54
2.7
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
The Principle of Local Reflexivity
In this section we we proof a result of J. Lindenstrauss and H. Rosenthal
[LR] which states that for a Banach space X the finite dimensional subspaces
of the bidual X ⇤⇤ are in a certain sense similar to the finite dimensional
subspaces of X.
Theorem 2.7.1. [LR] [The Principle of Local Reflexivity]
Let X be a Banach space and let F ⇢ X ⇤⇤ and G ⇢ X ⇤ be finite dimensional
subspaces of X ⇤⇤ and X ⇤ respectively.
Then, given " > 0, there is a subspace E of X containing F \ X (we
identify X with its image under the canonical embedding) with dim E =
dim F and an isomorphism T : F ! E with kT k · kT 1 k  1 + " such that
(2.11)
(2.12)
T (x) = x if x 2 F \ X and
hx⇤ , T (x⇤⇤ )i = hx⇤⇤ , x⇤ i if x⇤ 2 G, x⇤⇤ 2 F .
We need several Lemmas before we can prove Theorem 2.7.1. The first
one is a corollary the Geometric Hahn-Banach Theorem
Proposition 2.7.2. (Variation of Geometrical Version of the Theorem of
Hahn Banach)
Assume that X is a Banach space and C ⇢ X is convex with C 6= ; and
let x 2 X \ C (so x could be in the boundary of C). Then there exists an
x⇤ 2 X ⇤ so that
<hx⇤ , zi < 1 = hx⇤ , xi for all z 2 C 0 ,
and, if moreover C is absolutely convex (i.e. if ⇢x 2 C for all x 2 C and
⇢ 2 K, with |⇢|  1), then
|hx⇤ , zi| < 1 = hx⇤ , xi for all z 2 C 0 .
Lemma 2.7.3. Assume T : X ! Y is a bounded linear operator between
the Banach spaces X and Y and assume that T (X) is closed.
Suppose that for some y 2 Y there is an x⇤⇤ 2 X ⇤⇤ with kx⇤⇤ k < 1, so
that T ⇤⇤ (x⇤⇤ ) = y. Then there is an x 2 X, with kxk < 1 so that T (x) = y.
Proof. We first show that there is an x 2 X so that T (x) = y. Assume this
where not true, then we could find by the Hahn-Banach Theorem (Corollary
1.4.5) an element y ⇤ 2 Y ⇤ so that y ⇤ (z) = 0 for all z 2 T (X) and hy ⇤ , yi = 1
2.7. THE PRINCIPLE OF LOCAL REFLEXIVITY
55
(T (X) is closed). But this yields hT ⇤ (y ⇤ ), xi = hy ⇤ , T (x)i = 0, for all x 2 X,
and, thus, T ⇤ (y ⇤ ) = 0. Thus
0 = hx⇤⇤ , T ⇤ (y ⇤ )i = hT ⇤⇤ (x⇤⇤ ), y ⇤ i = hy, y ⇤ i = 1,
which is a contradiction.
Secondly assume that y 2 T (X) \ T (BX ). Since T is surjective onto its
(closed) image Z = T (X) it follows from the Open Mapping Theorem that
T (BX ) is open in Z, and we can use the geometric version of the HahnBanach Theorem 2.2.3 , to find z ⇤ 2 Z ⇤ so that hz ⇤ , T (x)i < 1 = hz ⇤ , yi for
all x 2 BX . Again by the Theorem of Hahn-Banach (Corollary 1.4.4) we
can extend z ⇤ to an element y ⇤ in Y ⇤ . It follows that
kT ⇤ (y ⇤ )k = sup hT ⇤ (y ⇤ ), xi = sup hz ⇤ , T (x)i  1,
x2BX
x2BX
and thus, since kx⇤⇤ k < 1, it follows that
|hy ⇤ , yi| = |hy ⇤ , T ⇤⇤ (x⇤⇤ )i| = |hx⇤⇤ , T ⇤ (y ⇤ )i| < 1,
which is a contradiction.
Lemma 2.7.4. Let T : X ! Y be a bounded linear operator between two
Banach spaces X and Y with closed range, and assume that F : X ! Y has
finite rank.
Then T + F also has closed range.
Proof. Assume the claim is not true. Put S = T + K and consider the map
S : X/N (S) ! Y,
x + N (S) ! S(x)
which is a well defined linear bounded Operator, which by Proposition 1.3.11
cannot be an isomorphism onto its image.
Therefore we can choose a sequence (zn ) in X/N (S) so that
we can choose a sequence (xn ) so that
lim (T + F )(xn ) = 0 and dist(xn , N (T + F ))
n!1
1.
Since the sequence (F (xn ) : n 2 N) is a bounded sequence in a finite
dimensional space, we can, after passing to a subsequence, assume that
(F (xn ) : n 2 N) converges to some y 2 Y and, hence,
lim T (xn ) =
n!1
y.
56
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
Since T has closed range there is an x 2 X, so that T (x) = y. Using again
the equivalences in Proposition 1.3.11 and the fact that T (xn ) ! y = T (x),
if n % 1, it follows for some constant C > 0 that
xn , N (T ))  lim CkT (x
lim dist(x
n!1
n!1
xn )k = 0,
and, thus,
y
F (x) 2 F (N (T )),
F (x) = lim F (xn )
n!1
so we can write y
F (x) as
F (x) = F (u), where u 2 N (T ).
y
Thus
lim dist(xn
lim kF (xn )
n!1
u, N (T )) = 0 and
x
n!1
F (x)
F (u)k = 0.
F |N (T ) has also closed range, Proposition 1.3.11 yields (C being some positive constant)
lim sup dist(xn x u, N (F )\N (T ))  lim sup CkF (xn ) F (x) F (u)k = 0.
n!1
n!1
Since T (x+u) = y = F (x+u) (by choice of u), and thus (T +F )(x+u) =
0 which means that x + u 2 N (T + F ). Therefore
lim sup dist(xn , N (T + F )) = lim sup dist(xn
x
u, N (T + F ))
 lim sup dist(xn
x
u, N (T ) \ N (F )) = 0.
n!1
n!1
n!1
But this contradicts our assumption on the sequence (xn ).
Lemma 2.7.5. Let X be a Banach space, A = (ai,j )im,jn an m be n
matrix and B = (bi,j )ip,jn a p by n matrix, and that B has only real
entries (even if K = C).
⇤⇤
Suppose that y1 , . . . ym 2 X, y1⇤ , . . . yp⇤ 2 X ⇤ , ⇠1 , . . . ⇠p 2 R, and x⇤⇤
1 , . . . xn 2
BX ⇤⇤ satisfy the following equations:
(2.13)
n
X
ai,j x⇤⇤
j = yi , for all i = 1, 2 . . . m, and
j=1
(2.14)
D
yi⇤ ,
n
X
j=1
E
bi,j x⇤⇤
= ⇠i , for all i = 1, 2 . . . p.
j
2.7. THE PRINCIPLE OF LOCAL REFLEXIVITY
57
Then there are vectors x1 , . . . xn 2 BX satisfying:
(2.15)
n
X
ai,j xj = yi , for all i = 1, 2 . . . m, and
j=1
(2.16)
D
yi⇤ ,
n
X
j=1
E
bi,j xj = ⇠i , for all i = 1, 2 . . . p.
Proof. Recall from Linear Algebra that we can write the matrix A as a
product A = U P V , where U and V are invertible and P is of the form
✓
◆
Ir 0
P =
,
0 0
where r is the rank of A and Ir the identity on Kr .
For a general s by t matrix C = (ci,j )is,jt consider the operator
TC : `t1 (X) ! `s1 (X),
(x1 , x2 , . . . xt ) 7!
t
⇣X
j=1
⌘
ci,j xj : i = 1, 2 . . . m .
If s = t and if C is invertible then TC is an isomorphism. Also if C (1) and
C (2) are two matrices so that the number of columns of C (1) is equal to the
number of rows of C (2) one easily computes that TC (1) C (2) = TC (1) TC (2) .
Secondly it is clear that TP is a closed operator (P defined as above), since
TP is simply the projection onto the first r coordinates in `n1 (X).
It follows therefore that TA = TU TP TV is an operator with closed
range. Secondly define the operator
SA : `n1 (X) ! `m
1 (X)
(x1 , . . . xn ) 7!
`p1 ,
TA (x1 , . . . xn ),
⇣D
yi⇤ ,
n
X
bi,j xj
j=1
E⌘p
i=1
!
.
SA can be written as the sum of SA and a finite rank operator and has
therefore also closed range by Lemma 2.7.4.
⇤⇤ is the operator
Since the second adjoint of SA
⇤⇤
⇤⇤
SA
: `n1 (X ⇤⇤ ) ! `m
1 (X )
⇤⇤
(x⇤⇤
1 , . . . xn )
7!
`p1 ,
⇤⇤
TA⇤⇤ (x⇤⇤
1 , . . . xn ),
⇣D
yi⇤ ,
n
X
j=1
bi,j x⇤⇤
j
E⌘p
i=1
!
58
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
with
TA⇤⇤
:
`n1 (X ⇤⇤ )
!
⇤⇤
`m
1 (X ),
⇤⇤
⇤⇤
(x⇤⇤
1 , x2 , . . . xn )
7!
t
⇣X
j=1
⌘
ai,j x⇤⇤
j : i = 1, 2 . . . m ,
our claim follows from Lemma 2.7.3.
Lemma 2.7.6. Let E be a finite dimensional space and (xi )N
i=1 is an "-net
of SE for some 0 < " < /3. If T : E ! E is a linear map so that
(1
Then
")  kT (xj )k  (1 + "), for all j = 1, 2 . . . N .
1 3"
1+"
kxk  kT (x)k 
kxk, for all x 2 E,
1 "
1 "
and thus
kT k · kT
1
k
(1 + ")2
.
(1 ")(1 3")
Proof. Let x 2 X. W.l.o.g. we can assume that kxk = 1. Pick j  N so
that kx xj k  ". Then
kT (x)k  kT (x)
T (xj )k + kT (xj )k  "kT k + 1 + ",
and thus
1+"
,
1 "
which implies the second inequality of our claim. We also have
kT k 
kT (x)k
kT (xj )k kT (x xj )k
1 " "
1+"
1
=
1 "
2" + "2 "
1 "
"2
=
1
1
which proves our claim.
We are now ready to proof Theorem 2.7.1.
Proof of Theorem 2.7.1. Let F ⇢ X ⇤⇤ and G ⇢ X ⇤ be finite dimensional
)2
subspaces, and let 0 < " < 1. Choose > 0, so that (1 (1+
)(1 3 ) < ", and a
N
⇤⇤ N
-net (x⇤⇤
j )j=1 of SF . It can be shown that (xj )j=1 generates all of F , but
we can also simply assume that without loss of generality, since we can add
a basis of F .
Let
N
X
N
S : R ! F,
(⇠1 , ⇠2 . . . ⇠N ) 7!
⇠j x⇤⇤
j ,
j=1
3"
,
2"
2.7. THE PRINCIPLE OF LOCAL REFLEXIVITY
59
and note that S is surjective.
Put H = S 1 (F \ X), and let (a(i) : i = 1, 2 . . . m) be a basis of H,
write a(i) as a(i) = (ai,1 , ai,2 , . . . ai,N ), and define A to be the m by N matrix
A = (ai,j )im,jN . For i = 1, 2 . . . m put
yi = S(a(i) ) =
N
X
j=1
ai,j x⇤⇤
j 2 F \ X,
⇤
choose x⇤1 , x⇤2 , . . . , x⇤N 2 SX ⇤ so that hx⇤⇤
, and pick basis
j , xj i > 1
⇤
⇤
⇤
{g1 , g2 , . . . g` } of G.
Consider the following system of equations in N unknowns z1⇤⇤ , z2⇤⇤ ,
⇤⇤ in X ⇤⇤ :
. . . , zN
N
X
ai,j zj⇤⇤ = yi for i = 1, 2 . . . m
j=1
hzj⇤⇤ , x⇤j i
hzj⇤⇤ , gk i
⇤
= hx⇤⇤
j , xj i for j = 1, 2 . . . N and
⇤
= hx⇤⇤
j , gk i for j = 1, 2 . . . N and k = 1, 2 . . . `.
By construction zj⇤⇤ = x⇤⇤
j , j = 1, 2 . . . N , is a solution to these equations.
Since kx⇤⇤
k
=
1
<
1
+
, for j = 1, 2 . . . N , we can use Lemma 2.7.5 and
j
find x1 , x2 , . . . xN 2 X, with kxj k = 1 < 1 + , for j = 1, 2 . . . N , which solve
above equations.
Define
N
X
N
S1 : R ! X,
(⇠1 , ⇠2 . . . ⇠N ) 7!
⇠j xj .
j=1
We claim that the null space of S isPcontained in the
space of S1 . Indeed
Pnull
N
⇤⇤
if we assumed that ⇠ 2 KN , and N
⇠
x
,
but
x
j
j
j=1
j=1 j 6= 0, then there
is an i 2 {1, 2, . . . N } so that
D
x⇤i ,
N
X
j=1
E
x⇤⇤
6 0,
=
j
but since hx⇤ ⇤j , x⇤i i = hxj , x⇤i i this is a contradiction.
It follows therefore that we can find a linear map T : F ! X so that
S1 = T S. Denoting the standard basis of RN by (ei )iN we deduce that
xi = S1 (ei ) = T S(ei ) = T (x⇤⇤
j ), and thus
1 + > kxj k = kT (x⇤⇤
j )k
⇤
|hx⇤j , xj i| = hx⇤⇤
j , xj i| > 1
.
60
CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
By Lemma 2.7.6 and the choice of it follows therefore that kT k · kT 1 k 
1 + ".
P
(i)
Note that for ⇠ 2 H = S 1 (F \ X), say ⇠ = m
i=1 i a , we compute
S1 (⇠) =
m
X
i=1
i S1 (a
(i)
)=
m
X
i
i=1
N
X
j=1
ai,j xj =
m
X
i=1
i
N
X
j=1
ai,j x⇤⇤
j
=
m
X
i S(a
(i)
i=1
We deduce therefore for x 2 F \ X, that T (x) = x.
Finally from the third part of the system of equations it follows, that
⇤
⇤
⇤
hx⇤ , T (x⇤⇤
j )i = hx , xj i = hx , xj i, for all j = 1, 2 . . . N and x 2 G ,
and, thus (since the x⇤⇤
j generate all of F ), that
hx⇤ , T (x⇤⇤ )i = hx⇤⇤ , x⇤ i, for all x⇤⇤ 2 F and x⇤ 2 G.
Exercises
1.
Prove Proposition 2.7.2.
) = S(⇠).