78
CHAPTER 3. BASES IN BANACH SPACES
Remark. Note that if aj = 1, for j 2 N, then in c0
sup
n2N
n
X
aj ej
j=1
c0
= 1,
P
but the series j2N aj ej does not converge in c0 .
Considering X as a subspace of X ⇤⇤ (via the canonical embedding) the
image of X under T is the space of sequences
1
n
o
X
Z := (ai ) 2 Y :
aj ej converges in X .
j=1
Proof of Proposition 3.3.6. Let K denote the basis constant of (en ), (e⇤n ) the
coordinate functionals, and (Pn ) the canonical projections. It is straightforward to check that Y is a vector space and that ||| · ||| is a norm on Y .
For x⇤ 2 X ⇤ and x⇤⇤ 2 X ⇤⇤ we have by Proposition 3.3.1
Pn⇤ (x⇤ ) =
n
X
j=1
hx⇤ , ej ie⇤j and
D
hPn⇤⇤ (x⇤⇤ ), x⇤ i = x⇤⇤ ,
n
X
j=1
n
E DX
E
hx⇤ , ej ie⇤j =
hx⇤⇤ , e⇤j iej , x⇤ ,
j=1
which implies that
(3.6)
⇤⇤
|||T (x )||| = sup
n2N
n
X
j=1
hx⇤⇤ , e⇤j iej = sup kPn⇤⇤ (x⇤⇤ )k  Kkx⇤⇤ k.
n2N
Thus T is bounded and kT k  K.
Assume that (an ) 2 Y . We want to find x⇤⇤ 2 X ⇤⇤ , so that T (x⇤⇤ ) =
(an ). Put
n
X
⇤⇤
xn =
aj ej , for n 2 N.
j=1
(where we identify X with its canonical image in X ⇤⇤ and, thus, ej with
(ej ) 2 X ⇤⇤ ) Since
kx⇤⇤
n kX ⇤⇤
=
n
X
j=1
aj ej
X
 |||(ai )|||, for all n 2 N,
3.3. SHRINKING, BOUNDEDLY COMPLETE BASES
79
and since X ⇤ is separable (and thus (BX ⇤⇤ , (X ⇤⇤ , X ⇤ )) is metrizable by
⇤
⇤⇤
Exercise 8 in Chapter 2 ) (x⇤⇤
n has a w -converging subsequence xnj to an
⇤⇤
element x with
kx⇤⇤ k  lim sup kx⇤⇤
n k  |||(aj )|||.
n!1
It follows for m 2 N that
⇤
hx⇤⇤ , e⇤m i = lim hx⇤⇤
n j , em i = a m ,
j!1
and thus it follows that T (x⇤⇤ ) = (aj ), and thus that T is surjective.
Finally, since (e⇤n ) is a basis for X ⇤ it follows for any x⇤⇤
n
X
|||T (x⇤⇤ )||| = sup
n2N
=
=
j=1
hx⇤⇤ , e⇤j iej
sup
n2N,x⇤ 2BX ⇤
⌦
n
X
j=1
⇤⇤
hx⇤⇤ , e⇤j ihx⇤ , ej i
sup sup x , Pn⇤ (x⇤ )
x⇤ 2BX ⇤ n2N
⇤⇤
↵
kx k (since Pn⇤ (x⇤ ) ! x⇤ if n ! 1),
which proves that T is an isomorphism, and, that |||T (x⇤ )|||
kx⇤⇤ k, for
x⇤⇤ 2 X ⇤⇤ . Together with (3.6) that shows T is an isometry if K = 1.
Lemma 3.3.7. Let X be a Banach space with a basis (en ), with basis constant K and let (e⇤n ) be its coordinate functionals. Let Z = span(e⇤n : n 2 N) ⇢
X ⇤ and define the operator
S : X ! Z ⇤,
x 7! (x)|Z
i.e. S(x)(z) = hz, xi, for z 2 Z and x 2 X.
Then S is an isomorphic embedding of X into Z ⇤ and for all x 2 X.
1
kxk  kS(x)k  kxk.
K
Moreover, the sequence (S(en )) ⇢ Z ⇤ are the coordinate functionals of (e⇤n )
(which by Proposition 3.3.1 is a basis of Z).
Proof. For x 2 X note that
kS(x)k =
sup
z2Z,kzkX ⇤ 1
|hz, xi|  sup |hx⇤ , xi| = kxk,
x⇤2BX ⇤
By Corollary 1.4.6 of the Hahn Banach Theorem.
80
CHAPTER 3. BASES IN BANACH SPACES
On the other hand, again by using that Corollary of the Hahn Banach
Theorem, we deduce that
kxk = sup |hw⇤ , xi|
⇤
w⇤ 2BX
= ⇤lim ⇤ lim |hw⇤ , Pn (x)i|
w 2BX n!1
= ⇤lim ⇤ lim |hPn⇤ (w⇤ ), xi|
w 2BX n!1
 sup sup |hPn⇤ (w⇤ ), xi
⇤
n2N w⇤ 2BX
 sup
sup
n2N z2span(e⇤j :jn),kzkK
|hz, xi = KkS(x)k.
Theorem 3.3.8. Let X be a Banach space with a basis (en ), and let (e⇤n )
be its coordinate functionals. Let Z = span(e⇤n : n 2 N) ⇢ X ⇤ . Then the
following are equivalent
a) X is isomorphic to Z ⇤ , via the map S as defined in Lemma 3.3.7
b) (e⇤n ) is a shrinking basis of Z.
c) If (aj ) ⇢ K, with the property that
sup
n2N
then
P1
j=1 aj ej
n
X
j=1
aj ej < 1,
converges.
In that case we call (en ) boundedly complete.
Proof. “(a))(b)” Assuming condition (a) we will verify condition (b) of
Theorem 3.3.4 for Z and its basis (e⇤n ). So let z ⇤ 2 Z ⇤ . By (a) we can write
z ⇤ = S(x) for some x 2 X. Since x = limn!1 Pn (x), where (Pn ) are the
canonical projection for (en ), we deduce that
sup
w2span(e⇤j :j>n),kwk1
hz ⇤ , wi =
=
=
sup
hS(x), wi
sup
hw, xi
sup
hw, (I
w2span(e⇤j :j>n),kwk1
w2span(e⇤j :j>n),kwk1
w2span(e⇤j :j>n),kwk1
Pn )(x)i
3.3. SHRINKING, BOUNDEDLY COMPLETE BASES
 k(I
81
Pn )(x)k !n!1 0.
It follows now from Theorem 3.3.4 that (e⇤j ) is a shrinking basis of Z.
“(b))(c)” Assume (b) and let (aj ) ⇢ K so that
n
X
|||(aj )||| = sup
n2N
aj ej = sup
n2N
j=1
n
X
j=1
aj (ej ) < 1.
As in Proposition 3.3.6 we can show that there is an x⇤⇤ 2 X ⇤⇤ , so that
hx⇤⇤ , e⇤n i = an for all n 2 N (for that part, the assumption that (en ) was
shrinking was not used). Let z ⇤ be the restriction of x⇤⇤ to the space Z.
Since (e⇤j ) is a shrinking basis of Z, since by Lemma 3.3.7 (S(en )) are the
coordinate functionals we can write (in a unique way)
z⇤ =
1
X
bj S(ej )
j=1
hx⇤⇤ , e⇤j i
But this means that aj =
= hz ⇤ , e⇤j i = bj for all j 2 N, P
and since
1
S is an isomorphism between X and its image it follows that
j=1 aj ej
converges in norm in X.
“(c)) (a)” By Lemma 3.3.7 it is left to show that the operator S is surjective. Thus, let z ⇤ 2 Z ⇤ . Since (e⇤n ) is a basis of Z and (S(en )) ⇢ Z ⇤ are
the coordinate functionals of (e⇤n ), it follows from Proposition 3.3.1 that z ⇤
is the w⇤ limit of (zn⇤ ) where
zn⇤ =
n
X
j=1
Since
w⇤ -converging
hz ⇤ , e⇤j iS(ej ).
sequences are bounded it follows that
n
X
sup
n2N
j=1
and, thus, by Lemma 3.3.7
sup
n2N
hz ⇤ , e⇤j iS(ej ) < 1
n
X
j=1
hz ⇤ , e⇤j iej < 1.
By our assumption (c) it follows therefore that x =
in X, and moreover
S(x) = lim
n!1
which proves our claim.
n
X
j=1
Pn
j=1 hz
hz ⇤ , e⇤j iS(ej ) = z ⇤ ,
⇤ , e⇤ ie
j j
converges
82
CHAPTER 3. BASES IN BANACH SPACES
Theorem 3.3.9. Let X be a Banach space with a basis (en ). Then X is reflexive if and only if (ej ) is shrinking and boundedly complete, or equivalently
if (ej ) and (e⇤j ) are shrinking.
Proof. Let (e⇤n ) be the coordinate functionals of (en ) and (Pn ) be the canonical projections for (en ).
“)” Assume that X is reflexive. By Proposition 3.3.1 it follows for every
x⇤ 2 X ⇤
n
X
⇤
⇤
x =w
lim
Pn⇤ (x⇤ ) = w
lim Pn⇤ (x⇤ ),
n!1
n!1
j=1
w
which implies that x⇤ 2 span(e⇤n : n 2 N) , and thus, by Proposition 2.2.5
k·k
k·k
x⇤ 2 span(e⇤n : n 2 N) . It follows therefore that x⇤ = span(e⇤n : n 2 N)
and thus that (ej ) is shrinking (by Proposition 3.3.1).
Thus X ⇤ is a Banach space with a basis (e⇤j ) which is also reflexive. We
can therefore apply to X ⇤ what we just proved for X and deduce that (e⇤n )
is a shrinking basis for X ⇤ . But, by Theorem 3.3.8 (in this case Z = X ⇤ )
this means that (en ) is boundedly complete.
“(” Assume that (en ) is shrinking and boundedly complete, and let x⇤⇤ 2
X ⇤⇤ . Then
X ⇤⇤ = (X ⇤⇤ , X ⇤ )
"
n!1
n
X
j=1
hx⇤⇤ , e⇤j i (ej )
By Proposition 3.3.1 and the fact that X ⇤ = span(e⇤j : j 2 N)
has (e⇤j ) as a basis, since (ej ) is shrinking
=k·k

lim
lim
n!1
n
X
j=1
#
hx⇤⇤ , e⇤j i (ej ) 2 (X)
P
Since supn2N k nj=1 hP ⇤⇤ (x⇤⇤ ), e⇤j iej k < 1, by Exercise 9
in Chapter 2,and since (ej ) is boundedly complete
which proves our claim.
The last Theorem in this section describes by how much one can perturb
a basis of a Banach space X and still have a basis of X.
Theorem 3.3.10. (The small Perturbation Lemma)
Let (xn ) be a basic sequence in a Banach space X, and let (x⇤n ) be the
⇤
coordinate functionals (they are elements of span(xj : j 2 N) ) and assume
3.3. SHRINKING, BOUNDEDLY COMPLETE BASES
83
that (yn ) is a sequence in X such that
(3.7)
c=
1
X
n=1
yn k · kx⇤n k < 1.
kxn
Then
a) (yn ) is also basic in X and isomorphically equivalent to (xn ), more
precisely
(1
c)
1
X
n=1
1
X
an xn 
n=1
for all in X converging series x =
an yn  (1 + c)
P
1
X
an xn ,
n=1
n2N an xn .
b) If span(xj : j 2 N) is complemented in X, then so is span(yj : j 2 N).
c) If (xn ) is a Schauder basis of all of X, then (yn ) is also a Schauder
basis of X and it follows for the coordinate functionals (yn⇤ ) of (yn ),
that yn⇤ 2 span(x⇤j : j 2 N), for n 2 N.
Proof. By Corollary 1.4.4 of the Hahn Banach Theorem we extend the functionals x⇤n to functionals x̃⇤n 2 X ⇤ , with kx̃⇤n k = kx⇤n k, for all n 2 N.
Consider the operator:
T : X ! X,
P1
x 7!
yn k · kx⇤n k
1
X
n=1
hx̃⇤n , xi(xn
yn ).
Since n=1 kxn
< 1, T is well defined, linear and bounded and
kT k  c < 1. It follows S = Id T is an isomorphism between X and it
self. Indeed, for x 2 X
kS(x)k kxk kT k · kxk (1 c)kxk and
P we have,
n (y) (T 0 = Id) then
if y 2 X, define x = 1
T
n=0
(Id
T )(x) =
1
X
n=0
T n (y)
T
1
⇣X
n=0
1
⌘ X
T n (y) =
T n (y)
n=0
1
X
T n (y) = y.
n=1
Thus Id T is surjective, and, it follows form Corollary 1.3.6 that Id T
is an isomorphism between X and itself.
(a) We have (I T )(xn ) = yn , for n 2 N, this means in particular that (yn )
is basic and (xn ) and (yn ) are isomorphically equivalent.
(b) Let P : X ! span(xn : n 2 N) be a bounded linear projection onto
span(xn : n 2 N). Then it is easily checked that
Q : X ! span(yn : n 2 N),
x 7! (Id
T) P
(Id
T)
1
(x),
84
CHAPTER 3. BASES IN BANACH SPACES
is a linear projection onto span(yn : n 2 N).
(c) If X = span(xn : n 2 N), then, since I T is an isomorphism, (yn ) =
((I T )(xn )) is also a Schauder basis of X.
Moreover define for k and i in N,
⇤
y(i,k)
=
k
X
j=1
hyi⇤ , xj ix⇤j =
n
X
j=1
h (xj ), yi⇤ ix⇤j 2 span(x⇤j : j 2 N).
⇤
It follows from Proposition 3.3.1, part (b), that w⇤ limk!1 y(i,k)
= yi⇤ ,
P
⇤
⇤
which implies that yi⇤ (x) = 1
j=1 hyi , xj ihxj , xi, for all x 2 X, and thus for
k i
kyi⇤
⇤
y(i,k)
k = sup |hyi⇤
⇤
y(i,k)
, xi|
x2BX
= sup
x2BX
= sup
x2BX
 kyi⇤ k
1
X
j=k+1
1
X
j=k+1
1
X
j=k+1
hyi⇤ , xj ihx⇤j , xi
hyi⇤ , xj
kxj
yj ihx⇤j , xi
yj k · kx⇤j k ! 0, if k ! 1.
⇤
so it follows that yi⇤ = k · k limk!1 y(k,i)
2 span(x⇤j : j 2 N) for every i 2 N,
which finishes the proof of our claim (c).
Exercises
1.
Prove that Y with ||| · |||, as defined in Proposition 3.3.6 is a normed
linear space.
2.
A Banach space X is said to have the Approximation Property if for
every compact set K ⇢ X and every " > 0 there is a finite rank
operator T so that kx T (x)k < " for all x 2 K.
Show that every Banach space with a Schauder basis has the approximation property.
3.3. SHRINKING, BOUNDEDLY COMPLETE BASES
3.
85
Show that (ei ) is a shrinking basis of a Banach space X, then the
coordinate functionals (e⇤i ) are boundedly complete basis of X ⇤ .
4.⇤ A Banach space is called L(p, ) -space, for some 1  p  1 and some
1, if for every finite dimensional subspace F of X and every " > 0
there is a finite dimensional subspace E of X which contains F and so
dim(E)
that dBM (E, `p
) < + ".
Show that Lp [0, 1], 1  p < 1 is a L(p,1) -space.
Hint: Firstly, he span of the first n elements of the Haar basis is isometrically isomorphically to `np (why?), secondly consider Small Perturbation Lemma.