2.5. THE THEOREM OF EBERLEIN SMULIAN
2.5
47
The Theorem of Eberlein Smulian
For infinite dimensional Banach spaces the weak topology is not metrizable
(see Exercise 1). Nevertheless compactness in the weak topology can be
characterized by sequences.
Theorem 2.5.1. (The Theorem of Eberlein- Smulian)
Let X be a Banach space. For subset K the following are equivalent.
a) K is relatively (X, X ⇤ ) compact, i.e. K
(X,X ⇤ )
is compact.
b) Every sequence in K contains a (X, X ⇤ )-convergent subsequence.
c) Every sequence in K has a (X, X ⇤ )-accumulation point.
We will need the following Lemma.
Lemma 2.5.2. Let X be a Banach space and assume that there is a countable set C = {x⇤n : n 2 N} ⇢ BX ⇤ , so that C? = {0}. We say that C is total
for X.
Consider for x, y
d(x, y) =
1
X
n=1
2
n
|hx⇤n , x
yi|.
Then d is a metric on X, and for any (X, X ⇤ )-compact set K,
coincides on K with the metric generated by d.
(X, X ⇤ )
Proof. It is clear that d is a metric on X. Assume that K ⇢ X is weak
compact. By the Theorem of Banach Steinhaus 2.3.6 K is therefore norm
bounded and we consider the identity I as map from the space (K, (X, X ⇤ )\
K) to (K, Td \ K) (Td being the topology generated by d). Then I is continuous: Indeed, if (xi : i 2 I) is a net which is converging in (X, X ⇤ ) to
some x 2 K and if " > 0 is arbitrary, we first use the boundedness of K to
find n 2 N so that
1
X
j=n+1
2j
1
|hx⇤j , xi
xi|  2
n+1
sup kxk < "/2,
x2K
P
and then we choose i0 2 I so that nj=1 2j 1 |hx⇤j , xi xi| < "/2 for all i 2 I,
with i i0 ,. It follows that d(xi , x) < ".
Since images of compact sets under continuous functions are compact,
and thus (by bijectivity of I) images of (X, X ⇤ )-open sets in K under I
are open in (K, T ) it follows that I is a homeomorphism.
48
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
Lemma 2.5.3. Assume that X is separable. Then there is a countable total
set C ⇢ X ⇤ .
Proof. Let D ⇢ X be dense, and choose by the Corollary 1.4.6 of the Theorem of Hahn Banach for each element x 2 D, an element yx⇤ 2 SX ⇤ so
that hyx⇤ , xi = kxk. Put C = {yx⇤ : x 2 D}. If x 2 X, x 6= 0, is arbitrary then there is a sequence (xk ) ⇢ D, so that limk!1 xk = x, and thus
limk!1 hyx⇤k , xi = kxk > 0. Thus there is a x⇤ 2 C so that hx⇤ , xi =
6 0,
which implies that C is total.
Proof of Theorem 2.5.1. “(a))(b)” Assume that K is (X, X ⇤ )-compact
(if necessary, pass to the closure) and let (xn ) ⇢ K be a sequence, and put
X0 = span(xn : n 2 N). X0 is a separable Banach space. By Proposition
2.2.5 the topology (X0 , X0⇤ ) coincides with the restriction of (X, X ⇤ ) to
X0 . It follows therefore that K0 = K \ X0 is (X0 , X0⇤ )-compact. Since X0
is separable, by Lemma 2.5.3 there exists a countable set C ⇢ BX0⇤ , so that
C? = {0}.
It follows therefore from Lemma 2.5.2 that (K0 , (X0 , X0⇤ ) \ K0 ) is
metrizable and thus (xn ) has a convergent subsequence in K0 . Again, using
the fact that on X0 the weak topology coincides with the weak topology on
X, we deduce our claim.
“(b))(c)” clear.
“(c))(a)” Assume K ⇢ X satisfies (c). We first observe that K is (norm)
bounded. Indeed, for x⇤ 2 X ⇤ , the set Ax⇤ = {hx⇤ , xi : x 2 K} ⇢ K is
the continuous image of A (under x⇤ ) and thus has the property that every
sequence has an accumulation point in K. This implies that Ax⇤ is bounded
in K for all x⇤ 2 X ⇤ , but this implies by the Banach Steinhaus Theorem
1.3.8 that A ⇢ X must be bounded.
Let
: X ,! X ⇤⇤ be the canonical embedding. By the Theorem of
(X ⇤⇤ ,X ⇤ )
Alaoglu 2.3.2, it follows that (K)
is (X ⇤⇤ , X ⇤ )-compact. There(X ⇤⇤ ,X ⇤ )
fore it will be enough to show that (K)
⇢ (x) (because this
would imply that every net ( (xi ) : i 2 I) ⇢ (K) has a subnet which
( (X), X ⇤ )) converges to some element (x) 2 (X)).
(X ⇤⇤ ,X ⇤ )
So fix x⇤⇤
. Recursively we will choose for each k 2 N,
0 2 (K)
⇤
xk 2 K, finite sets Ak ⇢ SX ⇤ , if k 2 N0 , so that
(2.4) hx⇤⇤
0
(xk ), x⇤ i <
[
1
for all x⇤ 2
A⇤j ,
k
0j<k
2.5. THE THEOREM OF EBERLEIN SMULIAN
(2.5) 8 x⇤⇤ 2 span(x⇤⇤
0 , (xj ), 1  j  k)
kx⇤⇤ k
49
max
|hx⇤⇤ , x⇤ i|
⇤
⇤
x 2Ak
kx⇤⇤ k
.
2
For k = 0 choose A⇤0 = {x⇤ }, x⇤ 2 SX ⇤ , with |x⇤ (x⇤⇤
kx⇤⇤
0 )|
0 k/2, then
condition (2.5) is satisfied, while condition (2.4) is vacuous.
Assuming that x1 , x2 , . . . xk 1 and A⇤0 , A⇤1 , . . . , A⇤k 1 have been chosen for
some k > 1, we can first choose xk 2 K so that (2.4) is satisfied (since A⇤j is
finite for j = 1, 2, . . . k 1), and then, since span(x⇤⇤
0 , (xj ), j  k) is a finite
⇤
dimensional space we can choose Ak ⇢ SX ⇤ so that (2.5) holds.
By our assumption (c) the sequence (xk ) has an (X, X ⇤ )- accumulation
point x0 . By Proposition 2.3.1 it follows that x0 2 Y = span(xk : k 2 N)
k·k
(X,X ⇤ )
=
span(xk : k 2 N)
.
We will show S
that x⇤⇤
0 = (x0 ) (which will finish the proof ). First note
that for any x⇤ 2 j2N A⇤j
hx⇤⇤
0
(x0 ), x⇤ i  lim inf
k!1
hx⇤⇤
0
(xk ), x⇤ i + hx⇤ , xk
x0 i
= 0.
k·k
⇤⇤
Secondly consider the space Z = span(x⇤⇤
0 , (xk ), k 2 N)
S1 ⇢⇤X it follows
from (2.5) that the set of restrictions of elements of k=1 Ak to Y is total
in Z and thus that
1
⇣[
⌘
⇤⇤
x0
(x0 ) 2 Z \
A⇤k
= {0},
k=1
?
which implies our claim.
Exercises
1.
Prove that if X is a separable Banach space (BX ⇤ , (X ⇤ , X)) is metrizable.
2.⇤ For an infinite dimensional Banach space prove that (X, (X, X ⇤ )) is
not metrizable.
Hint: Exercise 4 in Section 2.1
3.
Prove that for two Banach spaces X and Y , the adjoint of a linear bounded operator T : X ! Y is w⇤ continuous (i.e (Y ⇤ , Y )
(X ⇤ , X)continuous).
4.
Show that `1 isometric to a subspace of C[0, 1].
5.⇤ Show that `1 is not complemented in C[0, 1].
50
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
2.6
Characterizations of Reflexivity by Ptak
We present several characterization of the reflexivity of a Banach space,
due to Pták [Ptak]. We assume in this section that our Banach spaces are
defined over the real firel R
Theorem 2.6.1. The following conditions for a Banach space X are equivalent
1. X is not reflexive.
⇤ 1
2. For each ✓ 2 (0, 1) there are sequences (xi )1
i=1 ⇢ BX and (xi )i=1 ⇢
BX ⇤ , so that
(
✓ if j  i, and
⇤
(2.6)
xj (xi ) =
0 if j > i.
⇤ 1
3. For some ✓ > 0 there are sequences (xi )1
i=1 ⇢ BX and (xi )i=1 ⇢ BX ⇤ ,
for which (2.6) holds.
4. For each ✓ 2 (0, 1) there is a sequence (xi )1
i=1 ⇢ BX , so that
(2.7)
dist conv(x1 , x2 , . . . xn ), conv(xn+1 , xn+2 , . . .)
✓
5. For some ✓ > 0 there is a sequence (xi )1
i=1 ⇢ BX , so that (2.7) holds.
For the proof we will need Helly’s Lemma.
Lemma 2.6.2.
Let Y be an infinite-dimensional normed linear space y1⇤ , y2⇤ , . . . , yn⇤ 2 Y ⇤ ,
M > 0 and let c1 , c2 , . . . cn be scalars. The following are equivalent
(M) The Moment Condition
For all " > 0 there exists y 2 Y with
kyk = M + " and yk⇤ (y) = ck for k = 1, 2 . . . n.
(H) Helly’s Condition
n
X
j=1
aj cj  M
n
X
j=1
aj yj⇤ for any sequence (aj )nj=1 of scalars.
2.6. CHARACTERIZATIONS OF REFLEXIVITY BY PTAK
51
Proof. “(M ) ) (H)”. Let " > 0 and assume y 2 Y satisfies the condition
in (M ). Then
n
X
j=1
aj cj =
n
X
aj yj⇤ (y)
j=1
 kyk ·
n
X
aj yj⇤
= (M + ")
j=1
n
X
aj yj⇤ ,
j=1
which implies (H), since " > 0 was arbitrary.
“(H) ) (M )” Assume (H) and let " > 0. We can assume that not all
the yk⇤ are vanishing (otherwise also all the ck have to be equal to 0, and
any y 2 Y with kyk = M + " will satisfy the conditions in (M )). Secondly
we can assume, after reordering the yj⇤ , that for some k 2 {1, 2 . . . n} the
⇤ , y ⇤ , . . . y ⇤ 2 span(y ⇤ :
sequence (yj⇤ )kj=1 is linear independent and yk+1
n
j
k+2
j = 1, 2, . . . , k). This implies that if we have a y 2 Y , with kyk = M + "
and yj⇤ (y) = cj , for j = 1, 2 . . . k, then it also follows that yj⇤ (y) = cj ,
for j = k + 1, k + 2 . . . n. Indeed, for j = k + 1, k + 2, . . ., choose scalars
P
(j)
(j)
(ai : i = 1, 2 . . . k) so that yj⇤ = ki=1 ai yi⇤ , for j = k + 1, k + 2, . . . n. Now,
P
(j)
the inequality in (H) implies that cj = ki=1 ai ci , for j = k + 1, k + 2, . . . n
(j)
(choose aj = 1, aj = 0, if i 2 {k + 1, k + 2, . . . n} \ {j} and ai = ai , if
i 2 {1, 2 . . . k}). This yields that yj⇤ (y) = cj , for j = k + 1, k + 2, . . . , n.
We can therefore restrict ourselves to satisfy the second condition in (M )
for all j = 1, 2 . . . k. Define for j = 1, 2 . . . k the affine subspace Hj = {y 2
T
Y : yj⇤ (y) = cj }. Then G = kj=1 Hj is an affine subspace of codimension
k. Note that if we pick y 2 G, then G = y + G0 , where G0 is the closed
subspace
k
\
G0 =
{y 2 Y : yj⇤ (y) = 0}.
j=1
We need to show that
(2.8)
N := inf{kyk : y 2 G}  M.
Then our claim would follow, since the Intermediate Value implies that there
must be some y in G for which N < kyk = M + " < 1. Without loss of
generality we can assume that N > 0. We choose a functional g ⇤ in the dual
of the span of G so that g ⇤ (y) = N , for all y 2 G. This can be done by
⇤
picking a point y 2 G, and choosing by Hahn Banach g ⇤ 2 spanG , with
g ⇤ (y) = N and which vanishes on the linear closed subspace G0 .
Note that
(2.9)
span(G) = {ry : y 2 G and r 2 R}.
52
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
We note that kg ⇤ k
1 (otherwise there is a y 2 G and an r > 0 so
⇤
that kryk < |g (ry)| = rN , and thus kyk < N , which is a contradiction to
definition of N ), and secondly that kg ⇤ k  1, because we can use (2.9) and
obtain a y 2 G, kyk < N , so that |g ⇤ (y)| = N , which also contradicts the
definition of N ). Thus kg ⇤ k = 1. We let y ⇤ be a Hahn Banach extension of
g ⇤ to a functional defined on all of Y .
Since for all y 2 Y , we have that if yj⇤ (y) = cj , for j = 1, 2 . . . k it
follows that y ⇤ (y) = N , we have for all y 2 Y if yj⇤ (y) = 0, for j = 1, 2 . . . k,
then y ⇤ (y) = 0, in other words, the intersection of the null spaces of the yj⇤ ,
⇤
j = 1, 2 . . . k, is a subset of the null space of y ⇤ . This
Pkmeans ⇤that y is a
⇤
⇤
linear combination of the yj , j = 1, 2 . . . k, say y = j=1 aj yj . This also
P
P
implies that N = y ⇤ (y) = kj=1 aj yj⇤ (y) = kj=1 aj cj , for y 2 G.
Thus by our assumption (H)
N
N= ⇤ =
ky k
Pk
j=1 aj cj
Pk
⇤
j=1 aj yj
 M,
which proves our claim (2.8) and finishes the proof of the Lemma.
Proof of Theorem 2.6.1. “(i))(ii)” Assume that X is not reflexive and that
✓ 2 (0, 1). Since X is a closed subspace of X ⇤⇤ (as usual we are identifying X
with its image under the canonical embedding into X ⇤⇤ ), there is a functional
x⇤⇤⇤ 2 X ⇤⇤⇤ that kx⇤⇤⇤ k = 1, x⇤⇤⇤ |X ⌘ 0 and x⇤⇤⇤ (x⇤⇤ ) > ✓ for some
x⇤⇤ 2 X ⇤⇤ , with kx⇤⇤ k < 1.
Now we will choose inductively xn 2 BX and x⇤n 2 BX ⇤ , n 2 N, at each
step assuming that the condition (2.6) holds up to n, and additionally, that
x⇤⇤ (x⇤n ) = ✓.
For n = 1 we simply choose x⇤1 2 SX ⇤ so that x⇤⇤ (x⇤1 ) > ✓ and then we
choose x1 2 BX so that x⇤1 (x1 ) = ✓. Assuming we have chosen x1 , x2 , . . . xn
and x⇤1 , x⇤2 , . . . x⇤n so that
(
✓ if j  i  n, and
⇤
(2.10)
xj (xi ) =
0 if i < j  n.
Since x⇤⇤⇤ (xj ) = 0 for j = 1, 2 . . . n and x⇤⇤⇤ (x⇤⇤ ) > ✓, the elements
x1 , x2 , . . . xn , x⇤⇤ , seen as functionals on X ⇤ , together with the numbers
0, 0, . . . 0, ✓ and M = x⇤⇤⇤✓(x⇤⇤ ) < 1 satisfy Helly’s condition (H). Indeed,
for scalars a1 , . . . an+1 we have
|an+1 |✓ = M |an+1 x⇤⇤⇤ (x⇤⇤ )|
2.6. CHARACTERIZATIONS OF REFLEXIVITY BY PTAK
= M x⇤⇤⇤
n
⇣X
aj xj + an+1 x⇤⇤
j=1
⌘
M
n
X
53
aj x⇤⇤⇤ (xj ) + an+1 x⇤⇤ .
j=1
We can therefore choose an x⇤n+1 2 X ⇤ , kx⇤n+1 k  1 so that x⇤n+1 (xj ) = 0 for
all j = 1, 2 . . . n and x⇤⇤ (x⇤n+1 ) = ✓.
Secondly we note that the functionals x⇤1 , x⇤2 , . . . x⇤n+1 , the numbers ✓, ✓, . . . , ✓,
and the number M = kx⇤⇤ k < 1 satisfy Helly’s condition. Indeed, for scalars
a1 , . . . , an+1 we have
n+1
X
ai ✓ =
i=1
n+1
X
j=1
aj x⇤⇤ (x⇤j )  M
n+1
X
aj x⇤j .
j=1
We can therefore find xn+1 2 BX ⇤ , so that x⇤j (xn+1 ) = ✓, for all j =
1, 2, . . . , n.
“(ii))(iv)” and “(iii))(v)” Fix a ✓ 2 (0, 1) for which there are
P sequences
(xj ) ⇢ BX and (x⇤j ) ⇢ BX ⇤Pfor which (2.6) holds. Let x = nj=1 aj xj 2
conv(x1 , x2 , . . . xn ) and z = 1
j=n+1 bj xj 2 conv(xn+1 , xn+2 , . . .) then
kz
xk
x⇤n+1 (z
x) = x⇤n+1 (y) = ✓,
which implies our claim.
“(iv))(v)” obvious.
“(v))(i)” Assume that for ✓ > 0 and the sequence (xj ) ⇢ BX satisfies (2.7).
Now assume that our claim were wrong and X is reflexive.
Define Cn = conv(xj : j n + 1), for n 2 N, then the CnT, n 2 N, are
weakly compact, C1 C2 . . .. Thus there is an element v 2 n2N Cn . We
can approximate v by some u 2 conv(xj : j 2 N), with ku vk < ✓/2. There
is some n so that v 2 conv(x1 , . . . xn ). But now it follows, since u 2 Cn+1 ,
that dist(conv(x1 , . . . xn ), conv(xn+1 , xn2 , . . .))  kv uk < ✓/2, which is a
contradiction and finishes the proof.