x, y)| = λB(x, y) = B(λx, y) = u(λx) ≤ 1

because λx ∈ D, so y ∈ D ⊆ A. However, we have

|B(a, y)| ≥ |u(a)| > 1

so a 6∈ A.

We denote the absolutely convex σ(E,F )-closed hull of a subset A ⊆ E by acxA.

Proposition 11.4. Let (Ai)i be a nonempty family of absolutely convex nonempty σ(E,F )-closed subsets of E. Then (⋂

i

Ai

)= acx

(⋃i

Ai

).

Proof. We have ⋂i

Ai =⋂i

Ai =

(⋃i

Ai

)and hence (⋂

i

Ai

)=

(⋃i

Ai

)= acx

(⋃i

Ai

).

If M ⊆ E is a linear subspace then M = M⊥, defined as follows:

Definition 11.5. Let E,F be paired with respect to B. Given any subset M ⊆ E, thelinear subspace of F defined by M⊥ := y ∈ F : B(x, y) = 0 ∀x ∈ M is called thesubspace orthogonal to M . Similarly, one defines N⊥ ⊆ E for N ⊆ F .

If (Mi)i is a family of subsets of E we denote by∨iMi the smallest σ(E,F )-closed linear

subspace of E containing all sets Mi, equal to the σ(E,F )-closure of the linear span of⋃iMi.

Proposition 11.6. Let E,F be paired with respect to B.

(i) M1 ⊆M2 ⇒ M⊥1 ⊇M⊥

2 ,(ii) M ⊆M⊥⊥ := (M⊥)⊥,

(iii) M⊥ = M⊥⊥⊥,(iv) M⊥ is σ(F,E)-closed,(v) M⊥⊥ is the σ(E,F )-closed subspace of E generated by M ,

(vi) (⋃iMi)

⊥ = (∨iMi)

⊥ =⋂iM

⊥i , and

(vii) if each Mi is a σ(E,F )-closed subspace of E then (⋂iMi)

⊥ =∨iM

⊥i .

Proof. (i)–(iv) are clear (as in Proposition 11.2).

(v): by (iv) M⊥⊥ is a σ(E,F )-closed subspace of E and M ⊆ M⊥⊥ by (ii). Let L ⊆M⊥⊥ be the σ(E,F )-closed subspace generated by M . Supposing that there exists somea ∈ M⊥⊥ \ L, Proposition 9.6 (i) gives some u ∈ E ′ which is of the form u(x) = B(x, y)for some y ∈ F , such that u(x) = B(x, y) = 0 for x ∈ L ⊇ M but B(a, y) = 1, whichwould imply y ∈M⊥ but y 6∈M⊥⊥⊥, contradicting (iii).

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(vi): this is seen from (⋃i

Mi

)⊥=⋂i

M⊥i ,(⋃

i

Mi

)⊥=

(⋃i

Mi

)⊥⊥⊥,

(⋃i

Mi

)⊥⊥=∨i

Mi.

(vii): we have ⋂i

Mi =⋂i

M⊥⊥i = (

∨i

M⊥i )⊥,

(⋂i

Mi

)⊥=∨i

M⊥i .

12. S-topologies

Let E,F be vector spaces paired with respect to a bilinear form B and S be a collectionof σ(E,F )-bounded subsets of E. Then the polars A (A ∈ S), which are absorbentand absolutely convex, by Corollary 3.8 define a locally convex topology on F , calledthe S-topology or the topology of uniform convergence on sets of S. A 0-basis of thistopology is given by finite intersections of sets of the form λA (λ > 0, A ∈ S), andit is defined by the family of seminorms qA(y) = supx∈A |B(x, y)| for A ∈ S becausey ∈ λA ⇔ qA(y) ≤ λ.

Proposition 12.1. The S-topology is Hausdorff if and only if⋃A : A ∈ S is total in

E for σ(E,F ) (i.e., if its linear span is dense) and the pairing separates points of F .

Proof. Suppose that qA(y) = 0 for all A ∈ S, which is equivalent to

y ∈⋂A∈S

A⊥ =

(∨A∈S

A

)⊥.

If now⋃A∈SA is total then

∨A∈SA = E and B(x, y) = 0 for all x ∈ E, which implies

that y = 0, so the S-topology is Hausdorff.

Conversely, if∨A∈SA 6= E or B does not separate points of F , then there is y ∈ F \ 0

such that y ∈ A⊥ for all A ∈ S, i.e., such that qA(y) = 0 for all A ∈ S, which meansthat the topology is not Hausdorff.

Proposition 12.2. The S-topology on F is unchanged if we replace S by any of thefollowing:

(i) all subsets of sets in S,(ii) finite unions of sets in S,

(iii) sets λA with λ ∈ K and A ∈ S,(iv) balanced hulls of sets in S,(v) σ(E,F )-closures of sets in S,

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(vi) absolutely convex σ(E,F )-closed hulls of sets in S.

Proof. (i): If A1 ⊆ A ∈ S then A ⊆ A1, so A1 is a 0-neighborhood.

(ii): (⋃ni=1Ai)

=⋂ni=1A

i is a 0-neighborhood.

(iii): For λ 6= 0, (λA) = λ−1A is a 0-neighborhood; for λ = 0, (λA) = F .

(iv): (D · A) = A.

(v): A ⊆ A ⊆ A, so A ⊇ (A) ⊇ A = A, so (A) = A.

(vi): for A ∈ S, its absolutely convex σ(E,F )-closed hull is A, but A = A.

Example. The weak topology σ(E,F ) is the S-topology for S the collection of allsingletons of F .

Definition 12.3. If E,F form a pairing separating points of F , then a locally convextopology T on E is said to be compatible with the pairing if (E,T )′ = F .

Proposition 12.4. If E,F form a pairing separating points of F , then the closed convexsets of E are the same for all locally convex topologies on E compatible with the pairing.

Proof. We may assume that E,F are real vector spaces. A closed convex set is givenby the intersection of all half-spaces f(x) ≤ α containing it (f ∈ E ′, α ∈ R), and thecontinuous linear forms are the same for all topologies compatible with the pairing.

Definition 12.5. Let E be a topological space, F a TVS and H a set of mappingsE → F . H is called equicontinuous at a ∈ E if for every 0-neighborhood V in Fthere is a neighborhood U of a such that f(U) ⊆ f(a) + V for all f ∈ H. H is calledequicontinuous on E if it is equicontinuous at every point of E.

If also E is a TVS then H is called uniformly equicontinuous if for every 0-neighborhoodV in F there is a 0-neighborhood U in E such that x − y ∈ U implies f(x) − f(y) ∈ Vfor all f ∈ H.

If H is uniformly equicontinuous then it is equicontinuous and each f ∈ H is uniformlycontinuous.

Proposition 12.6. If E,F are TVS and H a set of linear maps E → F then H isuniformly equicontinuous if it equicontinuous at the origin.

Proof. If V is a 0-neighborhood in F there is a 0-neighborhood U in E with H(U) ⊆ V .Then, x− y ∈ U implies f(x)− f(y) = f(x− y) ∈ V for all f ∈ H.

Proposition 12.7. If E is a TVS then a subset A ⊆ E ′ is equicontinuous if and only ifA ⊆ U for some 0-neighborhood U in E.

Proof. If A is equicontinuous there is a 0-neighborhood U in E such that |〈x, u〉| ≤ 1for x ∈ U and u ∈ A, so A ⊆ U. Conversely, if A ⊆ U for some 0-neighborhoodU then |〈x, u〉| ≤ ε for x ∈ εU and u ∈ A, so A is equicontinuous at 0 and henceequicontinuous.

Proposition 12.8. The topology T of any LCS E coincides with the topology of uniformconvergence on the family of equicontinuous subsets of E ′.

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Proof. There is a 0-basis of T consisting of absolutely convex closed sets V , which byProposition 12.4 are also σ(E,E ′)-closed, so by Theorem 11.3 we have V = V . ByProposition 12.7 V is equicontinuous, so V is a 0-neighborhood in the S-topology, whichhence is finer than T .

Conversely, if M ⊆ E ′ is equicontinuous then M ⊆ V where V is an absolutely convexσ(E,E ′)-closed 0-neighborhood in T . Because M ⊇ V = V , M is a 0-neighborhoodin T and the topologies are equal.

[12.5.]

Theorem 12.9 (Alaoglu-Bourbaki). Let E be a TVS. Any equicontinuous subset of E ′

is relatively σ(E ′, E)-compact.

Proof. Let H ⊆ E ′ be equicontinuous. By Proposition 12.7, H ⊆ U for some 0-neighborhood U in E and it suffices to show that U is σ(E ′, E)-compact.

Note that(E ′, σ(E ′, E))→ (E∗, σ(E∗, E))→ KE

are topological embeddings.

Denoting by U• the polar of U in E∗, U• is σ(E∗, E)-closed (Proposition 11.2 (v)) andσ(E∗, E)-bounded (U absorbent ⇒ U•• absorbent ⇒ U• σ(E∗, E)-bounded by Proposi-tion 11.2 (vii)).

U• is bounded in KE, so each projection πx(U•) ⊆ K (x ∈ E) is bounded and U• ⊆∏

x∈E πx(U•) is relatively compact.

For u ∈ U•, u(εU) ⊆ εD gives u ∈ E ′, so U• = U ⊆ E ′.

We now have that V ⊆ KE is compact. By Lemma Lemma 12.10, the closure of V in KE

is contained in E ′ and hence equal to its σ(E ′, E)-closure, which therefore is compact.

Lemma 12.10. Let E be a TVS and H ⊆ E ′ equicontinuous. Then the closure of H inKE is contained in E ′ and equicontinuous.

Proof. Let u ∈ H ′, x, y ∈ E, ρ ∈ K, ε > 0. Then ∃f ∈ H such that

|u(z)− f(z)| ≤ minε3,

ε

3(|ρ|+ 1) (z ∈ x, y, ρx+ y),

so

|u(ρx+ y)− ρu(x)− ρ(y)| ≤ |u(ρx+ y)− f(ρx+ y)|+ |ρu(x)− ρf(x)|+ |u(y)− f(y)| ≤ ε

so u is linear. Moreover, H(U) ⊆ D for some 0-neighborhood U in E, soH ⊆⋂x∈U π

−1x (D),

which is a closed subset of KE mapping u to D. Because H(U) ⊆ D, H ⊆ E ′ and it isequicontinuous.

Corollary 12.11. If E is a normed space, the closed unit ball B′1 := x′ ∈ E ′ | ‖x′‖ ≤ 1is σ(E ′, E)-compact.

Proof. ‖x′‖ = sup‖x‖≤1 |〈x, x′〉|, so B′1 is the polar of B1 := x ∈ E | ‖x‖ ≤ 1, henceequicontinuous, and the claim follows from Theorem 12.9.

Definition 12.12. Let E,F be paired vector spaces. If S is the set of all σ(E,F )-boundedsubsets of E then the corresponding S-topology on F is denoted by β(F,E) and called thestrong topology or topology of uniform convergence on bounded subsets of E.

Example. Let E be a normed space and E ′ its dual.

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• A ⊆ E is σ(E,E ′)-bounded if for all x′ ∈ E ′ there is λ > 0 such that |〈x, x′〉| ≤ λfor all x ∈ A.• if A ⊆ E is bounded (w.r.t. the norm) it is bounded for σ(E,E ′): ‖x‖ ≤ M for

all x ∈ A implies |〈x, x′〉| ≤M‖x′‖ for all x ∈ A.• Recall: x 7→ [x′ 7→ 〈x, x′〉] is an isometric embedding of E into (E ′)′, i.e., ‖x‖ =

sup|〈x, x′〉| /‖x‖ | x′ ∈ E \ 0.• If A ⊆ E is σ(E,E ′)-bounded then for all x ∈ E ′, the set |〈x, x′〉| | x ∈ A is

bounded, i.e., A is pointwise bounded in (E ′)′. By the Banach-Steinhaus theoremit is uniformly bounded, i.e., there is M such that for all x ∈ A we have |〈x, x′〉| ≤M‖x′‖, so ‖x‖ ≤M for all x ∈ A and A is norm-bounded.

Hence, the σ(E,E ′)-bounded subsets of E are exactly the norm-bounded ones.• A ⊆ E is norm-bounded if and only if there is λ > 0 such that A ⊆ λB1, whereB1 is the closed unit ball in E. Hence, β(E ′, E) is the S-topology with S = B1.Because (B1)

= B′1 (unit ball in E ′), the β(E ′, E)-topology is the norm-topologyon E ′.• In particular, β(E ′, E) is not compatible with the duality of E is not reflexive.

13. The Mackey Topology

Let E,F form a pairing separating points of E (i.e., we have E ⊆ F ∗). For whichtopologies T on F do we have (F,T )′ = E – which topologies on F are compatible withthe pairing?

Proposition 13.1. Let S be a collection of absolutely convex σ(E,F )-closed σ(E,F )-bounded subsets of E such that

(i) (Ai)i finite family in S ⇒ acx⋃iAi ∈ S,

(ii) A ∈ S, λ 6= 0 ⇒ λA ∈ S.

Then, if TS is the corresponding S-topology on F , we have

(F,TS)′ =⋃clσ(F ∗,F )A | A ∈ S

where clT denote the closure with respect to a topology T .

Proof. Let u ∈ (F,TS)′. Then u(U) ⊆ D for some 0-neighborhood U in F which is of theform

U =n⋂i=1

λiAi (λi ∈ K \ 0, Ai ∈ S)

=n⋂i=1

(λ−1i Ai) = (

n⋃i=1

λ−1i Ai) = (

n⋃i=1

λ−1i Ai) = (acx

n⋃i=1

λ−1i Ai) = A

with A := acx⋃ni=1 λ

−1i Ai ∈ S. Now u(A) ⊆ D, so u ∈ A = clσ(F ∗,F )A.

Conversely, let u ∈ clσ(F ∗,F )A = A for some A ∈ S, then u(εA) ⊆ εD and εA =(ε−1A) is a 0-neighborhood in the S-topology.

Lemma 13.2. Let E be a LCS and A1, . . . , An ⊆ E convex and compact. Then theconvex and the absolutely convex hulls of

⋃ni=1Ai are compact.

Proof. L := (λ1, . . . , λn) ∈ Kn | λi ≥ 0,∑λi = 1 is compact, so K := L × A1 · · ·An

is compact in Kn × En. The mapping f : K → E, (λ1, . . . , λn, x1, . . . , xn) 7→∑λixi is

continuous, so f(K) (which is the convex hull of⋃iAi) is compact. For the absolutely

convex hull use L = (λ1, . . . , λn) |∑|λi| ≤ 1.

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Proposition 13.3. Let S0 be a family of σ(E,F )-bounded subsets of E. Then

(F,TS0)′ = spanacxσ(F

∗,F )B | B ∈ S0) =: H.

Proof. The set

S := acxσ(e,F )

n⋃i=1

λiBi | n ∈ N, λi ∈ K, Bi ∈ S0

satisfies (i) and (ii) of Proposition 13.1; to see this, set

σ1 := λB | λ 6= 0, B ∈ S0σ2 := B1 ∪ . . . ∪Bn | n ∈ N, Bi ∈ S1σ = acxB | B ∈ S2.

These define the same topologies as S0.

For (i), take Ai = (⋃j λijB

ij ∈ S. Then we have

(⋃i

Ai) = (

⋃i

(⋃j

λijBij)) = (

⋃i

(⋂j

λijBij))

= (⋂i

(⋃j

λijBij)) = (

⋃ij

λijBij) ∈ S.

For (ii), we have λ(⋃j λjBj)

= (⋃j λλjBj)

∈ S.

Each A ∈ S is σ(E,F )-bounded. We then have

(F,TS0)′ = (F,Tσ)′ =

⋃A∈S

Aσ(F ∗,F )

.

For B ∈ S0, acxσ(F∗,F )B = clσ(F ∗,F ) acxB ⊆ (F,Tσ0)

′, so H ⊆ (F,TS0)′.

Conversely, take A = acx(⋃i λiBi) ∈ S (Bi ∈ S0) and set Ai := acxσ(F

∗,F )Bi = Bi =(Bi )

⊆ H (polars taken with respect to (F ∗, F )), thenBi is a 0-neighborhood in (F,Tσ0),so (Bi )

⊆ (F,Tσ0)′ is equicontinuous, by Theorem 12.9 σ((F,Tσ0)

′, F )-compact andhence σ(F ∗, F )-compact. By Lemma 13.2, A0 := acx

⋃λiAi ⊆ H is σ(F ∗, F )-compact,

hence σ(F ∗, F )-closed in F ∗. Because⋃i λiBi ⊆ A0 we have

clσ(F ∗,F ) acx⋃i

λiBi ⊆ A0 ⊆ H,

so (F,Tσ0)′ ⊆ H.

Corollary 13.4. Let E,F form a pairing separating points of E and let S be a collectionof σ(E,F )-bounded subsets of E. Then

(i) (F,TS)′ ⊆ E ⇐⇒ ∀B ∈ S: acxB is σ(E,F )-compact.(ii) (F,TS)′ ⊇ E ⇐⇒ spanacxB | B ∈ S = E.

Proof. (i) “⇐”: acxB σ(E,F )-compact ⇒ σ(F ∗, F )-compact ⇒ σ(F ∗, F )-closed ⇒acxσ(F

∗,F )B ⊆ E ⇒ (F,TS)′ ⊆ E by Proposition 13.3.

“⇒”: for B ∈ S, acxσ(F∗,F )B ⊆ (F,TfS)′ ⊆ E by Proposition 13.3, so acxB is σ(F ∗, F )-

closed. As acxB = (B) and B is a 0-neighborhood in TS, (B) is relatively σ(F ′, F )-compact (Theorem 12.9), hence relatively σ(F ∗, F )-compact ⇒ σ(F ∗, F )-compact ⇒σ(E,F )-compact.

(ii) “⇐”: E ⊆ spanacxB | B ∈ S ⊆ spanacxσ(F∗,F )B | B ∈ S ⊆ (F,TS)′.

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“⇒”: E ⊆ E ∩ spanacxσ(F∗,F )B | B ∈ S = span(acxσ(F

∗,F )B) ∩ E | B ∈ S =spanacxB | B ∈ S. To see this, take x ∈ clσ(F ∗,F )A ∩E = clσ(E,F )A with A =acx

⋃i λiBi. As above, we have thatAi := acxBi is σ(E,F )-compact, soA0 := acx

⋃i λiAi

is σ(E,F )-compact, hence σ(E,F )-closed, and acx⋃i λiBi ⊆ A0 ⊆ spanacxB | B ∈

S.

Theorem 13.5 (Mackey-Arens). Let E,F form a pairing separating points of F . Atopology T on E is compatible with the pairing if and only if T is a S-topology for acollection of absolutely convex σ(F,E)-compact subsets of F which cover F .

Proof. Let N be the set of all absolutely convex closed 0-neighborhoods in E; these arealso σ(E,F )-closed. Set S := V | V ∈ N. Because V = V for V ∈ N , T = TS. ByCorollary 13.4 (i), acxV = V is σ(F,E)-compact. For y ∈ F , 〈., y〉 ∈ E ′, so ∃V ∈ N :|〈x, y〉| ≤ 1 for all x ∈ V , and y ∈ V .The converse follows directly from Corollary 13.4.

Definition 13.6. Let E,F form a pairing separating points of F . For S the collectionof all absolutely convex, σ(F,E)-compact subsets of F , the S-topology on E is called theMackey topology and denoted by τ(E,F ).

Proposition 13.7. Let E,F form a pairing separating points of F . A locally convextopology T on E is compatible with the pairing if and only if it is finer than σ(E,F ) andcoarser than τ(E,F ).

Proof. “⇒”: By Theorem 13.5, T = TS for some collection S of absolutely convexσ(F,E)-compact subsets of F , so T ≤ τ(E,F ). A 0-basis in σ(E,F ) is given by sets ofthe form

Uy1,...,yk,ε := x ∈ E | |〈x, yi〉| ≤ ε ∀i = 1 . . . kfor y1, . . . , yk ∈ F and ε > 0. Because (E,T )′ = F , there is a 0-neighborhood V in Tsuch that |〈V, yi〉| ∈ εD for i = 1 . . . k, i.e., V ⊆ Uy1,...,yk,ε and σ(E,F ) ≤ T .

“⇐”: σ(E,F ) ≤ T means that F ⊆ (E, σ(E,F ))′ ⊆ (E,T )′, and T ≤ τ(E,F ) meansthat (E,T )′ ≤ (E, τ(E,F ))′ = F .

Let now E be a LCS, A ⊆ E nonempty and absolutely convex, and set

EA := spanA =n⋃i=1

nA.

The gauge qA of A is a seminorm on this space.

Lemma 13.8. Let A 6= ∅ be an absolutely convex and bounded subset of the LCS E.

(i) If E is Hausdorff, EA is a normed space.(ii) If in addition A is complete, EA is a Banach space.

Proof. (i): x ∈ EA, x 6= 0 ⇒ ∃U 0-neighborhood in E such that x 6∈ U . With λ suchthat λA ⊆ U we have x 6∈ λ, hence qA(x) ≥ λ.

(ii): Let xk be Cauchy sequence in EA. Because it is bounded it is contained in someλA, which is complete for the topology induced by E, so xk → x ∈ λA in this topology,and by Lemma 13.9 also in EA, so EA is complete. Note that the topology of EA isfiner than the induced topology because give a 0-neighborhood U , we have λB ⊆ U forsome λ. Moreover, each λA is closed with respect to the induced topology because A iscomplete.

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Lemma 13.9. Let T ,T ′ be linear topologies on a vector space E such that T ′ is finerthan T . Suppose that T ′ has a 0-basis consisting of T -closed sets. If A ⊆ E and F isa T ′-Cauchy filter on A which T -converges to x ∈ A, then F → x in T ′.

Proof. This is contained in the proof of Proposition 6.16.

Recall that a barrel in a locally convex space is an absolutely convex, absorbent andclosed set.

Proposition 13.10. Let E be a LCS. A subset M ⊂ E ′ is σ(E ′, E)-bounded if and onlyif M ⊆ T for a barrel T in E.

Proof. “⇒”: T := M is absolutely convex, σ(E,E ′)-closed, absorbent, hence also closedfor the topology of E and a barrel; we have T = M ⊇M .

“⇐”: if T is a barrel, T = T , so T is σ(E ′, E)-bounded because its polar is absorbent.

Theorem 13.11 (Banach-Mackey). Let E be a Hausdorff LCS and T a barrel in E.Then T absorbs every absolutely convex, bounded, complete subset A of E.

Proof. EA is a Banach space; S = T ∩ EA is a barrel in EA because the topology of EAis finer, so S is σ((EA)′, EA)-bounded in (EA)′: for x ∈ EA there is λ > 0 such that|〈x, x′〉| < λ for x′ ∈ S, so by the Banach-Steinhaus theorem for normed spaces there isµ > 0 such that |〈x, x′〉| ≤ µ‖x‖ for x ∈ EA, x′ ∈ S, hence x | ‖x‖ ≤ 1

µ ⊆ S = S.

Because A ⊆ (qA)≤1 we have 1µA ⊆ S ⊆ T and T absorbs A.

[19.5.]

Theorem 13.12 (Mackey). Let E,F form a dual system (i.e., separating points of bothE and F ). The bounded subsets of E are the same for all locally convex topologies on Ecompatible with the pairing of E and F .

Proof. Because a bounded subset of E is also bounded for a coarser topology we onlyhave to show thach each σ(E,F )-bounded subset B of E is τ(E,F )-bounded. If we equipF with the topology σ(F,E) it is clear that B is also σ(F ′, F )-bounded in F , hence byProposition 13.10 contained in T for a barrel T in (F, σ(F,E)).

Let A be an absolutely convex σ(F,E)-compact subset of F . The polars of all such Aform a 0-basis of the Mackey topology τ(E,F ), so we need to show that A absorbsT . Because A is also σ(F,E)-complete and σ(F,E)-bounded and σ(F,E) is Hausdorff,Theorem 13.11 gives that λA ⊆ T for some λ > 0 and hence B ⊆ T ⊆ λ−1A.

Consequently, if E is a LCS then its bounded subsets are exactly the σ(E,E ′)-boundedones and the topology β(E,E ′) is the topology of uniform convergence on bounded sets.The σ(E ′, E)-bounded subsets of E ′ are furthermore characterized as follows:

Proposition 13.13. Let E be a LCS. Then M ⊆ E ′ is β(E ′, E)-bounded if and only ifM ⊆ T for a bornivorous barrel T in E (i.e., a barrel absorbing every bounded set).

Proof. If M is β(E ′, E)-bounded, its polar M is absolutely convex and σ(E,E ′)-closed,hence also closed. Because M is also σ(E ′, E)-bounded, M is absorbent, so M is abarrel in E. Given a bounded subset A ⊆ E, A is a 0-neighborhood in β(E ′, E) so thereis λ > 0 with M ⊆ λA and hence A ⊆ A ⊆ λM, so M is a bornivorous barrel andM ⊆ (M).

Conversely, let T be a bornivorous barrel and A ⊆ E bounded. Then A ⊆ λT for someλ > 0, i.e., T ⊆ λA which means that T is β(E ′, E)-bounded.

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Theorem 13.14. If a LCS E is quasicomplete then every σ(E ′, E)-bounded subset of E ′

is β(E ′, E)-bounded.

Proof. Using Proposition 13.10 and Proposition 13.13 we only have to show that everybarrel T is bornivorous. Give a bounded set B in E, also its absolutely convex closedhull is bounded and by assumption complete, so T absorbs this hull and also B.

14. Barrelled spaces

Proposition 14.1. Let E be a LCS and A ⊆ E ′. Then

A equicontinuous ⇒ A β(E ′, E)-bounded ⇒ A σ(E ′, E)-bounded .

Proof. If A is equicontinuous then A ⊆ V for an absolutely convex closed 0-neighborhoodV in E. V is a bornivorous barrel, so Proposition 13.13 gives the first claim. The secondclaim is clear because β(E ′, E) ≥ σ(E ′, E).

Example. Let E be a normed vector space and E ′ its dual. Then ρB1 | ρ > 0 is afundamental system of 0-neighborhoods and also of bounded sets.

• M ⊆ E ′ is equicontinuous if and only if M ⊆ (ρB1) = ρ−1B1 , i.e., if and only if

it is bounded in E ′ (B1 is the unit ball in E ′).• M is bounded in (E ′, β(E ′, E)) if and only if it is norm-bounded (the β(E ′, E)-

topology is the norm topology).• Hence, in E ′ the equicontinuous sets are exactly the strongly bounded sets.• If E is a Banach space and M ⊆ E ′ is σ(E ′, E)-bounded, then by the Banach-

Steinhaus theorem M is norm-bounded. Hence, in E ′ the equicontinuous, theβ(E ′, E)-bounded and the σ(E ′, E)-bounded sets coincide.

Definition 14.2. A LCS E is called barrelled if every barrel in E is a 0-neighborhood.

Proposition 14.3. A LCS E is barrelled if and only if every weakly bounded subset ofE ′ if equicontinuous.

Proof. Suppose that E is a barrelled and let A ⊆ E ′ be σ(E ′, E)-bounded. By Proposi-tion 13.10, A ⊆ T for a barrel in E which by assumption is also a 0-neighborhood, so Ais equicontinuous by Proposition 12.7.

Conversely, if T is a barrel then T is σ(E ′, E)-bounded and hence equicontinous byassumption, so by Proposition 12.8 T = T is a 0-neighborhood.

Corollary 14.4. If E is barrelled then for A ⊆ E ′ we have the equivalences

A equicontinuous

⇐⇒ A relatively σ(E ′, E)-compact

⇐⇒ A β(E ′, E)-bounded

⇐⇒ A σ(E ′, E)-bounded.

Examples of barrelled LCS are obtained by Baire’s theorem. Recall that if a Baire spaceis the union of a sequence of closed sets, then at least one of those sets has an interiorpoint.

Proposition 14.5. If a LCS E is a Baire space, it is barrelled.

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Proof. If T is a barrel in E then E =⋃∞n=1 nT , so one of the closed sets nT and hence

also T has to have an interior point x0 ∈ T . If U is a balanced 0-neighborhood such thatx0 + U ⊆ T we have −x0 + U ⊆ T and U 3 x = 1

2(x0 + x) + 1

2(−x0 + x) ∈ T because T

is convex, so T is a 0-neighborhood.

By Baire’s theorem, every complete metrizable topological space is a Baire space, so wehave:

Corollary 14.6. Every Frechet space is barrelled.

Proposition 14.7 (Banach-Steinhaus Theorem). Let E be a barrelled LCS and F afilter basis on E ′ containing a σ(E ′, E)-bounded set. If for every x ∈ E F (x) convergesto some f(x) then [f : x 7→ f(x)] ∈ E ′.

Proof. Let A ∈ F be σ(E ′, E)-bounded and hence equicontinuous. Then for each x ∈ E,FA(x)→ f(x), which means that FA → f in KE, so f is in the closure of A in KE andhence continuous by Lemma 12.10.

Corollary 14.8. If E is a barrelled LCS and (fn)n is a sequence in E ′ converging tosome f ∈ KE, then f ∈ E ′.

A slightly weaker notion is the following:

Definition 14.9. A LCS E is called infrabarrelled or quasibarrelled if every bornivorousbarrel in E is a 0-neighborhood.

Proposition 14.10. A LCS E is infrabarrelled if and only if every β(E ′, E)-boundedsubset of E ′ is equicontinuous.

Proof. “⇒”: if M if β(E ′, E)-bounded then M ⊆ T for a bornivorous barrel T , whichis a 0-neighborhood by assumption, so M is equicontinuous.

“⇐”: if T is a bornivorous barrel in E, then T is β(E ′, E)-bounded and hence equicon-tinuous, so by Proposition 12.8 T = T is a 0-neighborhood in E.

Theorem 13.14 implies:

Corollary 14.11. A quasicomplete Hausdorff LCS is infrabarrelled if and only if it isbarrelled.

Proposition 14.12. Let E be a Hausdorff LCS. If a convex subset A ⊆ E ′ is relativelyσ(E ′, E)-compact it is β(E ′, E)-bounded.

Proof. N := acx(A) is absolutely convex and weakly compact by Lemma 13.2, so N isa 0-neighborhood in the Mackey topology. If B ⊆ E is bounded then it is so for theMackey topology, so B ⊆ λN for some λ > 0, which implies A ⊆ N = N ⊆ λB, so Ais β(E ′, E)-bounded.

Proposition 14.13. If E is an infrabarrelled Hausdorff LCS, its topology coincides withthe Mackey topology τ(E,E ′).

Proof. Let U = A be a 0-neighborhood in E, where A ⊆ E ′ is equicontinuous. ThenacxA and hence also acxA are equicontinuous (Lemma 12.10) and by Theorem 12.9σ(E ′, E)-compact, hence (acxA) ⊆ U , which shows that the Mackey topology is finer.

Conversely, let U = B be a 0-neighborhood in the Mackey topology, i.e., B is absolutelyconvex and σ(E ′, E)-compact. Then it is β(E ′, E)-bounded by Proposition 14.12 andhence equicontinuous by Proposition 14.10, so U is a 0-neighborhood in the topology ofE and the topologies are equal.

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Proposition 14.14. Let E be a vector space, (Eι)ι a family of (infra-)barrelled locallyconvex spaces, and (fι)ι a family of linear mappings Eι → E. The inductive locally convextopology on E with respect to (fι)ι is (infra-)barrelled.

Proof. If T is a barrel in E then each f−1ι (T ) is a barrel in Ei: absolute convexity andclosedness are clear; for x ∈ Eι, λfι(x) ∈ T for some λ and hence λx ∈ f−1ι (T ) andf−1ι (T ) is absorbent. If T is bornivorous, then for a bounded set B ⊆ Eι, λfι(B) ⊆ Tfor some λ > 0 and λB ⊆ f−1ι (T ), so f−1ι (T ) is bornivorous. Because f−1ι (T ) then is a0-neighborhood of Eι, T is a 0-neighborhood in E by Proposition 6.8 (ii).

Corollary 14.15. (i) If E is an (infra-)barrelled LCS and M ⊆ E a subspace thenE/M is (infra-)barrelled.

(ii) If (Eι)ι is a family of (infra-)barrelled LCS, the locally convex direct sum⊕

ιEιis (infra-)barrelled.

(iii) Inductive limits of (infra-)barrelled LCS are (infra-)barrelled.

Examples. Normed spaces are infrabarrelled, Banach spaces are barrelled. The spacesEm(Ω), Dm(Ω) are barrelled because they are Frechet spaces. (LF)-spaces (e.g., D(K))are barrelled.

15. Bornological Spaces

Definition 15.1. A LCS E is called bornological if every absolutely convex bornivoroussubset of E is a 0-neighborhood.

Proposition 15.2. A LCS E is bornological if and only if for all locally convex spacesF , every linear map E → F which maps bounded sets to bounded sets is continuous. Itsuffices to know this for normed spaces F .

Proof. “⇒”: let V be an absolutely convex 0-neighborhood in F , then f−1(V ) is abso-lutely convex and bornivorous: given A ⊆ E bounded, f(A) is bounded and ∃λ > 0s.t. f(A) ⊆ λV and hence A ⊆ λf−1(V ), so f−1(V ) is a 0-neighborhood.

“⇐”: If U ⊆ E is absolutely convex and bornivorous and qU the gauge of U , considerthe quotient space EU := E/q−1U (0) with canonical surjection ϕ. Then ‖ϕ(x)‖ := qU(x)is a norm on EU . If A ⊆ E is bounded, A ⊆ λU for some λ > 0, so ‖ϕ(x)‖ ≤ λ forx ∈ A which means that ϕ(A) is bounded and hence ϕ is continuous by assumption.Consequently, ϕ−1(x | ‖x‖ < 1) = x ∈ E | ‖ϕ(x)‖ < 1 = (qU)<1 ⊆ U and U is a0-neighborhood.

Proposition 15.3. A LCS with a countable 0-basis is bornological.

Proof. Let (Un)n be a decreasing sequence of balanced 0-neighborhoods forming a 0-basisand let U ⊆ E be absolutely convex and bornivorous. If U contains a set of the formn−1Un it is a 0-neighborhood. Assuming the contrary we would have a sequence (xn)nwith xn ∈ n−1Un but xn 6∈ U . Then nxn → 0, hence A := nxn | n is bounded. Ucannot absorb A because nxn ∈ λU would imply xn ∈ λ

nU ⊆ U for n ≥ λ, which gives a

contradiction.

Examples. The spaces Em(Ω), Dm(K) are bornological.

Proposition 15.4. Let E be a vector space, (Eι)ι a family of bornological locally convexspaces, and (fι)ι a family of linear mappings Eι → E. The inductive locally convextopology on E with respect to (fι)ι is bornological.

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Proof. If U ⊆ E is absolutely convex and bornivorous, each f−1ι (U) is absolutely convex.If B ⊆ Eι is bounded, fι(B) is bounded and hence contained in some λU , so B ⊆λf−1ι (U) and f−1ι (U) is bornivorous, hence a 0-neighborhood, which proves that U is a0-neighborhood.

Corollary 15.5. (i) If E is a bornological LCS and M ⊆ E a subspace, then E/Mis bornological.

(ii) The direct sum of a family of bornological LCS is bornological.(iii) The inductive limit of a family of bornological LCS is bornological.

Example. The space Dm(Ω) is bornological.

We finish this section with two important results. The first one is an analogoue toProposition 7.4:

Proposition 15.6. Every (complete) Hausdorff bornological LCS can be represented asthe inductive limit of a family of normed (Banach) spaces.

Proof. For each absolutely convex closed bounded subset A ⊆ E, EA with the seminormqA is a normed (Banach) space by Lemma 13.8. Let fA : EA → E be the canonicalinjection. Denoting by T the original topology of E and by T ′ the inductive locallyconvex topology with respect to the mappings fA, we will show that T = T ′.

If U is an absolutely convex 0-neighborhood in T then λA ⊆ U for some λ > 0, i.e.,λA ⊆ f−1A (U), so each f−1A (U) is a 0-neighborhood in EA and U is a 0-neighborhood inT ′ by Proposition 6.8 (ii).

If U is an absolutely convex 0-neighborhood in T ′ then λA ⊆ f−1A (U) for some λ, so λA ⊆U which means that U is bornivorous and hence a 0-neighborhood in T . Proposition 7.9now gives the claim because E =

⋃EA | A ⊆ E absolutely convex, closed, bounded

and for A ⊆ B, EA ⊆ EB and qB ≤ qA, so (E,T ) = (E,Ti) = lim−→EA.

Proposition 15.7. If E is bornological then (E ′, β(E ′, E)) is complete.

Proof. Let F be a Cauchy filter on E ′, the it is also a Cauchy filter with respect toσ(E ′, E) and hence converges to some f in KE, and even in E∗ as seen in the proof ofLemma 12.10. To show continuity of f let B ⊆ E be bounded; then for some A ∈ F ,A − A ⊆ B. Fixing any v ∈ A there is α > 0 such that for all u ∈ A and x ∈ B,|u(x)| ≤ |u(x)− v(x)| + |v(x)| ≤ α. Because also FA → f we can for each x ∈ B findsome ux ∈ A such that |ux(x)− f(x)| ≤ α. Then |f(x)| ≤ |f(x)− ux(x)|+ |ux(x)| ≤ 2α,so f is bounded and hence continuous. Because β(E ′, E) has a 0-basis of σ(E ′, E)-closedsets, Lemma 13.9 gives the claim.

Example. The space D ′(Ω) := (D(Ω)′, β(D(Ω)′,D(Ω))) of distributions on Ω iscomplete.

16. Reflexivity

Definition 16.1. Let E be a Hausdorff LCS. E ′′ := (E ′, β(E ′, E))′ is called the bidualof E.

Definition 16.2. A Hausdorff LCS E is called semireflexive if the canonical embeddingE → E ′′ is surjective, and reflexive if it is an isomorphism onto (E ′′, β(E ′′, E ′)).

Lemma 16.3. The following statements are equivalent:

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(i) E is semireflexive.(ii) Every element of (E ′, β(E ′, E))′ is of the form x′ 7→ 〈x, x′〉 for some x ∈ E.

(iii) Every β(E ′, E)-continuous linear form on E ′ is σ(E ′, E)-continuous.(iv) (E ′, β(E ′, E))′ = E.

The proof is clear. Moreover, by (iv) and Theorem 13.5 E is semireflexive if and only ifβ(E ′, E) = τ(E ′, E).

Proposition 16.4. A LCS E is semireflexive if and only if every bounded σ(E,E ′)-closedsubset of E is σ(E,E ′)-compact.

Proof. If E is semireflexive we have E ′′ = (E ′,TS)′ ⊆ E where S is the family of allσ(E,E ′)-bounded subsets of E, hence by Corollary 13.4 (i), for every B ∈ S we havethat acxB is σ(E,E ′)-compact. If now A is bounded and σ(E,E ′)-closed it is σ(E,E ′)-bounded, so acxA is σ(E,E ′)-compact, hence A is σ(E,E ′)-compact.

Conversely, β(E ′, E) is the topology of uniform convergence on all absolutely convexσ(E,E ′)-closed σ(E,E ′)-bounded subsets of E, which by assumption are exactly theabsolutely convex σ(E,E ′)-compact subsets of E, so β(E ′, E) is the Mackey topology.

Proposition 16.5. The strong dual of a semireflexive Hausdorff LCS is barrelled.

Proof. If T is a barrel in E ′ then T is bounded in E ′′ = E for σ(E ′′, E ′) = σ(E,E ′) byProposition 13.10, so T = T is a 0-neighborhood in β(E ′, E).

Proposition 16.6. A Hausdorff LCS E is reflexive if and only if it is semireflexive andinfrabarrelled.

Proof. If E is reflexive then it clearly is semireflexive. Moreover, if B ⊆ E ′ is β(E ′, E)-bounded its polar B• is a 0-neighborhood in E ′′. As ι : E → E ′′ is a homeomorphism,ι−1(B•) = B is a 0-neighborhood in E, and B ⊆ (B), shows that B is equicontinuousand by Proposition 14.10 E is infrabarrelled.

If E is semireflexive then ι is surjective. Given a 0-neighborhood U = B• in E ′′ for someβ(E ′, E)-bounded subset B ⊆ E ′ , B is equicontinuous because E is infrabarrelled andhence B ⊆ V for some 0-neighborhood V in E. Hence, we have

U = B• ⊇ (V )• = ι(V ) ⊇ ι(V )

so ι is continuous. Conversely, if V = V is a 0-neighborhood in E then ι(V ) =ι(V ) = (V )• is a 0-neighborhood in E ′′ because V is equicontinuous and henceβ(E ′, E)-bounded in E ′.

[2.6]

Corollary 16.7. A reflexive space is barrelled.

Proof. Let M ⊆ E ′ be σ(E ′, E)-bounded. If E is semireflexive then M is also β(E ′, E)-bounded1, hence equicontinuous because E is infrabarrelled, so E is barrelled.

Proposition 16.8. The strong dual of a reflexive space is reflexive.

Proof. E ′ is barrelled by Proposition 16.5. If M ⊆ E ′ is bounded and σ(E ′, E ′′)-closedit is also σ(E ′, E)-closed because E is reflexive, and hence it is σ(E ′, E)-compact byCorollary 14.4, which shows that E ′ is also semireflexive by Proposition 16.4.

1the β-topology is compatible with the pairing

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Proposition 16.9. A normed vector space is reflexive if and only if its closed unit ballis weakly compact.

Proof. Suppose E is reflexive. The closed unit ball is bounded and convex, hence σ(E,E ′)-closed by Proposition 12.4 and σ(E,E ′)-compact by Proposition 16.4.

Conversely, each bounded σ(E,E ′)-closed subset of E is contained in some closed ballBρ = ρB1 and therefore is σ(E,E ′)-compact, so E is reflexive by Proposition 16.4.

17. Montel spaces

Recall: semireflexive spaces are those where every bounded set is weakly relatively com-pact (Proposition 16.4). When is this the case for the original topology instead of theweak one?

Definition 17.1. A Hausdorff LCS E is called semi-Montel if every bounded subset isrelatively compact. An infrabarrelled semi-Montel space is called a Montel space.

By Theorem 8.4, if a normed space if semi-Montel it is finite-dimensional.

Proposition 17.2. Every semi-Montel space is quasicomplete and semireflexive.

Proof. If B ⊆ E is bounded and F a Cauchy filter on B, F has a cluster point in E andhence converges. Moreover, if B is σ(E,E ′)-closed it is also σ(E,E ′)-compact and E issemi-reflexive by Proposition 16.4.

Proposition 16.6 gives:

Corollary 17.3. Every Montel space is reflexive.

In particular, taking into account Corollary 16.7 one sees that for a semi-Montel spacethe properties of being infrabarrelled, reflexive or barrelled are equivalent.

Definition 17.4. A subset A of a Hausdorff LCS E is called precompact or totallybounded if for every 0-neighborhood U in E there is a finite subset H ⊆ E such thatA ⊆ H + U .

Lemma 17.5. Let E be a Hausdorff LCS. Then for a subset A ⊆ E, the followingassertions are equivalent:

(i) Every ultrafilter on A is a Cauchy filter.

(ii) A is relatively compact in the completion E.(iii) A is totally bounded.

Proof. (i) ⇒ (ii): suppose that B := i(A) is not compact, where i : E → E is thecanonical injection. Then there is a filter F on B which has no adherent point. The

collection X + V | X ∈ F , V balanced 0-neighborhood in E is the basis of a filter G

on E. Because (X + V ) ∩ A 6= ∅ for all X, V , the trace GA is a filter on A and hencecontained in an ultrafilter U on A. Because U is a Cauchy filter, the filter generatedby it on B would be a Cauchy filter which would converge to some x ∈ B; x would bean adherent point of U , hence of GA, and of G . For F ∈ F and a 0-neighborhood U ,let V be a balanced 0-neighborhood with V + V ⊆ U . Then (F + V ) ∩ (x + V ) 6= ∅, sof + v1 = x+ v2 for some f, v1, x, v2, and f = x+ v2− v1 and F ∩ (x+U) 6= ∅, so x wouldbe an adherent point of F .

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(ii) ⇒ (iii): Given a 0-neighborhood U , let V ⊆ U be a closed 0-neighborhood and V its

closure in E. Because B is compact, B ⊆⋃ni=1(i(xi) + V ) for a finite family (xi)i in A

and A ⊆ ι(A) ∩ E ⊆⋃i(xi + V ) because V ∩ E = V .

(iii) ⇒ (i): Let U be an ultrafilter on A, U a 0-neighborhood and V a balanced 0-neighborhood such that V + V ⊆ U . For A ⊆

⋃(xi + V ), A =

⋃(xi + V ) ∩ A ∈ U , so

for some j, X := (xj + V ) ∩ A ∈ U , and we have X −X ⊆ V − V ⊆ U .

Definition 17.6. If E is a Hausdorff LCS, λ(E ′, E) denotes the topology of uniformconvergence on precompact subsets of E and κ(E ′, E) the topology of uniform convergenceon absolutely convex compact subsets of E.

Consider the following picture:

τ≤

σ ≤ κ≤

≤β

λ≤

Every precompact set is bounded: A ⊆ H + U ⊆ λU + U = (1 + λ)U if U is a convex0-neighborhood and λ chosen such that H ⊆ λU . Consequently, λ(E ′, E) ≤ β(E ′, E).

Lemma 17.7. If E is quasicomplete then κ = λ. If E is semi-Montel then λ = β andhence κ = τ = λ = β.

Proof. If E is quasicomplete and a 0-neighborhood A in λ is given with A precompact,acxA is precompact by Lemma 17.8 and hence relatively compact because it is bounded,so acxA is absolutely convex, compact and we have (acxA) ⊆ A.

For the second claim, by Lemma 17.5 in a semi-Montel space each bounded set is relativelycompact and hence precompact.

Lemma 17.8. If E is a LCS and A ⊆ E precompact, then acxA is precompact.

Proof. Set B := acxA = x =∑n

i=1 λixi |∑n

i=1 |λi| ≤ 1, xi ∈ A, n ∈ N. If U isan absolutely convex 0-neighborhood in E we have to show that B ⊆

⋃pj=1(aj + U)

for some aj ∈ A. Because A is precompact we know that A ⊆⋃qk=1(bk + 1

2U). Set

L := λ = (λ1, . . . , λq) |∑q

k=1 |λk| ≤ 1 ⊆ Kq compact. Then for each δ > 0 there are

µ1, . . . , µp ∈ Kq such that for all λ ∈ L one can find j with∑q

k=1

∣∣λk − µjk∣∣ < δ. Choose

δ with δA ⊆ 12U ; this implies δB ⊆ 1

2U because U is absolutely convex. We then claim

that B ⊆⋃pj=1(

∑qk=1 µ

(j)k bk + U). In fact, take x =

∑ni=1 λixi ∈ B. Then xi = bki + yi

with yi ∈ 12U , and x =

∑ni=1 λibki +

∑ni=1 λiyi =

∑qk=1 νkbk + y with

∑|νk| ≤ 1 and

y ∈ 12U . Consequently, x =

∑qk=1 µ

jkbk +

∑qk=1(νk − µ

jk)bk + y ∈

∑qk=1 µ

jkbk + U .

Proposition 17.9. If E a is Hausdorff LCS and H ⊆ E ′ is equicontinuous, the topologiesλ(E ′, E) and σ(E ′, E) coincide on H.

Proof. σ(E ′, E) is coarser than λ(E ′, E), so we only have to show the converse. LetH ⊆ E ′ be equicontinuous. It suffices to show that for u0 in H and A ⊆ E precompactthere is a finite subsetM ⊆ E such that (u0+M

)∩H ⊆ u0+A. Choose a 0-neighborhood

U such that H(U) ⊆ 12D, then there is a finite subset K ⊆ E such that A ⊆ K + 1

2U ; set

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M := 2K. Then for u ∈ (u0 + M) ∩ H, u(U) ∈ 12D, and u − u0 ∈ M = 1

2K implies

(u − u0)(K) ⊆ 12D. Any x ∈ A can be written as x = y + 1

2z with y ∈ K and z ∈ U ,

which gives |(u− u0)(x)| ≤ 1 and hence u− u0 ∈ A.

Corollary 17.10. If (E,T ) is a semi-Montel space and B ⊆ E bounded, then thetopology induced on B by T coincides with the topology induced by σ(E,E ′).

Corollary 17.11. Let E be semi-Montel. If (xn)n converges weakly to x ∈ E then xn → xin E.

Proof. The set of all xn together with x is weakly bounded and hence bounded, so thetopologies induced by T and σ(E,E ′) coincide.

Proposition 17.12. Closed subspaces and products of semi-Montel spaces are semi-Montel. Projective limits of semi-Montel spaces are (semi-)Montel. Regular inductivelimits of (semi-)Montel spaces are (semi-)Montel.

Proof. Let M ⊆ E be closed and A ⊆ M bounded. Then A is bounded in E and hencerelatively compact, because M is closed also relatively compact in M .

Let a family (Eι)ι of semi-Montel spaces be given and A ⊆ E be bounded. Each πι(A) isbounded and hence relatively compact, then the product

∏ι πι(A) is relatively compact.

Because it contains A, A also is relatively compact.

Proposition 17.13. The strong dual of a Montel space is a Montel space.

Proof. By Corollary 17.3 E is reflexive, so E ′ is barrelled by Proposition 16.5.

Let M ⊆ E ′ be bounded for β(E ′, E), then M is contained in an absolutely convexβ(E ′, E)-closed β(E ′, E)-bounded set N , which is equicontinuous by Proposition 14.10because E is infrabarrelled. N is σ(E ′, E)-closed because β(E ′, E) is compatible with theduality, and hence σ(E ′, E)-compact by Theorem 12.9. On N , the topologies σ(E ′, E) andλ(E ′, E) coincide, so N is λ(E ′, E)-compact and hence β(E ′, E)-compact (Lemma 17.7),and M is relatively compact.

Examples. One can show that the spaces E (Ω), S (Rn), D(Ω) and their strong dualsare Montel spaces.

18. The transpose of a linear map

We recall that the transpose (or adjoint) of a linear map u : E1 → E2 is defined as themapping u′ : E∗2 → E∗1 , y

′ 7→ y′ u.

Proposition 18.1. Let (E1, F1) and (E2, F2) be two dual systems and u : E1 → E2

linear. Then u′(F2) ⊆ F1 if and only if u is (σ(E1, F1), σ(E2, F2))-continuous. In thiscase, u′ : F2 → F1 is (σ(F2, E2), σ(F1, E1))-continuous and u = (u′)′.

Proof. Using Proposition 10.5, the first claim is a direct consequence of 〈x, u′(y)〉 =〈u(x), y〉.The second claim is equivalent to σ(F2, E1)-continuity of y 7→ 〈x, u′(y)〉 = 〈u(x), y〉 forall x ∈ E1, which is clear from E2 = (F2, σ(F2, E2))

′, and T = (T ′)′ is clear from thedefinition.

Proposition 18.2. Let (E1, F1) and (E2, F2) be two dual systems and u : E1 → E2 weaklycontinuous, A ⊆ E1 and B ⊆ E2. Then:

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(i) (u(A)) = (u′)−1(A).(ii) u(A) ⊆ B ⇒ u′(B) ⊆ A

(iii) If A,B are nonempty, absolutely convex and weakly closed then u(A) ⊆ B ⇔u′(B) ⊆ A.

(iv) u′ is injective if and only if the image of u is σ(E2, F2)-dense.

Proof. (i): y ∈ (u(A)) ⇔ ∀x ∈ u(A): |〈x, y〉| ≤ 1 ⇔ ∀x ∈ A : |〈u(x), y〉| = |〈x, u′(y)〉| ≤1 ⇔ u′(y) ∈ A ⇔ y ∈ (u′)−1(A).

(ii): for y ∈ B and x ∈ A, |〈x, u′(y)〉| = |〈u(x), y〉| ≤ 1.

(iii): u(A) = u′′(A) ⊆ B = B.

(iv): keru′ = (u′)−1(0) = (u′)−1(E1) = (u(E1)) = u(E1)

⊥. Now u(E1)⊥ = 0 if

and only if u(E1) = E2 (weak closure) (“⇒”: u(E1)⊥⊥ = u(E1) = 0⊥ = E2; “⇐”:

u(E1)⊥ = u(E1)

⊥⊥⊥ = (u(E1))⊥ = E⊥2 = 0).

Proposition 18.3. Let E,F be LCS and u : E → F linear.

(i) If u is continuous it is weakly continuous.(ii) u is weakly continuous if and only if it is ((τ(E,E ′), τ(F, F ′))-continuous.

(iii) If u is weakly continuous then it is (β(E,E ′), β(F, F ′))-continuous.

Proof. (i) If u : E → F is continuous, y u ∈ E ′ for all y ∈ F ′, so u′(F ′) ⊆ E ′ and u isweakly continuous by Proposition 18.1.

By (i), continuity w.r.t. the Mackey topologies implies weak continuity. Now supposethat u is weakly continuous. If B ⊆ F ′ is absolutely convex and weakly compact, thenalso u′(B) ⊂ E ′ is so, and we have u((u′(B))) = u((u′′)−1(B)) = u(u−1(B)) ⊆ B, so uis continuous w.r.t. the Mackey topologies, which shows (ii). The same argument shows(iii).

Corollary 18.4. If E has the Mackey topology τ(E,E ′) then every weakly continuouslinear map E → F is continuous.

This applies in particular to Frechet spaces.

19. Topological tensor products

Let L(E,F ;G) be the vector space of all bilinear mappings E × F → G (where E,F,Gare vector spaces) and set B(E,F ) := L(E,F ;K) (the space of bilinear forms). Forx, y ∈ E × F define ux,y ∈ B(E,F )∗ by ux,y(b) := b(x, y) for all b ∈ B(E,F ). Then themapping

⊗ : E × F → B(E,F )∗,

(x, y) 7→ ux,y

is bilinear and E ⊗ F := span(⊗(E × F )) is called the tensor product of E and F . Wewrite x⊗ y instead of x⊗ y.

Proposition 19.1. If (ei)i is a basis for E and (fj)j a basis for F then (ei ⊗ fj)i,j is abasis for E ⊗ F .

Proof. We only have to show linear independence. Let z =∑zijei ⊗ fj = 0 (finite

sum). Given i0, j0 there is b ∈ B(E,F ) such that b(ei0 , fj0) = 1 and b(ei, fj) = 0 fori 6= i0, j 6= j0, and z(b) = 0 gives zi0j0 = 0.

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Hence, every z ∈ (E ⊗ F ) \ 0 can be written as z =∑n

i=1 xi ⊗ yi with xi ∈ E linearlyindependent and yi ∈ F linearly independent. In fact, let z =

∑ri=1 xi⊗yi with r minimal.

For r = 1, clearly xi 6= 0 and yi 6= 0. For r ≥ 2, assume xr =∑r−1

i=1 λixi. Then

z =r−1∑i=1

xi ⊗ (yi + λiyr) =r−1∑i=1

xi ⊗ yi,

contradicting minimality of r.

The map

θ : L(E ⊗ F,G)→ L(E,F ;G),

T 7→ T ⊗is linear and injective. Given b ∈ L(E,F ;G) we define T ∈ L(E ⊗F,G) by T (ei⊗ fj) :=b(ei, fj); then (ΘT )(ei, fj) = T (ei ⊗ fj) = b(ei, fj), so Θ is an isomorphism; T is calledthe linearization of b.

Theorem 19.2. L(E,F ;G) ∼= L(E ⊗ F,G).

In particular, B(E,F ) ∼= (E ⊗ F )∗.

Let Ei, Fi be vector spaces and Ti ∈ L(Ei, Fi). The linearization T1 ⊗ T2 : E1 ⊗ E2 →F1 ⊗ F2 of the bilinear map

E1 × E2 → F1 ⊗ F2

(x1, x2) 7→ T1(x1)⊗ T2(x2)is called the tensor product of T1 and T2.

Proposition 19.3. If T1 and T2 are injective (surjective), then so is T1 ⊗ T2.

Proof. Clear from Proposition 19.1 because a linear map is injective (surjective) if andonly if the image of a given basis is linearly independent (spans the codomain).

For vector spaces E and F , define χ : E∗ ⊗ F → L(E,F ) as the linearization of

E∗ × F → L(E,F )

(u, y) 7→ [x 7→ u(x) · y].

Proposition 19.4. χ is injective and imχ = T ∈ L(E,F ) | dim imT <∞.

Proof. Let z =∑

ui⊗yi with yi linearly independent. Then χ(z) = 0 implies that∑ui(x)yi = 0 for all x, hence ui = 0 and z = 0.

T ∈ imχ has finite rank. Conversely, if T ∈ L(E,F ) has finite rank, choose a basis(y1, . . . , yn) of imT and extend it to a basis C of F . Define vi ∈ F ∗ by vi(yi) = 1, vi(y) = 0for y ∈ C \ yi. With ui := vi T ∈ E∗, χ(

∑ni=1 ui ⊗ yi) =

∑vi(T (x))yi = T (x).

Let E,F now be locally convex spaces. Let Tπ be the finest locally convex topology onE⊗F such that ⊗ : E×F → E⊗F is continuous; it is given by the projective topologywith respect to all mappings

id : E ⊗ F → (E ⊗ F,T )

where T is any locally convex topology such that ⊗ : E×F → (E⊗F,T ) is continuous.

Let U , V be 0-basis in E and F , respectively, consisting of absolutely convex sets. ForA ⊆ E and B ⊆ F we set A⊗B = ⊗(A,B). We claim that

acx(U ⊗ V ) | U ∈ U , V ∈ V

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is a 0-basis of Tπ.

First, we show that acx(U ⊗V ) is absorbent. For z =∑n

i=1 xi⊗ yi ∈ E⊗F with xi ∈ E,yi ∈ F , we choose ρi > 0 and σi > 0 with xi ∈ ρiU , yi ∈ σiV , and obtain

(3) z =∑

ρiσixiρi⊗ yiσi∈ (∑

ρiσi) · acx(U ⊗ V ).

Obvisouly acx(U ⊗ V )‖ U ∈ U , V ∈ V is a filter basis on E ⊗ F , and for ρ > 0 andgiven U, V there are U1, U2 such that acx(U1 ⊗ U2) ⊆ ρ · acx(U ⊗ V ) for some U1, V1. ByTheorem 2.10 there is a unique locally convex topology T on E ⊗ F having this familyas a 0-basis. [9.6.]

(1) T ≤ Tπ: we need that ⊗ : E × F → (E ⊗ F,T ) is continuous. This is the casebecause U ⊗ V ⊆ acx(U ⊗ V ).

(2) Tπ ≤ T : every 0-neighborhood in Tπ contains a set of the form W1 ∩ . . . ∩Wk,where each Wi is a 0-neighborhood in E ⊗ F for some topology into which ⊗ iscontinuous. Hence, Ui ⊗ Vi ⊆ Wi for some Ui, Vi, and as we can assume Wi tobe absolutely convex, acx(Ui ⊗ Vi) ⊆ Wi so Wi and hence W1 ∩ . . . ∩ Wk is a0-neighborhood in T .

Definition 19.5. Tπ is called the projective tensor topology or π-topology on E ⊗ F ,and E ⊗π F := (E ⊗ F,Tπ) is called the projective tensor product or π-tensor productof E and F . In case it is Hausdorff, its completion is denoted by E⊗πF .

For LCS E,F,G let L(E,F ;G) be the set of all continuous bilinear mappings E⊗F → Gand L(E,F ) the set of all linear continuous mappings E → F .

Proposition 19.6. Let E,F,G be LCS. Then the map θ : L(E ⊗ F,G) → L(E,F ;G),θT = T ⊗, induces an isomorphism

(4) L(E ⊗π F,G)→ L(E,F ;G).

Proof. As ⊗ is continuous, we have an injection (4). Let b ∈ L(E,F ;G) and an absolutelyconvex 0-neighborhood W in G be given. Choose U, V such that b(U, V ) ⊆ W . ThenT := Θ−1(b) ∈ L(E ⊗ F,G) satisfies T (acx(U × V )) ⊆ acxT (U ⊗ V ) = acx b(U, V ) ⊆acxW , so T ∈ L(E ⊗π F,G).

In particular, we have (E ⊗π F )′ = B(E,F ) := L(E,F ;K).

We will now show that for Hausdorff LCS E,F we have E ′ ⊗ F ′ ⊆ (E ⊗π F )′. In fact,for u ∈ E ′, v ∈ F ′ define bu,v ∈ B(E,F ) by bu,v(x, y) := 〈u, x〉 · 〈v, y〉 for x, y ∈ E × F .Let ψ : E ′ ⊗ F ′ → B(E,F ) be the linearization of the biliner map

E ′ × F ′ → B(E,F )

(u, v) 7→ bu,v.

Suppose now that ψ(z) = 0, where z =∑ui ⊗ vi; we can assume the vi to be linearly

independent. Choose yj ∈ F such that 〈vi, yj〉 = δij. Then for all x ∈ E, ψ(z)(x, yj) =∑〈ui, x〉〈vi, yj〉 = 〈uj, x〉 = 0, so uj = 0 for all j and z = 0, so ψ is injective.

Identifying E ′ ⊗ F ′ with a subspace of (E ⊗π F )′, for u, v we denote by u ⊗ v also thecorresponding linear or bilinear form.

Proposition 19.7. If E and F are Hausdorff LCS then E ⊗π F is Hausdorff.

Proof. Let z 6= 0, then z =∑xi⊗yi with (xi)i and (yi)i linearly independent. Let u ∈ E ′,

v ∈ F ′ be such that u(xi) = δ1i = v(yi). Then u⊗ v ∈ (E ⊗π F )′ and (u⊗ v)(z) = 1.

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We will now examine a family of seminorms defining the projective tensor topology.

Definition 19.8. Given two vector spaces E and F and seminorms p on E and q on F ,we define a seminorm p⊗ q on E ⊗ F by

(p⊗ q)(z) := inf n∑i=1

p(xi)q(yi) | z =n∑i=1

xi ⊗ yi

(z ∈ E ⊗ F ).

Proposition 19.9. Let U ⊆ E and V ⊆ F be absolutely convex and absorbent, andqU , qV their gauges. If πU,V is the gauge of acx(U ⊗ V ) then πU,V = qU ⊗ qV . Moreover,for x ∈ E and y ∈ F we have πU,V (x⊗ y) = qU(x)qV (y).

Proof. Recall (3): we can write any z =∑n

i=1 xi ⊗ yi as

z =n∑i=1

ρiσixiρi⊗ yiσi∈ (∑

ρiσi) acx(U ⊗ V )

for any ρi > 0, σi > 0 with xi ∈ ρiU , yi ∈ σiV . For any ε > 0 it is possible tochoose ρi < qU(xi) + ε, σi < qV (yi) + ε, so πU,V (z) ≤

∑(qU(xi) + ε)(qV (yi) + ε) and

πU,V (z) ≤ qU ⊗ qV .

Conversely, let z ∈ ρ · acx(U ⊗V ) for some ρ > 0. Then z =∑λixi⊗ yi with

∑|λi| ≤ ρ,

xi ∈ U , yi ∈ V , so we have (qU ⊗ qV )(z) ≤ ρ and hence (qU ⊗ qV )(z) ≤ infρ > 0 | z ∈acx(U ⊗ V ) = πU,V .

For the last claim let z = x ⊗ y. Choose u ∈ E ′, v ∈ F ′ such that |u| ≤ qU , |v| ≤ qV ,〈u, x〉 = qU(x) and 〈v, y〉 = qV (y) (Theorem 9.4). For any representation z =

∑xi ⊗ yi

then

πU,V (z) ≤ qU(x) · qV (y) = 〈u, x〉 · 〈v, y〉 = 〈u⊗ v, x⊗ y〉

= 〈u⊗ v, z〉 ≤∑|〈u, xi〉| · |〈v, yi〉| ≤

∑qU(xi) · qV (yi).

Taking the infimum over all representations of z gives the claim.

Corollary 19.10. Let P and Q be directed families of seminorms generating the topolo-gies of E and F ; then p⊗q | p ∈P, q ∈ Q is a directed family of seminorms generatingthe topology of E ⊗π F .

Corollary 19.11. If E,F are metrizable (normable) LCS then E ⊗π F is metrizable(normable).

Proposition 19.12. Let Ei, Fi be LCS and Ti ∈ L(Ei, Fi). Then T1 ⊗ T2 ∈ L(E1 ⊗πE2, F1 ⊗π F2).

Proof. Given absolutely convex 0-neighborhoods Ui in Fi, choose absolutely convex 0-neighborhoods Vi in Ei such that Ti(Vi) ⊆ Ui; then (T1 ⊗ T2)(acx(V1 ⊗ V2)) ⊆ acx(T1 ⊗T2)(V1 ⊗ V2) ⊆ acx(U1 ⊗ U2).

To state an important result about projective tensor products of Frechet spaces we needthe following terminology.

Definition 19.13. Let E be a LCS.

A sequence (xn)n in E is called summable if for each 0-neighborhood V in E there isN ∈ N such that for all finite subsets J ⊆ n ∈ N | n ≥ N we have

∑n∈J xn ∈ V .

A sequence (xn)n is called absolutely summable if for each continuous seminorm p on Ethe sequence (p(xn))n is summable.

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Remark 19.14. An absolutely summable sequence is summable. If (xn)n is summable,

the sequence of partial sums (∑N

n=1 xn)N is a Cauchy sequence. If the limit exists it isdenoted by

∑∞n=1 xn is called an (absolutely) convergent series.

Theorem 19.15. Let E,F be Frechet spaces. Any element u ∈ E⊗πF can be representedby an absolutely convergent series

u =∞∑n=1

λnxn ⊗ yn

with (λn)n ∈ `1, xn → 0 in E and yn → 0 in F .

Proof. Let (pn)n and (qn)n be increasing sequences of seminorms defining the topologiesof E and F , respectively, and set rn := pn ⊗ qn. Denote by rn the continuous extensionof rn to E⊗πF . There is a sequence (un)n in E ⊗ F such that rn(u− un) < n−22−(n+1).

Set vn = un+1 − un for n ∈ N and let u1 =∑i1

i=1 λixi ⊗ yi be any representative of u1.Then

rn(vn) ≤ rn(u− un) + rn(u− un+1) ≤ rn(u− un) + rn+1(u− un+1)

< n−22−(n+1) + (n+ 1)−22−(n+2) ≤ n−22−n.

Because rn = πUn,Vn where Un and Vn are the unit balls of pn and qn Proposition 19.9 thismeans that vn ∈ n−22−n acx(Un ⊗ Vn) = 2−n acx(Un

n⊗ Vn

n), so there is a representation

vn =∑in+1

i=in+1 λixi ⊗ yi such that pn(xi) ≤ n−1, qn(xi) ≤ n−1 for in < i ≤ in+1 and∑in+1

i=in+1 |λi| ≤ 2−n. Thus, we can write

u = u1 +∞∑n=1

vn =∞∑i=1

λixi ⊗ yi

as desired. This series is absolutely convergent because

rn(λixi ⊗ yi) ≤ |λi|

for i large enough.

We now move to the other important topology on the tensor product. Because E ′⊗F ′ ⊆B(E,F ) we have

E ⊗ F = (E ′σ)′ ⊗ (F ′σ)′ ⊆ B(E ′σ, F′σ) ⊆ Bs(E ′σ, F

′σ)

where Bs(E ′σ, F′σ) denotes the set of separately continous bilinear mappings E ′σ×F ′σ → K,

and E ′σ = (E ′, σ(E ′, E)).

Definition 19.16. The topology of bi-equicontinuous convergence on Bs(E ′σ, F′σ) is the

unique locally convex topology having as 0-basis all sets of the form

WA,B := φ ∈ Bs(E ′σ, F′σ) | φ(A,B) ⊆ D

for A ⊆ E ′ and B ⊆ F ′ equicontinous.

For this to be well-defined by Theorem 2.10 we have to verify:

(1) WA,B is absolutely convex (this is clear) and absorbent (see below).(2) WA∪A′,B∪B′ ⊆ WA,B ∩WA′,B′ , so these sets form a filter basis.(3) ∀ρ,A,B ∃A′, B′ : WA′,B′ ⊆ ρWA,B: in fact, WρA,B ⊆ ρWA,B.

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For WA,B to be absorbent we note that φ ∈ λWA,B ⇔ φ(A,B) ⊆ λD, so we need thateach φ ∈ Bs(E ′σ, F

′σ) is bounded on products of equicontinuous sets. We can assume that

A,B are absolutely convex and weakly closed (Lemma 12.10), hence weakly compact(Theorem 12.9) and thus bounded and complete. The following result then gives theclaim:

Lemma 19.17. Let E,F be LCS and φ : E×F → K bilinear and separately continuous.If A ⊆ E and B ⊆ F are absolutely convex, bounded and complete then φ(A × B) isbounded.

Proof. Set φx := φ(x, .) ∈ F ′. The family φx : x ∈ A is σ(F ′, F )-bounded, so U :=⋂x∈A φ

−1x (D) ⊆ F is absolutely convex, closed and absorbent (y ∈ F ⇒ φ(A, y) ⊆ λD⇒

λ−1y ∈ U), i.e., a barrel. By Theorem 13.11, U absorbs B, and B ⊆ λU means thatφ(A,B) ⊆ λD.

We denote by Bsε(E

′σ, F

′σ) the space Bs(E ′σ, F

′σ) with the topology of bi-equicontinuous

convergence, and by E⊗ε F the space E⊗F with the induced topology. E⊗ε F is calledthe ε-tensor product of E and F , its topology the ε-tensor topology (or: injective tensorproduct, injective tensor topology).

Proposition 19.18. E ⊗π F → E ⊗ε F is continuous.

Proof. Let WA,B be given, then A ⊆ U, B ⊆ V for some 0-neighborhoods U, V . Forx ∈ U , y ∈ V we have (x⊗y)(a, b) = 〈a, x〉 ·〈b, y〉 ∈ D, so x⊗y ∈ WA,B and acx(U⊗V ) ⊆WA,B.

Proposition 19.19. For Ti ∈ L(Ei, Fi), T1 ⊗ T2 ∈ L(E1 ⊗ε E2, F1 ⊗ε F2).

Proof. Given WA,B, T ′1(A) and T ′2(B) are equicontinuous and (T1 ⊗ T2)(WT ′1(A),T′2(B)) ⊆

WA,B.

We now move to nuclear mappings. Note that the canonical mapping χ : E∗ ⊗ F →L(E,F ) maps E ′ ⊗ F into L(E,F ), as ‖χ(x′ ⊗ y)(x)‖ = ‖x′(x) · y‖ ≤ ‖x′‖ · ‖y‖ · ‖x‖.Lemma 19.20. If E,F are Banach spaces and L(E,F ) is endowed with the operatornorm, then χ : E ′ ⊗π F → L(E,F ) is continous.

Proof. For u ∈ E ′ ⊗ F , given any representation u =∑x′i ⊗ yi we have

‖χ(u)‖ = sup‖x‖≤1

∑|x′i(x)| · ‖yi‖ ≤

∑‖xi‖‖yi‖

and the infimum over all representations of u gives ‖u‖ on the right side.

Definition 19.21. Let E,F be Banach spaces. A linear map T ∈ L(E,F ) is callednuclear T = χ(u) for some u ∈ E ′⊗πF , where χ is the extension of χ to the completion.

Using Theorem 19.15 we obtain the following characterization:

Proposition 19.22. Let E,F be Banach spaces and T ∈ L(E,F ). The following areequivalent:

(i) T is nuclear.(ii) T is given by

T (x) =∞∑n=1

λnx′n(x)yn

with∑|λn| <∞, x′n → 0 and yn → 0.

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(iii) T is given by

T (x) =∞∑n=1

x′n(x)yn

with∑‖x′n‖‖yn‖ <∞.

Proof. (i) ⇒ (ii): if T is nuclear then T = χ(u) for some u ∈ E ′⊗πF , and u =∑∞n=1 λnx

′n ⊗ yn with (λn) ∈ `1, xn → 0 in E and yn → 0 in F . Because χ is continuous

and u = limN→∞ uN with uN :=∑N

i=1 λnxn ⊗ yn we have χ(u)(x) = limχ(uN)(x) =∑∞n=1 λnx

′n(x)yn.

(ii) ⇒ (iii): set x′n := λnx′n, and we have

∑‖χ′n‖ · ‖yn‖ =

∑|λn| ‖x′n‖‖yn‖ <∞.

(iii) ⇒ (i): define u :=∑∞

n=1 x′n ⊗ yn ∈ E ′⊗πF , which is absolutely convergent by

assumption. Then χ(u)(x) = χ(limN→∞ uN)(x) = limN→∞ χ(uN)(x) =∑∞

n=1 x′n(x)yn =

T (x).

We now generalize the definition of nuclearity to arbitrary LCS E,F , which we suppose tobe Hausdorff. We call a linear map u ∈ L(E,F ) bounded if it maps some 0-neighborhoodU to a bounded set, i.e., u(U) is bounded in F . In this case it also is continuous. Wecan suppose that U and B are absolutely convex. First, we note that u(E) ⊆ FB: x ∈ Eimplies x ∈ λU for some λ > 0, so u(x) ∈ λB and u(x) ∈ FB. If pU(x) = 0 then u(x) = 0by Lemma 13.8 (i), so the map v ∈ L(E,FB) defined by u factors as v = u0 φU forsome u0 ∈ L(Eu, FB). Because u0(φU(U)) = u(U) ⊆ B and φU(U) is a 0-neighborhoodin EU , u0 is continuous. Hence, we can write u = ψB u0 φU , where φU ∈ L(E,EU)and ψB ∈ L(FB, F ) are the canonical mappings and u0 ∈ L(Eu, FB). If FB is complete,

u0 has a continuous extension u0 ∈ L(EU , FB) for which u = ψB u0 φV .

Definition 19.23. Let E,F be Hausdorff LCS. u ∈ L(E,F ) is called nuclear if thereis an absolutely convex 0-neighborhood U in E and an absolutely convex bounded subsetB ⊆ F which is infracomplete (i.e., FB is complete) and such that the induced mapping

u0 ∈ L(EU , FB) is nuclear.

The following characterization is very useful:

Theorem 19.24. u ∈ L(E,F ) is nuclear if and only if it is of the form

u(x) =∞∑n=1

λnfn(x)yn

where∑∞

n=1 |λn| <∞, fn : n ∈ N ⊆ E ′ is equicontinuous, and yn : n ∈ N is boundedin FB for some absolutely convex bounded infracomplete set B ⊆ F .

If u is given like this we write u =∑∞

n=1 λnfn ⊗ yn.

Proof. If u is nuclear then u = ψB u0 φU with U an absolutely convex 0-neighborhood,

B absolutely convex bounded infrabarrelled, and u0 ∈ L(EU , FB) nuclear. We know

(Proposition 19.22) that u0(x) =∑∞

n=1 λngn(x)yn with∑|λn| <∞, gn ∈ (EU)′, gn → 0

and yn ∈ FB with yn → 0. Define fn := gn φU ∈ E ′. Because (gn)n is bounded in (EU)′

it is equicontinuous and (fn)n is equicontinuous. Hence, we have

u(x) =∞∑n=1

λnfn(x)yn

which is of the desired form.

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Conversely, if u(x) =∑∞

n=1 λnfn(x)yn, set U :=⋂n f−1n (D). The series converges abso-

lutely in FB, which is complete:∑‖λfn(x)yn‖ ≤

∑|λn| · sup

n|fn(x)| · ‖yn‖

and in fact u(x) = ψB(∑∞

n=1 λnfn(x)yn).

As u(U) is bounded in FB we have pU(x) = 0, hence u(x) = 0, so we can factoru : E → fB as u = ψB u0 φU , with u0(φU(x)) = u(x) for all x ∈ E; Because fn ≡ 0on p−1U (0), there is hn ∈ (EU)′ with fn = hnφU . We have ‖hn‖ = supx∈EU ,‖x‖≤1 |hn(x)| =supp(x)≤1,x∈E |hn(φU(x))| = supp(x)≤1,x∈E |fn(x)| ≤ 1. Hene, u0(φU(x)) =

∑∞n=1 λnhn(φU(x))yn,

and because φU is surjective, u0(x) =∑∞

n=1 λnhn(x)yn. Because u0(x) =∑∞

n=1 λnhn(x)yn,we are finished.

Thus,

u0(x) =∞∑n=1

λnhn(x)yn

and because

u0(x) =∞∑n=1

λnhn(x)yn

(use absolute convergence of the series) we are finished (using Proposition 19.22).

We have seen in the proof that we can have fn → 0 in (EU)′ and yn → 0 in FB. Hence,assuming that

∑|λn| ≤ 1, fn(U) with U =

⋂n f−1n (D) is contained in the absolutely

convex closed hull C of yn in FB; C is compact in FB and hence also in F . This proves:

Corollary 19.25. Every nuclear map is compact (i.e., maps a 0-neighborhood to a rela-tively compact set).

We note the following ideal property:

Proposition 19.26. Let u ∈ L(E,F ), v ∈ L(F,G) and w ∈ L(G,H). If v is nuclearthen v u and w v are nuclear.

Proof. First, we have (v u)(x) =∑λn(fn u)(x)yn with fn u equicontinuous.

Second, there is an absolutely convex 0-neighborhood V such that v(V ) = B is compact.Hence, B1 = w(B) is compact and HB1 is complete, so we have

(w u)(x) =∑

λnfn(x)w(yn)

which shows nuclearity.

Definition 19.27. A LCS E is called nuclear if it has a 0-basis U consisting of absolutely

convex 0-neighborhoods such that for each U ∈ U the canonical mapping E → EU isnuclear.

Proposition 19.28. If a normed space E is nuclear then it is finite dimensional.

Proof. If U is an absolutely convex bounded 0-neighborhood then E → EV is a homeo-

morphism; if E → EV is nuclear it is compact and hence E is locally compact and finitedimensional by Theorem 8.4.

We give the following characterization:

Proposition 19.29. For a LCS E, the following assertions are equivalent:

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(i) E is nuclear(ii) Every continuous linear map of E into any Banach space is nuclear.

(iii) For U ⊆ E an absolutely convex 0-neighborhood there is V ⊆ U absolutely convex

0-neighborhood such that EV → EU is nuclear.

Proof. (i) ⇒ (ii): if F is any Banach space and u ∈ L(E,F ), there is V such that

φV : E → EV is nuclear and u(V ) is bounded in F . As u vanishes on p−1V (0) and φV (E) =

EV there is a map v ∈ L(EV , F ) such that u = v φV , hence u is nuclear.

(ii) ⇒ (iii): given U , φU : E → EU is nuclear, hence given by φU =∑λnfn ⊗ yn. We set

V := U∩⋂n f−1n (D), which is an absolutely convex 0-neighborhood, and each fn induces a

a continuous linear form of norm ≤ 1 on EV , which has a continuous extension hn to EV .

As V ⊆ U there is a canonical map φU,V : EV → EU given by (the continuous extension of)φU,V (φV (x)) = φU(x). Now φU,V (φV (x)) = φU(x) =

∑λnfn(x)yn =

∑λnhn(φV (x))yn,

hence φU,V =∑λnhn ⊗ yn, which is nuclear.

(iii)⇒ (i): given U there is V with φU,V nuclear; because φU = φU,V φV , φU is nuclear. [23.6.]

Lemma 19.30. If E is nuclear, every bounded subset of E is precompact.

Proof. We use the representation E = lim←− EU , where U runs through a 0-basis of abso-

lutely convex 0-neighborhoods. Each φU : E → EU is nuclear, so if B ⊆ E is bounded,

KU := φU(B) is compact (Corollary 19.25). The injection f : E → lim←− EU is given

by f(x) = (φU(x))U , so f(B) = (KU)U ⊆ f(E) is compact because f(E) is closed in∏EU .

Corollary 19.31. If E is nuclear and quasicomplete, τ(E ′, E) = β(E ′, E).

We come back to the injective tensor product in more detail now.

Definition 19.32. Let E,F be vector spaces. We call a topology T on E⊗F compatiblewith the tensor product if the following conditions are satisfied:

(a) The canonical mapping E×F → E⊗τ F := (E⊗F,Tτ ) is separately continuous.(b) u ∈ E ′, v ∈ F ′ ⇒ u⊗ v ∈ (E ⊗τ F )′.(c) If G1 ⊆ E ′ is equicontinuous and G2 ⊆ F ′ is equicontinuous then G1 ⊗ G2 ⊆

E ′ ⊗ F ′ ⊆ (E ⊗τ F )′ is equicontinuous.

We remark that (a) implies (E ⊗τ F )′ ⊆ Bs(E,F ): u ∈ (E ⊗τ F )′, ε > 0 and W a0-neighborhood in E ⊗τ F with u(W ) ⊆ εD. Given x0 ∈ E there is a 0-neighborhood Vin F with x0⊗ V ⊆ W , so (θu)(x0, V ) = u(x0⊗ V ) ⊆ u(W ) ⊆ εD, so θu is continuous iny; similarly one sees continuity in x.

Moreover, (a) and (b) imply that E ′ ⊗ F ′ ⊆ (E ⊗τ F )′ ⊆ Bs(E,F ).

Lemma 19.33. Tτ is compatible with the tensor product if and only if T is the S-topology for a class S of subsets of Bs(E,F ) ⊂ (E ⊗ F )∗ such that

(1) each A ∈ S is separately equicontinuous,(2) for all equicontinuous subsets G1 ⊆ E ′ and G2 ⊆ F ′, G1 ⊗G2 ∈M .

Proof. Suppose Tτ is compatible and let S be the class of all equicontinuous subsetsof (E ⊗τ F )′. If G1, G2 are equicontinuous then G1 ⊗ G2 is equicontinuous by Defi-nition 19.32 (c), hence an element of S. Moreover, for A ∈ S, then A(W ) ⊆ D for

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some 0-neighborhood W in E ⊗τ F . For given x0 ∈ E there is V with x0 ⊗ V ⊆ W ,hence A(x0 ⊗ V ) ⊆ D and (as the same works for the second variable) A is separatelyequicontinuous.

Conversely, if these conditions hold, let u ∈ E ′ and v′ ∈ F ′. Then A := u⊗ v is in S,hence A is a 0-neighborhood in E ⊗τ F with (u ⊗ v)(A) ⊆ D, so u ⊗ v ∈ (E ⊗τ F )′.This implies Definition 19.32 (b), and (c) is seen the same way.

Given W and x0 we now want V with x0 ⊗ V ⊆ W . We can assume W = M for someM which is absolutely convex and separately equicontinuous, hence M(x0, V ) ⊆ D forsome V , which means x0 ⊗ V ⊆M = W .

This motivates to take for S the family of all G1 ⊗ G2 where G1 ⊆ E ′ and G2 ⊆ F ′

are equicontinuous; this gives the coarsest topology compatible with the tensor prod-uct. Is is evident that this topology coincides with the ε-tensor topology introduced inDefinition 19.16.

For the seminorms we recall the following result (see also Section 12).

Lemma 19.34. Let (E,F ) be paired vector spaces and A ⊆ F be σ(F,E)-bounded. Thenthe gauge of A is given by

qA(x) = supy∈A|〈x, y〉|

Proof. Suppose r = supy∈A |〈x, y〉| > 0. Then |〈x, y〉| ≤ r for all y ∈ A, hence x/r ∈ A,x ∈ rA and qA(x) ≤ r. For any t < r, x/t 6∈ A and hence x 6∈ tA, which gives theclaim. For r = 0, x ∈ λA for all λ > 0 and hence qA(x) = 0.

The seminorms of the ε-tensor topology hence are given by

εG1,G2(z) = supu∈G1,v∈G2

|〈u⊗ v, z〉| sup(u,v)∈G1×G2

∣∣∣∣∣n∑i=1

〈u, xi〉〈v, yi〉

∣∣∣∣∣for z =

∑ni=1 xi ⊗ yi, where G1 ⊆ E ′ and G2 ⊆ F ′ are equicontinuous.

If E and F are normed spaces then E ⊗ε F is a normed space with norm

ε(z) = ‖z‖ε = sup‖u‖≤1,‖v‖≤1

|(u⊗ v)(z)| = sup‖u‖≤1,‖v‖≤1

∣∣∣∑〈u, xi〉〈v, yi〉∣∣∣ .With this norm, if U and V are the closed unit balls of E and F , then (U ⊗ V ) is theclosed unit ball of E ⊗ε F .

Proposition 19.35. If E,F are LCS and E is nuclear, E ⊗ F is dense in Bsε(E

′σ, F

′σ).

For this, we need some preparations.

Let F be a topological vector space and T any set, S a family of subsets of T directedunder inclusion. Then the family

WA,V := f : T → F | f(A) ⊆ V with A ∈ S and V running through all 0-neighborhoods in F , is a 0-basis of a uniquetranslation-invariant topology on F T , called the S-Topology or topology of uniform con-vergence on sets A ∈ S. In fact, if V3 ⊆ V1 ∩ V2 and A1 ∪ A2 ⊆ A3, then WA3,V3 ⊆WA1,V1 ∩WA2,V2 , so these sets form a filter basis. Moreover, we have

V + V ⊆ U ⇒ WA,V +WA,V ⊆ WA,U .

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Note that we can replcae S with a fundamental subfamily S1, i.a., a family such thateach A ∈ S is contained in some A1 ∈ S1, an we can take the 0-neighborhoods of V onlyfrom some 0-basis, without changing the S-topology.

Lemma 19.36. A linear subspace G ⊆ F T is a topological vector space with respect tothe S-topology if and only if ∀f ∈ G ∀A ∈ S, f(A) is bounded in F .

Proof. One only has to show that WA,V is absorbent, but this onle is the case if the statedcondition holds.

If pα is a family of seminorms generating the topology of F , the seminorms

pA,α(u) := supx∈A

pα(u)

generate the S-topology on L(E,F ).

Lemma 19.37. Bsε(E

′σ, F

′σ) ∼= Lε(E ′τ , F ), where E ′τ = (E ′, τ(E ′, E)) is E ′ with the

Mackey topology.

Proof. Algebraically we have B(E,F ) ∼= L(E,F ∗), where b ∈ B(E,F ) corresponds to

b ∈ L(E,F ∗) via b(x, y) = b(x)(y). Topologically, if b ∈ Bs(E,F ) then obviously b ∈L(E,F ′). Moreover, if M with M ⊆ F finite is a 0-neighborhood in F ′σ, there is a

0-neighborhood U in E with b(U,M) ⊆ D and hence b(U) ⊆ M, which means that

b ∈ L(E,F ′σ). Conversely, if b ∈ L(E,F ′σ) then b ∈ B(E,F ) and B(x., ) ∈ F ′ for fixedx. For y ∈ F , y is a 0-neighborhood in F ′σ, so there is a 0-neighborhood U in E with

b(U) ⊆ y, which means that B(U, y) ⊆ D. This shows that Bs(E,F ) ∼= L(E,F ′σ), so

Bs(E ′σ, F′σ) ∼= L(E ′σ, (F

′σ)′σ) ∼= L(E ′σ, Fσ).

Moreover, if u ∈ L(E ′σ, Fσ) then u ∈ L(E ′τ , Fτ ) by Proposition 18.3 (ii), hence u ∈L(E ′τ , F ), which in turn implies u ∈ L(E ′σ, Fσ) by Proposition 18.3 (i).

For the topological isomorphism we note that for equicontinuous subsets A ⊆ E ′ andB ⊆ F ′,

b ∈ WA,B ⇔ b(A,B) ⊆ D⇔ b(A)(B) ⊆ D⇔ b(A) ⊆ B ⇔ b ∈ WA,B .

Proposition 19.38. Lε(E ′τ , F ) is complete if E and F are complete.

Proof. Let F be a Cauchy filter on Lε(E ′τ , F ). The evaluation map Lε(E ′µ, F ) → F iscontinuous, hence for each x ∈ E ′, F (x) is a Cauchy filter basis converging to someu(x) ∈ F , which defines a map u : E ′τ → F . u is linear because F (x) +F (y) ⊆ F (x+ y)(for F,G ∈ F there is H ∈ F with H(x) ⊆ F (x)+G(x), hence H(x+y) ⊆ F (x)+G(y))implies that F (x + y)→ u(x) + u(y); similarly, F (λx) = λF (x)→ λu(x). So, F → uin FE′ .

Now because the 0-neighborhoods WA,V of Lε(E ′µ, F ) are closed w.r.t. the topology of

pointwise convergence (WA,V =⋂x∈A ev−1x (V ) is closed in Lσ(E ′τ , F )), Lemma 13.9 (or

rather: its variant for topological groups) applies and the convergence of F is uniform onequicontinuous sets. It suffices to show that u ∈ L(E ′, F ) is weakly continuous becausethen it is continuous w.r.t. the Mackey topologies and hence in L(E ′τ , Fτ ) ⊆ L(E ′τ , F ).Now u is weakly continuous if and only if its transpose u′ ∈ L(F ′, (E ′)∗) maps F ′ into(E ′τ )

′ = E. Transposition is a mapping

ad: L(E ′, F )→ L(F∗, (E ′)∗),ad: L(E ′τ , F )→ L(F ′σ, Eσ)

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F is a filter in L(E ′τ , F ), so ad(F ) is a filter in L(F ′σ, Eσ), and ad(F )(v) a filter inE = (E ′τ )

′. ad(u)(v) is an element of (E ′)∗. Let ε > 0, M ⊆ E ′ equicontinuous and V ⊆ Fa 0-neighborhood such that v(V ) ⊆ εD. Then there is F ∈ F with F (M)−u(M) ⊆ V , so(ad(F )(u)−ad(u)(v))(M) = v(F (M)−u(M)) ⊆ v(V ) ⊆ εD, hence ad(F )(v)→ ad(u)(v)uniformly on equicontinuous subsets of E ′, which implies that ad(u)(v) = u′(v) ∈ Ebecause (E ′τ )

′ε = E is complete.

Moreover, we need the following result.

Lemma 19.39. Let U ⊆ E be an absolutely convex 0-neighborhood and set A = U.

Then the adjoint φ′U : (EU)′ → E ′ of the quotient mapping φU : E → EU maps into E ′Aand defines a topological isomorphism (EU)′ ∼= E ′A.

Proof. Let u ∈ (EU)′ and set v := u φU ; then clearly v ∈ E ′ and because v(U) isbounded, v is absorbed by A.

Conversely, let v ∈ E ′A be given. Then v(U) is bounded, hence v vanishes on q−1U (0) andthere is u ∈ (EU)′ such that u φU = v; extending u to the completion gives the desired

element of (EU)′.

Proof of Proposition 19.35. Let f ∈ L(E ′τ , F ), A ⊆ E ′ absolutely convex and equiconti-nous, and V ⊆ F an absolutely convex 0-neighborhood. We have to find f0 ∈ E⊗F suchthat (f − f0)(A) ⊆ V .

Set U := A, which is an absolutely convex 0-neighborhood in E. By Proposition 19.29(iii) there is an absolutely convex 0-neighborhood W ⊆ U such that the canonical map-

ping φU,W : EW → EU is nuclear, hence given by

φU,W =∑

λiyi ⊗ zi

for (λi) ∈ `1, (yi) bounded in (EW )′, and (zi) bounded in EU . Set B := W . BecauseW ⊆ U we have A ⊆ B, so there is a canonical inclusion φB,A : E ′A → E ′B. We now claimthat the following diagram commutes:

E ′AφB,A //

(φ′U )−1

E ′B

(EU)′φ′U,W

// (EW )′

φ′W

OO

In fact, for u ∈ E ′A we have

φ′W (φ′U,W ((φ′U)−1(u)))(x) = (φ′U)−1(u)(φU,WφW (x)) = (φ′U)−1(u)(φU(x)) = u(x)

for all x ∈ E. It follows that we can write (with zi ∈ Eu ⊆ (E ′W )′)

φBA =∞∑i=1

λ(zi (φ′U)−1)⊗ φ′W (yi).

and given any ε > 0 we can choose n such that

‖n∑i=1

λi(zi (φ′U)−1)⊗ φ′W (yi)− φBA‖ ≤ ε/2.

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where the norm is the operator norm in L(E ′A, E′B). Moreover, for any given κi > 0 we

can choose zi ∈ EU such that ‖zi − zi‖ ≤ κi in ((EU)′)′. It follows that for x′ ∈ A,

‖n∑i=1

(zi (φ′U)−1)(x′)φ′W (yi)− x′‖ ≤ ‖n∑i=1

((zi − zi) (φ′U)−1)(x′)φ′W (yi)‖+ ε/2 ≤ ε

(norm in E ′B) for the κi chosen small enough, i.e.,n∑i=1

(zi (φ′U)−1)(x′)φ′W (yi)− x′ ∈ εB.

Choose now ε such that εf(B) ⊆ V . Thenn∑i=1

(zi (φ′U)−1)(x′)f(φ′W (yi))− f(x′) ∈ V

and if we take z′i ∈ E such that zi = φU(z′i) = z′i φ′U then

f0 :=n∑i=1

zi ⊗ f(φ′W (yi))

is the desired element of E ⊗ F .

Proposition 19.40. If E,F are LCS and E is nuclear then E ⊗π F = E ⊗ε F .

Proof. The map E ⊗π F → Bsε(E

′σ, F

′σ) is continuous and has dense image, hence by

Proposition 18.2 its adjoint (Bsε(E

′σ, F

′σ))′ → B(E,F ) is injective. In order to show that

the identity id : E⊗πF → E⊗εE is open, let U ⊆ E⊗πF be an absolutely convex closed0-neighborhood. We need to find a 0-neighborhood W in Bs

ε(E′σ, F

′σ) such that W ∩ (E⊗

F ) ⊆ U . The polar A := U is an equicontinuous subset of B(E,F ), and we can assumethat A = u ∈ B(E,F ) | u(U, V ) ∈ D for some absolutely convex 0-neighborhoods Uin E and V in F . We claim that A is contained and equicontinuous in (Bs

ε(E′σ, F

′σ))′,

because then its polar W is a 0-neighborhood in Bsε(E

′σ, F

′σ) and W ∩ (E ⊗ F ) is the

polar in E ⊗ F , which equals U = U .

Because E is nuclear the map φU : E → EU is nuclear, hence of the form

φU =∑

λifi ⊗ xi

with (xi) bounded in EU and (fi) ⊆ E ′ equicontinous as well as∑|λi| ≤ 1. Let u ∈ A;

because u(U, V ) is bounded there is v ∈ B(EU , F ) such that u(x, y) = v(φU(x), y), and

v(B, V ) ⊆ D if B is the unit ball in EU . So, we have

u(x, y) =∞∑i=1

λifi(x)v(xi, y) =∑

λifi(x)gi(y)

if we set gi := v(xi, .). The family (gi)i is equicontinuous, hence u ∈ acx(G1 ⊗ G2) forequicontinuous sets G1 ⊆ E ′, G2 ⊆ F ′. Because (acx(G1 ⊗G2))

is a 0-neighborhood inBsε(E

′σ, F

′σ), and A is contained in its polar, A is equicontinuous as desired.

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REFERENCES 66

References

[D] J. A. Dieudonne. “Recent developments in the theory of locally convex vectorspaces”. In: Bull. Am. Math. Soc. 59 (1953), pp. 495–512. issn: 0002-9904. doi:10.1090/S0002-9904-1953-09752-X.

[S] H. H. Schaefer. Topological Vector Spaces. New York: Springer-Verlag, 1971.isbn: 978-0-387-05380-6.

https://doi.org/10.1090/S0002-9904-1953-09752-X

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