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CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
2.4
Annihilators, Complemented Subspaces
Definition 2.4.1. (Annihilators, Pre-Annihilators)
Assume X is a Banach space. Let M ⇢ X and N ⇢ X ⇤ . We call
M ? = {x⇤ 2 X ⇤ : 8x 2 M hx⇤ , xi = 0} ⇢ X ⇤ ,
the annihilator of M and
N? = {x 2 X : 8x⇤ 2 N hx⇤ , xi = 0} ⇢ X,
the pre-annihilator of N .
Proposition 2.4.2. Let X be a Banach space, and assume M ⇢ X and
N ⇢ X ⇤.
a) M ? is a closed subspace of X ⇤ , M ? = (span(M ))? , and (M ? )? =
span(M ),
b) N? is a closed subspace of X, N? = (span(N ))? , and span(N ) ⇢
(N? )? .
c) span(M ) = X () M ? = {0}.
Proposition 2.4.3. If X is Banach space and Y ⇢ X is a closed subspace
then (X/Y )⇤ is isometrically isomorphic to Y ? via the operator
: (X/Y )⇤ ! Y ? , with
(z ⇤ )(x) = z ⇤ (x).
(recall x := x + Y 2 X/Y for x 2 X).
Proof. Let Q : X ! X/Y be the quotient map.
For z ⇤ 2 (X/Y )⇤ , (z ⇤ ), as defined above, can be written as (z ⇤ ) =
z ⇤ Q. Thus (z ⇤ ) 2 X ⇤ . Since Q(Y ) = {0} it follows that (z ⇤ ) 2 Y ? .
For z ⇤ 2 (X/Y )⇤ we have
k (z ⇤ )k = sup hz ⇤ , Q(x)i =
x2BX
sup hz ⇤ , xi = kz ⇤ k(X/Y )⇤ ,
x2BX/Y
where the second equality follows on the one hand from the fact that kQ(x)k 
kxk, for x 2 X, and on the other hand, from the fact that for any x = x+Y 2
X/Y there is a sequence (yn ) ⇢ Y so that lim supn!1 kx + yn k = kx̃k.
Thus is an isometric embedding. If x⇤ 2 Y ? ⇢ X ⇤ , we define
z ⇤ : X/Y ! K,
x + Y 7! hx⇤ , xi.
2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES
41
First note that this map is well defined (since hx⇤ , x + y1 i = hx⇤ , x + y2 i for
y1 , y2 2 Y ). Since x⇤ is linear, z ⇤ is also linear, and |hz ⇤ , xi| = |hx⇤ , xi|, for all
x 2 X, and thus kz ⇤ k(X/Y )⇤ = kx⇤ k. Finally, since h (z ⇤ ), xi = hz ⇤ , Q(x)i =
hx⇤ , xi, it follows that (z ⇤ ) = x⇤ , and thus that is surjective.
Proposition 2.4.4. Assume X and Y are Banach spaces and T 2 L(X, Y ).
Then
(2.2)
(2.3)
T (X)? = N (T ⇤ ) and T ⇤ (Y ⇤ ) ⇢ N (T )?
T (X) = N (T ⇤ )? and T ⇤ (Y ⇤ )? = N (T ).
Proof. We only prove (2.2). The verification of (2.3) is similar. For y ⇤ 2 Y ⇤
y ⇤ 2 T (X)? () 8x 2 X
() 8x 2 X
hy ⇤ , T (x)i = 0
hT ⇤ (y ⇤ ), xi = 0
() T ⇤ (y ⇤ ) = 0 () y ⇤ 2 N (T ⇤ ),
which proves the first part of (2.2), and for y ⇤ 2 Y ⇤ and all x 2 N (T ),
it follows that hT ⇤ (y ⇤ ), xi = hy ⇤ , T (x)i = 0, which implies that T ⇤ (Y ⇤ ) ⇢
N (T )? , and, thus, T ⇤ (X ⇤ ) ⇢ N (T ).
Definition 2.4.5. Let X be a Banach space and let U and V be two closed
subspaces of X. We say that X is the complemented sum of U and V and
we write X = U V , if for every x 2 X there are u 2 U and v 2 V , so that
x = u + v and so that this representation of x as sum of an element of U
and an element of V is unique.
We say that a closed subspace Y of X is complemented in X if there is
a closed subspace Z of X so that X = Y Z.
Remark. Assume that the Banach space X is the complemented sum of the
two closed subspaces U and V . We note that this implies that U \ V = {0}.
We can define two maps
P : X ! U and Q : X ! V
where we define P (x) 2 U and Q(x) 2 V by the equation x = P (x) + Q(y),
with P (x) 2 U and Q(x) 2 V (which, by assumption, has a unique solution).
Note that P and Q are linear. Indeed if P (x1 ) = u1 , P (x2 ) = u2 , Q(x1 ) = v1 ,
Q(x2 ) = v2 , then for , µ 2 K we have x1 +µx2 = u1 +µu2 + v1 +µv2 , and
thus, by uniqueness P ( x1 +µx2 ) = u1 +µu2 , and Q( x1 +µx2 ) = v1 +µv2 .
Secondly it follows that P P = P , and Q Q = Q. Indeed, for any
x 2 X we we write P (x) = P (x) + 0 2 U + V , and since this representation
42
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
of P (x) is unique it follows that P (P (x)) = P (x). The argument for Q is
the same.
Finally it follows that, again using the uniqueness argument, that P is
the identity on U and Q is the identity on V .
We therefore proved that
a) P is linear,
b) the image of P is U
c) P is idempotent, i.e. P 2 = P
We say in that case that P is a linear projection onto U . Similarly Q is a
a linear projection onto V , and P and Q are complementary to each other,
meaning that P (X)\Q(X) = {0} and P +Q = Id. A linear map P : X ! X
with the properties (a) and (c) is called projection.
The next Proposition will show that P and Q as defined in above remark
are actually bounded.
Lemma 2.4.6. Assume that X is the complemented sum of two closed subspaces U and V . Then the projections P and Q as defined in above remark
are bounded.
Proof. Consider the norm ||| · ||| on X defined by
|||x||| = kP (x)k + kQ(x)k, for x 2 X.
We claim that (X, ||| · |||) is also a Banach space. Indeed if (xn ) ⇢ X with
1
X
n=1
|||xn ||| =
1
X
n=1
kP (xn )k +
1
X
n=1
kQ(xn )k < 1.
P
P1
Then u = 1
n=1 P (xn ) 2 U , v =
n=1 Q(xn ) 2 V (U and V are assumed
to be closed) converge in U
Pand V (with respect to k · k, respectively, and
since k · k  ||| · ||| also x = 1
n=1 xn converges and
x=
1
X
xn = lim
n!1
n=1
n
X
P (xj )+Q(xj ) = lim
n!1
j=1
n
X
P (xj )+ lim
n!1
j=1
and
x
n
X
j=1
xn = u
n
X
j=1
P (xn ) + v
n
X
j=1
Q(xn )
n
X
j=1
Q(xj ) = u+v,
2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES
= u
n
X
P (xn ) + v
j=1
n
X
j=1
43
Q(xn ) !n!1 0,
(here all series are meant to converge with respect to k · k) which proves that
(X, ||| · |||) is complete.
Since the identity is a bijective linear bounded operator from (X, ||| · |||) to
(X, k·k) it has by Corollary 1.3.6 of the Closed Graph Theorem a continuous
inverse and is thus an isomorphy. Since kP (x)k  |||x||| and kQ(x)k  |||x|||
we deduce our claim.
Proposition 2.4.7. Assume that X is a Banach space and that P : X ! X,
is a bounded projection onto a closed subspace of X.
Then X = P (X) N (P ).
Theorem 2.4.8. There is no linear bounded operator T : `1 ! `1 so that
the kernel of T equals to c0 .
Corollary 2.4.9. c0 is not complemented in `1 .
Proof of Theorem 2.4.8. For n 2 N we let e⇤n be the n-th coordinate functional on `1 , i.e.
e⇤n : `1 ! K,
x = (xj ) 7! xn .
Step 1. If T : `1 ! `1 is bounded and linear, then
N (T ) =
1
\
n=1
N (e⇤n T ).
Indeed, note that
x 2 N (T ) () 8n 2 N
e⇤n (T (x)) = he⇤n , T (x)i = 0.
In order to prove our claim we will show that c0 cannot be the intersection
of the kernel of countably many functionals in `⇤1 .
Step 2. There is an uncountable family (N↵ : ↵ 2 I) of infinite subsets of N
for which N↵ \ N is finite whenever ↵ 6= are in I.
Write the rational numbers Q as a sequence (qj : j 2 N), and choose for
each r 2 R a sequence (nk (r) : k 2 N), so that (qnk (r) : k 2 N) converges to r.
Then, for r 2 R let Nr = {nk (r) : k 2 N}.
For i 2 I, put x↵ = 1N↵ 2 `1 , i.e.
(
1 if k 2 N↵
(↵)
(↵)
x↵ = (⇠k : k 2 N) with ⇠k =
0 if k 62 N↵ .
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CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
Step 3. If f 2 `⇤1 and c0 ⇢ N (f ) then {↵ 2 I : f (x↵ ) 6= 0} is countable.
In order to verify Step 3 let An = {↵ : |f (x↵ )|
1/n}, for n 2 N.
It is enough to show that for n 2 N the set An is P
finite. To do so, let
↵1 , ↵2 , . . . , ↵k be distinct elements of An and put x = kj=1 sign f (x↵j ) x↵j
(for a 2 C we put sign(a)
= a/|a|) and deduce that f (x)
k/n. Now
S
consider Mj = N↵j \ i6=j N↵i . Then N↵j \ Mj is finite, and thus it follows
for
k
X
x̃ =
sign(f (x↵j ))1Mj
j=1
that f (x) = f (x̃) (since x x̃ 2 c0 ). Since the Mj , j = 1, 2 . . . k are pairwise
disjoint, it follows that kx̃k1 = 1, and thus
k
 f (x) = f (x̃)  kf k.
n
Which implies that that An can have at most nkf k elements.
T1
⇤
Step 4. If c0 ⇢
T1n=1 N (fn ), for a sequence (fn ) ⇢ `1 , then there
T is an ↵ 2 I
so that x↵ 2 n=1 N (fn ). In particular this implies that c0 6= n2N N (fn ).
Indeed, Step 3 yields that
[
C = {↵ 2 I : fn (x↵ ) 6= 0 for some n 2 N} =
{↵ 2 I : fn (x↵ ) 6= 0},
n2N
is countable, and thus I \ C is not empty.
Remark. Assume that Z is any subspace of `1 which is isomorphic to c0 ,
then Z is not complemented. The proof of that is a bit harder.
Theorem 2.4.10. [So] Assume Y is a subspace of a separable Banach space
X and T : Y ! c0 is linear and bounded. Then T can be extended to a
linear and bounded operator T̃ : X ! c0 . Moreover, T̃ can be chosen so that
kT̃ k  2kT k.
Corollary 2.4.11. Assume that X is a separable Banach space which contains a subspace Y which is isomorphic to c0 . Then Y is complemented in
X.
Proof. Let T : Y ! c0 be an isomorphism. Then extend T to T̃ : X ! c0
and put P = T 1 T̃ .
2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES
45
Proof of Theorem 2.4.10. Note that an operator T : Y ! c0 is defined by a
(Y ⇤ , Y ) null sequence (yn⇤ ) ⇢ Y ⇤ , i.e.
T : Y ! c0 ,
y 7! (hyn⇤ , yi : n 2 N).
We would like to use the Hahn Banach theorem and extend each yn⇤ to an
element x⇤n 2 Xn⇤ , with kyn⇤ k = kx⇤n k, and define
T̃ (x) := (hx⇤n , xi : n 2 N),
x 2 X.
But the problem is that (x⇤n ) might not be (X ⇤ , X) convergent to 0, and
thus we can only say that (hx⇤n , xi : n 2 N) 2 `1 , but not necessarily in c0 .
Thus we will need to change the x⇤n somehow so that they are still extensions
of the yn⇤ but also (X ⇤ , X) null.
Let B = kT kBX ⇤ . B is (X ⇤ , X)-compact and metrizable (since X
is separable). Denote the metric which generates the (X ⇤ , X)-topology
by d(·, ·). Put K = B \ Y ? . Since Y ? ⇢ X ⇤ is (X ⇤ , X)-closed, K is
compact and every (X ⇤ , X)-accumulation point of (x⇤n ) lies in K. Indeed, this follows from the fact that x⇤n (y) = yn⇤ (y) !n!1 0. This implies that limn!1 d(x⇤n , K) = 0, thus we can choose (zn⇤ ) ⇢ K so that
limn!1 d(x⇤n , zn⇤ ) = 0, and thus (x⇤n zn⇤ ) is (X ⇤ , X)-null and for y 2 Y it
follows that hx⇤n zn⇤ , yi = hx⇤n , yi, n 2 N. Choosing therefore
T̃ : X ! c0 ,
x 7! (hx⇤n
zn⇤ , xi : n 2 N),
yields our claim.
Remark. Zippin [Zi] proved the converse of Theorem: if Z is an infinitedimensional separable Banach space admitting a projection from any separable Banach space X containing it, then Z is isomorphic to c0 .
Exercises
1.
2.
Prove Proposition 2.4.2.
a) Assume that `1 isomorphic to a subspace Y of some Banach
space X, then Y is complemented in X.
b) Assume Z is a closed subspace of a Banach space X, and T : Z !
`1 is linear and bounded. Then T can be extended to a linear
and bounded operator T̃ : X ! `1 , with kT̃ k = kT k.
3.
Show that for a Banach space X, the dual space X ⇤ is isometrically
isomorphic to complemented subspace of X ⇤⇤⇤ , via the canonical embedding.
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CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
4.
Prove Proposition 2.4.7.
5.
Prove (2.3) in Proposition 2.4.4.