Math. Log. Quart. 44 (1998) 545 - 552

A Definitive Constructive Open Mapping Theorem?

Douglas Bridges” and Hajime Ishiharab

” Department of Mathematics, University of Waikato, Hamilton, New Zealand’) School of Information Science, Japan Advanced Institute of Science and Technology, Tatsunokuchi, Ishikawa 923- 12, J span')

Abstract. It is proved, within Bishop’s constructive mathematics (BISH), that, in the context of a Hilbert space, the Open Mapping Theorem is equivalent to a principle that holds in intuitionistic mathematics and recursive constructive mathematics but is unlikely to be provable within BISH.

Mathematics Subject Classification: 03F60, 25340, 46830, 47830.

Keywords: Open mapping theorem, Constructive mathematics, Linear mapping, Linear Operator, Hilbert space, Banach space.

1 Introduction

The search for constructive analogues of the Open Mapping Theorem, one of the cornerstones of classical functional analysis, appears to have begun with the work of STOLZENBERG [13] and was continued in [6], where there appeared a number of such theorems, each requiring strong, but commonly satisfied, locatedness hypo the~es .~ ) The main such results in [6] were the following, referred to as the Unopen Mapping Theorem and the Open Mapping Theorem, re~pectively.~)

T h e o r e m 1 . Le t T be a bounded l inear mapp ing of a B a n a c h space X o n t o a normed space Y such tha t T (Bx (1 ) ) i s located in Y . T h e n T i s u n o p e n i f and on ly i f Y is incomplete .

T h e o r e m 2. Le t T be a bounded l inear mapp ing of a B a n a c h space X on to a n o r m e d space Y such t h a t T(Bx(1)) i s bilocated in Y . T h e n T i s open i f and on ly i f Y is complete .

On the other hand, by weakening the concept of openness and restricting to op- erators on a Hilbert space, we were able to establish the following result in [5].

T h e o r e m 3. Le t T be a bounded l inear operator o n a Halbert space s u c h tha t T* exists. T h e n T is sequentially open i f and only if its range is complete , in wh ich case both the range and the k e r n e i of T are located.

‘)e-mail: douglas@waikato.ac.nz 2, e-mail: ishihara@jaist .ac.jp 3)For related results see [ S ] and [l]. 4)We use B ~ ( T ) (respectively, B ~ ( T ) ) to denote the open (respectively, closed) ball with centre 0

and radius T in a normed space.

546 Douglas Bridges and Hajime Ishihara

(To understand these results fully, we need a number of definitions which will be given shortly.)

Building on the work of “61 and [5], in this paper we prove that the classical Open Mapping Theorem for operators on a Hilbert space is equivalent, within BISHOP’S constructive mathematics (BISH - see [l], [Z], [7], and [14]), to a certain principle about bounded subsets of the set N+ of positive integers. That principle, and hence the classical Open Mapping Theorem, holds in each of the two standard models of BISH - namely, BROUWER’S intuitionistic mathematics (INT) and the recursive constructive mathematics (RUSS) of the MARKOV School; but it appears unlikely to be provable in BISH. Nevertheless, our main theorem has a fair claim to being the definitive constructive Open Mapping Theorem in the context of a Hilbert space.

It will be convenient to gather together a few of the definitions that have particular importance for the present work.

A subset S of a metric space (X, e ) has complemen t

-S = {z E X : (Vs E S ) ( e ( x , s ) > 0))

and m e t r i c complemen t

-s = {z E x : ( 3 c > O)(Vs E S ) (e(x, s) 2 c)}

We say that S is

2 E x; - located (in X ) if the distance p(z,S) inf{e(z,s) : s E S } exists for each

colocated if -S is located; - bilocated if S is both located and colocated.

Let T be a linear mapping of a normed space X into a normed space Y. We denote the range and kernel of T by ran(T) and ker(T), respectively. We say that T is

sequence (Tzn) converges to Tx;

that B y ( r ) C T(Bx(1));

yn # T z whenever 3: E X and llzll 5 1;

a sequence (yn) in ker(T) such that (5 , - y,) + 0;

- sequentially con t inuous if for each sequence (2,) converging to x in X, the

- open if it maps open sets to open sets - equivalently, if there exists T > 0 such

9 u n o p e n if there exists a sequence (9,) of vectors in Y converging to 0 such that

sequentially open if for each sequence (z,) in X such that Txn -+ 0 there exists

- a lmos t open if there exists r > 0 such that llyll 2 T for all y E -T(Bx(l)) .

It is trivial that openness implies both sequential openness and almost openness; and that openness, sequential openness, and almost openness are classically equivalent notions.

A metric space ( X , e ) is incomplete if it contains a Cauchy sequence (z,) that is eventually bounded away from each element of X , in the sense that for each z E X there exist c > 0 and a positive integer N such that ~ ( z , xn) 2 c for all n 2 N .

A Definitive Constructive Open Mapping Theorem? 547

Under classical logic, our notions of “unopen” and “incomplete” are equivalent to

Additional background material in constructive mathematics can be found in [a] the negations of “open” and “complete”, respectively.

and [7], and will be used without further comment.

2 An Open Mapping Theorem

Our aim in this section is to prove another version of the Open Mapping Theorem within BISH. For this we need three lemmas, the first two of which are used to rule out an unwanted alternative in the proof of the main theorem and refer to the limited principle of omniscience:

LPO

This principle is known to be essentially nonconstructive (it is false in both INT and RUSS).

L e m m a 1. If there exists a n unopen sequentially con t inuous l inear mapp ing of a normed space on to a Banach space, t h e n LPO holds.

P r o o f . Let T be an unopen sequentially continuous linear mapping of a normed space X onto a Banach space Y , and choose a sequence (2,) in X such that for each n , T I , E --T(Bx(n)) and ((Tz,(( 5 l /n . Given a binary sequence (u,), construct an increasing binary sequence (A,) such that

If (a,) is a binary sequence, t h e n ei ther Vn ( a , = 0) or 3n (a, = 1).

A, = 0 * ( V k 5 n) (uk = O ) , A, = 1 3 (3k 5 78) (Uk = 1).

We may assume that A 1 = 0. If A, = 0, put z, Z 0; if A, = 1 - A n - l , put zk I 2,-1 for all k 2 n. Then (Tz,) is a Cauchy sequence in Y and so converges to T z , for some t, E X. Choose a positive integer N > ) ( z o o ( / . If AN = 1, then a, = 1 for some n 5 N . So we may assume that AN = 0. Then, supposing that A, = 1 - A,-1 for some m > N , we have T z , = T z , E -T(Bx(rn)), which is absurd, as z , E Bx(m).

0 Thus A, = 0 for all n > N and therefore for all n, whence V n (a, = 0). C o r o l l a r y 1.

RUSS/INT I- There does no t e t i s t a n unopen sequentially cont inuous l inear mapp ing of a normed space on to a B a n a c h space.

P r o o f . LPO is false in both RUSS and INT. 0

We conclude from Corollary 1 that within BISH we will never find an example of

L e m m a 2. If LPO holds, t h e n every separable subset of a m e t r i c space is located. P r o o f . This is a simple application of the constructive least-upper-bound prin-

0

L e m m a 3. If T is a l inear mapp ing of a normed space X on to a B a n a c h space

an unopen sequentially continuous linear mapping onto a Banach space.

ciple ([2, Ch. 2, (4.4.3)]).

Y such that ker(T) is located, t h e n

-T(BX(l)) = T ( { 2 E : [1211X/ker(T) 2 l)), where ( 1 . I l , y / k e r ( ~ ) denotes t h e quotient n o r m o n X / ker(T).

548 Douglas Bridges and Hajime Ishihara

P r o o f . Let y = Tx E Y . If IIxllx/ker(~) < 1, then, choosing 2’ E ker(T) such It follows that if that 1 1 % - 1/11 < 1, we have y = T ( z - x’), so y E T(Bx(1) ) .

Y E w T ( B X ( l ) ) , then IlxllX/ker(T) 2 1. Now suppose, conversely, that Ilxllx/kerp) 2 1. If 1 1 ~ ~ 1 1 < 1, then for each 1 in

ker(T) we have

II(. - x’) - 111 2 IIx - 111 - llxlll > e(., ker(T)) - 1 2 0. It follows from Theorem 1 in [3] that T(x - z‘) # 0 and hence that y = T x # Tz’.

L e m m a 4. Let T be a n almost open l inear mapp ing of a normed space X on to a

P r o o f . Choose T > 0 such that y E -T(Bx( l ) ) implies that llyll 2 T . If y = T x

Thus y E -T(Bx(l)). 0

Banach space Y such that ker(T) is located. T h e n T as open.

lies in B y ( r ) , then y $-T(Bx(l)) and so, by Lemma 3,

ker(T)) = IIxllX/ ker(T) 5 1.

Choosing t E ker(T) such that IIx - z11 < 2, we see that T(x - t ) = T x = y. Hence 0

Our next result uses sequential continuity, rather than boundedness, of the linear mapping. Constructively, sequential continuity is a weaker property than bounded- ness for linear mappings; but as any linear mapping on a Banach space is bounded within both RUSS and INT, we cannot expect t o find, within BISH, an example of a linear map that fails to be bounded on a Banach space.

Although the results of [6] were proved for bounded linear mappings, rather than ones that are sequentially continuous, it is not hard to adapt the proofs so that they apply to the latter type of mapping; see also the proof of [ l l , Lemma 31). We shall assume without further comment that those results hold in that more general situation.

T h e o r e m 4. Le t T be a sequentially cont inuous l inear mapp ing of a separable B a n a c h space X on to a Banach space Y such tha t ker(T) and -T(Bx(l)) are located. T h e n T zs open.

By(.) c T(Bx(2)) and therefore T is open.

P r o o f . Construct an increasing binary sequence (A,) such that

An = 0 + e(0, mT(Bx(1)) < l /n2, A n = 1 3 e(O,-T(Bx(I)) > ~ / ( n + I ) ~ .

In view of Lemma 4, it will suffice to prove that there exists N such that AN = 1, so we may assume that XI = 0. If A, = 0, choose [, such that TFn E - T ( B x ( l ) ) and IITEnII < l / n 2 , and set 2, = ne,; if An = 1 - A n - l , set Xk = zn-l for all Ic 2 n. Then (Txn) is a Cauchy sequence in Y : in fact, llTxm - TxnII < 2/72 for all m and n. Choose in turn an element x, E X such that T z , -+ Txm, and a positive integer N > 1 1 ~ ~ 1 1 . Suppose that AN = 0. If there exists m > N such that Am+l = 1 - A,, then I n = 2, for all n 2 m, so Tx, = Tz, E -T(Bx(rn)); this is absurd, as x, E Bx(m). Hence An = 0 for all n 2 N and therefore for all n. It follows that for each n, Tx, E -T(Bx(n)) and IITxnll 5 1/71. Thus T is unopen. By Lemmas 1 and 2, T (Bx (1)) is located in Y. It now follows from Theorem 1 that Y is incomplete, which is a contradiction. Hence, in fact, AN = 1. 0

A Definitive Constructive Open Mapping Theorem? 549

We end this section with a result that substantially improves upon the “only if” part of Theorem 2 in [S].

P r o p o s i t i o n 1. Banach space into a normed linear space, then ran(T) is complete.

P r o o f . Let T map the Banach space X onto the normed linear space Y , choose T > 0 such that B y ( r ) C T(Bx( l ) ) , and set = 0 for all n. Let (Txn) be a Cauchy sequence in Y , and compute n1 such that (ITxj - Txk(( < 2-’T for all j, k 2 nl. For each k 2 n1 there exists y1,k E ker(T) such that (lxn, - X k - yl,k(l < 2 - l . Now compute n2 > n1 such that (1Tzj - TxkII < 2-’r for all j, k 2 n2. For each k 2 n2

there exists y2,k E ker(T) such that ll(xnz + ~ 1 , ~ ~ ) - (xk + yz,t)(l < 2-2. Carrying on in this way, we construct a strictly increasing sequence (nk)rzl of positive integers and a sequence such that 11(%k + Yk-i,nk) - (%,+I + yk,nk+l)II < 2-k for each k. Thus (zn, + yk-l,nk)F=l is a Cauchy sequence in X and so converges to a limit z E X. The sequential continuity of T now ensures that

I f T as a sequentially continuous open linear mapping of a

Txn, = T(xnk + ~ k - 1 , ~ ~ ) + TZ as k + 00.

Since (Ttn) is a Cauchy sequence, it follows that Tx, + Tz as n -+ 00. Hence Y is complete. 0

3 BD-N and the Open Mapping Theorem

A subset S of Nt is pseudobounded if limn..+m n-ls,, = 0 for each sequence (sn) in S . In this section we show how the following principle sheds light on the open mapping theorem.

BD-N Every nonempty5)countable pseudobounded subset of N+ is bounded.

This principle is trivially true in classical mathematics, holds in both INT and RUSS, and appears not to be provable in BISH. Moreover, any theorem that we can prove in BISH + BD-N will hold in both INT and RUSS (see [lo]).

Let T be a nonzero sequentially continuous linear mapping of a normed space X onto a Banach space Y such that -T(Bx(l)) is separable, and let (Tzn) be a dense sequence in -T(Bx(l)). Construct a binary double sequence am,n such that

is a countable set. Moreover, a positive integer n belongs to AT if and only if there exists m such that ((Tz,(( 5 l /n2 .

5)By nonempty we mean inhabited, in the sense that we can construct an element of the set in question.

550 Douglas Bridges and Hajime Ishihara

L e m m a 5. Let (a,) be a sequence i n A T . T h e n for all posit ive numbers a , /3 wi th

P r o o f . Construct an increasing binary sequence (A,) such that (Y < p, ei ther n - la , > a for s o m e n or else n - la , < p for all n.

A n = 0 + (Vk 5 n) (k - ' ak< p), A, = 1 - A , - ~ 3 n- la , > a.

We may assume that A1 = 0. If A, = 0, put [, = 0. If A, = 1 - A n - l , then there exists m such that llTzmll < l / a t ; setting [k = n x , for all k 2 n, we see that llT&iI < 1 / a 2 n . It is easy to see that (T&) is a Cauchy sequence in Y and so converges to T& for some too E X. Choose an integer N > Il[M]I. If AN = 1, then there exists n 5 N such that n-la, > a . Consider the case where AN = 0, and suppose that An+l = 1 - A, for some n 2 N . Then Ttm = T& = T(nx,) for some m. Since too E Bx(n ) and T(nz,) E -T(Bx(n)) , we have a contradiction; so

0 A, = 0 for all n 2 N and therefore for all n. Thus n- la , < /3 for all n. L e m m a 6. U n d e r t h e hypotheses of L e m m a 5, AT i s pseudobounded. P r o o f . Given E > 0 and using Lemma 5, construct an increasing binary sequence

(A,) such that

A, = 0 * (3k 2 n) (k - ' ak > ~ / 2 ) , A, = 1 + (Vk 2 n)(k- 'ak < E ) .

We may assume that A 1 = 0. If A, = 0, choose k 2 n such that k-'ak > ~ / 2 ; choose also m such that llTzmII < l / a i , and set [, = nz,. Then ((T[,(I < 4/c2n. On the other hand, if A, = 1 - A,-1, set ( j = <,-I for all j 2 n. Then (TC,) is a Cauchy sequence in Y and so converges to T& for some tm E X. Choosing an integer N > l l [oo/ / , suppose that AN = 0. Suppose also that there exists n 2 N such that X n + l = 1 - A,. Then T& = T<, = T(nz,) for some m; but T(nz,) E -T(Bx(n)) and Ttoo E T ( B x ( n ) ) , which is absurd. Hence A, = 0 for all n 2 N and therefore for all n. Thus <, E-T(Bx(n)) and I(TC,II 5 4 / 2 n for each n, whence T is unopen and therefore, by Lemmas 1 and 2, T(Bx(1)) is located in Y . It follows from Theorem 1 that Y is incomplete. This contradiction ensures that AN = 1 and therefore that

0

P r o p o s i t i o n 2. n - l a , < E for all n 2 N.

BD-N I- Let T be a sequentially cont inuous l inear mapp ing of a B a n a c h space X on to a B a n a c h space Y such that -T(Bx( l ) ) is separable and ker(T) is located. T h e n T is open.

P r o o f . In view of Lemma 4, it is enough to prove that T is almost open. By Lemma 6 and BD-N, AT is bounded; so there exists a positive integer N such that for all m, a m , N - 1 = 1 and therefore //Tz,// > 1 / N 2 . Since T is sequentially continuous and the sequence (TZ,)~=~ is dense in -T(Bx(l)), it follows that IITxlI 2 1,"' for

0 all 2 E -T( Bx (1)). Hence T is almost open. In order to progress towards our main theorem, we need a simple lemma L e m m a 7. Let T be a nonzero sequentially cont inuous l inear mapp ing of a sep-

T h e n

P r o o f . The canonical mapping of X onto X/ ker(T) is bounded, so X/ ker(T) is X : 1 1 2 1 1 x / k e r ( ~ ) 2 l}, being located in X/ker(T) , is a

arable n o r m e d space X on to a Banach space Y such t h a t ker(T) is located. -T( Bx (1)) is separable.

separable. Hence S 5 {z

A Definitive Constructive Open Mapping Theorem? 55 1

separable subset of X/ ker(T), as is its image under the sequentially continuous map 0

Here, at last, is our version of the Open Mapping Theorem for operators on a

T h e o r e m 5. T h e fol lowing proposit ions are equivalent:

T : X/ ker(T) - Y But T(S) = -T(Bx(l)) , by Lemma 3.

Hii ber t space.

(i) BD-N. (ii) If T is a nonzero bounded linear mapp ing of a separable Halbert space H i n t o

itself such that T* exists and ran(T) is complete, t h e n T i s open.

P r o o f . First assume BD-N and let T satisfy the hypotheses of (ii). By Theorem 3, T is sequentially open and ker(T) is located. It follows from Lemma 7 that -T(Bx( 1)) is separable, and then from Proposition 2 that T is open. Thus (i) implies (ii).

To prove the converse, assume (ii) and let A = { a l , a2, . . .} be a countable pseudo- bounded subset of N+. Let H be an infinite-dimensional Hilbert space, and ( e n ) an orthonormal basis of H . We first prove that for each x E H,

Sx C r ' i a n ( z , e n ) e n

is well defined. Let ( N , ) be an increasing sequence of positive integers such that

CTzN, I ( x > e k ) I 2 < 2-n-1n-2, and construct a binary sequence (A,) such that

Define a sequence (b,) in A as follows. If A, = 0, we set b, = a l . If A, = 1, then

2-"-1 -2 Nn+1-1 2 mZ{U; : Nn 5 k < Nn+l} 2 x k = N n ak1(x,ek)12 > 2-,-',

so ak > n for some k with N,, _< k < Nn+1; in this case we set b, = a k . Since A is pseudobounded, there exists m such that bnn-l < 1 for all n 2 m. If A, = 1 for some n 2 m, then 1 5 b,n-l < 1, a contradiction. Hence A, = 0 for all n 2 m, and therefore the series defining Sx converges.

It is clear that S is a one-one selfadjoint linear mapping of H onto itself. So T S-l is also a one-one selfadjoint linear mapping of H onto itself. By the Hellinger- Toeplitz theorem ([4, Theorem 3]), T is sequentially continuous. But if T is open, then S is bounded and therefore A is a bounded set. 0

Taken with [4, Theorem 41, the second part of the proof of the last theorem leads to an interesting corollary.

C o r o 1 1 a r y 2. T h e fol lowing proposit ions are equivalent:

(i) BD-N. (ii) E v e r y one-one selfadjoint sequentially cont inuous l inear mapp ing f r o m a Hilbert

0 space on to itself as bounded.

552 Douglas Bridges and Hajime Ishihara

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(Received: July 10, 1997)

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