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Chapter 2
Weak Topologies, Reflexivity,
Adjoint operators
2.1
Topological vector spaces and locally convex spaces
Definition 2.1.1. [Topological Vector Spaces and Locally convex Spaces]
Let E be a vector space over K, with K = R or K = C and let T be a
topology on E. We call (E, T ) (or simply E, if their cannot be a confusion),
a topological vector space, if the addition:
+ : E × E → E,
(x, y) #→ x + y,
and the multiplication by scalars
· : K × E → E,
(λ, x) #→ λx,
are continuous functions. A topological vector space is called locally convex
if 0 (and thus any point x ∈ E) has a neighbourhood basis consisting of
convex sets.
Remark. Topological vector spaces are in general not metrizable. Thus,
continuity, closeness, and compactness etc, cannot be described by sequences.
We will need nets.
Assume that (I, ≤) is a directed set . This means
• (reflexivity) i ≤ i, for all i ∈ I,
27
28CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS
• (transitivity) if for i, jk ∈ I we have i ≤ j and j ≤ k, then i ≤ k, and
• (existence of upper bound) for any i, j ∈ I there is a k ∈ I, so that
i ≤ k and j ≤ k).
A net is a family (xi : i ∈ I) indexed over a directed set (I, ≤).
A subnet of a net (xi : i ∈ I) is a net (yj : j ∈ J), together with a map
j →
# ij from J to I, so that xij = yj for all j ∈ J for all i0 ∈ I there is a
j0 ∈ J, so that ij ≥ i0 for all j ≥ j0 .
Note: A subnet of a sequence is not necessarily a subsequence.
In a topological space (T, T ), we say that a net (xi : i ∈ I) converges to
x, if for all open sets U with x ∈ U there is a i0 ∈ I, so that xi ∈ U for all
i ≥ i0 . If (T, T ) is Hausdorff x is unique and we denote it by limi∈I xi .
Using nets we can describe continuity, closeness, and compactness in arbitrary topological spaces:
a) A map between two topological spaces is continuous if and only if the
image of converging nets are converging.
b) A subset A of a topological space S is close if and only if the limit point
of every converging net in A is in A.
c) A topological space S is compact if and only if every net has a converging subnet.
In order to define a topology on a vector space E which turns E into a
topological vector space we need (only) to define an appropriate neighborhood basis of 0.
Proposition 2.1.2. Assume that (E, T ) is a topological vector space. And
let
U0 = {U ∈ T , 0 ∈ U }.
Then
a) For all x ∈ X, x + U0 = {x + U : U ∈ U} is a neighborhood basis of x,
b) for all U ∈ U0 there is a V ∈ U0 so that V + V ⊂ U ,
c) for all U ∈ U0 there is an ε > 0 and V ∈ U0 so that
{λ ∈ K : |λ| < ε} · V ⊂ U,
2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 29
d) for all U ∈ U0 and x ∈ E there is an ε > 0, so that λx ∈ U , for all
λ ∈ K with |λ| < ε ,
e) if (E, T ) is Hausdorff, then for every x ∈ E, x (= 0, there is a U ∈ U0
with x (∈ U ,
f ) if E is locally convex, then there is for all U ∈ U0 a convex V ∈ U0 ,
with V ⊂ U , i.e. 0 has a neighborhood basis consisting of convex sets.
Conversely if E is a vector space over K, K = R or K = C and
U0 ⊂ {U ∈ P(E) : 0 ∈ U }
is non empty and is downwards directed, i.e. if for any U, V ∈ U0 , there is a
W ∈ U0 , with W ⊂ U ∩ V and satisfies (b) (c) and (d), then
T = {V ⊂ E : ∀x ∈ V ∃ ∈ U : x + U ⊂ V },
defines a topological vector space for which U0 is the neighborhood basis of 0.
(E, T ) is Hausdorff if U also satisfies (e) and locally convex if it satisfies (f ).
Proof. Assume (E, T ) is a topological vector space and U is defined as above.
We observe that for all x ∈ E the linear operator Tx : E → E, z #→ z + x
is continuous. Since also Tx ◦ T−x = T−x ◦ Tx = Id, it follows that Tx
is an homeomorphism, which implies (a). Property (b) follows from the
continuity at 0 of addition and (c) and (d) follow from the continuity of
scalar multiplication at 0. If E is Hausdorff then U0 clearly satisfies (e) and,
by definition, U0 satisfies (f) if we assume that E is locally convex.
Now assume that U0 ⊂ {U ∈ P(E) : 0 ∈ U } is non empty and downwards
directed, that for any U, V ∈ U0 , there is a W ∈ U0 , with W ⊂ U ∩ V and
satisfies (b), (c) and (d). Then
T = {V ⊂ E : ∀x ∈ V ∃ ∈ U : x + U ⊂ V },
is finitely intersection stable and stable by taking (arbitrary) unions. Also
∅, X ∈ T . Thus T is a topology. Also note that for x ∈ E,
Ux = {x + U : U ∈ U0 }
is a neighborhood basis of x.
30CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS
We need to show that addition and multiplication by scalars is continuous.
Assume (xi : i ∈ I) and (yi : i ∈ I) converge in E to x ∈ E and y ∈ E, and
if U ∈ U0 , respectively. By (b) there is a V ∈ U0 with V + V ⊂ U . We can
therefore choose i0 so that xi ∈ x + V and yi ∈ x + V , for i ≥ i0 , and, thus,
xi + yi ∈ x + y + V + V ⊂ x + y + U , for i ≥ i0 . This proves the continuity
of the addition in E.
Assume (xi : i ∈ I) converges in E to x , (λi : i ∈ I) converges in K to λ
and let U ∈ U0 . Then
choose V ∈ U0 so that V + V ⊂ U ,
choose ε1 > 0 so that µx ∈ V , whenever |µ| ≤ ε1 ,
choose ε2 > 0 and W ∈ U0 , so that {µ ∈ K : |µ| < ε2 } · W ⊂ V , and
finally
choose i0 ∈ I, so that |λ − λi | < min(ε1 , ε2 ) and xi ∈ x + W , whenever
i ≥ i0 .
Then it follows for i ≥ i0
λi xi = (λi −λ)(xi −x)+(λi−λ)x+λx ∈ λx+(λi −λ)W+V ⊂ λx+V +V ⊂ λx+U.
If U0 satisfies (e) and if x (= y are in E, then we can choose U ∈ U0 so
that y − x (∈ U and then, using the already proven fact that addition and
multiplication by scalars is continuous, there is V so that V − V ⊂ U . It
follows that x + V and y + V are disjoint. Indeed, if x + v1 = y + v2 , for some
v1 , v2 ∈ V it would follows that y − x = v2 − v1 ∈ U , which is a contradiction.
If (f) is satisfied then E is locally convex since we observed before that
Ux = {x + U : U ∈ U0 } is a neighborhood basis of x, for each x ∈ E.
Let E be a vector space over K, K = R or K = C, and let F be a subspace
of
E # = {f : E → K linear}.
Assume that for each x ∈ E there is an x∗ ∈ F so that x∗ (x) (= 0, we say in
that case that F is separating the elements of E from 0. Consider
U0 =
!
"
∩nj=1 {x ∈ E : |x∗i (x)| < εi } : n ∈ N, x∗i ,∈ F, and εi > 0 for i = 1, 2ṅ .
U0 is finitely intersection stable and it is easily checked that U0 satisfies
that for assumptions (b)-(f). It follows therefore that U0 is the neighborhood
basis of a topology which turns E into local convex Hausdorff space.
2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 31
Definition 2.1.3. [The Topology σ(E, F )]
Let E be a vector space and let F be a separating subspace of E # .
Then we denote the local convex Hausdorff topology generated by
!
"
U0 = ∩nj=1 {x ∈ E : |x∗i (x)| < εi } : n ∈ N, x∗i ∈ F, and εi > 0 for i = 1, 2, . . . n ,
by σ(E, F ).
If E is a Banach space X and F = X ∗ we call σ(X, X ∗ ) the Weak Topology
on X and denote it also by w. If E is the dualspace X ∗ of a Banach space X
and we consider X (via the canonical map χ : X %→ X ∗∗ ) a subspace of X ∗∗
we call σ(X ∗ , X) the Weak∗ Topology on X ∗ , which is also denoted by w∗ .
Proposition 2.1.4. Let E be a vector space and let F be a separating subspace of E # .
For a net (xi )i∈I ⊂ E and x ∈ E
lim xi = x in σ(E, F ) ⇐⇒ ∀x∗ ∈ F
i∈I
lim1x∗ , xi 2 = 1x∗ , x2.
i∈I
An easy consequence of the geometrical version of the Hahn-Banach Theorem 1.4.12 is the following two observation.
Proposition 2.1.5. If A is a convex subset of a Banach space X then
w
$·$
A =A .
If a representation of the dual space of a Banach space X is not known,
it might be hard to verify weak convergence of sequence directly. The following Corollary of Proposition 2.1.5 states an equivalent criterium condition a
sequence to be weakly null without using the dual space of X.
Corollary 2.1.6. For a bounded sequence (xn ) in Banach space X it follows
that (xn ) is weakly null if and only if for
# all subsequences (zn ), all ε > 0
and there is a convex combination z = kj=1 λj zj of (zj ) (i.e. λi ≥ 0, for
#
i = 1, 2, . . . k, and lj=1 λj = 1) so that 3z3 ≤ ε .
Proposition 2.1.7. If X is a Banach space and Y is a closed subspace of
X the σ(Y, Y ∗ ) = σ(X, X ∗ ) ∩ Y , i.e. the weak topology on Y is the the weak
topology on X restricted to Y .
Theorem 2.1.8. [Theorem of Alaoglu]
∗
BX
is w∗ compact for any Banach space X.
32CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS
Sketch of a proof. Consider the map
$
∗
Φ : BX
→
{λ ∈ K : |λ| ≤ 3x3},
x∈X
x∗ #→ (x∗ (x) : x ∈ X).
Then we check that Φ is continuous
with respect to w∗ topology on BX ∗ and
%
the product topology on x∈X {λ ∈ K : |λ| ≤ 3x3}, has a closed image, and
∗
is a homeorphism from BX
onto its image.
%
Since by the Theorem of Tychanoff x∈X {λ ∈ K : |λ| ≤3 x3} is compact,
Φ(BX ∗ ) is a compact subset, which yields (via the homeomorphism Φ−1 ) that
BX ∗ is compact in the w∗ topology.
Theorem 2.1.9. [Theorem of Goldstein]
∗∗
BX is (via the canonical embedding) w∗ dense in B X .
The proof follows immediately from the following Lemma.
Lemma 2.1.10. Let X be a Banach space and let x∗∗ ∈ X ∗∗ and x∗1 ,
x∗2 , . . . , x∗n ∈ X ∗ Then
inf
$x$≤1
n
&
i=1
|1x∗∗ , x∗i 2 − 1x∗i , x2|2 = 0.
#
Proof. For x ∈ X put φ(x) = ni=1 |1x∗∗ , x∗i 2−1x∗i , x2|2 and β = inf x∈BX φ(x),
and choose a sequence (xj ) ⊂ BX so that φ(xj ) 4 β, if k 5 ∞. W.l.o.g we
can also assume that ξi = limk→∞ 1x∗i , xk 2 exists for all i = 1, 2, . . . , n.
For any t ∈ [0, 1] and any x ∈ BX we note for k ∈ N
φ((1 − t)xk + tx)
n
&
=
|1x∗∗ , x∗i 2 − (1 − t)1x∗i , xk 2 − t1x∗i , x2|2
=
i=1
n
&
i=1
=
n
&
i=1
|1x∗∗ , x∗i 2 − 1x∗i , xk 2 + t1x∗i , xk − x2|2
|1x∗∗ , x∗i 2 − 1x∗i , xk 2|2
+2t7
n
'&
i=1
1x∗∗ , x∗i 2−
(
1x∗i , xk 21x∗i , xk −x2
2
+t
n
&
i=1
|1x∗i , xk 2−1x∗i , x2|2
2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 33
→k→∞ β + 2t7
n
'&
i=1
n
(
&
(1x∗∗ , x∗i 2−ξi )(ξi − 1x∗i , x2) + t2
|ξi −1x∗i , x2|2 .
)
*+
,
i=1
=:λi
From the minimality of β it follows that for all x ∈ BX
7
n
'&
i=1
n
(
&
λi ξi ≥ 7(1x∗ , x2) with x∗ :=
λi x∗i ,
i=1
and thus
n
'&
(
3x 3 ≤ 7
λi ξi .
∗
i=1
Indeed, write 1x∗ , x2 = reia , then
∗
−ia
|1x , x2| = e
∗
∗
−ia
1x , x2 = 1x , e
On the other hand limk→∞ 1x∗ , xk 2 =
#n
i=1
i=1
λi ξi and thus
n
n
-&
'&
(
3x 3 ≥ λi ξi - ≥ 7
λi ξi ,
∗
i=1
which implies that
i=1
3x∗ 3 = 7
So
n
'&
i=1
(
λi ξi .
β = lim φ(xk )
=
=
k→∞
n
&
i=1
n
&
|λi |2
λi λi
i=1
=
n
'&
(
x2 ≤ 7
λi ξ .
n
&
i=1
=7
λi (1x∗∗ , x∗i 2− ξi )
n
'&
i=1
λi (1x
∗∗
(
, x∗i 2−ξi )
34CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS
= 71x∗∗ , x∗ 2 − 7
∗∗
∗
n
'&
i=1
∗
λi ξi
(
≤ 3x 3 ·3 x 3 − 3x 3 ≤ 3x∗ 3 − 3x∗ 3 = 0.
Thus β = 0 which proves our claim.
Theorem 2.1.11. Let X be a Banach space. Then X is reflexive if and only
if BX is compact in the weak topology.
Proof. Let χ : X %→ X ∗∗ be the canonical embedding.
“⇒” If X is reflexive and thus χ is onto it follows that χ is an homeomorphism
between (BX , σ(X, X ∗ )) and (BX ∗∗ , σ(X ∗∗ , X ∗ )). But by the Theorem of
Alaoglu Bourbaki 2.1.8 (BX ∗∗ , σ(X ∗∗ , X ∗ )) is compact.
“⇐” Assume that (BX , σ(X, X ∗ ) is compact, and assume that x∗∗ ∈ BX ∗∗
we need to show that there is an x ∈ BX so that χ(x) = x∗∗ , or equivalently
that 1x∗ , x2 = 1x∗∗ , x∗ 2 for all x∗ ∈ X ∗ .
For any finite set A = {x∗1 , . . . x∗n } ⊂ X ∗ and for any ε > 0 we can,
according to Lemma 2.1.10, choose an x(A,ε) ∈ BX so that
n
&
i=1
|1x∗∗ − x(A,ε) , x∗i 2|2 ≤ ε.
The set
I = {(A, ε) : A ⊂ X ∗ finite and ε > 0},
(
(
is directed via (A,.ε) ≤ (A( , ε( ) : ⇐⇒
/ A ⊂ A and ε ≤ ε. Thus, by compactness, the net x(A,ε) : (A, ε) ∈ I must have a subnet (zj : j ∈ J) which
converges weakly to some element x ∈ BX .
We claim that 1x∗ , x2 = 1x∗∗ , x∗ 2, for all x∗ ∈ X ∗ . Indeed, let j #→ ij
be the map from J to I, so that zj = xij , for all j ∈ J, and so that, for
any j0 there is a i0 with ji ≥ j0 , for i ≥ i0 . Let x∗ ∈ X ∗ and ε > 0. Put
i0 = ({x∗ }, ε) ∈ I, choose j0 , so that ij ≥ i0 , for all j ≥ j0 , and choose j1 ∈ J,
j1 ≥ j0 , so that |1x − zj∗ , x∗ 2| < ε, for all j ≥ j1 . It follows therefore that
(note that for ij1 = (A, ε( ) it follows that x∗ ∈ A and ε( ≤ ε)
- ∗∗
- - -1x − x, x∗ 2- ≤ -1x∗∗ − xi , x∗ 2- + -1zj1 − x, x∗ 2- ≤ ε + ε = 2ε.
j1
Since ε > 0 and x∗ ∈ X ∗ where arbitrary we deduce our claim.
Theorem 2.1.12. For a Banach space X the following are equivalent.
2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 35
a) X is reflexive,
b) X ∗ is reflexive,
c) every closed subspace of X is reflexive
Proof. “(a)⇒(c)” Assume Y ⊂ X is a closed subspace. Proposition 2.1.5
yields that BY = BX ∩ Y is a σ(X, X ∗ )-closed and, thus, σ(X, X ∗ )-compact
subset of BX . Since, by the Theorem of Hahn-Banach (Corollary 1.4.4), every
y ∗ ∈ Y ∗ can be extended to an element in X ∗ , it follows that σ(Y, Y ∗ ) is the
restriction of σ(X, X ∗ ) to the subspace Y . Thus, BY is σ(Y, Y ∗ )-compact,
which implies, by Theorem 2.1.11 that Y is reflexive.
“(a)⇒(b)” If X is reflexive then σ(X ∗ , X ∗∗ ) = σ(X ∗ , X). Since by the Theorem of Alaoglu Bourbaki 2.1.8 BX ∗ is σ(X ∗ , X)-compact the claim follows
from Theorem 2.1.11.
“(c)⇒(a)” clear.
“(b)⇒(a)” If X ∗ is reflexive, then, “(a)⇒(b) ” X ∗∗ is also reflexive and thus,
the implication “(a)⇒(c)” yields that X is reflexive.
An important consequence of the Uniform Boundedness Principle is the
following
Theorem 2.1.13. [Theorem of Banach-Steinhaus]
a) If A ⊂ X, and supx∈A |1x∗ , x2| < ∞, for all x∗ ∈ X ∗ , then A is (norm)
bounded.
b) If A ⊂ X ∗ , and supx∗ ∈A |1x∗ , x2| < ∞, for all x ∈ X, then A is (norm)
bounded.
In particular weak compact subsets of X and weak∗ compact subsets of X ∗
are norm bounded.
Exercises
1. Show Theorem 2.1.9 using Lemma 2.1.10.
2. Prove Proposition 2.1.7 and Corollary 2.1.6.
36CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS
3. Show that B#∗∞ is not sequentially compact in the w∗ -topology.
Hint: Consider the unit vector basis of )1 seen as subsequence of B#∗1 .
4. Prove that for a Banach space X every w∗ converging sequence in X ∗
is bounded, but that if X is infinite dimensional, X ∗ contains nets
(x∗i : i ∈) which converge to 0, but so that for every c > 0 and all i ∈ I
there is a j0 ≥ i, with 3xj 3 ≥ c, whenever j ≥ j0 .
5. Show that in each infinite dimensional Banach space X there is a weakly
null net in SX .
6.∗ Prove that every weakly null sequence in )1 is norm null.
Hint: Assume that (xn ) ⊂ S#1 is weakly null. Then there is a subsequence xnk and a block sequence (zk ) so that limk→∞ 3xnk − zk 3 = 0.