Inventiones math. 13, 284-294 (1971) �9 by Springer-Verlag 1971

On Simultaneous Extensions

MURALI RAO (Aarhus)

Introduction and Summary

Let X be a compact metric space, F a closed subset of X and C(X), C(F) the spaces of continuous complex functions on X and F respectively. The Borsuk-Dugundji theorem [2, p. 36] asserts the existence of a linear operator T on C(F) into C(X) such that

1) (Tf)(x)=f(x) for all xeF and f e C(F);

2) Tf>O if f>O, feC(F); 3) T ] = 1, i.e. constants are carried by T into constants.

Any linear operator T on a subspace E of C(F) into C(X) satisfying (Tf)(x)=f(x) for all xeF and f e e will be called a linear extension map on E. In this paper we shall prove that every linear extension map on a finite dimensional subspace E of C(F) can be extended to a linear extension map on C(F) without increase in norm i.e. that in the Borsuk- Dugundji theorem the values of a linear extension map may be prescribed on a finite dimensional subspace E. I f there is a projection of norm one f rom C(F) onto E this can be deduced from Theorem 1.5 of [4]. Our p roof is elementary. As an application we will use this result to imitate a separation argument, used for instance as in [3] to deduce a linear extension theorem, of Michael and Pelczynski namely Theorem 1.5 of [4].

Notations and Definitions

X will always denote a fixed compact Hausdorff space and F a fixed closed subset of X. C (X) and C (F) complex valued continuous functions on X and F respectively with the usual sup norms. E will always be a fixed finite dimensional subspace of C (F).

Definition l. A linear operator T on a subspace HcC(F) into C(X) will be called a linear extension map if (Tf)(x)=f(x) for all f e l l and xeF.

From now on u will be a fixed linear extension map on E.

Definition 2. Again let H be a subspace of C(F). A function p(x,f) defined on the product X x H will be called a seminorm function with

On Simultaneous Extensions 285

domain H if

a) p is continuous on X • H;

b) for each xEX, p(x, .) is a seminorm on H;

c) JeH, ][f]] = 1 implies p(x,f)>O for all xeX.

With these definitions and notations we have the following

Lemma 1. Let go6C(F) and H the subspace of C(F) generated by E and go. Let p(x,f) be a seminorm Jhnction with domain H. Suppose G is a closed subset of X containing F and vl: H--* C(X) a linear extension map which extends u: u f =vljJbr f cE . I f we have

puJ'(x)[<p(x.f), J~E, drfl[=l, x ~ X (1)

Jvlf(x)[ <p(x,f), f ~ H , IIfH = I, xEG

then there exists a linear extensi6"n map v: H--* C(X) such that

v f = uJ; f~ E

vf(x)=vlf(x) , f ~ H , xeG (2)

[vf(x)]<p(x,f), f e H , x e X .

Proof The reader will soon notice that this is indeed a selection theorem. By an appeal to Michael's selection theorem it is possible to shorten the proof a little. For x~X, uf(x) is a linear functional on E dominated by the seminorm p (x , f ) and by Hahn Banach theorem we can extend it to a linear functional on H still dominated by p. The set of all such extensions is a compact convex set. This way we associate to each x ~ X a compact convex set in the dual H* and the whole proof depends on showing that this set valued map is lower semi continuous. One can then appeal to Michael's well known selection [1] or simply bypass it as we do below.

Now every extension of the linear map f ~ uf(x) to H is uniquely determined by its value on the element go. Thus the set of all such extensions can be identified with a subset V(x) of the complex plane:

V(x)= {Lgo: LEH*, [gfl<p(x,f) f o r f ~ H and Lf=uf (x ) for f e E } . (3)

Clearly V(x) is a compact convex set. We will show how we can pick continuously one element out of each V(x).

Now vlgo(x) is continuous on X and for xeG, vlgo(x) belongs to V(x). Since V(x) is a compact convex set there is a unique element g(x) in V(x) closest to vl go(x):

[g (x ) - Vl go (x)] = inf Vl go(x)- z[. (4) z ~ V x)

286 M. Rao:

Since for x~G, Vlgo(X ) belongs to V(x) we see that for x~G, g(x}= L'I go(X). Now we proceed in steps.

Step 1. The function D(x, y) defined on X x C (C = complex plane) by

D(x, y)= sup Re(yz) z~V(x)

is continuous. To show this note that the definition of V(x) and the proof of the Hahn Banach theorem imply that

D (x, y) = ~nf (p (x, f + y g o ) - Re uf(x)) (5)

= inf inf (p (x, r f+ y g0)- Re r uf(x)}. I1$11=1,$~E r>=o

Now we have

By (1) p(x , f ) - Re uf(x) > 0 whenever I[ f II = 1. By continuity the same is true in a neighbourhood of the compact set (f: Ilfll = 1, feE) . There- fore for lyl<R and all r large enough (depending only on R and go)

p(x , f + ~ ) - R e u f ( x ) > e

for some positive e uniformly for xeX, [yl<R and f e E , I!fll = 1. Thus as r--* ~ the quantity

tends to oo uniformly in (x, Y, fl. Therefore we may write for some Ro (depending on go and R)

in~(p (x, r f + y go) - R e r uf(xo)) = i n f o ( p (x, r f + y' go) - R e r uf(Xo))

or from (5) that

D(x,y)= infs~E(p(x,f+ygo)-Reuf(x) ). I!fll

The right side is continuous since the ball of radius Ro is compact. The continuity of D has thus been established.

Step 2. The function d on X defined by

Iv, go(x)-=r

On Simullaneous Extensions 287

is continuous. To prove this we need the following fact: For each open set U in the complex plane the set {x: V(x)c~ U#:O} is open in X i.e. that the set valued map x---, V(x) is lower semi continuous. Without loss in generality assume U is convex. Supposing the claim were false there would exist a net (x.) in X converging to a point xo in X such that V(x,)c~U=O and V(xo)c~U+O. Since the compact convex set V(x,) is disjoint from the open convex set U there would exist (by Hahn Banach theorem) complex numbers y, such that PY, I = 1 and

sup Re(y,z)<infRe(y,z). (61 z ~ V ( x ~ ) z ~ U

By choosing a subnet if necessary we may assume that y,--+ Yo. F rom Step 1 the left side of (6) is cont inuous function of (x, y). It follows that

sup Re(voZ)<_infRe(YoZ). z ~ V ( x o ) ~ - - z E U

This is absurd since V(xo)c~ U+-O. Using this we now prove that d(x) is upper semi continuous. Let (x,) be a net converging to x o. We must show that lim supd(x,)<d(xo). There is a unique point Zoe V(xo) such that

d(xo) = Iv, go (Xo)- zol-

Let U denote the open disc of radius a with centre z o. By what we have already shown the set of x such that V{x)c~ U 4 = 0 is an open set contain- ing Xo. Since (.x~) converges to xo, for all sufficiently far n we have V(x,)~ U+O, i.e. there exists z,~V(x,,) with pzn-zl<a. It follows that

d(x.)<= [t', go(x,)-z.I < It', go (Xo)- t'l go (x.){

+ Iv1 go(Xo)-Zo]-r [z,,-z,,I

i.e. that lim sup d(x,)<= d(xo)+ a. Since a is arbitrary we are done.

Lower semi continuity of d is easy. Indeed if (x.) converges to xo and z.e V(x.) satisfies

a (x.)= lvl go (x~ z.I

we may by taking subnets assume that (z,) converges to z o. Then z o necessarily belongs to V(xo). Hence

lira inf d (x , )= lim infl vl go ( x , ) - z , I (7)

= I Vl go (Xo)- Zo[ >= d(xo)

i.e. d is lower semi continuous. Finally we have

Step 3. g is cont inuous in x. Indeed let (x,,) be a net converging to Xo. By choosing a subnet if necessary we may assume that g(x.) converges to Zo. By Step 2 the right side of (4) is cont inuous in x. We

288 M. Rao:

thus have taking limits [Zo - 1)1 go (Xo)[ ~-- d (Xo).

By uniqueness we must have zo=g(xo). Thus every limit point of the net (g(x,)) coincides with g(x0) i.e. g is continuous.

It is very simple now to show that if we define the operator v on the element go by v g o=g we get the required linear extension map. That proves the lemma.

We now remove some restrictions in Lemma 1. The argument in the next lemma follows one due to Pelczynski [1, p. 23].

In the following lemma H will be a finite dimensional subspace of C(F) containing E. As in Lemma 1 v~ will be a linear extension map extending u to H and G a closed subset of X containing F.

Lemma 2. The fi)llowing are equivalent.

(a) For every seminorm function p (x , f ) with domain H satisfying

]uf(x) l<p(x, f ) , f e E , [If[] =1 , x e X

Ivl f (x)r<p(x, f ) , f e l l , H i l t = l , x e G

there exists a linear extension map v on H, which extends u such that

v j ' (x)=vl f (x) , f e l l , x e G

[vf(x)] <p(x,j ') , f e l l , x e X .

(b) For every seminorm function p(x , f ) with domain H satisJ);ing

[uf(x)[<p(x, f ) , f e E , x ~ X

Ivl.f(x)[<p(x,f), f E H , xeG

there exists a linear extension map v on H which extends u such that

/ ) f (x)=vl f (X) , f e l l , x 6 a

Ivf(x)[ <p(x , f ) , f e l l , x e X .

Before proving the lemma let us note the following: Given a semi- norm p and a linear functional S (on some linear space) with p dominating S the function q defined by

q(x) = sup Re (p(a x) - S(a x)) la l = 1

can be verified to be a seminorm. Given two seminorms ql and q2 the function q defined by

q(x)= inf (ql(xl)+q2(x2)) X14, -X2=X

is also a seminorm, q is the largest seminorm less or equal to the minimum of ql and q2; we denote this by ql Aq2.

On Simultaneous Extensions 289

Proof of Lemma 2. We show (a) implies (b). A sequence pl, P2 . . . . of seminorm functions and a sequence T1, Tz .... of linear extension maps T~ on H will be constructed by induction satisfying:

(i) pl(x,f)=3p(x,f), x~X, f~H. (ii) T ~ f = u f f e E

Tif(x)=v I f (x ) , f ~H, x~G. (iii) IT/f (x)[ <2'pi(x,f), x~X, f~H, [l/I[ = l.

(iv) [uf(x)l <2ipi(x,f), x~X,jEE, [I./'ll = 1 Iv, f(x)[<2ip,(x,f), x ~ X , f ~H, I]fH = 1.

(v) For all II.fll#:0, xeX, pi(x,f)>o; p i+ l (x , f ) is constructed as follows: define

pi+ l (x,f )= sup Re ( p(x, a f ) - a ~ 2- ' Tj l and put 1

Pi+l =q~+l A (~ 2 - i -1P) .

(iv) is valid for Pl. By (a) there is a T~ on H satisfying (ii) and (iii). Assuming that Pl . . . . . Pn, T~ . . . . . T, have been chosen, define P,+I as outlined in (v). We will show that p,+l(x,f)>O, x~X, Hfl] + 0 and verify (iv) is valid for p,+~. We can then use (a) to construct T,+I satis- fying (ii) and (iii).

Now for rlfl] = 1, 12-"T,f(x)J<p,(x,f)=(~2-"p)Aq,. Therefore for x~X, I l fH#0, q,+l(x,f)>O. Also p(x, f )~oc as I r J ' l l ~ , uniformly in x. We deduce that

p,+~(x,f)= inf (~2-"-lp(x, fO+q,+~(x,f)) f l + f 2 = f

II f l tl ~ R

where R depends on f but not on x. This establishes the continuity of p,+~(x,f). Clearly p.+~(x,f) is at least equal to the minimum of the

quantities inf ( ~ 2 - " - l p ( x , g)), [[gll ~�89

inf q. l(x,g)

and both are strictly positive. Thus p ,+L(x , f )>O, xeX, Ilfll #0 . We need only show the validity of (iv) for q.+l (this is because p . + ~ ( x , f ) = ~2-"-~p(x, fO+q,+~(x,f2) for some Z + f z = f ) . For feE, I lf l l --1 we have

2"+~Re p(x,J)- 2-.JTjf(x 1

> 2p(x,f) > luf(x) l.

290 M. Rao: o0

Finally define v = ~ 2-" T,. v is the desired linear extension map. This finishes the proof.

Combining Lemma 1 and Lemma 2 we get the following:

Theorem 3. Let E ~ C ( F ) be .finite dimensional and H a separable subspace of C(F) and p ( x , f ) a seminorm fimetion with domain H. E~'ery linear extension map u: E--+ C(X) such that

luf(x)[ < p ( x , f ) . . f~E, x ~ X

can be extended to a [i~Tear extension map t': H ~ CrX ~ sati,~f~ving

IvJ'(x)l <=p(x,f), f c H , x ~ X .

Further (f G is any closed set containing F, t'~ a linear extension map on H extending u, Ivaf(x)l<=p(x,j), .f'~H, x~G then it is possible to choose ~~ so that ~ ' f ( x ) = v l f ( x ) f o r aft x c G and f ~ H .

Taking p(x, f ) = Jl f I[F we get the most interesting case of Theorem 3.

Corollary 4. Every linear extension map u: E - + C ( X ) defined or a .finite dimensional subspaee E ~" C(F} ean be extemted without increa,se in norm to a linear extension map l~: H ~ C(X) where H is any separable subspace of C(F) eontoining E.

Remarks. It does not seem likely that in Corollary 4 we can replace E by an arbitrary separable subspace. It would clearly be of interest to know under what conditions on E the corollary is valid. It goes without saying that Corol lary4 is a considerable strengthening of Borsuk- Dugundji theorem (see [-2]).

An Application of Lemma 1

The basic data is as follows: X is a compact metric space and F a closed subset of X. B is a closed subspace of the space C(X) of all complex valued continuous functions on X. We assume that B separates points of X and contains constants. B • will denote the set of measures which annihilate B. B is further supposed to satisfy the following

Condition G. For any m~B ~, m[~. (the restriction of m to F) is in B s.

Under condition G it is well known [3, pp. 280-283] that BII: is closed in C(F) and every function in BI~. can be extended to a function in B without increase in norm.

Lemma 5 below states a condition equivalent to condition (G) and is useful. Let K(F) denote the subspace of B which annihilates F i.e. K(F) = {f: f ~B , . f (F )=0} .

On Simultancous Extensions 291

Lemma 5. The following conditions are equivalent.

1) m~B l implies the restriction of m to F is also in B a.

2) m~K(F) • implies that the restriction of m to X'-, F is in B •

Proofi 2) implies l) is easy�9 Let us show that 1) implies 2). 1) implies that Bt~. is closed in C(F). If reeK(F) • we may define a linear functional L on BIV by saying Ll&)=m(f). Continui ty of L is a consequence of the open mapping theorem. Hence there exists a measure l on F such that [ ( f ) = L ( f t r )=m( f ) for all f e B . m - I thus annihilates B. Condi t ion I. then gives that m - t restricted to X \ F also annihilates B. Since 1 lives on F the proof is finished.

Let B, X, F, K(F) be as above, p a strictly positive function on X, p (x )> 1 on F. Le t j l . . . . , j . be functions in B such that for some 1 < k < n

~ a~;i (x)<p(x) ~ a~;i , (s)

k for all compIex al . . . . . a k with ~ f a ~ I = l . Let U and 1 /be open sets (in C(X) x . . - • C(X)) defined by 1

V= ( (hk+, , . . . , h,): ~ aifi+ k+, ~ ai(�9

<p(x aifi for all F

LII . . . . . (Jn with ~lai[=l) !

U = ( ~ ' k + l . . . . . gn): atj;-~ 2ai~i k + l

n

<p(x) ~ air r for all

cq . . . . . a , with ~ [ a i , = l ) . 1

Note that U and V are convex open sets. Clearly (gk+~ . . . . . g , ) e U if and only if(Jk+~-~:k+l . . . . . . / s E That U is non empty follows from Theorem 3. Lemma 6 below shows how big K(F) is in B under Con- dit ion (G).

Lemma 6. Retain the abot:e notation. Under Condition (G), B • ... • B ca U+ 0 (/'aJ~d only (f K(F)• ... • K(F)c~ V+O.

292 M. Rao :

Proof Assuming K(F) x ... x K(F) c~ V = ~ by H a h n Banach theorem there would exist measures m,+~, ..., m, in K(F) • such that

R e ~ m ~ h ~ > 0 fo ra l l (,hk+l,...,h,)~V. (9) k + l

Let S be any compac t subset of X \ F and ( h k + 1 . . . . . h,)eV. The vector equal to (fk+j--hk+~ . . . . . J~,--h,) on S and equal to (.f~+~ . . . . . L ) on F can be extended by T h e o r e m 3 to a vector in U i.e. the vector equal to (hk+ 1 . . . . . h,) on S and (0, 0 . . . . ,0). on F can be extended to an element in V. Since V is a bounded set we deduce from (9) that

Re ~ ,~ hi dmi>=O for all (hk+, . . . . . h,)~ V. (10) k + l X ' - F

For any (gk+l . . . . . g,)E U, we thus have from (I0)

n

Re Z S (J;-gO dm,=>O. (11) k + l X \ F

Since mi annihilate K(F), by L e m m a 5 and (11)

n

R e ~ S g, dmi <0 f o r a l l (gk+l . . . . . g , )~U. (12) k + l X - . F

Since U is open the inequali ty in (12) must be strict unless all the measures m~ are concent ra ted on F. However , as we have seen the vector equal to (0, 0 . . . . . 0) on F can be extended to a vector in V and (9) cannot be valid if all the measures ml were concentra ted on F. Since m~ restricted to X'- , F do annihilate B we have thus shown that U cannot intersect B. Tha t completes the proof.

Corol lary 7. Let f l , ... , f , , in B be such that fil~ are linearly independent, satisfy (8) and

~ bis >= blJl Jbr all complex bl . . . . . b,. (13) F F

Then there exist functions gk+l . . . . . g, EB such that g~(x)=Ji(x), k+ 1 < i<n, x EF and

aig i (x)<p(x) 1 k + l

n

for all complex a~ . . . . . a. with ~ [ a i [ = 1 and all xEX. 1

On Simul taneous Extens ions 293

Proq[2 Define the open sets U and V as before. Then (0, 0 . . . . . 0)e U.

Indeed let aa . . . . . an be such that [a,I--1. If ~ l a i l = 0 the inequality 1 1

k

is trivial; on the other hand if ~lai[ =t:0 we have from (8) 1

~1 aifi (x)~p(x) ~1 ai,'i F<:P(X) ~1 aiji F (the last inequality follows from (13)).

An appeal to Lemma 6 concludes the proof.

Note that (13) says that the natural projection from the space spanned (in BIF ) bYfllF . . . . . s onto the space spanned byfJlF . . . . . fi, lv has norm one. Now we will define n~-spaces.

Definition 3. A separable Banach space B is called a n~-space if it has an increasing sequence F~, F 2 ... of finite dimensional subspaces whose union is dense in B, such that there is a projection of norm one from B onto each F,.

Step by step application of Corollary 7 then gives

Theorem 8. l f B satisfies Condition (G), BIF is a hi-space, 0 < p e C(X) with p(x)> 1Jbr x 6 F then there is a linear extension map u: Bit---' B such that (uf)(x)=f(x) for all x cF, f~BtF and

fufl(x)<p(x)lrfJlF, x~X, f~BIF. In Theorem 8 we have the condition that p(x)> 1 for x~F, It is not

clear how this directly implies the existence of a linear extension map of norm one from BIF into B. We can however get a linear extension map of norm one in the following way. We will be brief since we have already encountered these ideas.

Supposef~ . . . . . . ]~, belong to B, are linearly independent when restricted to F and satisfy

I~ aif~(x) <(1 +b)

whenever [l~aif~llv=l. Let K denote the compact set of vectors

for which . ~ a j , ~=1. Let 0 < c < b and let U denote the ? I

(a, an)

open convex set of vectors (hi . . . . . h , ) eC(X) • x C(X) for which

~ ai ( f i -h i ) (x ) <(1 +c)

21 Inventiones math,Vol 13

294 M. Rao: On Simultaneous Extensions

and

~ aihi(x) < b - c

for all (al . . . . . a,)eK. As in L e m m a 6 if Ur~K(F)• xK(F)= 0 there would exist measures mi, 1 < i < n , in K(F) • with

n

R e ~ rnihi>0, (hi . . . . . h,)eg. 1

Again as in Lemma 6 we conclude that

n

R e ~ .[ hidmi>O, (h I .. . . . h,)~U. 1 X \ F

b - c Since ~ (fl . . . . . . s clearly belongs to U, the last inequality contra-

dicts Lemma 5. What we have shown is simply that given a linear extension map u: E--*B, with E finite dimensional and Ilull < 1 + b and c < b there exists a linear extension map r : E ~ B such that II v - u I[ < b - c and II vii < 1 +c . Repeti t ion of this a rgument gives a Cauchy sequence and its limit v will satisfy Hvll = 1, Ilv-uH <b. N o w the p roof is exactly as in [4] but we give it for Reader 's convenience. Start with F 1 and let ul" F1 ~ B be a linear extension map of norm one. Extend Ul using Corol lary 7 to a linear extension map u~" F2--' B, Ilu2 II < 1 + �89 Find u 2" Fz-+B , 11u2]]=l , ]IUZ--U21]< 1. And so on. This gives a Cauchy sequence of linear extensions maps of F, and their limit will be a linear extension map on Q)F,, of norm one. Extend this to BIv.

n

The author records his indebtedness to Karl Pedersen and W. Slowikowskl for valuablc discussions.

References

1. Semadenx, Z.: Simultaneous extensions and projections in spaces of continuous func- tions. Lecture notes, No. 4. Mathematics Institute, Aarhus University.

2. Bade, W,G.: The Banach space C(S). Lecture notes, No. 26. Mathematics lnstitutc, Aarhus University.

3. Gamehn, T.W.: Restrictions ofsubspaces ol C(X). Trans. Amer. Math. Soc. 112, 278 286 /1964).

4. Michael, E., Pelczynskl, A.: A linear extensmn theorem, i11. J. Math. II, 563-579 (1967).

Murah Rao Matemat~sk lnstitut Universitetsparken Ny Munkegade DK-8000 Aarhus C, Denmark

(Received February 8, 1971/Apri115, I971)