A Constructive Treatment of Open and Unopen Mapping Theorems

Zci t sd~r . i. math. Log& und Grundlagen d. Malh. Bd. 35, S. 29-43 (1989)

A CONSTRUCTIVE TREATMENT OF OPEN AND UNOPEN MAPPING THEOREMS

by DOUGLAS BRIDGES in Buckingham (England) and WILLIAM JULIAN and RAY MINES in Las Cruces, New Mexico (U.S.A.)

1. Introduction

The work below originated in our search, within BISHOP’S constructive mathematics [l, 21, for a proof of appropriate versions of the classical open mapp ing theorem:

(OMT) A bounded linear m a p of a Banach s p c e into a normed linear space i s operr if and only if its range i s complete.

Here we recall that a mapping between normed linear spaces is open if it maps open subsets of its domain onto open subsets of its range.

A detailed constructive analysis of the open mapping theorem has been carried out by STOLZENBERG [9], but is inconclusive about the possibility of a constructive theorem classically equivalent to (OMT) .

From a constructive viewpoint, the standard argument used to prove (OMT) (see, for example, [8, 5.91) breaks down at its application of BAIRE’S Theorem, which, in the form used there, has no known constructive proof. However, a constructive anal- ogue of that form of BAIRE’S Theorem can be proved [5, Ch. 2, § 21, and, taken with certain results below, leads to a proof of (OMT) under the additional, classically super- fluous, hypothesis that the image of the unit ball under the bounded linear map is bilocated in the range (Theorem 4.8). (A subset S of a metric space ( X , e) is located in X if Q ( X , S ) = inf{e(x, y ) : y E X> exists for each IC in X ; S is colocated in X if its metric complement X - S, the set of points of X that are bounded away from S, is located in X ; and X is bilocated in X if it is both located and colocated in X . Hypo- theses of locatedness, colocatedness, or bilocatedness, which are often used in constructive theorems, are classically redundant.)

By classical logic, (OMT) is equivalent to the following contrapositive version :

(OMT’) A bounded linear m a p of a Banach space into a normed linear space is unopen if and only if its range i s incomplete.

From a constructive point of view, the investigation of (OMT’), which we also under- take in this paper, is an entirely separate problem from that of (OMT), because of the affirmative interpretation of such terms as “unopen” and “incomplete” in constructive mathematics.

A bounded linear map u between normed linear spaces is unopen if each ball in the range of u contains a point which is distinct from each point in the image of the unit

30 D. BRIDGES, W. JULIAN, Ih’U R. IIlh’ES

ball. A normed linear space X is incomplete if it contains a Cauchy sequence that is eventually bounded away from any given vector in X . (In constructive mathematics, it is typical that concepts are defined affirmatively in pairs, like open/unopen, in which each part is classically equivalent to the negation of the other. Likewise, theorems frequently occur in pairs, in which each part is classically equivalent to the contrapositive of the other.)

The main result of this paper (Theorem 4.5) is (OMT’), with the additional hypothesis that the image of the unit ball under the bounded linear map is located in the range. A corollary (4.6) of that theorem supports STOLZENBERG’S suggestion in [ Q ] that the open mapping theorem, suitably adapted to fit into the constructive framework, will provide nonsurjectivity conditions on a bounded linear map between Banach spaces.

We also prove a constructive version (Theorem 4.9) of the following proposition, which is classically equivalent to (OMT) :

If u i s a bounded linear m a p of a normed linear space X onto a normed linear space Y , such that either (i) u is unopen and Y is complete, or (ii) u i s open and Y is incomplete, then X i s incomplete.

It is interesting to note that although we apparently have to supplement hypothesis (i) for our constructive proof, hypothesis (ii) can be used as it stands.

The standard open mapping theorem (OMT) can be proved under the additional assumption that the bounded lincar map is compact : in Section 5 we prove that the following conditions on a compact linear map u are equivalent:

(a) u is open;

(b) the range of u is finite-dimensional;

(c) the range of u is complete.

We also prove a natural constructive contrapositive of that theorern, using an affirmative definition of ‘‘ infinite-dimensional” : a normed linear space X is infinite-dimensional if for each finite-dimensional subspace F of X there exists (we can construct) an element x of X such that p(x, F ) > 0, where e is the distance function as- sociated with the norm.

Much of our work depends on elementary, but nontrivial, results about finite- and infinite-dimensional normed linear spaces, which we gather together in section 2. For example, we prove the classically trivial proposition that a normed linear space over R or C is infinite-dimensional if i t is incomplete.

In section 3 we prove some results, in particular characterizations of open and unopen linear maps, which are of interest, but are not applied in the present context.

We assume familiarity with the relevant portions of [2]; the reader may also be interested to consult [3] and [7].

Notation: We write B,(r) for the open ball of radius r in a normed linear space X , ran(u) for the range of a mapping u, and ker(u) for the kernel (null space) of u.

A CONSTRUCTIVE TREATMENT OF OPEN AND UNOPEN MAPPING THEOREMS 3 1

2 . Preliminary results on metric and normed spaces

Let ( X , e ) be a metric space. A sequence (x,) in X is said to be eventually bounded away from the subset K of X if there exists an cx > 0 such that e(xn, K ) 2 cx-that is, e(xn, y ) 2 a for all y in K-for all sufficiently large n. If this holds and K E {x) is a singleton, we say that the sequence (x,,) is eventually bounded awuy from X .

The metric space X is incomplete if there exists a Cauchy sequence in X that is eventually bounded away from each point of X .

The proposition,

(*)

entails Markov’s P r inc ip l e :

(MP) If (a,,) is a binary sequence such that iVn (a , = O), then 3 n ( ~ , ~ = 1). To see this, let 8, be the set of all elements of P with finitely many nonzero terms, and consider any sequence (a,,) in (0, 1) such that iVn(a , , = 0). Then

if a normed lineur space i s not complete, then i t i s incomplete,

V = (x E Po: 3n(an = I)> u {O),

taken with the ll-norm, is a normed linear space. Suppose that V is complete. If there exists an n such that a, = 1, then V = V o , which is incomplete; thus a,, = 0 for all n, a contradiction. Hence V is not complete. On the other hand, if V is incomplete, then there is a Cauchy sequence in V that is eventually bounded away from 0 ; so that V contains a nonzero element, and therefore there exists an n such that a,, = 1. Thus (*) entails (MP).

As we know of no proof of (MP) within BISHOP’S constructive mathematics, we are careful to distinguish between “not complete” and “incomplete”.

2.1. Propos i t ion . A metric space X is incomplete if and only if it contains a Cauchy sequence that i s eventually bounded away from each complete, located subset of X .

Proof. Assume that X is incomplete, and embed it isometrically, as a dense subset, in its completion s. Let (x,) be a Cauchy sequence in X that is eventually bounded away from each point of X . Then (x,,) converges to a limit x, in X. Let K be a complete, located subset of X . Then K is complete and located as a subset of s. By [ 2 , Ch. 4, 3.81, there exists y in K such that e(xm, y ) > 0 entails e(xm, K ) > 0. Since (a,,) is eventually bounded away from y, we have e(xm, y ) > 0 and therefore e(xm, K ) > 0. Hence e(xn, K ) 2 +e(xm, K ) > 0 for all sufficiently large n. The converse is trivial. 0

We shall need the following result later.

2.2. Corol lary. An incomplete normed linear space i s infinite-dimerLsio.1. Proof. Let X be an incomplete normed linear space, and F a finite-dimensional

subspace of X . Then I? is complete and located [2 , p. 3071. By Proposition 2.1, there exists an x in X such that e (x , F) > 0. Hence X is infinite-dimensional.

Our next proposition, whose proof is left to the reader, is classically equivalent to Proposition 2.1. Although we shall not use it in this paper, we mention it here for the sake of completeness, and to illustrate the constructive distinction between a theorem and its contrapositive; that distinction will appear several times in the course of our work below.

32 U. URIDGCS, W. JULIAN, AND E. MINES

2.3. Propos i t ion . A metric space X i s complete if and only if for each Cauchy sequence (xn) in X there exists a complete, located subset K of X such that for each E > 0, e(xn, K ) < E for all sufficiently large n.

In the rest of this section we gather together some results about finite- and infinite- dimensional normed linear spaces that will be needed later in the paper.

2.4. Lemma. If F i s a finite-dimensional subspace of a normed linear space E, r > 0, and x is an element of BE(r) such that e(x, BF(r)) > 0, then e(x , F ) > 0.

Proof . Clearly, we may take r = 1. Let 6 = e(x , B F ( l ) ) . Either e(z, F ) > 0 or e(x , F ) < 6/2. In the latter case, choosing y in B with (Ix - yll < 6/2, we must have 1 S llyll 6 1 + 6/2. Thus Ijy1j-l y E B F ( l ) , and so

11% - llYll-lYll 6 IIZ - YII + (1 - llYll-l) llvll < 6j2 + (1 - ( 1 + 6/2)-1) (1 + 6/2) = 6 ,

which contradicts the definition of 6. Hence e(x , F ) > 0. j-J

2.5. Lemma. Let F be a finite-dimensional subspace of a normed linear space E , and y an element of E such that e ( y , F ) > 0. Then for each I with 0 < r < 1 there exists a unit vector z in the span of F v { y ) such that ~ ( x , F ) > r .

Proof . Construct x in F such that

e (Yt F ) I IIY - 41 i

112 - x’ll = IIY - 41-111Y - x - IIY - 41 2’11

e(y9 F ) >

and write z = IIy - x11-l (y - z). Then for each x‘ in F we have

2 IIY - 41-1 @(Y, F ) .

Hence e(2, F ) h Ily - 41-1 e(Y, F ) > re D Vectors y l , . . ., y, in a normed linear space Y are linearly independent if

C Aiyt + 0 whenever A,, . . .,A,, are scalars with C lLil + 0. A linearly independent

sequence of vectors in Y is a sequence (y,) in Y such that yl , . . . , yn are linearly independent for each n.

2.6. Lemma. Let E be an infinite-dimensional normed linear space, and 0 < r < 1. T h e n there i s a linearly independent sequence (y,) of unit vectors in E such that

n n

f = 1 i = 1

(2.6.1) e(yn+lt span(y,, . . ., yn}) > ( 7 ~ = 1, 2 , . . .). Proof. The construction of a sequence (y,) of unit vectors satisfing (2.6.1) is car-

ried out inductively, using Lemma 2.5. An induction argument using (2.6.1) then shows that (y,) is linearly independent. j-J

2.7. Lemma. Let E be an infinite-dimensional normed linear space, and B a totally bounded subset of E. Then for each r with 0 < r < 1 there exists a uni t vector x in E’ such that e(x, B ) > r .

Proof. Let E = (1 - r ) / 4 , and construct a finite sapproximation {xl, . . . , xm> to B. By [2 , Ch. 7, 2.51, there exists a finite-dimensional subspace F of E such that e ( x j , F ) < E (1 i 5 m). Using Lemma 2.5, construct a unit vector x in E such that

.\ (‘OHSTBUCTIVE TREdTMENT OF OPEN AND UNOPEN MAPPINO THEOREMS 3 3

p(.r. E’) > 1 - t. Given any b in B, choose i in ( 1 , . . . ,m> such that IJb - xtl/ < E ,

and then choose y in F such that ((y - X I ( ( < B. Then

I/z - b// >= I ( x - y(I - (Iy - x,// - \ /xi - b(l > 1 - 38.

Henee e(z, B ) >= 1 - 3s > r .

A subset S of a metric space ( X , e ) is noncompact if there exist an a > 0 and a se- qiience (2,) in S such that e(x, , , , x,,) 2 OL whenever m =j= n.

It follows from Lemma 2.6 that any ball in an infinite-dimensional normed linear space is noncompact. Conversely, we have

2.8. Propos i t ion . A normed linear space that contains a bounded noncompact subset is. Lnjiil.ite-dirnen.siona1.

Proof . Let X he a normed linear space containing a bounded noiicompact subset AS. Choose r > 0 such that /lxl/ < r for each x in S . Choose also an OL > 0 and a sequence (s,) in S such that - Y,,)) >= OL whenever m + n. Given a finite-dimensional sub- spacc F of X , let A 3 (xl, . . . . xh) be a finite a/8-approximation to the compact ball B, ( r ) . Either e(s,, A ) > ar/4 for some i with 1 5 i 5 N + I, or else e(si, A ) < ~ / 2 for all i in { 1, . . . , N + 13. In the latter case, there exist i, j , k such that 1 i < j 5 - - < N + 1, 1 5 k s N , (IsI - z k l / < 4 2 , and 118, - xkll < a/2; whence I(si - s,I( < a, a contradiction. Thus e(s , , A ) > a/4 for some i with 1 6 i N + 1. For this i we have /IsI - x/1 >= B / S whenever z ~ B ~ ( r ) . It now follows from Lemma 2.4 that p(s,. P ) > 0. Hence X is infinite-dimensional.

3. Open and unopen linear maps

Let u : X + Y be a linear map between normed linear spaces X and Y. We say that tc is open if it maps open subsets of X onto open subsets of ran(u). It is easy to prow that u is open if and only if Brancu,(r) c u(B,(l)) for some r > 0. On the other hand. we say that u is unopen if for each r > 0 there exists a y in BranCu)(r) such that y + u(i)-that is, 1Iy - u(x)/I > 0-whenever x ~ B , ( l ) . We may replace the unit halls in these two definitions by balls of any positive radius.

I t is interesting to have criteria for openness and unopenness. A simple lemma leads to such criteria.

3 . 1 . Lemma. Let S be a nonvoid subset of a normed linear space X and let c be a constant in (0, 1) such that for each x in X, either llxll < 1 or there exists an x’ in S with j1x’/) s c/(xI/ . Then there exists an element x in X with (1x11 < 1.

3.2. Propos i t ion . Let u : X 3 Y be a linear map between normed linear spaces. Then u is open if and only if it satisfies the condition:

(4.2.1) There exists an r > 0 such that for each unit vector x in X with /lu(x)II < r , there exists a y in ker(u) such that 11x - yII < 1/2.

Proof. Assume that u is open, and choose r > 0 so that Brancu,(r) c u ( B X ( l / 2 ) ) . Given any vector x: in X with I/u(x)II < T , choose 5‘ such that I(x’(1 < 1/2 and u(x’) = u(x). Then z - x‘ belongs to ker(u), and IIx - ( x - x‘)(I < 1/2. Thus (3.2.1) holds. Conversely, assume that condition (3.2.1) is satisfied. We prove that

3 Ztschr. f . math. Logik

34 U. BRIDGES, W. JULIAN, AND R. MINES

BrBn(u)(r/2) c u(Bx(l)) . To this end, consider any x with /lu(x)ll < rj2. Either I1xII < 1 or, as we may assume, IJx11 > 1/2. Thus 11 1 1 ~ ) ) - ~ zll = 1 and

IIu(ll~ll-l~)II = 1 1 ~ 1 1 - 1 1 1 ~ ~ ~ ~ 1 1 < 211U(~)Il < r .

By (3.2.1), there exists a z in ker(u) such that 1 ) 1 1 ~ 1 1 - ~ x - zIJ < 1/2. Setting x‘ = x - 11x1) z , we have lIx’II < llxll/Z and u(x’) = u(x) - 11x11 u(z) = u(x) . The result now follows by taking S = u-l(u(x)) and c = 1/2 in Lemma 3.1.

A linear map u between normed linear spaces is said to be well-behaved if (i) ker(u) is located in the domain of u, and (ii) IIu(x)II > 0 whenever e(z, ker(u)) > 0.

The assertion that any linear map whose kernel is located in its domain is well- behaved is equivalent to the following statement, which, in turn, is equivalent to Markov’s Principle [2, p. 63, Problem 131:

(MP’) To see this, suppose that each linear map with located kernel is well-behaved. Let 1 (a = 0) , and consider the map u : R --f R taking x to ax. Then ker(u) = (0) is located, and e(1, ker(u)) = 1. Thus la1 = IIu(1)IJ > 0; whence (MP’) obtains. Con- versely, assume (MP’), and consider a linear map u whose kernel is located. Let x be in the domain of u with e(x, ker(u)) > 0. If IIu(x)II = 0, then e(x, ker(u)) = 0, a contradiction. Thus by (MP’), we have )[u(z)I( > 0. Therefore the map u is well-behaved.

As the constructive status of (MP’) is unclear, and as we shall need property (ii) in the proof of Proposition 3.4, we have included that property in our definition of a well-behaved linear map.

The following proposition is of interest in view of the last Brouwerian example.

3.3. Propos i t ion . If u is a bounded linear map of a normed linear space X onto a finite-dimensional space Y , such that ker(u) is located, then u is well-behaved.

Proof . Since Y is finite-dimensional, it contains a finite linearly independent span- ning subset { e l , . . ., e,,). Write e, = u(x,) for i in (1, . . ., n). Consider any x in X

with e(z, ker(u)) > 0, and choose scalars A , , . . ., A,, such that u(z) = 2 A,e,. Then

x - C Lixi is in ker(u), so that

If a € R and i ( a = 0 ) , then a + 0.

n

n i = 1

i = I n n n

i = 1 i = I I = 1 0 < IIX - (x - c ArxAll = I1 c A P i I l 5 c 141 l l~ i l l .

Hence lAi l > 0 for some i in ( 1 , . . ., n}, so that ~ ( z ) + 0. 0 3.4. Propos i t ion . Let u : X .--* Y be a bounded linear map between normed linear

spaces, such that ker(u) is located in X . If u is unopen, then it satisfies the following condition :

(3.4.1) Por each r > 0 there exists a unit vector E in X such that

Ilu(t)II < r and e(E, k e W ) 2 1/2.

Conversely, if u is well-behaved and satisfies (3.4.1), then u is unopen.

Proof . Assume that u is unopen. To prove (3.4.1), let r > 0 and put

S = ( x ~ X : Ilu(x)II < r ]2, and Vy~B, (1 /2 ) ( u ( x ) =k u ( y ) ) ) .

A CONSTRUCTIVE TREATMENT OF OPEN AND UNOPEN M-@PING THEOREMS 35

As u is unopen, there exists an x in S. Note that llxll 2 1/2. P u t 5 = 1 1 ~ 1 1 - ~ x; then //[I / = 1 and llu([)// < r. Either p([, ker(u)) > 1/2, in which case (5.4.1) holds; or, as we assume, @([, ker(u)) < 314. Choosing z in ker(u) such that 115 - zll < 3/4, write x’ = x - (IxJI z . Then llx’ll < (3/4) 1 1 ~ 1 1 and u(x’) = u(x). Thus, taking c = 3/4, we may assume by Lemma 3.1 that )1x)) < 1. We claim that e([, ker(u)) 2 1/2. If t ~ k e r ( u ) and \I[ - 511 < lj2, then 11% - 1 1 ~ 1 1 (11 = llxll 116 - ( 1 1 < 1/2; so that, as x E s,

and therefore llu([)II > 0, a contradiction. Thus e ( [ , ker(u)) 2 l/2, and so condition (3.4.1) holds.

Conversely, assume that (3.4.1) holds. and consider any r > 0. Choose a unit vector 5 in X such that IIu(6)II < r and @([, ker(u)) 2 1/2. Given x in B,(1/4), we have

0 < IlNx - (x - llxll mil = 1141 Ilu(5‘)II >

e(5 - 2, k e W ) 2 e(5, kertu)) - llrll 2 V4. Hence Ilu([ - x)II > 0, as u is well-behaved. Since x and r are arbitrary, it follows that u is unopen.

An obvious generalization of the proof of [3, Lemma 31 gives a weak version of the open mapping theorem.

3.5. Propos i t ion . If a bounded linear m a p between normed linear spaces has finite- dimensional range, then the m a p is open.

Not surprisingly, we can also prove a related weak unopen mapping theorem.

3.6. Propos i t ion . If a bounded linear m a p between normed linear spaces i s unopen, then the m a p has infinite-dimensional range.

Proof . Let u : X + Y be an unopen bounded linear map, and consider any finite dimensional subspace F of ran(u). The restriction of u to u-l(P) is a bounded linear map of u-l(F) onto P, and hence, by Proposition 3.6, is open. Thus there exists an r > 0 such that

BF(r) c u(u-’(P) A B,( 1)).

As u is unopen, there exists a y in ran(u) such that llyll < r and IIy - u(x)II > 0 for each x in B x ( l ) . Hence y y’ for each y’ in BF(r) . Since P is finite-dimensional, B,(r) is compact and so is complete and located in ran(u). Hence e(y ,B , (r ) ) > 0, by [a, Ch. 4, 3.81; so that e (y , P ) > 0, by Lemma 2.4. Hence Y is infinite-dimensional. 0

In Section 4 we will prove a generalization of Proposition 3.6, under appropriate conditions, by replacing ‘‘ infinite-dimensional ” by “incomplete ”.

4. The unopen mapping theorem

Let u: X + Y be a linear map between normed linear spaces X and Y. We say that u is weakly open if Brancu,(r) c u(B,(l)) for some r > 0. On the other hand, we say that u is strongly unopen if for each r > 0 there exist a y in Brancu,(r) and an OL > 0 such that IIy - u(x)II > a whenever x E Bx(l) .

The first part of the next proposition is well known in classical mathematics (see [6, proof of 1.8.21).

3*

3 6 D. BRIDGES, W. JULIAN, AND 1%. M I N E S

4.1. Proposi t ion. Let u be a weakly open bounded linear m a p on a n o r m d linear space X . Then

(i) if X is complete, u i s open; (ii) if u i s unopen, X i s incomplete.

and n is a positive integer, then there exist xl, . . . , x, in B,( 1) such that Proof . Since u is weakly open, it is straightforward to prove that if Y E Bran(",(r/2)

n

(**) IIy - u( c 2-+c1)l1 < 2-n- l r . i = I

To prove (i), assume that X is complete, consider any y in Brancu,(r/2), and construct

a sequence (J,) in B,(l) such that (**) holds for each n. As X is complete, c 2-'x1

converges to a vector x in B,( 1). As u is continuous, it follows from (a*) that y = u(x). Hence Bra,(,,,(r/2) c u(B,(l)), and u is open. To prove (ii), assume that u is unopen, and choose y in Bran(u)(r/S) such that y $: u(x) for each x in B,( 1). Construct a sequence

(x,) such that (**) holds for each n. Then ( 2 2-'x,);= is a Cauchy sequence in B,( 1).

We prove that this sequence is eventually bounded away from any x in X. To thls end, we may assume that 11x11 < 1. Let c > 0 be a bound for u. Choose N such that 2-N-1r < i l l y - u(x)I/. Then for all n 2 N we have

a,

i= 1

n

I = I

n n

i = I i = 1 112 - c 2-'2,/1 >= c-'llu(x) - u( c 2-'x,)J/

2 "-'(IIY - u(x)II - t l l Y - u(x)ll) = +c-1lly - u(x))) > 0 .

This completes the proof of (ii).

I n order to characterize strongly unopen bounded linear maps, we need a lemma, whose proof is left to the reader.

4.2. Lemma. Let u be a strongly unopen bounded linear m a p o n a normed linear space X , and let y E ran(u). Then for all r , R > 0 there exist a y' in ran(u) and an a > 0 such that lly - y'II 5 r and lly' - u(x)ll 2 a for each x in B,(R). 0 4.3. Propos i t ion . The following are equivalent conditions on a bounded linear map u

on a normed linear space X : (i) u is strongly unopen; (ii) there exists a Cauchy sequeicw (y,,) i r ~ ran(u) that i s eventually bounded away from

u ( B x ( r ) ) for each r > 0; (iii) there exists a bounded sequence (y,) in ran(u) such that for each n, y,, is bounded

away from u(B,(n)). Proof . Let u: X + Y be a bounded linear map. Assume first that u is strongly

unopen. Using Lemma 4.2 inductively, construct a sequence (y,) in ran(u) and a sequence (a,,) of positive numbers, such that for each n,

(4 0 < &+1 < &"/4, (b) I IYn+l - Ynll 5 a n / 2

A CONSTRUCTIVE TREATMENT O F OPEN AND UNOPBN nlAPPIN43 THEOREXS 37

and

(c) IIy, - u(x))I 2 01, for all x in B,(n). Then for m > n we have

m - 1 m - 1

IIYlll - YnII 5 c IIYi+l - YiII I .z (a,/2) 1 = f l I = ,

m < (a,/2) z 4-' = 2arfl/3,

i = O

so that (y,) is a Cauchy sequence in ran(u). Consider any positive integer N and any x in B,(N). If n > N , then

IIYn - 4411 1 IIYN - U ( 4 l I - 11% - YNII 2 OLN - 2a,/3 = a,/3.

Hence (i) implies (ii).

Since a Cauchy sequence is bounded, it is easy to show that (ii) implies (iii). Now assume that (iii) holds, and choose c > 0 so that ()y,,l) 5 c for each n. Consider any r > 0, and let n be an integer greater than 2clr. Then there exists an 01 > 0 such that lJy, - u(2)Jj 2 a for all x in B,(n). Noting that ]I?/,,// 2 a , write y = (2l]y,,l/)-' ry,,. Then y E Bran(,,,(r). Moreover, for each x in B,(1) we have

ll~lly,ll ~ - ' 4 l 5 2c/r < n ,

IIY - u(x)lI = (2llYnIl)-' r l lYn - U(21IYnII r-'4Il

1 (211Ynll)-1 > 0 -

so that

Since r is arbitrary, i t follows that ZL is strongly unopen. Hence (iii) implies (i).

The next lemma makes possible the extension of Proposition 4.3 to an unopen mapping theorem.

4.4. Lemma. Let X be a Banach space, Y a normed linear spare, u : X -+ Y n bounded linear mapping whose range i s dense in Y . such that u(B,(l)) i s located in Y , ond r a positive number. Then for each y i n BY(r) there exists a n z in Bx(3) such that if y =k u(z), then e(y', u(B,(l)) > 0 for sornc $1' in Brancu,(r).

Proof . Fix y in BY(r), and define an increasing sequence (A,) in (0, 1) and a sequence (2,) in B,(2), as follows. Either e(?/, u(B,(l))) > 0, in which case we set A, = 1 and x, = 0 for all n ; or e(y, u(B,(l))) < r / 2 . In the latter case. set A1 = 0 and choose z1 in B,(2) such that I1y - u($zl)ll < 1.12. Then, trivially, /I2y - u(xl)ll < r. Either e(2y - u(x,), u(B,(l))) > 0, in which case we set A, E 1 and zfl = 0 for all i~ 2 2 ; or ~ ( 2 y - u(x,), u(B,(l))) < r / 2 . In the latter case, set I , = 0 and choose x, in B,(2) such that )I2y - u ( x l ) - u(fxn)ll < rj2. Then 1(22y - u(2xl) - u(xz)ll < r. The construction of the sequences (A,) and (r,) is continued in this manner. As X is

complete, the series C 2-'si converges to an element z of Bx(2). Suppose that y + u(x). With c > 0 a bound for u, choose the integer rt so that r) 2 2 and

4,

i = 1

2-"r + P + ' r < I/y - u(s)!I.

If A,, = 0, then n

IJy - u( C 2-izl)Jl < 2-"r i = I and so

n m m

i = I i = n + l i = n + l 1Iy - U(Z) I ! s jly - u( C 2-izL)ll + Ilu( C 2Pzi)l1 =< 2-"r + 2c C 2-'

- - < 2-"r + 2-"+'c,

which contradicts our choice of n. Hence A, = I , and e ( z , u(B,(l))) > 0, where

n-1

i = I z 2"-' y + u( C 2"-i-'zi) E B y ( r ) .

AsB,,,,~,,(~) is dense in By(r) . we can now choose y' in Brancu,(r) such that JJy' - z / / < < e(z , 4B,(l)))> whence @(Y', u(B,(I))) > 0.

We now prove the unopen mapping theorem:

4.5. Theorem. Let u be a bounded linear map on u Banach space X , such that u(B,(l)) i s located in ran(u). Then u is mopen if and only if ran(u) is incomplete.

Proof . If u is unopen, then for each r > 0 we can find y in Brancu,(r) such that y + u(x) whenever x ~ B , ( 2 ) . It follows from Lemma 4.4 that u is strongly unopen; whence, by Proposit.ion 4.3, ran(u) is incomplete. Conversely, assume that ran(u) is incomplete, and let Y he the completion of t'he riormed linear space ran(u). Choose a Cauchy sequence (y,) in ran(u) that is eventually bounded away from each point of ran(u). As Y is complete, (y,) converges t,o a point y of Y such that y =/= u(z) for each x in X . Consider any r > 0, and choose N so that lly - yNI( < r. For each x in X we have yN + u(x) E ran(u); whence y -+ yN + u(z), and therefore y - yN + u(z). Hence by Lemma 4.4, e(y', u(B,(l))) > 0 for some y' in Brancu,(r). As r is arbit'rary, i t follows that u is strongly unopen, and hence unopen.

The following corollary is classically equivalent to the open mapping theorem.

4.6. Corol lary. Let u be an unopen bounded linear map of a Banach space X into a Banach space Y , such that u(B,(l)) is located ,in ran(u). Then there ezists a point y in Y such that y $. u(z) for each x in X.

Proof . By Theorem 4.5, ran(u) is incomplete. Hence there exists a Caiichy sequence (y,) in Y that is eventually bounded away from each point of ran(u). The limit of t'his sequence in Y is the required point y.

The follou~irig constriict.ive version of BAIRE'S Theorem is discussed in [ 5 , Chap. 2, ( 2 . 5 ) ] :

If (C,) is a sequence of closed, bilocated subsets of a complete metric space 9,

such that X = U C , , then C, has nonvoid interior for some n. m

n = l

4.7. Lemma. Let u be a bounded lineur map of a normed linear space X onto a Banach

Proof . Apply the above version of BAIRE'S Theorem, with C, the closure of

space Y , such that u(B,(l)) is bilocated in, Y . Then u is weakly open.

{nu(z): z E B,( I ) } in Y . 0

A CONSTRUCTIVE TREATMENT OF OPEN AND UNOPEN MAPPINQ THEOREMS 39

Lemmas 4.4 and 4.7 enable us to prove the constructive open mapping theorem:

4.8. Theorem, Let u be a bounded linear m a p of a Banach space X onto a normed linear space Y , such that u(B,(l)) i s bilocated in Y . Then u is open if and only if Y i s complete.

Proof . If Y is complete, then u is weakly open, by (4.7), whence u is open, by (4.1 (i)). Conversely, assume that u is open, and choose r > 0 such that B y ( r ) c u(B,(l)). To prove Y complete, i t will suffice to show that any Cauchy sequence in B y ( r ) has a limit in Y . To this end, let (y,,) be a Cauchy sequence in B y ( r ) , and ym its limit in the completion of Y . By (4.4), there exists an x in B x ( 2 ) such that if (Iym - u(x)II > 0, then e(y', u(B,(l))) > 0 for some y' in B y ( r ) . As the last condition contradicts our choice of r , we must have ym = u(z) E Y . Hence Y is complete. 0

We now prove yet another result classically equivalent to (OMT).

4.9. Theorem. If u i s a bounded linear map of a normed linear space X onto a normed linear space Y , such that either (i) u i s unopen, u(B,(l)) is bilocated in Y , and Y i s complete, or (ii) u i s open and Y i s incomplete, then X i s incomplete.

Proof. Under hypotheses (i), the conclusion follows immediately from (4.7) and (4.1(ii)). Now assume hypotheses (ii), and choose r > 0 so that B y ( r ) c u(Bx(l)) . Construct a Cauchy sequence (y,,) in Y that is eventually bounded away from each element of Y . We may assume that y1 = 0, and that for each n, Ily,,+l - ynll < 2 " r . With x1 = 0, we construct inductively a sequence (x,,) in X such that, for each n , y,, = u(x,,) and llx,,+l - 511 < 2-"; indeed, having constructed x l , . . . , x,,, we choose x in BX(2-") such that y,,+, - y,, = u(x), and set z,,+~ E x,, + x. Then (x,,) is a Cauchy sequence in X ; we show that it is eventually bounded away from each element of X . To this end, let c > 0 be a bound for u, let x be any point of X , and choose an cx > 0 such that jiu(x) - ynli 2 M for all sufficiently large n. Then for such n we have

For the sake of completeness, we now state three theorems classically equivalent to BANACH'S inverse mapping theorem (a bijective bounded h e a r map between Banach spaces has a bounded inverse); the proofs of these theorems are left to the reader.

112 - X,ll 2 c- la . 0

A map u: X + Y is injective if z = y whenever u(x) = u ( y ) .

4.10. Theorem. If u i s an injective bounded linear m a p of a Banach space X onto (I Banach space Y , such that u(B,(l)) is bilocated in Y , then u-l i s bounded.

4.11. Theorem. Let u be a bounded linear m a p of a Banach space X onto a Banach space Y , such that u(B,(l)) i s bilocated in Y , and such that for each r > 0 there exists an 5 in X with IIu(x)II < r and llxll > 1. Then there exist x, x' in X such that llxll < 1, 112'1[ > 1, and u(x) = u(x'). 0

A linear map v : X -+ Y between normed linear spaces X , Y is unbounded if for each r > 0 there exists an x in X such that llxll < r and IIw(x)II > 1. A map u : X -+ Y between normed spaces is strongly injective if ~ ( x ) + u ( y ) whenever x + y.

4.12. Theorem. Let u be a strongly injective bounded linear m a p of a normed linear space X onto a normed linear space Y , such that u-' i s unbounded. Then

(i) if u(B,(l)) i s located and X i s complete, then Y is incomplete; ( i i ) if u(B,(l)) i s bilocated and Y is complete, then X i s incomplete. 0

40 D. BRIDGES, W. SULIAN, AND R. MINES

5. Compact and noncompact linear maps

A linear map u : X -+ Y between normed linear spaces X and Y is compact if u(B,( 1)) is totally bounded. In that case, u is not only bounded, but normable.

We now prove a strong version of the open mapping theorem for compact linear maps. In this connection, note that a bounded linear map with finite-dimensional range need not be constructively compact: i t is so if and only if its kernel is located in its domain [3, Ex. 1 and Thm. 13.

5.1. Theorem. The following are equivalent conditions on a compact linear map u between normed linear spaces:

(i) u i s open; (ii) ran(u) i s finite-dimensional; (iii) ranfu) is complete.

Proof. Let u : X -+ Y be a compact linear map. Assume, t o begin with, that u is open, and choose an r > 0 so that Brancu,(r) c u(Bx(l)). Since Bran(,,,(r) is located in ran(u), and u(Bx(l)) is totally bounded, we see that Bran(,,)(r) is totally bounded [ 2 , Ch. 4, 4.51. Hence ran(u) is finite-dimensional, by [2, Ch. 7, 2.61, and so (i) implies (ii). We see from Proposition 3.5 that (ii) implies (i). Clearly, (ii) implies (iii). Finally the Corollary in [7] shows that (iii) implies (ii). 0

Sext. we have a strong version of the unopen mapping theorem for compact linear

5.2. Theorem. The following are equivalent conditions on a compact linear niap u

maps.

between normed linear spaces: (i) u i s unopen; (ii) ran(u) i s infinite-dimensional; (iii) ran(u) is incomplete.

Proof . We see from Proposition 3.6 that (i) implies (ii). Conversely. assume that ran(u) is infinite-dimensional, and let r > 0. As ~ ( B ~ ( 2 r - ~ ) ) is totally bounded, we see from Lemma 2.7 that there exists a unit vector in ran(u) such that ~ ( y , u ( B x ( W 1 ) ) > l j2. Thentry E Bran(,,)(r) ande(+ry, u(Bx(l))) > r/4. Thusu is htrongly nnopen, and hence unopen. Moreover, by Proposition 4.3, ran(u) is incomplete. Hence (i i ) entails both (i) and (iii). Finally, we see from Corollary 2.2 that (iii) implies ( i i ] .

Theorems 5.1 and 5.2 suggest the possibility of related results about bounded Intear maps that are noncompact, in some appropriate sense.

A nonconipact linear m a p is a linear map u: X -+ Y between normed linear spaces such that u(B,(I)) is a noncompact subset of Y .

5.3. Propos i t ion . The following are equivalent conditions on an open bounded l i i i m r niap u between normed linear spaces: ( i ) u is nonconzpact; (i i) ran(u) is infinite-dimensional.

Proof . Let u : X -+ Y be an open bounded linear map. It follows from Proposi- tion 2.8 that (i) implies (ii). Next assume that ran(u) is infinite-dimensional. As u is

A COXSTHL'CTIVE TREATMENT O F OPEN AND UNOPEN MAPPING THEOREMS 41

open, there exists an R > 0 such that Bran<")(2) c u(B,(R)). B y Lemma 2.6, there exists a sequence (y,,) of unit vectors in ran(u) such that llym - y,,/nll 2 l j2 whenever m =+ n. For each rL, choose tn in B,(R) such that ~ ( 5 , ) = y,,, and write x,, E R-lt,,. Then ~lz,,~~ c I , and if m =+ n,

llu(xnJ - 4 X n ) I l = R-'IIym - ~ n l l 1 (2R)-'

Hence u is noncompact, and (ii) implies (i). 0

incomplete range, then the mup is noncompact. 5.4. Corollary. If an open bounded linear m p between normed linear spaces has

Proof. This follows from Corollary 2.2 and Proposition 5.3. Theorem 5.1 also suggests the conjecture that a bounded linear map onto an infinite-

dimensional Banach space is noncompact (cf. [7]). Here is a proof of a restricted case of that conjecture.

5.5. Theorem. If u i s a bounded linear imzp of a normed linear space X onto an infinite-dimensional Banach space Y , such fhat u(B,(l)) is bilocated in Y , then u is n o w compact.

Proof . By (4.7), u is weakly open-that is, u(B,(l)) contains a ball of positive radius. -4s Y is infinite-dimensional, this ball, and therefore u(B,(l)), is noncompact. The conclusion now follows because q~(B,(l)) is dense in its closure.

6. Limiting examples

Although the locatedness and bilocatedness of u(B,( l ) ) , which play an essential role in the proofs of our main results, are classically insignificant conditions, they need not hold in the constructive setting.

6.1. A Brouwerian example of a normnble linear map ZL of X = 1' onto R2 stich that u(B,(l)) is not located in ran(u).

Let (a,,),"=, be a sequence in (0, 1: and define m

u(2) = (3x,. x2 + 2 c a,x,) n = 3

for all x in X . If u(B,(l)) is located in R2, then either s E e((0, 2), u(B,(l))) < 1 , in \r.Iiich case there exists an n such that a, = 1, or else s > 0, and a, = 0 for all n.

It is an interesting, and apparently nontrivial, problem to find conditions on u, a bounded linear mapping between normed linear spaces X and Y , that ensure that u(B,( l ) ) is located in ran(u). Among those bounded linear maps that have the latter property are the compact ones and those introduced by the following definition.

Let H be a Hilbert space, with scalar product (...). A bounded linear operator u on H is a weighted multiple shift if there exist an orthonormal basis (en) of H , a nonnegative integer 7, and R bounded sequence (A,,) of complex numbers such that

u ( x ) = c

A,(.c. e , ) en+,. for all x in H . 11 = I

32 D. BRIDGES, W. JULIAE, AND R. MINES

6.2. Propos i t ion . If u is a weighted multiple shift on the separable Hilbert s p e H ,

Proof . Choose an orthonormal basis (en) of H , a nonnegative integer r , and a se-

quence (An) in C such that U ( X ) = C jl,,(x, en) en+r for each x in H . Consider any finite

linear combination x = C xnen of the elements en, and let P be the projection of I1

onto the finite-dimensional subspace spanned by ( e , , . . . , eN+r}. Then u commritts with P. For each y in B,(l) we have

(1)

then u (B , ( l ) ) is located in H .

m

N n = l

n = 1

Ilx - U(Y)I12 = IIX - PU(Y)Il2 + IlU(Y) - PU(Y)IIZ

2 IIx - UP(Y)Il2. Now, P(BH(1)) = BPcH,(l) is totally bounded, as P(H) is finite-dimensional; whence, as u is continuous, uP(B,(l)) is totally bounded and therefore located. As P(BJj( l ) ) c BH(l ) , i t follows from (1) that e(x , u(BH(l))) exists and equals @(z,uP(BH(l))) . Since the set of finite linear combinations of the en is dense in H , we conclude that u(BH(l)) is located in H . 0

6.3. A Brouwerian example of an open normable linear mapping u of X = l 2 onto 1' such that u(B,(l)) is located but not colocated in ran(u).

Let (an):=, be a binary sequence with a t most one term equal to 1, let (en) be the standard orthonormal basis of 1 2 , and define the linear map u : l 2 + l 2 by

m m

4 C s e n ) E C ( 1 - $an) %"en* n = l n = l

Then u is normable, with norm l,,and clearly maps Z2 onto itself.

((yII 5 112, choose x = Cq,en in l 2 such that y = u(z). Then m

n = l

m m

1/4 2 IIYII' = C ( 1 - L Cxn2/4, n = 1 n = I

Given y in 1' with

so ))x1/2 _I 1. Hence Brantu,(l/2) c u(B,(l)), and u is open. By Proposition 6.2, S E u(B,(l)) is located in ran(u). Suppose that T = 1' - S is located. Then either e(0, T ) > 112 or e(0, T ) < 1 . In the first case, a,, = 0 for all n: for if a, = 1, then ( 2 - E ) - ~ en E T for each E E (0, 2 ) ; whence e(0, T ) _I l j2. I n the second case, choose

y in T such that llyll < 1. Then y = U ( X ) for some x 3 C x,en with e(x , B,(l)) > 0 ; so that x + u(x), and therefore aN = 1 for some N . n = l

We believe that for any bounded linear map u that occurs in important practical cases, the image under u of the unit hall will be bilocated in ran(u). This belief is jus- tified in the case where ran(u) is finite-dimensional:

6.4. Propos i t ion . The following are equivalent conditions on a bounded linear mapping u from a normed linear space X onto R": (i) u(B,(l)) is located;

( i i ) u(B,( 1)) is colocated;

m

A CONSTRUCTIVE TREATMENT OF OPEN AND UNOPEN MAPPING THEOREMS 43

(iii) u(B,( 1)) i s bilocated;

(iv) u is compact.

Proof . This follows from [a, (2.1)] and [5 , Ch. 2, 4.111. 0 There now arises a second open problem: find conditions which ensure that if u is

a bounded linear mapping between normed linear spaces X and Y such that u(B,(l)) is located in ran(u), then u(B,(l)) is bilocated in ran(u). Example 6.1 shows that even when X is complete, ran(u) is finite-dimensional, and u is normable, u(B,(l)) need not be located. Example 6.3 shows that even when both X and ran(u) are Hilbert spaces, u is open and normable, and u(B,( 1)) is located, u(B,( 1)) need not be bilocated.

References

[l] BISHOP, E., Foundations of Constructive Analysis. NcGraw-Hill, New York 1967. [2] BISHOP, E., and D. BRIDGES, Constructive Aniilysis. Springer-Verlag, New York-Heidelberg-

Berlin 1985. 1 3 1 BRIDGES, D. S., CALDER, A., JULIAN, W., MINES, K., and F. RICHMAN, Bounded linrar

maps of finite rank. J. Functional Analysis 43 (1981), 143-148. [4] BRIDGES, D. S., CALDER, A., JULIAN, W., MINES, R., and F. R r c ~ a r a ~ , Locating metric comple-

ments in R". In: Constructive Mathematics (F. RICHMAN, ed.), Springer Lectures Notes in Mathematics 873 (1981), pp. 241 -249.

151 BRIDGES, D. S., and F. RICHMAN, Varieties of Constructive Mathematics. London Math. Soc. Lecture Notes 97 (1987), Cambridge University Press.

161 KADISON, R. V., and J. R. RINGROSR, Fundamentals of the Theory of Operator Algebras, Vol. 1 . Academic Press, New York 1983.

[7] RICHMAN, F., BRIDGES, D., CALDER, A., JULIAN, \V., and R. MINES, Compactly generated Banarh spaces. Arch. Math. 36 (1981), 239-243.

181 RUDIN, W., Real and Complex Analysis. McGraw-Hill, New York 1970. 191 STOLZENBERG, G., A Critical Analysis of Banach's Open Mtipping Theorem. Northeastern Uni-

versity, Boston, Mass. 1971.

D. Bridges The University of Buckingham Buckingham, MK18 1EG England

W. Julian, R. Mines New Mexico State University Llts Cruces, New Mexico 88003 U.S.A.

(Eingegangen am 9. November 1987)

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A Constructive Treatment of Open and Unopen Mapping Theorems