1Functional analysis tools

1. Banach space: weak topologies

Let X, Y be Banach spaces (normed vector spaces in which every Cauchy se-

quence has a limit).

Let A be linear X ! Y . Then A is continuous i↵ there exists c > 0 such that

(1.1) kAxkY ckxkX , for all x 2 X.

The space L(X, Y ) of linear mappings X ! Y , is a Banach space when endowed

with the norm

(1.2) kAkL(X,Y ) := sup{kAxkY ; kxkX 1}.1.0.1. Unique extension of linear mappings. Let X, Y be Banach spaces, and E

be a dense subspace of X.

Let A : E ! Y be linear and such that for some c > 0:

(1.3) kAekY ckekX , for all e 2 E.

Then A has a unique extension to L(X, Y ), i.e. there exists a uniqueA 2 L(X, Y ),

such that Ae = Ae, for all e 2 E, and for x 2 X (the limit below exists):

(1.4) Ax := lim

k{Aek, ek 2 E, ek ! x}.

0Chapter from the lecture notes on ’Optimal control of partial di↵erential equations’ by J.F.Bonnans. Version of April 11, 2020. Updates and additional material on the page

http://www.cmap.polytechnique.fr/⇠bonnans/notes/cedp/cedp.html

9

10 1. FUNCTIONAL ANALYSIS TOOLS

1.0.2. Multilinear mappings. Let X := X1 ⇥ · · · ⇥ Xn, a : X ! Y multilinear,

the Xi and Y being Banach spaces. Then a is continuous i↵ there exists c > 0

such that

(1.5) ka(x1, . . . , xn)kY ckx1kX1 · · · kxnkXn

for all x = (x1, . . . , xn) 2 X.

If E is a dense subset of X, a : E ! Y is multilinear, and the above inequality holds

for all x 2 E, then a has a unique continuous multilinear extension to X.

Remark 1.1. If the above multilinear mapping a is continuous, then it is of

class C1, with derivative at x 2 X in direction h 2 X given by

(1.6) Da(x)h =

nX

i=1

a(x1, . . . , xi�1, hi, xi+1, . . . , xn).

Example 1.2. Let p, q, r belong to (1,1), such that 1/r = 1/p + 1/q, and let

⌦ be an open subset of Rn. By Holder’s inequality, if f 2 Lp

(⌦) and g 2 Lq(⌦),

then fg 2 Lr(⌦) and

(1.7) kfgkLr(⌦) kfkLp(⌦)kgkLp(⌦).

It follows that the mapping Lp(⌦)⇥ Lq

(⌦) ! Lr(⌦), (f, g) 7! fg is of class C1

.

1.0.3. Topological dual. Let X be a Banach space. A linear form over X is a

linear mapping X ! R.We call topological dual (or, in short, dual) and denote by X⇤

, the set of

continuous linear forms over X.

The action (duality product) of x⇤ 2 X⇤over x 2 X is denoted by hx⇤, xiX . The

dual X⇤is a Banach space, endowed with the dual norm

(1.8) kx⇤kX⇤:= sup{|hx⇤, xiX |; kxkX 1}.

1.0.4. Bidual; reflexive spaces. The bidual (dual of the dual) of X is denoted by

X⇤⇤.

With x 2 X associate the linear form over X⇤:

(1.9) `x(x⇤) := hx⇤, xiX .

The mapping x ! X⇤⇤, x 7! `x is isometric: k`xkX⇤⇤

= kxkX .So, we can identify X with the image of `, which is a closed subspace of X⇤⇤

.

We say that X is reflexive if ` is onto: in that case we can identify X and X⇤⇤,

i.e., X is the dual of X⇤.

1.0.5. Transpose operator. Given A 2 L(X, Y ), fix y⇤ 2 Y ⇤. The mapping

(1.10) `y⇤ : X ! R, x 7! hy⇤, AxiYis continuous since |`y⇤(x)| ky⇤kkAkkxk, so that

(1.11) k`y⇤kX⇤ ky⇤kkAk.

1. BANACH SPACE: WEAK TOPOLOGIES 11

Also y⇤ 7! `y⇤ is linear and (by the above inequality) continuous: so, we may denote

it as A>y⇤, with A> 2 L(Y ⇤, X⇤), and:

(1.12) hA>y⇤, xiX = hy⇤, AxiY , for all x 2 X, y⇤ 2 Y ⇤.

1.0.6. Examples of Banach spaces.

• Hilbert space X: scalar product denoted by (·, ·)X .Associated norm kxkX := (x, x)1/2X . Such spaces are reflexive.

Riesz theorem: if x⇤ 2 X⇤, there exists y 2 X such that

hx⇤, xiX = (y, x)X , for all x 2 X.

• ⌦ open subset of Rn, Lp

(⌦) reflexive if p 2 (1,1) and

(1.13) Lp(⌦)

⇤= Lq

(⌦), 1/p+ 1/q = 1, p 2 [1,1),

• ⌦ as above: L1(⌦) and L1

(⌦) not reflexive.

1.0.7. Punctual convergence by a density argument.

Lemma 1.3. Let X, Y be Banach spaces and Ak be a bounded sequence inL(X, Y ). Let Akx ! 0 for all x in a dense subset E of X. Then Akx ! 0,for all x 2 X.

Proof. Let x 2 X. Fix " > 0, and e 2 E such that

(1.14) sup

`kA`kkx� ek ".

Then

(1.15) kAkxk kAk(x� e)k+ kAkek sup

`kA`kkx� ek+ kAkek.

It follows that

(1.16) lim sup

kkAkxk ".

The result follows. ⇤1.0.8. Weak convergence. Given a sequence {xk} in a Banach space X, and

x 2 X, we say that xk weakly converges to x, and write xk * x if

(1.17) hx⇤, xkiX ! hx⇤, xiX , for all x⇤ 2 X⇤.

Lemma 1.4. Any weakly convergent sequence is bounded and, if X is reflexive,any bounded sequence has a weakly convergent subsequence.

Proof. See Brezis [10]. ⇤1.0.9. Testing over a dense subset.

Lemma 1.5. We have that xk * x in the Banach space X i↵ xk is bounded and,for some dense subset E of X⇤:

(1.18) hx⇤, xkiX ! hx⇤, xiX , for all x⇤ 2 E.

Proof. Apply lemma 1.3. ⇤

12 1. FUNCTIONAL ANALYSIS TOOLS

1.0.10. About unbounded sequences. A counterexample is as follows. In X = `2

(space of summable square sequence) identified with its dual, let xk := kek (ek is

the kth element of natural basis) and E be the subspace of elements with finitely

many nonzero coordinates.

The hypothesis (1.18) of the lemma holds with x = 0, but xk does not weakly

converge since it is unbounded. A direct argument for the impossibility of weak

convergence is: let z :=

Pk ek/k

2/3, then z 2 X and (z, xk)X = k1/3

is unbounded.

1.0.11. Transportation of weak convergence by linear operators. If A 2 L(X, Y )

(X, Y Banach spaces), then

(1.19) xk * x in X implies Axk * Ax in Y .

Indeed, for any y⇤ 2 Y ⇤:

(1.20) hy⇤, AxkiY = hA>y⇤, xkiX ! hA>y⇤, xiX = hy⇤, AxiY .1.0.12. Weak versus strong convergence. Strong convergence implies weak con-

vergence. When does the converse hold ? Easy answer in the case of an Hilbertspace.

Lemma 1.6. Let X be a Hilbert space and xk * x in X. Then

(1.21) kxkX lim inf

kkxkkX lim sup

kkxkkX ,

with equality in both inequalities i↵ xk ! x.

Proof. Use

(1.22)

lim sup kxk � xk2 = lim sup kxkk2X � 2 lim(xk, x)X + kxk2X ,= lim sup kxkk2X � kxk2X .

⇤

Example: X = L2(0, ⇡), xk(t) = sin kt.

Sequence of constant norm, weakly (but not strongly) converging to 0.

Remark 1.7. In the case of a Lpspace, see the related Brezis-Lieb theorem 1.17.

1.0.13. Weak⇤ convergence. Let X be a Banach space. We say that the sequence

x⇤k in X⇤

weakly⇤ converges to x⇤ 2 X⇤if

(1.23) hx⇤k, xi ! hx⇤, xi, for all x 2 X.

Note that, if X is reflexive, then the weak and weak⇤ convergence coincide. In

general, we can only say that the former implies the later. Then, see Brezis [10],ch. 3, Coro. 3.30:

Lemma 1.8. Let X be a separable Banach space (i.e., there exists a dense se-quence). Then any bounded sequence in X⇤ has a weakly⇤ converging subsequence.

1. BANACH SPACE: WEAK TOPOLOGIES 13

Proof. Let x⇤k be a bounded sequence in X⇤

and x` be a dense sequence in X.

It su�ces to check that there exists x⇤ 2 X⇤such that up to the extraction of a

subsequence of x⇤k:

(1.24) hx⇤k, x`i ! hx⇤, x`i, for all ` 2 N .

We obtain this by the following diagonal extraction argument. Indeed, since

hx⇤k, x0i is bounded there exists a subsequence k0

i , for i 2 N, and ↵0 2 R such that

(1.25) |hx⇤k0i

, x0i � ↵0| 2

�i.

By induction we can construct, for any j 2 N⇤, a subsequence i 7! kji of kj�1

i such

that

(1.26) |hx⇤kji

, xii � ↵j| 2

�i.

Then the sequence j 7! x⇤j := x⇤

kjj

satisfies

(1.27) hx⇤j , x`i ! ↵`, for all ` 2 N.

For x 2 X, define

(1.28) hx⇤, xi := lim

jhx⇤, x`

j

i,

where `j is a subsequence such that x`j

! x. One can easily check that the limit

does not depend on the subsequence, and that this defines a continuous linear form

that is a weak⇤ limit of a subsequence of x⇤k. ⇤

Example 1.9. Let ⌦ be an open subset of Rn. Then X = L1

(⌦) is separable.

So, any bounded sequence in X⇤= L1

(⌦) has a weakly⇤ converging subsequence.

1.0.14. Closed convex sets are weakly sequentially closed. Indeed, let X be a

Banach space and K ⇢ X be convex and closed. Let {xk} ⇢ K weakly converge to

x. We need to prove that x 2 K.

If this is not true, by the Hahn Banach theorem (see Brezis [10], ch.1), thereexists x⇤ 2 X⇤ strictly separating x from K:

(1.29) hx⇤, xi < inf{hx⇤, xi; x 2 K}.then we get a contradiction since

(1.30) hx⇤, xi < lim

khx⇤, xki = hx⇤, xi.

1.0.15. What about weakly⇤ closed convex sets ? While for dual spaces we can-

not use the above separation argument, in practice we often find sets which are

intersection of “weakly⇤ closed” halfspaces, i.e., for some E ⇢ X ⇥ R:

(1.31) K := {x⇤ 2 X⇤; hx⇤, xiX ↵, for all (x,↵) 2 E}.

Clearly such a set is weak⇤ sequentially closed.

14 1. FUNCTIONAL ANALYSIS TOOLS

Example 1.10. Let X = L1(Rn

), f 2 X and f ⇤ 2 X⇤= L1

(Rn). We say that

f � 0 if f(x) � 0 a.e., and that f ⇤ � 0 if hf ⇤, fiX � 0, for all f � 0. Thus, a weak⇤limit of a nonnegative sequence in X⇤

is nonnegative.

1.0.16. Case of probability measures. Let ⌦ ⇢ Rnbe compact, and X = C(⌦)

be the Banach space of continuous functions over ⌦, endowed with the norm

(1.32) kfk := max{|f(x)|; x 2 ⌦}.The dual space is X⇤

:= M(⌦), space of bounded Borel measures over ⌦.

The set P(⌦) of Borel probability measures over ⌦ is a sequentially weakly⇤ closed

subset of M(⌦).

Since X is separable it follows that a (necessarily) bounded sequence of prob-

ability measures over ⌦ has a subsequence, weakly⇤ converging to a probability

measure.

Remark 1.11. The set ⌦ being as before, we can identify L1(⌦) with the subset

of M(⌦) of measures having a density. So, a bounded sequence in L1(⌦) will have a

weakly⇤ converging subsequence in M(⌦), but not necessarily a weakly converging

subsequence in L1(⌦).

1.0.17. Jensen’s inequality. Let � : R ! R [ {+1} be convex and lowersemi continuous (l.s.c.):

(1.33) �(x) lim inf

x!x�(x), for all x 2 R.

Let X(!) be a scalar random variable over the probability space (⌦,F , µ), withfinite expectation. Then

(1.34) �(IEX) IE�(X).

In particular if �(x) = |x|p, 1 p < 1:

(1.35) |IEX|p IE|X|p = kXkpLp(⌦).

2. Integration

2.1. Classical results. Three basic theorems

(1) Monotone convergence

(2) Dominated convergence

(3) Fatou lemma

2.1.1. Vitali’s theorem.

Definition 1.12 (Uniform integrability). Let (⌦,F ,P) be a probability space.

We say that a set E of measurable functions is uniformly integrable if, for all

" > 0, there exists M" > 0 such that

(1.36) IE|f |1{|f |>M"

} ", for all f 2 E.

2. INTEGRATION 15

Theorem 1.13. Let (⌦,F ,P) be a probability space, and fk be a uniformly inte-grable sequence in L1

(⌦), with a.s. finite limit f . Then f 2 L1(⌦), and fk ! f in

L1(⌦).

Remark 1.14. The uniform integrability of the sequence fk means that for all

" > 0, there exists M" > 0 such that

(1.37) IE|fk|1{|fk

|>M"

} ", for all k 2 N.Observe that this hypothesis is weaker than the one for the dominated convergence

theorem.

Counter example: ⌦ = R, Lebesgue measure, fk(x) := f(x+k) where f nonzero,

continuous with compact support. So, in general the conclusion does not hold when

meas(⌦) = 1.

2.2. Weak and a.e. convergence.

Lemma 1.15. Let ⌦ be an open subset of Rn, and for 1 < q < 1, a boundedsequence gk in Lq

(⌦), converging a.e. to some g. Then g 2 Lq(⌦) and gk weakly

converges to g in Lq(⌦).

Proof. We have that g 2 Lq(⌦), since by Fatou’s lemma,

(1.38) kgkqLq(⌦) =

Z

⌦

|g(!)|qd! lim inf

k

Z

⌦

|gk(!)|qd!.

For any nonzero N 2 N, set ⌦N := ⌦ \ B(0, 1/N), and

(1.39) EN := {x 2 ⌦N ; |gk(x)� g(x)| 1, for all k � N}.Let h 2 Lp

(⌦), 1/p+ 1/q = 1. Set hN := h1⌦N

and 'k(x) := gk(x)hN(x). Then

(1.40) |'k(x)| (|g(x)|+ 1)hN(x); |g(x)|+ 1 2 Lq(⌦N),

since the measure of ⌦N is finite. By Holder’s inequality:

(1.41) (|g(x)|+ 1)hN(x) 2 L1(⌦N).

By the dominated convergence theorem,

(1.42) lim

k

Z

⌦

gk(x)hN(x)dx = lim

k

Z

⌦N

gk(x)h(x)dx =

Z

⌦N

g(x)h(x)dx.

On the other hand, since [NEN = ⌦ up to a null set, hN ! h a.e. By the dominated

convergence theorem, hN ! h in Lp(⌦). So,

(1.43)

Z

⌦

gk(x)f(x)dx !Z

⌦

g(x)f(x)dx

for f in a dense subset of Lp(⌦). Since gk is bounded in Lq

(⌦), the conclusion

follows. ⇤Remark 1.16. The above result was obtained in J.L. Lions [20, Lemma 1.3,

p.12], in the case of a bounded domain ⌦.

16 1. FUNCTIONAL ANALYSIS TOOLS

2.2.1. The Brezis-Lieb theorem.

Theorem 1.17. Let (⌦,F , µ) be a measured space, and fk be a bounded sequencein Lp

(⌦), p 2 [1,1[, converging a.e. to some f . Then f 2 Lp(⌦), and

(1.44) kfkpp = lim

k

�kfkkpp � kf � fkkpp�.

And so, fk ! f strongly i↵ kfkkp ! kfkp.Proof. See Brezis and Lieb [11]. ⇤Remark 1.18. This improves the classical consequence of Fatou’s lemma: for

p 2 [1,1), convergence a.e. and convergence of the norm implies the convergence

in norm. See T. Gallouet, R. Herbin [16], Ex. 6.17, p. 315 of the Hal 2018 version.

2.2.2. Compactness in Lp. We follow T. Gallouet, R. Herbin [16], Ex. 6.19, p.

317 of the Hal 2018 version.

Lemma 1.19. Let ⌦ have finite measure, and 1 p < q < 1. Let fk be abounded sequence in Lq

(⌦), converging a.e. to f . Then fk ! f in Lp(⌦).

Proof. By Fatou’s lemma, f 2 Lq(⌦). We have that gk := |fk � f |p converges

a.e. to 0, and is bounded in Lr(⌦), with r = q/p > 1. By the lemma below,

a bounded sequence in Lr(⌦) is uniformly integrable. We conclude by applying

Vitali’s theorem to the sequence gk. ⇤Lemma 1.20. A bounded subset of Lr

(⌦), for 1 < r < 1, is uniformly integrable.

Proof. Iet E be a bounded subset in Lr(⌦). By Jensen’s inequality, for h 2 E

and " > 0, setting

(1.45) AM := {x 2 ⌦; |h(x)| > M},we have that

(1.46)

✓1

|AM |Z

AM

|h(x)|dx◆r

1

|AM |Z

AM

|h(x)|rdx 1

|AM |khkrr,

so that, since E ⇢ B(0, R) for some R > 0:

(1.47)

Z

AM

|h(x)|dx |AM |1�1/rkhkr |AM |1�1/rR.

On the other hand

(1.48) |AM | M�r

Z

AM

|h|r M�r

Z

⌦

|h|r M�rRr,

so that |AM | ! 0 when M " 1, uniformy over h 2 E. Combining with (1.47), the

conclusion follows. ⇤

3. MOLLIFICATION 17

3. Mollification

Let be C1: Rn ! R+,

RRn

(x)dx = 1, have support on the unit ball B. The

associated family of mollifiers, for " > 0, where x 2 Rn, is:

(1.49) "(x) := "�n (x/").

It has support on "B and unit integral.

The mollification of f 2 Lp(Rn

), 1 p < 1, is (⇤: convolution product)

(1.50) f"(x) := f ⇤ "(x) =

Z

Rn

f(x� y) "(y)dy.

3.0.1. Regularizing kernel and Jensen’s inequality. Interpret f"(x) as an expec-tation of f(x� ·):

(1.51) f"(x) =

Z

Rn

f(x� y) "(y)dy.

By Jensen’s inequality:

(1.52) |f"(x)|p Z

Rn

|f(x� y)|p "(y)dy.

Integrating over Rn, by Fubini’s theorem, denoting by k ·kp the norm in Lp

(Rn), get

(1.53) kf"kpp kfkpp.So, f 7! f" is linear and nonexpansive in Lp

(Rn).

Remark 1.21. We will use later the fact that this actually holds for an arbitrary

probability measure over Rn.

3.0.2. Regularizing kernel: convergence.

Lemma 1.22. Let f 2 Lp(Rn

), 1 p < 1. Then f" ! f in Lp(Rn

).

Proof. (a) One easily checks that this holds if f 2 C00(Rn), the set of continu-

ous functions with compact support, by the dominated convergence theorem applied

to the relation f"(x) =Rf(x� "y) (y)dy.

(b) C00(Rn) is a dense subset of Lp

(Rn) (enough to approximate simple function or

even characteristic ones, with the argument of regularity of the Lebesgue measure,

used in section 4).

(c) Given ↵ > 0, there exists g 2 C00(Rn) such that kf � gkp < ↵, then when " # 0,

since the regularization is nonexpansive:

(1.54) kf" � fkp kf" � g"kp + kg" � gkp + kg � fkp 2↵ + kg" � gkp,so that lim sup kf" � fkp 2↵. The conclusion follows with ↵ # 0. ⇤

Remark 1.23. We can as well deduce the result from (1.53) and the Brezis Lieb

theorem 1.17.

18 1. FUNCTIONAL ANALYSIS TOOLS

3.0.3. Generalized regularizing kernel: convergence. Let f 2 Lp(Rn

), 1 p < 1,

and let µ be a probability measure over Rn (arbitrary support). Set, for any

Borelian A ⇢ Rnand " > 0:

(1.55) µ"(A) := µ("�1A); f ⇤ µ(x) :=Z

Rn

f(x� y)dµ(y; f"(x) := f ⇤ µ"(x).

By similar arguments, we get:

Lemma 1.24. We have that f ⇤ µ" 2 Lp(Rn

), and f ⇤ µ" ! f in Lp(Rn

).

4. Weak derivatives

We call domain a nonempty open subset ⌦ of Rn. Let L1

loc(⌦) be the set of

measurable functions over ⌦, whose restriction over any compact subset of ⌦ is

integrable. By D(⌦) we denote the set of C1functions over ⌦, with compact

support. Let f 2 L1loc(⌦), and 1 i n. We say that g 2 L1

loc(⌦) is the weakpartial derivative of f w.r.t. xi, if for all ' 2 D(⌦):

(1.56)

Z

⌦

f(x)Dxi

'(x)dx+

Z

⌦

g(x)'(x)dx = 0.

4.0.1. Uniqueness of the weak derivative. The weak derivative is unique.Indeed if for some nonzero g 2 L1

loc(⌦):

(1.57)

Z

⌦

g(x)'(x)dx = 0, for all ' 2 D(⌦),

for ⌘ > 0 small, the following set has positive measure:

(1.58) E := {x 2 ⌦; g(x) > ⌘; |x| 1/⌘}.The Lebesgue measure being regular (e.g. Malliavin [25], Thm. 3.2, p. 76), there

exists K ⇢ E, compact of positive measure. Set, for ↵ > 0:

(1.59) ↵(x) := (1� dist(x,K)/↵)+.

It converges punctually when ↵ # 0 to the characteristic function of K (equal to 1

over K and 0 outside). By the dominated convergence theorem,

(1.60)

Z

⌦

g(x) ↵(x)dx !Z

K

g(x)dx > ⌘meas(K) > 0.

So, fix ↵ such that

R⌦ g(x) ↵(x)dx > 0. Observe that ↵ is continuous with compact

support. Let '" 2 D(⌦) be obtained by convolution of ↵ with a regularizing kernel.

Then by the dominated convergence theorem

(1.61) lim

"#0

Z

⌦

g(x)'"(x)dx =

Z

⌦

g(x) ↵(x)dx > 0.

But this is in contradiction with (1.57).

4. WEAK DERIVATIVES 19

4.0.2. Derivatives of mollification = mollification of derivative.

Lemma 1.25. If f , g are locally integrable, g being the weak derivative of f w.r.t.xi, with mollifications denoted resp. g" and f", then g"(x) = Dx

i

f"(x) (punctually).

Indeed, since y 7! "(x� y) belongs to D(Rn):

(1.62) g"(x) =

Z

Rn

g(y) "(x� y) =

Z

Rn

f(y)Dxi

"(x� y) = Dxi

f"(x).

Second equality: definition of weak derivative

Third equality: dominated convergence.

4.0.3. Derivatives in strong/weak Lp sense. Let f , g belong to Lp(Rn

); 1 p <1. Set, for given i 2 {1, · · · , n}:

(1.63) �"f(x) :=1

"(f(x+ "ei)� f(x)).

We say that g is the strong (resp. weak) partial derivative of f w.r.t. xi, in the

strong (resp. weak) Lp(Rn

) sense, if �"f ! g strongly (resp. weakly) in Lp(Rn

).

Since (Lp(Rn

))

⇤= Lq

(Rn), 1/p+1/q = 1, �"f ! g weakly in Lp

(Rn) means that,

for all h 2 Lq(Rn

):

(1.64)

Z

Rn

�"f(x)h(x)dx !Z

Rn

g(x)h(x)dx.

Lemma 1.26. Let f 2 Lp(Rn

) have a weak derivative g in Lp(Rn

) i w.r.t. xi.Then for all " > 0:

(1.65) �"f(x) =

Z 1

0

g(x+ t"ei)eidt a.e.

Proof. Set X := Lp(Rn

) and G(g)(x) :=

R 1

0 g(x + t"ei)eidt. By Jensen’s in-

equality, G is a nonexpansive mapping in X. So, the operator

(1.66) F (f, g) := �"f �G(g)

is linear and continuous L(X ⇥ X) ! X, and vanishes whenever f is C1with

derivative g w.r.t. xi. Now let (f, g) be as in the lemma, with mollifications denoted

by (f↵, g↵). Since (f↵, g↵) ! (f, g) in X ⇥X, we deduce that

(1.67) F (f, g) = limF (f↵, g↵) = 0.

⇤

4.0.4. Weak derivatives and weak derivatives in Lp(Rn

) coincide !!

Lemma 1.27. Given 1 p < 1, let f , g belong to Lp(Rn

). Then g is the weakpartial derivative of f (w.r.t. xi) i↵ it is the weak partial derivative in the Lp sense.

20 1. FUNCTIONAL ANALYSIS TOOLS

Proof. (a) Let f be as above, and ' 2 D(Rn). We have the discrete inte-

gration by parts formula

(1.68)

Z

Rn

�"f(x)'(x)dx = �Z

Rn

f(x)'(x)� '(x� "ei)

"dx.

The dominated convergence theorem, applied to the r.h.s., implies that

(1.69) lim

"#0

Z

Rn

�"f(x)'(x)dx = �Z

Rn

f(x)Dxi

'(x)dx.

(b) If g is a weak derivative of f w.r.t. xi:

(1.70) lim

Z

Rn

�"f(x)'(x)dx =

Z

Rn

g(x)'(x)dx.

Since D(Rn) is a dense subset of Lq

(Rn), �"f * g, so that the latter is a partial

derivative of f in the weak Lp(Rn

) sense.

(c) Conversely, let g be a partial derivative in the weak Lp(Rn

) sense.

Then the l.h.s. in (1.69) is equal to

RRn

g(x)'(x)dx.So, by (1.69), g is a weak partial derivative of f . ⇤

4.0.5. Weak and strong derivatives in Lp(Rn

) coincide !

Theorem 1.28. The weak di↵erentiability in the Lp(Rn

) sense, where p 2 [1,1),implies the strong one.

Proof. If f has derivative g in the weak Lpsense, by Jensen’s inequality:

(1.71) (i) �"f(x) =

Z 1

0

g(x+ "tei)eidt a.e.; (ii) k�"fkpp kgkpp.

Denoting by f↵, g↵ the mollification of f , g:

(1.72) k�"f � gkp k�"f � �"f↵kp + k�"f↵ � g↵kp + kg↵ � gkp.First term in r.h.s. dominated by last one in view of (1.71)(ii). Last term less than

12"0 for small ↵ > 0. So

(1.73) k�"f � gkp "0 +R"; R" := k�"f↵ � g↵kp.Being smooth, (f↵, g↵) satisfies (1.71), so that

(1.74) Rp" :=

Z

Rp

����Z 1

0

g↵(x)(x+ "tei)eidt� g↵(x)

����p

dx.

The inner integral is equal to g↵ ⇤ µ"(x), where µ is the uniform probability on the

interval [�ei, 0]. By lemma 1.24, R" ! 0 when " # 0. The conclusion follows. ⇤

On this subject, see also Theorem 3.3.2, p. 138 in Malliavin [25].

5. FOURIER TRANSFORM 21

5. Fourier transform

The Fourier transform of f 2 L1(Rn

) is defined as

1 F(f) : Rn ! C, such that

(1.75) F(f)(⇠) = ˆf(⇠) =

Z

Rn

f(x)e�2⇡ix·⇠dx,

where “·” denotes the scalar product in Rn. By the dominated convergence theorem,

ˆf is continuous and bounded, and

(1.76) k ˆfk1 kfk1.By Malliavin [25], thm 2.4.0.2 p. 127, we also have

ˆf(⇠) ! 0 when |⇠| ! 1.

5.0.1. Space ˆC0(Rn). Let

ˆC0(Rn) be the space of continuous complex valued

functions over Rn, with limit 0 at infinity, endowed with the uniform norm.

We have proved that f ! ˆf is nonexpansive L1(Rn

) ! ˆC0(Rn).

Dual space:

ˆM(Rn), space of complex-valued bounded Borel measures over Rn

,

and the duality product reads

(1.77) hµ, hiC0(Rn) =

Z

Rn

h(⇠)dµ(x) =

Z

Rn

(hR(⇠)dµR(x) + hI(⇠)dµI(x))

and for a real-valued measure �: � = �+ � ��, |�| = �+ + ��; the measure is

bounded if

RRn

d|�|(x) < 1.

5.0.2. Fourier transform in L2. When f 2 L1(Rn

) \ L2(Rn

) we have that

ˆf 2L2

(Rn), and the Plancherel formula holds:

(1.78) k ˆfk2 = kfk2.Since L1

(Rn) \ L2

(Rn) is a dense subset of L2

(Rn), the Fourier transform has a

continuous extension to L2(Rn

), called the Fourier transform in L2(Rn

). The

latter is isometric, i.e., (1.78) holds for all f 2 L2(Rn

).

5.0.3. Fourier transform of derivatives in L1 sense. Let f 2 L1(Rn

), such that

�"f ! g in L1(Rn

) where for " > 0:

(1.79) �"f(x) := (f(x+ "ej)� f(x))/".

Lemma 1.29. We have that g(⇠) = 2⇡i⇠j ˆf(⇠) (equality in ˆC0(Rn)), or in other

words

(1.80)

d@f@xj

(⇠) = 2⇡i⇠j ˆf(⇠).

Proof. That ˆf(⇠) =RRn

f(x)e�2⇡ix·⇠dx implies

(1.81)

\f(·+ "ej)(⇠) = e2⇡i"⇠j ˆf(⇠).

1Several possible conventions: we follow [15].

22 1. FUNCTIONAL ANALYSIS TOOLS

Consequently

(1.82)

c�"f = a"(⇠) ˆf(⇠), where a"(⇠) :=e2⇡i"⇠j � 1

".

By the definition of derivative in L1sense:

(1.83) �"f ! g in L1(Rn

) ) a"(⇠) ˆf(⇠) ! g(⇠) in

ˆC0(Rn).

So, for any ⇠ 2 Rn, as was to be proved:

(1.84) g(⇠) = lim

"#0a"(⇠) ˆf(⇠) = 2⇡i⇠j ˆf(⇠).

⇤

5.0.4. Fourier transform of derivatives in the L2 sense.

Theorem 1.30. Let f 2 L2(Rn

) has a partial derivative g with respect to xj,in the L2 sense. Then g(⇠) = 2⇡i⇠j ˆf(⇠) (equality in the space ˆL2

(Rn) of complex

valued square summable functions).

Proof. Since formula valid in L1framework:

Enough to construct (fk, gk) in L1 \ L2, such that gk = Dx

j

fk and

(fk, gk) ! (f, g) in L2(implies convergence a.e. of subsequence).

Cut o↵ functions: k(x) 2 D(Rn) with values in [0, 1], gradient with value in

B(0, 1), equal to 1 on B(0, k) and to 0 if |x| > k + 2.

Easy to check: fk := f k has weak derivative gk = g k + fDxi

k.

Then (fk, gk) (i) is integrable as L2and null out of compact set, and (ii) converges

in L2by dominated convergence. Details left to reader. ⇤

5.0.5. High order derivatives. With ↵ = (↵1, . . . ,↵n) multiindex (vector of Nn),

of order |↵| := Pi ↵i, associate the monomial (symbol) ⇠↵ := ⇠↵1

1 · · · ⇠↵n

n , and the

di↵erential operator:

(1.85) D↵f(x) :=D|↵|f(x)

D↵1x1 · · ·D↵nxn

.

If f and g are locally integrable, we say that g = D↵f in the weak sense if for all

' 2 D(Rn):

(1.86) (�1)

|↵|Z

Rn

f(x)D↵'(x)dx =

Z

Rn

g(x)'(x)dx.

5. FOURIER TRANSFORM 23

5.0.6. Constant di↵erential operators. Consider the polynomial over Rn:

(1.87) P (⇠) =X

↵

a↵⇠↵,

↵ = (↵1, . . . ,↵n) multiindex, finitely many nonzero a↵ 2 R, and ⇠↵ := ⇠↵11 · · · ⇠↵n

n .

Associated di↵erential operator:

(1.88) DPf =

X

↵

a↵D|↵|f

D↵1x1 · · ·D↵nxn

.

If f and g are locally integrable, we say that g = DPf in the weak sense if for all

' 2 D(Rn):

(1.89)

Z

Rn

f(x)X

↵

(�1)

|↵|a↵D↵'(x)dx =

Z

Rn

g(x)'(x)dx.

Adjoint di↵erential operator:

(1.90) D>P g :=

X

↵

(�1)

|↵|a↵D|↵|g

D↵1x1 · · ·D↵nxn

.

Then (1.89) can be restated as

(1.91)

Z

Rn

f(x)D>P'(x)dx =

Z

Rn

g(x)'(x)dx.

Example 1.31. Laplace operator:

(1.92) P (⇠) =X

i

⇠2i , DPf(x) = D>P f(x) =

X

i

D2f(x)

@x2i

.

5.0.7. Spaces Wm,p(⌦). For ⌦ domain (= open subset) of Rn

, and p 2 [1,1),

the Sobolev space of function over ⌦, with weak partial derivatives of order at most

m 2 N in Lp(⌦), is denoted by Wm,p

(⌦) and endowed with the norm

(1.93) kfkWm,p(⌦) :=

0

@X

|↵|m

kD↵fkpLp(⌦)

1

A1/p

.

When p = 2 we redefine it as the Hilbert space Hm(⌦). We define in a similar way

Wm,1(⌦), endowed with the norm

(1.94) kfkWm,1(⌦) :=

X

|↵|m

kD↵fkL1(⌦).

24 1. FUNCTIONAL ANALYSIS TOOLS

5.0.8. Fourier characterization of Hm(Rn

).

Theorem 1.32. Let f 2 L2(Rn

). Then

(1.95) f 2 Hm(Rn

) , (1 + |⇠|m) ˆf(⇠) 2 L2(Rn

).

Proof. An induction argument reduces the proof to the case when m = 1. By

theorem 1.30, f 2 H1(Rn

) implies that (1 + |⇠|) ˆf(⇠) belongs to L2(Rn

). For the

converse implication, see th. 3.5.1 p. 142 in Malliavin [25]. ⇤5.0.9. Spaces Hs

(Rn). Define, for s � 0:

(1.96) Hs(Rn

) := {f 2 L2(Rn

); |⇠|s ˆf(⇠) 2 L2(Rn

)},with natural norm

(1.97) kfkHs(Rn) := k(1 + |⇠|2)s/2 ˆf(⇠)k2.For s 2 N we recover the Hm

(Rn) spaces previously defined. Set for s > 0:

(1.98) H�s(Rn

) := Hs(Rn

)

⇤.

6. The Fourier approach to Poisson type equations

6.1. First results.

Theorem 1.33. Given f 2 L2(Rn

), the equation

(1.99) u(x)��u(x) = f(x), x 2 Rn,

has a unique solution u in H2(Rn

), and for all 1 i j n:

(1.100) kuijk2 kfk2.Proof. (a) If u 2 H2

(Rn) is solution, then

(1.101) (1 + 4⇡2|⇠|2)u =

ˆf.

Set

(1.102) µ(⇠) := 1/(1 + 4⇡2|⇠|2); u := µ(⇠) ˆf.

Then u 2 L2is the Fourier transform of some u 2 L2

, in addition u = u satisfies

(1 + |⇠2|)u 2 L2. This implies u 2 H2

(Rn) and

(1.103) kD2xi

xj

uk2 = k\D2xi

xj

uk2 = k4⇡2⇠i⇠jµ(⇠) ˆfk2 k ˆfk2.⇤

6.1.1. High order regularity. By the same arguments we can prove:

Theorem 1.34. Given f 2 Hs(Rn

) with s � 0, the equation u��u = f has aunique solution u 2 Hs+2

(Rn), and for some c depending only on (n, s):

(1.104) kukHs+2 ckfkHs .

6. THE FOURIER APPROACH TO POISSON TYPE EQUATIONS 25

6.1.2. Including first-order terms.

Theorem 1.35. Given f 2 L2(Rn

) and b 2 Rn, the equation,

(1.105) u+ b ·ru��u = f

has a unique solution u in H2(Rn

), and for all 1 i j n:

(1.106) kuijk2 kfk2.Proof. Use previous arguments; the solution u is such that

(1.107) u(⇠) = µ(⇠) ˆf(⇠); µ(⇠) := 1/(1 + 4⇡2|⇠|2 + 2i⇡b · ⇠).⇤

Observe that the Lax-Milgram variational theory (see the chapter on the Lax

Milgram theory, or Brezis [10], ch. 5) applies to the previous example only under

restrictive hypotheses, such as |b| small !

Nevertheless by the Fourier approach we were able to find the solution of this

PDE !

However, we needed to have a domain equal to Rnand a constant di↵erential

operator.

6.1.3. Interior estimate. Let ⌦ be an open subset of Rn, f 2 L2

(⌦), u 2 H1(⌦)

such that, in a weak sense (we give no boundary condition !)

(1.108) u(x)��u(x) = f(x), x 2 ⌦.Let v := u', ' 2 D(⌦). Then v 2 H1

(⌦) and we easily check that

(1.109) rv = ur'+ 'ru; �v = divrv = �u'+ u�'+ 2ru ·r'.So,

(1.110) v ��v = f'� u�'� 2ru ·r' =: g 2 L2(⌦).

Extending v, g by 0 over Rn \ ⌦, we see that v ��v = g over Rn, and so:

(1.111) kvijk2 kgk2 c'�kfk2 + kukH1(⌦)

�.

Let ⌦

0 ⇢ ⌦ be open, with compact closure in ⌦. Choosing ' with value 1 on ⌦

0, so

that u = v on ⌦

0, deduce that

(1.112) kuijkL2(⌦0) c',⌦0�kfk2 + kukH1(⌦)

�.

Remark 1.36. Typically, by variational methods, we can estimate kukH1(⌦"

),

under some boundary conditions. So, the above estimate makes sense.

Remark 1.37. Obtaining estimates near the boundary depends on the boundary

conditions and is much more involved, see e.g. Lions and Magenes [23].

6.2. Complement: strong solutions in Lp spaces.

26 1. FUNCTIONAL ANALYSIS TOOLS

6.2.1. The Poisson equation in Lp spaces. The Fourier approach is the key for

obtaining the following extension (see the hard proofs in Gilbarg and Trudinger

[17]):

Theorem 1.38. Let p 2 (1,1) and f 2 Lp(Rn

). Then the equation

(1.113) u��u = f in Rn

has a unique solution in W 2,p(Rn

), and for some cp > 0 not depending on f :

(1.114) kukW 2,p(Rn) cpkfkLp(Rn).

6.2.2. Fractional Sobolev space. Following Grisvard [18], for s = m + �, withm 2 N and � 2 (0, 1), and p 2 [1,1], set

(1.115) W s,p(Rn

) := {u 2 Wm,p(Rn

); Ns,p(u) < 1},where

(1.116) Ns,p(u) :=X

|↵|=m

Z

Rn

Z

Rn

D↵u(x)�D↵u(y)|p|x� y|n+�p

dxdy.

This space is endowed with the norm

(1.117) kukW s,p(Rn) :=

⇣kukpWm,p(Rn) +Ns,p(u)

⌘1/p

.

This is a Banach space (a Hilbert space if p = 2).

6.2.3. Lp spaces based on the Fourier transform. For s 2 R+, let Gs denote the

Bessel potential of order s, defined by its Fourier transform

(1.118) FGs(⇠) = (1 + |⇠|2)s/2.For p 2 (1,1), define

(1.119) Hs,p(Rn

) := {u 2 L1loc(Rn

); Gs ⇤ u 2 Lp(Rn

)}.The following holds, see Grisvard [18, Ch. 1]: Hs,p

(Rn) = W s,p

(Rn) if s is integer,

and also,

(1.120) W s0,p(Rn

) ⇢ Hs,p(Rn

) ⇢ W s00,p(Rn

), whenever s0 > s > s00.

7. Elliptic control over Rn

7.1. Linear-quadratic setting.

(1.121) y(x)��y(x) = f(x) + u(x), x 2 Rn.

with f 2 L2(Rn

) given and u 2 L2(Rn

). Cost function

(1.122) J(u, y) := 12

Z

R(y(x)� yd(x))

2dx+

12

Z

Ru(x)2dx.

Optimal control problem

(P ) Min

u,yJ(u, y) s.t. (1.121); u 2 KU ,

7. ELLIPTIC CONTROL OVER Rn 27

with KU nonempty closed convex subset of U := L2(Rn

). State space Y := H2(Rn

).

For each u 2 U , the state equation (1.121) has a unique solution denoted by y[u] inY , and u 7! y[u] is a�ne and continuous U ! Y . The reduced cost function is

(1.123) F (u) := J(u, y[u]).

Being continuous and strongly convex, it has a unique minimizer u over KU , and

each minimizing sequence uk in KU strongly converges to u.Reduction Lagrangian: sum of cost function and product of state equation

by a multiplier:

(1.124) L(u, y, p := J(u, y) +

Z

Rn

p(x) (f(x) + u(x) +�y(x)� y(x)) dx.

A priori the space for the state equation is L2(Rn

), so we should take the multiplier

in the dual space L2(Rn

), but we decide to search for it in the smaller space Y . Then

by Green’s theorem over Y ⇥ Y (valid over the dense subset D(Rn)

2and extended

by continuity):

(1.125)

Z

Rn

p(x)�y(x)dx =

Z

Rn

y(x)�p(x)dx,

so that for y = y[u] and z 2 Y :

(1.126) Lyz =

Z

Rn

(y(x)� yd(x) +�p(x)� p(x))z(x)dx.

For y = y[u], this gives the costate equation, with unique solution p[u] in Y :

(1.127) p(x)��p(x) = y(x)� yd(x), x 2 Rn.

The linearized state equation, with (v, z) 2 U ⇥ Y , is:

(1.128) z(x)��z(x) = v(x), x 2 Rn.

And we get as expected, writing z = z[v] and p = p[v]:

(1.129)

F 0(u)v =

Z

Rn

((y(x)� yd(x))z(x) + u(x)v(x))dx,

=

Z

Rn

((p(x)��p(x))z(x) + u(x)v(x))dx,

=

Z

Rn

((z(x)��z(x))p(x) + u(x)v(x))dx,

=

Z

Rn

(p(x) + u(x))v(x)dx.

Since U = L2(⌦), this means that the derivative of F is

(1.130) F 0(u) = p[u] + u.

28 1. FUNCTIONAL ANALYSIS TOOLS

Let u be solution of the optimal control problem. Set y = y[u], p = p[u]. if F is

strictly convex, (u, y, p) is the unique solution in U⇥Y ⇥Y of the optimality system

(1.131)

8>><

>>:

y ��y = f + u;

p��p = y � yd;Z

Rn

(p(x) + u(x))(v(x)� u(x))dx � 0, for all v 2 KU .

The above inequality is equivalent to u = PKU

(�p). Eliminating u, we obtain that

(y, p) is the unique solution in Y ⇥ Y of

(1.132)

⇢y ��y = f + PK

U

(�p);p��p = y � yd.

7.2. Case when KU = U . Then u = �p and (y, p) is unique solution in Y ⇥Yof:

(1.133)

⇢y ��y = f � p;p��p = y � yd.

This is a system of two coupled linear elliptic equations. We next see how

to study this system in a direct way, i.e., without reference to the optimal control

problem.

7.2.1. Direct Study based on the Fourier transform. Set µ := 1 + 4⇡2|⇠|2. Ap-

plying the Fourier transform to the above system we find that

(1.134) µy + p =

ˆf ; �y + µp = yd.

We find that the unique solution satisfies

(1.135) (1 + µ2)y = µ ˆf � yd; (1 + µ2

)p =

ˆf + µyd.

Therefore (µy, µp) 2 L2(Rn

), so that (as expected) y and p belong to Y .

7.2.2. Direct study based on the Lax-Milgram theorem. Here we anticipate on the

Lax-Milgram theory, presented in the next chapter. The variational formulation is,

denoting the test function by (y, p) 2 Y ⇥ Y :

(1.136)

Z

Rn

(yy +ry ·ry + py + pp+rp ·rp� yp)dx =

Z

Rn

(fy � pyd)dx.

Set V := H1(Rn

). The Lax-Milgram theorem applies since taking (y, p) = (y, p) weget after cancellation that the bilinear form on the r.h.s. is continuous over V ⇥ Vand satisfies

(1.137) a((y, p), (y, p)) = kyk2V + kpk2V .

7.3. Cheap control.

7. ELLIPTIC CONTROL OVER Rn 29

7.3.1. Zero cost control. Consider a variant of the previous linear quadratic set-

ting, where the control does not appear in the cost function

(1.138) J(u, y) := 12

Z

R(y(x)� yd(x))

2dx,

and there is no control constraints. Then a minimizing sequence could be un-

bounded. Yet we have that, since the problem is convex, a solution (u, y) is charac-terized by the existence of a solution p of the costate equation

(1.139) p��p = y � yd,

with derivative p of the reduced cost equal to 0. This means that y = yd. So, the

problem has a solution i↵ �yd 2 L2(Rn

), i.e. i↵ yd 2 H2(Rn

) and then the unique

solution is u := yd ��yd, y = yd.7.3.2. Small cost control. Assuming again that yd 2 H2

(Rn), let the cost function

be, for " > 0:

(1.140) J"(u, y) :=12"

Z

Rn

u(x)2dx+

12

Z

Rn

(y(x)� yd(x))2dx,

with yd 2 H2(Rn

) and again u := yd ��yd, y = yd. Then J" atttains its minimum

at some (u", y") and

(1.141) ku"k22 +1

"ky" � ydk22

2

"J"(u, y) = kuk22.

So y" ! y in L2(⌦) and u" is bounded, so it has at least one limit-point u. Passing

to the limit in the state equation we obtain that u = y, so that u" * u. By

lemma 1.6, lim sup ku"k2 � kuk2. Combining with the above display we deduce that

ku"k2 ! kuk2. By lemma 1.6 it follows that u" ! u in L2(Rn

).

Remark 1.39. On this ’singular perturbation’ problems (for which the limiting

problem is not of the same nature as the original one), see J.-L. Lions [21].