Next consider the weak topology. The most interesting results have to do with a reflexive
Banach space. The following lemma ties together the weak and weak ∗ topologies in the
case of a reflexive Banach space.
Lemma 15.5.7Let J : X → X^{′′}be the James map
Jx (f) ≡ f (x)
and let X be reflexive so that J is onto. Then J is a homeomorphism of
(X, weak topology)
and (X^{′′}, weak ∗ topology).This means J is one to one, onto, and both J and J^{−1}arecontinuous.
Proof:Let f ∈ X^{′} and let
Bf (x,r) ≡ {y : |f (x) − f (y)| < r}.
Thus B_{f}
(x,r)
is a subbasic set for the weak topology on X. I claim that
JB (x,r) = B (Jx,r)
f f
where B_{f}
(Jx, r)
is a subbasic set for the weak ∗ topology. If y ∈ B_{f}
(x,r)
, then
∥Jy− Jx∥
=
∥x − y∥
< r and so JB_{f}
(x,r)
⊆ B_{f}
(Jx,r)
. Now if x^{∗∗}∈ B_{f}
(Jx,r)
, then
since J is reflexive, there exists y ∈ X such that Jy = x^{∗∗} and so
∥y− x∥ = ∥Jy − Jx∥ < r
showing that JB_{f}
(x,r)
= B_{f}
(Jx,r)
. A typical subbasic set in the weak ∗ topology is of
the form B_{f}
(Jx,r)
. Thus J maps the subbasic sets of the weak topology to the
subbasic sets of the weak ∗ topology. Therefore, J is a homeomorphism as claimed.
■
The following is an easy corollary.
Corollary 15.5.8If X is a reflexive Banach space, then the closed unit ball isweakly compact.
Proof:Let B be the closed unit ball. Then B = J^{−1}
(B∗∗)
where B^{∗∗ }is the unit ball
in X^{′′} which is compact in the weak ∗ topology. Therefore B is weakly compact because
J^{−1} is continuous. ■
Corollary 15.5.9Let X be a reflexive Banach space. If K ⊆ X is compact in theweak topology and X^{′}is separable in the weak ∗ topology, then there exists a metricd, on K such that if τ_{d }is the topology on K induced by d and if τ is the topologyon K induced by the weak topology of X, then τ = τ_{d}. Thus one can consider Kwith the weak topology as a metric space.
Proof: This follows from Theorem 15.5.5 and Lemma 15.5.7. Lemma 15.5.7 implies
J
(K)
is compact in X^{′′}. Then since X^{′} is separable in the weak ∗ topology, X is
separable in the weak topology and so there is a metric, d^{′′} on J
Lemma 15.5.10Let Y be a closed subspace of a Banach space X and let y ∈ X∖Y.Then there exists x^{∗}∈ X^{′}such that x^{∗}
(Y)
= 0 but x^{∗}
(y)
≠0.
Proof: Define f
(x+ αy)
≡
∥y∥
α. Thus f is linear on Y ⊕ Fy. I claim that f is also
continuous on this subspace of X. If not, then there exists x_{n} + α_{n}y → 0 but
|f (xn + αny)|
≥ ε > 0 for all n. First suppose
|αn|
is bounded. Then, taking a further
subsequence, we can assume α_{n}→ α. It follows then that
{xn}
must also converge
to some x ∈ Y since Y is closed. Therefore, in this case, x + αy = 0 and so
α = 0 since otherwise, y ∈ Y . In the other case when α_{n} is unbounded, you have
(xn∕αn + y)
→ 0 and so it would require that y ∈Y which cannot happen
because Y is closed. Hence f is continuous as claimed. It follows that for some
k,
|f (x + αy)| ≤ k∥x+ αy ∥
Now apply the Hahn Banach theorem to extend f to x^{∗}∈ X^{′}. ■
Next is the Eberlein Smulian theorem which states that a Banach space is reflexive if
and only if the closed unit ball is weakly sequentially compact. Actually, only half the
theorem is proved here, the more useful only if part. The book by Yoshida [?] has
the complete theorem discussed. First here is an interesting lemma for its own
sake.
Lemma 15.5.11A closed subspace of a reflexive Banach space is reflexive.
Proof: Let Y be the closed subspace of the reflexive space, X. Consider the following
diagram
Y ′′ i∗∗→ 1-1 X ′′
′ i∗ onto ′
Y ←i X
Y → X
This diagram follows from Theorem 15.2.10 on Page 1282, the theorem on adjoints. Now
let y^{∗∗}∈ Y^{′′}. Then i^{∗∗}y^{∗∗} = J_{X}
(y)
because X is reflexive. I want to show that y ∈ Y . If
it is not in Y then since Y is closed, there exists x^{∗}∈ X^{′} such that x^{∗}
a contradiction. Hence y ∈ Y . Letting J_{Y } denote the James map from Y to Y^{′′} and
x^{∗}∈ X^{′},
y∗∗(i∗x∗) = i∗∗y∗∗(x∗) = JX (y)(x∗)
∗ ∗ ∗ ∗ ∗∗
= x (y) = x (iy) = ix (y) = JY (y)(ix )
Since i^{∗} is onto, this shows y^{∗∗} = J_{Y }
(y)
. ■
Theorem 15.5.12
(Eberlein Smulian )
Theclosed unit ball in a reflexive Banach spaceX, is weakly sequentially compact. By this is meant that if
{xn}
is contained in theclosed unit ball, there exists a subsequence,
{xnk}
and x ∈ X such that for allx^{∗}∈ X^{′},
∗ ∗
x (xnk) → x (x).
Proof: Let {x_{n}}⊆ B ≡B
(0,1)
. Let Y be the closure of the linear span of {x_{n}}.
Thus Y is a separable. It is reflexive because it is a closed subspace of a reflexive space so
the above lemma applies. By the Banach Alaoglu theorem, the closed unit ball B^{∗} in Y^{′}
is weak ∗ compact. Also by Theorem 15.5.5, B^{∗} is a metric space with a suitable
metric.
B ∗∗ Y′′ i∗∗→ 1- 1 X ′′
∗ ′ i∗ onto ′
weakly separable B Y ←i X
separable B Y → X
Thus B^{∗ }is complete and totally bounded with respect to this metric and it follows
that B^{∗ } with the weak ∗ topology is separable. This implies Y^{′} is also separable in the
weak ∗ topology. To see this, let
{y∗n}
≡ D be a weak ∗ dense set in B^{∗} and let y^{∗}∈ Y^{′}.
Let p be a large enough positive rational number that y^{∗}∕p ∈ B^{∗}. Then if A is any
finite set from Y, there exists y_{n}^{∗}∈ D such that ρ_{A}
(y∗∕p − y∗n)
<
ε
p
. It follows
py_{n}^{∗}∈ B_{A}
(y∗,ε)
showing that rational multiples of D are weak ∗ dense in Y^{′}.
Since Y is reflexive, the weak and weak ∗ topologies on Y^{′} coincide and so Y^{′} is
weakly separable. Since Y^{′} is weakly separable, Corollary 15.5.6 implies B^{∗∗}, the
closed unit ball in Y^{′′} is weak ∗ sequentially compact. Then by Lemma 15.5.7B, the unit ball in Y , is weakly sequentially compact. It follows there exists a
subsequence x_{nk}, of the sequence
{x }
n
and a point x ∈ Y , such that for all
f ∈ Y^{′},
f (xnk) → f (x) .
Now if x^{∗}∈ X^{′}, and i is the inclusion map of Y into X,
x∗(x ) = i∗x ∗(x ) → i∗x∗(x) = x∗(x).
nk nk
which shows x_{nk} converges weakly and this shows the unit ball in X is weakly
sequentially compact. ■
Corollary 15.5.13Let
{xn}
be any bounded sequence in a reflexive Banachspace X. Then there exists x ∈ X and a subsequence,
{xnk}
such that for allx^{∗}∈ X^{′},
lim x∗(xnk) = x∗ (x )
k→∞
Proof: If a subsequence, x_{nk} has
||xn ||
k
→ 0, then the conclusion follows.
Simply let x = 0. Suppose then that
||xn||
is bounded away from 0. That is,
||xn||
∈
[δ,C]
. Take a subsequence such that
||xn ||
k
→ a. Then consider x_{nk}∕
||xn ||
k
. By
the Eberlein Smulian theorem, this subsequence has a further subsequence,
x_{nk
j}∕
|||| ||||
||xnkj||
which converges weakly to x ∈ B where B is the closed unit ball. It
follows from routine considerations that x_{nk
j}→ ax weakly. This proves the
corollary.