By On the extension of Lipschitz maps

A R K I V F O R M A T E M A T I K Band 7 nr 15 1.66 16 2 Communicated 7 December 1966 by OTTO FROSTMAN

On the extension of Lipschitz maps By

STEIN OLOF SCHONBECK

Introduction

L e t X a n d Y be metric spaces. A m a p T f r o m X into Y is called a

Lipschitz map

if there is a constant M, such t h a t

d(Tx 1, Tx2) ~ Md(x 1, x2)

for all xl, x~ E X, a n d t h e n t h e n o r m of T, ]] TI], is defined as t h e least such constant. T is called a

contraction

if ]]TII ~<1. I f D is a subset of X a n d

T:D->

Y a Lipschitz map, it is n a t u r a l to ask w h e t h e r T m a y be e x t e n d e d to a Lipschitz m a p T : X - ~ Y so t h a t ]]TI] = HTII. This problem has been studied b y several authors. A short s u r v e y of results in this con- nection is given in the paper of Danzer, G r i i n b a u m a n d Klee [2]. I n t h e present n o t e we are concerned with some aspects of this problem in t h e case t h a t X a n d Y are real B a n a c h spaces. More precisely, we are interested in those pairs of B a n a c h spaces X a n d Y, for which the extension problem f o r m u l a t e d above always has a solution. I t is obvious that, in the case of B a n a c h spaces, it is sufficient to consider contractions, for if each c o n t r a c t i o n of a subset of X into Y m a y be e x t e n d e d to a c o n t r a c t i o n of X into Y, t h e n also each Lipschitz m a p of a subset of X into Y m a y be e x t e n d e d to a Lipschitz m a p of X into Y, w i t h o u t increasing t h e norm. I n order t o abbreviate, we will say t h a t (X, Y) has E P C (extension p r o p e r t y for contrac- tions) if for each subset D c X a n d each c o n t r a c t i o n

T:D-~ Y,

there exists a contrac- t i o n T : X - > Y, which extends T.

The fact t h a t (X, Y) has E P C m a y also be f o r m u l a t e d as a n intersection p r o p e r t y f o r cells in X a n d Y. A

cell

in a B a n a c h space X is a set of the f o r m

Sx(xo; r)=

{xEX: IIX-xoi] <~r}.

The

unit cell o / X

is the cell

Sx=Sx(O;

1). W h e n :~ a n d ~ are families of cells in X a n d Y respectively, we shall write :~ >-~ if there is a one-to- one correspondence between D: a n d 6 , such t h a t corresponding cells h a v e equal radii a n d the distance between t h e centers of a n y t w o cells in ~ is less t h a n or equal tO t h e distance between t h e centers of the corresponding cells in :~. If :~ is a f a m i l y of sets, t h e n ~:~ denotes t h e intersection of the sets belonging to :~. T h e n t h e fol- lowing two s t a t e m e n t s are equivalent (see [2]):

(i) (X, Y) has EPC,

(ii) whenever D: a n d ~ are families of cells in X a n d Y respectively, such t h a t : ~ N ~ a n d 7c:~ q=r t h e n ~ - ~ r

We will also use the following terminology, due to A r o n s z a j n a n d P a n i t c h p a k d i [1] (see also Klee [10]). F o r a cardinal n u m b e r y >~3, a space X is said to be

~-hyper- convex

if each collection {S~}, a E

A,

of m u t u a l l y intersecting cells in X with card A

< 7 , has a n o n e m p t y intersection. X is called

weakly ~-hyperconvex

if this condi- tion holds u n d e r the f u r t h e r a s s u m p t i o n t h a t all t h e S~ have a c o m m o n radius. X is

15:3 201

S. O. SCH6NBECK, Extension of Lipschitz maps

called (weakly) hyperconvex if it is (weakly) y - h y p e r c o n v e x for all 7" I t was p r o v e d b y H a r m e r [7, T h e o r e m 6.1] t h a t if 7 is finite, t h e n for finitedimensional spaces w e a k ~ - h y p e r c o n v e x i t y implies 7 - h y p e r c o n v e x i t y , and this result was e x t e n d e d to a r b i t r a r y B a n a c h spaces b y Lindenstrauss [11, T h e o r e m 4.3]. H a r m e r also p r o v e d that, for finitedimensional spaces, 5 - h y p e r c o n v e x i t y implies tC0-hyperconvexity, and this, too, was e x t e n d e d to a r b i t r a r y B a n a c h spaces b y Lindenstrauss [11, T h e o r e m 4.1].

I t is well-known ( A r o n s z a j n - P a n i t c h p a k d i [1], Goodner [5], K e l l e y [9], N a c h b i n [12]) t h a t the following properties of a real B a n a c h space are equivalent:

(a) X is h y p e r c o n v e x ,

(b) X is a ~)1 space, i.e. for a n y B a n a c h space Z containing X as a subspace, t h e r e is a projection of n o r m 1 from Z onto X,

(c) X is isometric to a space C(K), where K is an extremally disconnected c o m p a c t Hausdorff space.

R e t u r n i n g now to our object in this paper, we note t h a t the following facts are previously known (for the proper references we refer to [2]):

(1) (X, Y) has E P C for all X if and only if Y is a 01 space, (2) (X, Y) has E P C if X and Y are H i l b e r t spaces,

(3) If dim X = 2 , t h e n (X, X ) has E P C if and only if X is a Hilbert space or 01 space (Grtinbaum [6]).

I n section 1 we will prove t h a t if Y is strictly convex and dim Y > 1, t h e n (X, Y) has E P C only if X and Y are t t i l b e r t spaces (this result was a n n o u n c e d in [13]).

Section 2 contains a proof of the fact t h a t 3) above remains true if dim X = 2 is replaced b y dim X < o~. Finally, we give, in section 3, some results in the case t h a t X is a C(K) space; in particular, we prove t h a t (C(K), C(K)) has E P C only if K is e x t r e m a l l y disconnected.

1. Pairs (X, Y) with Y strictly convex I n this section we prove the following theorem:

Theorem 1.1. I] X and Y are Banach spaces such that Y is strictly convex, dim Y > 1 and (X, Y) has E P C then X and Y are Hilbert spaces.

T h e o r e m 1.1 follows from the following two lemmas.

L e m m a 1.2. I] X and Y are Banach spaces such that Y is strictly convex and (X, Y ) has EPC, then X and Y satis/y the ]ollowing condition:

(A) I / x x , x 2 E X , Yx, y~E Y and [[Xln = ]]YlH, ]]x2]] = ]]Y2H, ]]Xl x2]] ⁼ ]]Yl--Y~]] theu [[/~Xl -~ teX2][ ~ [[2Y1 § /or all ,~, te.

Proo/. L e t xl, x 2 e X , Yl, Y2 e Y a n d [[xl[ [ = [[y,][, [[x2[ [ = [[y~[[, [[xl-x~[[ = [[Yl-Y~[[ = d - Suppose first t h a t ~t ~> 0, # >~ 0, ~t + # = 1. Consider the cells S~x = Sx(xl; ted), S2x = Sx(x~;

2d), S3x=Sx(O; [[~tXl+tex2[[) in X and S ~ = S r ( Y l ; ted), S2r=Sr(y2; ~g), S ~ = S r ( 0 ; [[~txl +tex2[[) in Y. If we p u t :~={S~x, S~, Sax} and ~={S~y, S~, S3r}, we have Y > ~ and ~ =#r because 2x, +tex2 ez:~. Thus we get ~ =4=r and, since S~ N S~ = {~ty I +teY2}

ARKIV leOR MATEMATIK. B d 7 n r 15

(here we use t h e strict c o n v e x i t y of Y), we m u s t h a v e 4y, +/~Y2 e 83, i.e.

114y, +~y211

<

4x 1 +#x~ . T h e condition 4 + # = 1 m a y n o w be r e m o v e d a n d so we h a v e p r o v e d t h a t

4Xl+#X 2 >~ 4 y , + # y 2

for 4>~0, ~u~>0. I f we a p p l y this result to

x l - x ~, - x ~

a n d Y, -Y2, - Y ~ instead of Xl, x2 a n d Y,, Y2 we get 114(Xl --X2) --fiX211 ~ I[)t(yl --y~) --#Y~II for 4 >~ 0, # ~> 0, which is e q u i v a l e n t to 114x* + ~ Z 211 ~ 114Y1 + ~Y2 II f o r 2~ >//0, ~ ~ 0, 4 +/-~ ~ 0.

Similarly, replacing Xl, x2 a n d Y,, Y2 b y

x ~ - x l , - x 1

a n d

Y 2 - Y , , - Y , ,

we get 114x* +/~x2 II ~> 114Y* +#Y2 I I for 4 >~ 0,/~ ~< 00 4 + # >~ 0. T h e inequalities n o w p r o v e d t o g e t h e r i m p l y t h a t 114x, +~ux211 ~> 114y, +juy,~ll for all 4,/~, a n d t h e proof is complete.

L e m m a 1.3.

Suppose X and Y are two normed linear spaces such that dim Y > 1.

I / X and Y satis/y the condition (A) o] L e m m a

1.2,

then X and Y are euclidean (i.e.

prehilbert) spaces.

Proo/.

I n t h e proof we will use t h e concept of n o r m a l i t y defined as follows: L e t L be a real n o r m e d linear space. A n e l e m e n t x EL is said to be n o r m a l to a n e l e m e n t

y E L

if

IIx+4yl] >~ Ilxl[

for all 4. I f x is n o r m a l to y we write

xNy.

W e first p r o v e t h a t (A) implies:

(B) I f x I , X 2 E X ,

y , , y ~ E Y

a n d

IlXll]=llyill, I[x~ll=lly~ll, x , N x ~, y,Ny~

t h e n ll~xl+,x~ll ~> 114yl+#y~ll for all 4,

~

I t is sufficient to a s s u m e t h a t

IIx, H=Hx~II=Hy, H=IIy2H=I.

T h e r e exist two sequences (x~) a n d (x~') in t h e linear s p a n of x I a n d x2, such t h a t

H x~ H = IIx~'H = 1, xn'--x~ = HXn'-x~ Ilx2 =#0, lim X'n = l i m x~' = x r

T h e n we can find two sequences (y~) a n d (y~') in t h e linear s p a n of Y, a n d y2, such t h a t

Ily'll ^{n =} Ily"H ^{n = 1 ,}

Yn - Y n = xn --Xn

" ' l] " 'lly

3'

lim y~ = l i m y~ =Yl. '

"

F r o m (A) it follows t h a t H4x~ +#x~'ll/> 114yn +ttyn'll for all ~t, #, which gives

* H t *t z ~ 9 H 9 9

114~ +~,(xn - x~)/llz~ - x~ll II ~ 114y~ +z( y~ -y~/llx~ - x'nll H for all

~, , ,

or H4x~ +/tx2H ~> 114y~ +try211 for all 4,/t. L e t t i n g n t e n d to infinity we get H4Xl +#x211 >~

114y~ +~y~]l for all 4, ~, which was to be proved.

W e can n o w show t h a t n o r m a l i t y is a s y m m e t r i c relation in b o t h X a n d Y. Suppose first t h a t xl, x2 E X a n d

x,Nxe.

B y a result of D a y [3] we k n o w t h a t t h e r e exist two e l e m e n t s YD y2E Y such t h a t

IlY*II = IIXlll, Hy2H = IIx~ll, y,Ny2

a n d y 2 Y Y l . Hence we get, using (B),

II,~xi + x211 >~ ll4y, + y211 >~ IlY211 = IIx~H ,

which m e a n s t h a t

x~Nx,.

This p r o v e s t h a t n o r m a l i t y is s y m m e t r i c in X. N o w suppose t h a t y,, y2 E Y, y, # 0 , y~ # 0 a n d

y,Ny~.

Obviously, t h e r e exist two elements

x,, x ~ E i ,

such t h a t

llxill

⁼

By, l[, I1~11 = Ily~ll, **N.~

a n d

I I x ~ + ~ l l > IIx, ll for 4 + 0 .

L e t u be a n e l e m e n t in t h e linear span of y~ and y~ such that Ilull =IlYIII and y~Nu. If u = ~ y ~ + ~ y ~ and if we p u t z = ~ x I +fix2, t h e n b y (B), for each v,

IIx2 + ~ll

= H v~xi + (1 +

~)x~ll >~ II~y, +

(1 +

~)y~ll = Ily~+ ~ull >~ Ily~ll ~ Ilx211.

H e n c e we get

x2Nz

and, since n o r m a l i t y is s y m m e t r i c in

X , zNxz.

T o g e t h e r w i t h the condition

IIz~+~ll

> I1~111 for

4+0, ~Nx~

implies f i = 0 . T h u s we h a v e

u = ~ y 1

and, 203

s. o. scrtSNBECK, Extension of Lipschitz maps

since

Ilull- IlYall,

u - 2 [ - y 1. H e n c e y2Nya a n d it follows t h a t n o r m a l i t y is s y m m e t r i c in Y.

W e n o w use the following result which follows f r o m a construction of t w o d i m e n - sional spaces with s y m m e t r y of n o r m a l i t y given b y D a y [4, p. 330]:

L e t L a n d M be t w o d i m e n s i o n a l n o r m e d linear spaces, in b o t h of which n o r m a l i t y is s y m m e t r i c , a n d let Xl, x 2 e L , YD Y2 E M be two pairs of elements such t h a t

Ilxlll

=

IlYall, IIx~ll=llY,II and

aleX2,

ya~y2.

T h e n if

II~xl+flx~ll~>llXYa+flY~ll

for 2#~>0

we have IlXxl+flx, II < II2Yi+flY.II for ~ttt ~<0.

F r o m this result a n d w h a t we h a v e p r o v e d above, it follows t h a t if xi, x2EX, YI, Y2 E Y a n d

Ilxill =llYall, IIx~ll-IlY~II, XaNX~, y~Ny~, then we have II~xi+~x~ll =

libYa +~Y~II for all

~, ~

In particular, we see that if Z is a twodimensional subspace of either X or Y a n d if z, z' EZ,

I1~11

=

I1~'11-1,

t h e n there is a n i s o m e t r y of Z o n t o itself which carries z into z'. H o w e v e r , it is well-known t h a t this is possible o n l y if the u n i t circle in Z is a n ellipse, i.e. if Z is a n euclidean space. H e n c e all t w o d i m e n - sional subspaces of X a n d Y are euclidean, which m e a n s t h a t X and Y are euclidean spaces. T h u s the l e m m a is proved.

Remark. I t m a y be o b s e r v e d t h a t L e m m a 1.2 a n d therefore also T h e o r e m 1.1, is still valid u n d e r the w e a k e r h y p o t h e s i s t h a t (X, Y) has t h e following p r o p e r t y : I f a n d ~ are families of cells in X a n d Y respectively, such t h a t :~ ~ ~, ~ 9 : # 4 a n d card ~ = 3 , t h e n zt~ # 4 . This is a p p a r e n t f r o m the proof of L e m m a 1.3.

2. Pairs ( X , X ) with dim X <

W e first p r o v e two l e m m a s w i t h o u t a n y restriction on d i m X.

L e m m a 2.1. Let X be a Banach space such that (X, X) has EPC. Suppose that e 1, ee are extreme points o/ S z and Xa, x2EX are such that Hal+elf[ :]]x2+e2H, Hxl-elH =

Hxe-e211. Then it follows that H,~xi+fleall = II~x2+fle211 /or ])'1 >~ ]fl]"

Proof. I t is clearly sufficient to p r o v e t h a t llXa+fleall =llx2+fle~ll for 0 ~ < ~ < 1 . Consider t h e cells S 1 = S(X 1 -- el; 1 +/t~), S 2 = S(X 1 + el; 1 - - f l ) , S 3 = S ( 0 ; Ilxi + f i e 1H) a n d S;=S(x,z-%; l +fl), S 2 = S ( x e + e 2 ; 1 - # ) , S~=S 3. I f we p u t ~={Sa, S2, S3} a n d

={Si, S~, S~} t h e n we h a v e ~ 7 ~ a n d ~ z ~ # 4 (for X a + f l e l E T t ~ ). Since (X, X) h a s E1)C it follows t h a t ~ # 4 . Now, since e 2 is a n e x t r e m e p o i n t of S x t h e intersection S~ f3 S~ contains e x a c t l y one point, n a m e l y x 2 +fie 2. T h u s we m u s t h a v e x 2 + # e 2 E S~ =

$3, which m e a n s t h a t ]]x2+fle211 <]]Xl+flelH. B y s y m m e t r y we m u s t also h a v e IIx~ +~e~ll ~< II/~

+fle~ll

a n d hence fix 1 + f i e a[[ = fiX 2 +fle2[ [, which completes t h e proof.

L e m m a 2.2. Let X be a Banach space such that (X, X) has EPC. I f Y is a twodimen- sional subspace o/ X containing an extreme point e of Sx, there exists y E Y such that ]]y]] = 1 and ]]~y+#e H = H2y-tte]] for all ~, #. I~ Y' is another twodimensional sub- space of X containing an extreme point e' o / S x and i / y ' e Y' is such that IlY' II = 1 and ]],~y' +fie' H = ll2y'-/~e' H /or all ,~, fl, then the linear map T of Y onto Y' defined by T e = e', T y = y' is an isometry.

Proof. I t is easily seen t h a t for each n = l , 2, ..., we can find y~e Y w i t h ]]y~]] = 1 such t h a t Hy~+ne H = ]]y~-ne H. A p p l y i n g L e m m a 2.1 with c a - e , % = - e , Xa=X2=

AllKIV FOIl MATEMATIK. Bd 7 nr 15

y,,In we get II~(Ydn)+ffell

= IlX(Y./n)-ffell for I~1/> Iffl or IlXY~+ffell = II~Y:-ffell for nl;~l >~ Iffl" I f y is a limitpoint of the sequence (y~), it follows that IlYll-1 and

Ilay+ffell = IlXY-ffell

for all X, #. Thus t h e first s t a t e m e n t of the l e m m a is proved.

To p r o v e t h e second s t a t e m e n t , we observe t h a t 1 =

lie II <~

1/2(llXY

+ ell + IlXY- ell) = It~Y+ell

for all ~. T h u s for each n = l , 2 ... there is a n u m b e r Sn>~O such t h a t

II(y/n) +ell = II~=Y'+e'll.

T h e n we also h a v e

II(y/n)-ell = II~Y'-e'll. An

application of L e m m a 2.1 gives II~(Y/n)+~ell- II~=y' +~e'll for I~1 ~> I~1" P u t t i n g ff = 0 we get s ~ = l / n . T h u s for each n we h a v e

II~(Y/n)+~ell=ll~(Y'/n)+#e'll

for I~1/>

I~1

or

IlXY+ffell = IlXY'+ffe'll

^for

nlXl >~ Iffl"

H e n c e it follows t h a t

IlXY+ffell = IlXY' +ffe'll

for all X, if, which means t h a t T is an isometry.

L e m m a 2.3. Let X be a Banach space such that (X, X ) has E P C , d i m X < oo and X is not strictly convex. Then S x is a polyhedron. Moreover, there is a number a > 0 , such that i / e is an extreme point o / S x , i / x E X and i / t h e segment [e, x] between e and x is a maximal line segment on the boundary o / S x , then I l e - x l l = a .

Proo/. Since X is n o t strictly convex, it is clear t h a t the b o u n d a r y of S x m u s t contain a p r o p e r line segment, one e n d p o i n t of which is an e x t r e m e p o i n t of S x . I n other words, there exists an extreme point e of S x a n d a point x o with IIx011 = 1, such t h a t x 0 ~ _+ e a n d ]]~e + fix o II - '~ + # for 2 >~ 0,/x ~> 0. L e t Y be t h e twodimensional subspace of X d e t e r m i n e d b y e a n d x0, a n d let y be an element of Y such t h a t l i l y + r e H = H~y-ffe]l for all ~ , # ( L e m m a 2.2). T h e n we m a y write X o = ~ e + f l y . I f we p u t x0 = ~ e - f l y , t h e n ]]x0]] = 1 a n d I[~e +#x0]] = 2 + f f for ~t ~>0, ff >~0. Geometrically this means t h a t the two line segments [e, x0] a n d [e, x0] belong to the b o u n d a r y of St. The points x 0 a n d x0 determine t w o closed arcs of the u n i t circle in Y with end- points x 0 a n d Xo. One of these arcs, which we shall call F, does n o t contain e. L e t d be t h e distance f r o m e to F. T h e n d > 0 .

Now, if e I a n d e z are two a r b i t r a r y extreme points of Sx, el # +_ e2, a n d if Z is the twodimensional subspace of X d e t e r m i n e d b y e I a n d e2, there is b y L e m m a 2.2 an i s o m e t r y of Z onto Y, which carries e 1 into e. F r o m t h e c o n s t r u c t i o n of F it is obvious t h a t this i s o m e t r y carries e 2 into a point belonging to F. Hence we have lie1-e2[ [ >~d.

Since this holds for e v e r y pair of e x t r e m e points of S x a n d since dim X < ~ , it follows t h a t there are only a finite n u m b e r of e x t r e m e points of Sx, which means t h a t S x is a polyhedron.

I n order to prove the second s t a t e m e n t of the lemma, let el, e2 be e x t r e m e points of S x a n d xl, x 2 points w i t h HXlH = ]]x2] ] = 1, such t h a t [e, xl] a n d [e 2, x2] are m a x i m a l line segments belonging to t h e b o u n d a r y of Sx. W e t h e n h a v e to prove t h a t lie 1 - x l ] ] -

]]e2-x2H. L e t Yl a n d Y2 be the two-dimensional subspaces of X, d e t e r m i n e d b y e 1, x 1 a n d e2, xz respectively. According to L e m m a 2.2, we m a y t a k e yl E Yl SO t h a t []Yl[[ = 1 , [[Jtyl+ffei[ [ = ][~tyl--ffel] ] for all X,# a n d so t h a t Yl a n d x I lie in t h e same halfplane in ]71 d e t e r m i n e d b y t h e line t h r o u g h 0 a n d e x. W e take ysE Y2 in t h e corresponding way. If T is the i s o m e t r y of Y1 onto Y2 for which Te~ =%, T y l = y 2, t h e n it is obvious t h a t T x 1 = x 2. T h u s [[e I - X l ] ] = ne2--x2n a n d the proof is complete.

Theorem 2.4. Let X be a / i n i t e - d i m e n s i o ~ l Banach space such that (X, X ) has E P C . Then X is either a Hilbert space or a 01 space (i.e. the unit cell o] X is either a euclidean cell or a parallelotope).

Proo/. As we m e n t i o n e d in the introduction, this t h e o r e m was p r o v e d b y Griin- b a u m [6] in the case d i m X = 2. T h u s we m a y assume t h a t dim X >~ 3. If X is strictly 205

s. o. SCH6~BECK, Extension of Lipschitz maps

c o n v e x , i t follows f r o m T h e o r e m 1.1 t h a t X is a H i l b e r t space. I t r e m a i n s t o p r o v e t h a t if X is n o t s t r i c t l y c o n v e x t h e n X is a ~)x space. I n t h i s case we k n o w f r o m L e m m a 2.3 t h a t S x is a p o l y h e d r o n . L e t F b e a t w o d i m e n s i o n a l face of Sx. T h e n F is t h e c o n v e x h u l l of a c e r t a i n s e t {e I . . . era} , m ~> 3, of e x t r e m e p o i n t s of Sx. L e t a be t h e n u m b e r , whose e x i s t e n c e was p r o v e d in L e m m a 2.3. I f et, er d e t e r m i n e a side [e~, ej] of F a n d if % ~ [e~, ej] t h e n [ek, x] is a m a x i m a l line s e g m e n t on t h e b o u n d a r y of S x for e a c h xe[e~, e,]. C o n s e q u e n t l y ,

Ile -xll

= a for x e [ e , , e,]. W e n o w p r o v e t h a t e v e r y t w o d i s j o i n t sides of F a r e parallel. S u p p o s e t h a t el, e 2 a n d % e 4 d e t e r - m i n e t w o d i s j o i n t sides of iv. W i t h o u t loss of g e n e r a l i t y , we m a y a s s u m e t h a t e 1 a n d e 4 lie on d i f f e r e n t sides of t h e line j o i n i n g e z a n d e a. W e k n o w t h a t I l e i - x l [ = a for all x E [ % %]. F r o m t h i s it follows t h a t lies - xll = a for all x E [e~. + e s - % %1- Since we also h a v e I l e a - x l l = a for all x E [ q , ez], a n d since % + e a - e a a n d e I lie on t h e s a m e side of t h e line j o i n i n g e 3 a n d %, it follows t h a t [ e 2 + e 3 - e a , %] a n d [el, %] a r e parallel. T h u s [el, %] a n d [%, ed] a r e p a r a l l e l .

T h e p r o p e r t y of F j u s t p r o v e d i m p l i e s t h a t t h e b o u n d a r y of Y is e i t h e r a t r i a n g l e w i t h v e r t i c e s % % e a or a p a r a l l e l o g r a m w i t h v e r t i c e s el, % % % I n t h e first case p u t x 1 - e l , x ~ = e 2 § D x a = 2 e 2 - e a. T h e n IIx~-x~ll : IIx~-x~ll = IIx~-xall = 2 a a n d

Ilxl-e~ll

⁼

IIx -e ll

^:

IIx~-e~ll-a.

I n t h e s e c o n d case, a s s u m e t h a t [el, %] is a side of F a n d p u t x 1 = 2 q , x 2 = 2 % , x a = 2 e 3. T h e n we h a v e IlXl-x~ll = Ilx~-x~ll : IIx~

-Xlll

= 2a and IIxl-(el+e )ll =llx -(el+e )ll =llx -(e +e )ll = a In any case this shows t h a t X c o n t a i n s t h r e e cells S,=S(x,; a), i - l , 2, 3, such t h a t

Ilx,-x, II

= 2 a for i # ] a n d n ~ - i S~ 4 r Since (X, X) h a s E P C , i t follows t h a t X is w e a k l y 4 - h y p e r c o n v e x . H a n n e r [7, p. 75] h a s p r o v e d t h a t if d i m X = n < ~ a n d X is w e a k l y 4 - h y p e r c o n v e x , t h e n t h e u n i t cell of X is a f f i n e l y e q u i v a l e n t t o t h e c o n v e x h u l l of s o m e of t h e ver- tices of a n n - d i m e n s i o n a l cube. As is e a s i l y seen, t h i s i m p l i e s t h a t for a n y p a i r of e x t r e m e p o i n t s % % of S x w i t h e 1 # _ % we h a v e ]1 e 1 + e 2

II

= II - e~ll = 2. H e n c e , since t h e f o u r cells S(d__e 5 1), i = 1, 2, h a v e a n o n e m p t y i n t e r s e c t i o n a n d (X, X) h a s E P C it follows first t h a t P is w e a k l y 5 - h y p e r c o n v e x a n d f r o m t h i s t h a t X is 5 - h y p e r - convex. B u t t h i s i m p l i e s t h a t X is ~r a n d t h e r e f o r e X is h y p e r c o n v e x , b e c a u s e d i m X < c~. This c o m p l e t e s t h e p r o o f of T h e o r e m 2.4.

Remark. F r o m t h e p r o o f s we h a v e g i v e n i t is e v i d e n t t h a t T h e o r e m 2.4 r e m a i n s t r u e , if t h e h y p o t h e s i s t h a t (X, X) has E P C is r e p l a c e d b y t h e following w e a k e r h y p o t h e s i s : I f J a n d ~ a r e families of cells in X such t h a t D:N~, 7~D:#r a n d c a r d J~<4, t h e n 7e~ # r

3. Pairs (X, Y) with X = C ( K )

I n t h i s section we give some r e s u l t s c o n c e r n i n g p a i r s of B a n a e h s p a c e s X a n d Y, such t h a t (X, Y) h a s E P C a n d X is a C(K) space, i.e. t h e space of r e a l c o n t i n u o u s f u n c t i o n s on a c o m p a c t H a u s d o r f f s p a c e K . W e first p r o v e a s i m p l e l e m m a .

L e m m a 3.1. I / T is a topological space, there exists a dense subset V c T and a mapping #: V-+.,4( T), where .,4( T) is the/amily o/ closed subsets o/ T, such that

(i) v r /or each v E V,

(ii) I / v l , v2E V, v I # v 2, we have either vlE#(v2) or v~Ett(vj).

Proo/. L e t s d e n o t e t h e set of all p a i r s (U, v), where U = T a n d v is a m a p p i n g of U i n t o A ( T ) s a t i s f y i n g (i) a n d (ii). C l e a r l y s is n o t e m p t y . W e o r d e r ~ b y a rela-

ARKIV FOR MATEMATIK. B d 7 n r 15

tion ~<where (U1, vl)<~(U2, v2) means t h a t U 1 c U~ and vl(ul)= vz(Ul) for u 1 E U 1. I t is clear t h a t each linearly ordered subset of ~ has an u p p e r bound in ~ , and so b y Zorn's l e m m a there exists a m a x i m a l element (V,/x) in ~ . We assert t h a t V is dense in T. Indeed, if this were not the case, we could take a point p E T with p eV, a n d t h e n ( V ' , # ' ) E ~ , where V ' = V U { p } and /z'(v)=/u(v) for v E V , # ' ( p ) = V . Since (V, #) <(V',/x') this would contradict the m a x i m a l i t y of (V, ju). Hence V is dense in T a n d the l e m m a is proved.

Definition 3.2. I / T is a topological space we denote by d( T) the least cardinal number d, such that T contains a dense subset with d elements.

L e m m a 3.3. Let K be a compact Hausdor/[ space and {r~},

.eA,

a / a m i l y o / n o n - negative numbers with card A <d(K). Then there exists a /amily o/ cells S(x,; r~), o~EA, in C(K), such that

IIx -xpll

=r~+rB /or :r and f]~eA S(xa; r~) 4 r

Pro@ According to L e m m a 3.1 there exists a family {k~},

aeA,

of elements of K and a family {F~}, ~EA, of closed subsets of K, such t h a t k~ e F ~ a n d for a # f l we have k~EFp or k z E F ~. Since K is completely regular, there is, for each ~EA, a continuous realvalued function x~ on K, such t h a t x~(k~) = r:, x~(k) = - r ~ for k E F~

and

Ix (k)l

for k e g . Thus we have x~ =supk~x x~(k)] =r a a n d from the properties of {k~} and {F~} it follows t h a t x ~ - x p =r~ + r z for a 4/3. Consequently, the family of ceils {S(x~; r~)}, ~EA, in C(K) has the required properties.

])efinition 3.4. Let X be a Banach space. B y dim X we understand the least cardinal number 7 such that X is the closed linear hull o/ a subset with 7 elements.

Theorem 3.5. Let K be a compact Hausdor/] space and Y a Banach space such that d(K) >1 dim Y. I / ( C ( K ) , Y) has E P C then Y is a ~1 space.

Pro@ If dim Y < o~, it suffices to show t h a t Y is 5-hyperconvex. E x c e p t in the trivial case dim Y = 1, we have d(K)>~2. Then it is easy to see that, for a n y r, > 0 , i = 1 ... 4, C(K) contains four cells S(x 5 r~), such t h a t Hx~-xjll = r , + r s for i=#j and

f] ~=1 S(x,; r~) =4=r Since (C(K), Y) has E P C it follows t h a t Y is 5-hyperconvex.

Now assume t h a t dim Y is infinite. Since d(K)~>dim Y a n d (C(K), Y) has E P C it follows, with the aid of L e m m a 3.3, t h a t a n y collection {S~}, a EA, of m u t u a l l y intersecting cells in Y with card A = dim Y has a n o n e m p t y intersection. Aronszajn and Panitchpakdi (Theorem 1 of section 2 in [1]) have proved the following result:

I f a metric space E is 7-hyperconvex a n d at the same time y-separable, then E is hyperconvex (E is 7-separable means t h a t E contains a dense subset of cardinality

<7). Using this result we m a y now conclude t h a t Y is hyperconvex, i.e. Y is a ~)1 space. Hence Theorem 3.5 is proved.

I t is well-known t h a t there are no separable infinite-dimensional ~1 spaces. Thus we get

Corollary 3.6. I / K is an in]inite compact Hausdor// space and Y a separable Banach space such that (C(K), Y) has EPC, then Y is a ]initedimensional ~1 space.

The result of Aronszajn-Panitchpakdi used in the proof of Theorem 3.5 m a y be improved in the case of C(K) spaces.

L e m m a 3.7. Let 7 be a cardinal number and K a compact Hausdor// space containing a dense subset o/cardinality <7. I] C(K) is 7-hyperconvex then it is hyperconvex (and hence K is extremaUy disconnected).

207

S. O, scnSNBECK, Extension of Lipschitz maps

Proo/. I f ~ <~r there is n o t h i n g to prove, a n d so we m a y assume t h a t 7 > ~r L e t {S~)={S(x~; r~)}, ~EA, be a collection of m u t u a l l y intersecting cells in C(K).

We h a v e to prove t h a t N ~A S~ # r L e t D be a dense subset of K with card D < 7 a n d p u t / ( k ) = sup~ ~A (x~(k) -- r~), g(k) = inf~ ~A (x~(k) + r~). T h e n / ( k ) <~ g(k) for k E K a n d N ~A S~ = {x E C(K) :/(k) ~< x(k) <~ g(k) for k E K}. To each k E D a n d each n a t u r a l n u m - ber n we choose ~ = ~(k, n) EA so t h a t x~(k) + r~ < g(k) + (l/n) a n d fi =fl(k, n) E A so t h a t x,~(k)-rp>/(k)-(1/n). The cells S:(k,~) a n d S~(k,~) for k E D a n d n - l , 2, ..., are m u t u a l l y intersecting a n d - f o r m a family o f c a v d i n u l i t y ~ 2 ~ (card D) ~r Since C(K) is 7 - h y p e r c o n v e x it follows t h a t there exists xEC(K), such t h a t x E S~(k, ~) N SZ(k, ~) for k E D, n = l, 2 ... This means t h a t / ( k ) - (1/n) < x(k) < g(k) + (1/n) for kED, n - l , 2 ... f r o m which follows that/(k)<~x(k)<~g(k) for kED. Since / is lower semicontinuous a n d g is u p p e r semicontinuous a n d D is dense in K, this implies t h a t /(k)<~x(k)<~g(k) for all k E K , i.e. X E N ~ K S ~ a n d hence N ~ A S , # r which was to be proved.

W e can now prove an i m p r o v e d version of T h e o r e m 3.5 for the case t h a t Y is a C(K) space.

Theorem 3.8. Let K 1 and K 2 be compact Hausdor// spaces such that d(K1) >~ d(K2).

Then (C(K1) , C(K2)) has EPC i / a n d only i / K s is extremally disconnected.

Proo/. The if p a r t is clear. Assume then t h a t (C(K1), C(K2)) has EPC. L e t ~ be t h e least cardinal n u m b e r such t h a t 7 >d(K1)" Then, using L e m m a 3.3, we see t h a t C(K2) is 7 - h y p e r c o n v e x , a n d since d(K2)<7 it follows f r o m L e m m a 3.7 t h a t C(K2) is h y p e r c o n v e x a n d hence t h a t K s is e x t r e m a l l y disconnected.

Corollary 3.9. I / K is a compact Hausdor// space, then (C(K), C(K)) has EPC i/

and only i / K is extremally disconnected.

Remarks. 1. To justify the s t a t e m e n t t h a t L e m m a 3.7 a n d T h e o r e m 3.8 are im- p r o v e m e n t s of the earlier results, it should be n o t e d t h a t there are c o m p a c t Haus- dorff spaces for which d(K) < d i m C(K). I t is easy to verify t h a t t h e S t o n e - ~ e c h compactification of a countable discrete space has this p r o p e r t y . Less trivial examples m a y also be given. As a m a t t e r of fact, it is n o t difficult to prove t h a t for a n y c o m p a c t H a u s d o r f f space we h a v e dim C ( K ) - b(K), where b(K) is t h e least cardinal n u m b e r b, such t h a t the t o p o l o g y of K has a base with b elements. I t is well-known t h a t there are K for which d(K)<b(K). (See, for instance, E x a m p l e M on p. 164 of Kelley [8].) 2. If K is t o t a l l y disconnected, it is easy to show t h a t dim C(K)= card :H, where :H is t h e f a m i l y of all open a n d closed subsets of K. This fact m a y be used to show t h a t the conclusion of L e m m a 3.3 still holds for K, if the hypothesis card A <~d(K) is weakened to card A ~<dim C(K). This means that, if K is t o t a l l y disconnected, T h e o r e m 3.5 remains true, when t h e hypothesis d(K)~>dim Y is replaced b y d i m C(K) >~dim Y. Similarly, if K 1 is t o t a l l y disconnected, we m a y replace d(K1)>~d(K~) b y dim C(K1)>~d(K2) in T h e o r e m 3.8. We do n o t k n o w w h e t h e r these i m p r o v e m e n t s are a c t u a l l y valid for all c o m p a c t spaces.

University o] Stockholm, Sweden

208

ARKIV FOR MATEMATIKo B d 7 n r 1 5

R E F E R E N C E S

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Soc. Proc. S y m p . Pure Math., Vol. 7, Convexity, 101-77 (1963).

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Tryckt den 6 oktober 1967 Uppsala 1967. Almqvist & Wiksells B oktryckeri AB

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