Chapter 3
Bases in Banach Spaces
Like every vector space a Banach space X admits an algebraic or Hamel
basis, i.e. a subset B ⇢ X, so that every x 2 X is in a unique way the (finite)
linear combination of elements in B. This definition does not take into
account that we can take infinite sums in Banach spaces and that we might
want to represent elements x 2 X as converging series. Hamel bases are
also not very useful for Banach spaces, since (see Exercise 1), the coordinate
functionals might not be continuous.
3.1
Schauder Bases
Definition 3.1.1. (Schauder bases of Banach Spaces)
Let X be an infinite dimensional Banach space. A sequence (en ) ⇢ X is
called Schauder basis of X, or simply a basis of X, if for every x 2 X, there
is a unique sequence of scalars (an ) ⇢ K so that
x=
1
X
an en .
n=1
Examples 3.1.2. For n 2 N let
en = ( 0, . . . 0 , 1, 0, . . .) 2 KN
| {z }
n 1 times
Then (en ) is a basis of `p , 1  p < 1 and c0 . We call (en ) the unit vector
basis of `p and c0 , respectively.
Remarks. Assume that X is a Banach space and (en ) a basis of X.
61
62
CHAPTER 3. BASES IN BANACH SPACES
a) (en ) is linear independent.
b) span(en : n 2 N) is dense in X, in particular X is separable.
c) EveryPelement x is uniquely determined by the sequence (an ) so that
x = 1
. So we can identify X with a space of sequences in
j=1 an en
P
N
K , for which
an en converges in X.
Proposition 3.1.3. Let X be a normed linear space and assume that (en ) ⇢
X has the property that each x 2 X can be uniquely represented as a series
x=
1
X
n=1
an en , with (an ) ⇢ K
(we could call (en ) Schauder basis of X but we want to reserve this term
only for Banach spaces).
For n 2 N and x 2 X define e⇤n (x) 2 K to be the unique element in K,
so that
1
X
x=
e⇤n (x)en .
n=1
e⇤n
Then
: X ! K is linear.
For n 2 N let
Pn : X ! span(ej : j  n), x 7!
n
X
e⇤n (x)en .
j=1
Then Pn : X ! X are linear projections onto span(ej : j  n) and the
following properties hold:
a) dim(Pn (X)) = n,
b) Pn Pm = Pm Pn = Pmin(m,n) , for m, n 2 N,
c) limn!1 Pn (x) = x, for every x 2 X.
Conversely if (Pn : n 2 N) is a sequence of linear projections satisfying
(a), (b), and (c), and moreover are bounded, and if e1 2 P1 (X) \ {0} and
en 2 Pn (X) \ N (Pn 1 ), with en 6= 0, if n > 1, then each x 2 X can be
uniquely represented as a series
x=
1
X
n=1
an en , with (an ) ⇢ K,
so in particular (en ) is a Schauder basis of X in case X is a Banach space.
3.1. SCHAUDER BASES
63
⇤
Proof. The linearity
P1 of ⇤en follows from the unique representation of every
x 2 X as x =
j=1 en (x)en , which implies that for x and y in X and
↵, 2 K,
↵x + y = lim ↵
n!1
= lim
n!1
n
X
n
X
e⇤j (x)ej +
j=1
n
X
e⇤j (y)ej
j=1
(↵e⇤j (x)
+
e⇤j (y))ej
=
j=1
1
X
(↵e⇤j (x) + e⇤j (y))ej ,
j=1
and, on the other hand
↵x + y =
1
X
e⇤j (↵x + y)ej ,
j=1
thus, by uniqueness, e⇤j (↵x + y) = ↵e⇤j (x) + e⇤j (y), for all j 2 N. The
conditions (a), (b) and (c) are clear.
Conversely assume that (Pn ) is a sequence of bounded and linear projections satisfying (a), (b), and (c), and if e1 2 P1 (X) \ {0} and en 2
Pn (X) \ N (Pn 1 ), if n > 1, then for x 2 X, by (b)
Pn
1 (Pn (x)
Pn
1 (x))
= Pn
and thus Pn (x)
Pn (x)
Pn
1 (x)
1 (x)
Pn
= Pn (Pn (x)
Pn
1 (x)
Pn
1 (x)
2 N (Pn
1 (x))
= 0,
1)
and
2 Pn (X),
and therefore Pn (x) Pn 1 (x) 2 Pn (X) \ N (Pn 1 ). Thus we can write
Pn (x) Pn 1 (x) = an en , for n 2 N, and it follows from (c) that (letting
P0 = 0)
x = lim Pn (x) = lim
n!1
n!1
n
X
Pj (x)
Pj
1 (x) = lim
n!1
j=1
n
X
j=1
aj ej =
1
X
In order to show uniqueness of this representation of x assume x =
From the continuity of Pm Pm 1 , for m 2 N it follows that
am em = (Pm
and thus am = bm .
Pm
1 )(x) = lim (Pm
n!1
Pm
1)
n
⇣X
j=1
aj ej .
j=1
P1
j=1 bj ej .
⌘
b j e j = bm e m ,
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CHAPTER 3. BASES IN BANACH SPACES
Definition 3.1.4. (Canonical Projections and Coordinate functionals)
Let X be a normed linear space and assume that (ei ) satisfies the assumptions of Proposition 3.1.3. The linear functionals (e⇤n ) as defined in Proposition 3.1.3 are called the Coordinate Functionals for (en ) and the projections
Pn , n 2 N, are called the Canonical Projections for (en ).
Proposition 3.1.5. Suppose X is a normed linear space and assume that
(en ) ⇢ X has the property that each x 2 X can be uniquely represented as a
series
1
X
x=
an en , with (an ) ⇢ K.
n=1
If the canonical projections are bounded, and, moreover, supn2N kPn k < 1
(i.e. uniformly the Pn are bounded), then (ei ) is a Schauder basis of its
completion X̃.
Proof. Let P̃n : X̃ ! X̃, n 2 N, be the (by Proposition 1.1.5 and Exercise
1 in Section 1.2 uniquely existing) extensions of Pn . Since Pn has finite
dimensional range it follows that P̃n (X̃) = Pn (X) = span(ej : j  n) is
finite dimensional and, thus, closed. (P̃n ) satisfies therefore (a) of Proposition 3.1.3. Since the Pn are continuous, and satisfy the equalities in (b) of
Proposition 3.1.3 on a dense subset of X̃, (b) is satisfied on all of X̃. Finally,
(c) of Proposition 3.1.3 is satisfied on a dense subset of X̃, and we deduce
for x̃ 2 X̃, x̃ = limk!1 xk , with xk 2 X, for k 2 N, that
kx̃
P̃n (x̃)k  kx̃
xk k + sup kPj kkx̃
j2N
xk k + kxk
Pn (xk )k
and, since (Pn ) is uniformly bounded, we can find for given " > 0, k large
enough so that the first two summands do not exceed ", and then we choose
n 2 N large enough so that the third summand is smaller than ". It follows
therefore that also (c) is satisfied on all of X̃. Thus, our claim follows from
the second part of Proposition 3.1.3 applied to X̃.
We will now show that if (en ) is a basis for a Banach space X then the
coordinate functionals, and thus the canonical projections are bounded, and
moreover we will show that the canonical projections are uniformly bounded.
Theorem 3.1.6. Let X be a Banach space with a basis (en ) and let (e⇤n ) be
the corresponding coordinate functionals and (Pn ) the canonical projections.
Then Pn is bounded for every n 2 N and
b = sup ||Pn kL(X,X) < 1,
n2N
3.1. SCHAUDER BASES
65
and thus e⇤n 2 X ⇤ and
ke⇤n kX ⇤ =
kPn
Pn
ken k
1k

2b
.
ken k
We call b the basis constant of (ej ). If b = 1 we say that (ei ) is a monotone
basis.
Furthermore there is an equivalent renorming ||| · ||| of (X, k · k) for which
(en ) is a monotone basis for (X, ||| · |||).
Proof. For x 2 X we define
|||x||| = sup kPn (x)k,
n2N
since kxk = limn!1 kPn (x)k, it follows that kxk  |||x||| < 1 for x 2 X.
It is clear that ||| · ||| is a norm on X. Note that for n 2 N
|||Pn ||| =
=
sup
x2X,|||x|||1
sup
|||Pn (x)|||
sup kPm Pn (x)k
x2X,|||x|||1 m2N
=
sup
sup kPmin(m,n) (x)k  1.
x2X,|||x|||1 m2N
Thus the projections Pn are uniformly bounded on (X, ||| · |||). Let X̃
be the completion of X with respect to ||| · |||, P̃n , for n 2 N, the (unique)
extension of Pn to an operator on X̃. We note that the P̃n also satisfy the
conditions (a), (b) and (c) of Proposition 3.1.3. Indeed (a) and (b) are purely
algebraic properties which are satisfied by the first part of Proposition 3.1.3.
Moreover for x 2 X then
(3.1)
|||x
Pn (x)||| = sup kPm (x)
m2N
= sup kPm (x)
m n
Pmin(m,n) (x)k
Pn (x)k ! 0 if n ! 1,
which verifies condition (c). Thus, it follows therefore from the second part
of Proposition 3.1.3, the above proven fact that |||Pn |||  1, for n 2 N, and
Proposition 3.1.5, that (en ) is a Schauder basis of the completion of (X, ||| · |||)
which we denote by (X̃, ||| · |||).
We will now show that actually X̃ = X, and thus that, (X, ||| · |||) is
already complete. Then it would follow from Corollary 1.3.6 of the Closed
Graph Theorem that ||| · ||| is an equivalent norm, and thus that
C = sup sup kPn (x)k = sup |||x||| < 1.
n2N x2BX
x2BX
66
CHAPTER 3. BASES IN BANACH SPACES
P1
So, let x̃ 2 X̃ and write it (uniquely) as x̃ =
j=1 aj ej , where this
convergence
happens in ||| · |||. Since k · k  ||| · |||, and since X is complete the
P
series 1
a
j=1 j ej also converges with respect to k · k in X to say x 2 X.
Important
Side Note: This means that the sequence of partial sums
Pn
· k) to x, which means that (an ) is the unique
j=1 aj ej ) converges in (X, k
P
sequence in K, for which x = 1
j=1 aj ej . In particular this means that
Pn (x) =
n
X
j=1
aj ej = P̃n (x̃), for all n 2 N.
But now (3.1) yields that Pn (x) also converges in ||| · ||| to x.
This means (since (Pn (x) cannot converge to two di↵erent elements) that
x = x̃, which finishes our proof.
After reading the proof of Theorem 3.1.6 one might ask whether the last
part couldn’t be generalized and whether the following could be true: If k · k
and |||·||| are two norms on the same linear space X, so that k·k  |||·|||, and so
that (k · k, X) is complete, does it then follow that (X, ||| · |||) is also complete
(and thus k · k and ||| · ||| are equivalent norms). The answer is negative, as
the following example shows.
Example 3.1.7. Let X = `2 with its usual norm k·k2 and let (b : 2 ) ⇢ S`2
be a Hamel basis of (by Exercise 2, is necessarily uncountable). For
x 2 `2 define |||x|||,
X
|||x||| =
|x |,
2
P
where x =
2 x b is the unique representation of x as a finite linear
combination
of
elements of (b : 2 ). Since kb k2 , for 2 , it follows for
P
x=
x
b
2 `2 from the triangle inequality that
2
|||x||| =
X
2
|x | =
X
2
kx b k2
X
x b
2
2
= kxk2 .
Finally both norms k · k and ||| · |||, cannot be equivalent. Indeed, for arbitrary
" > 0, there is an uncountable set 0 ⇢ , so that kb
b 0 k2 < ", , 0 2 0 ,
( is uncountable but S`2 is in the k · k2 -norm separable). for any two
di↵erent elements , 0 2 0 it follows that
kb
b 0 k < " < 2 = |||b
b 0 |||.
Since " > 0 was arbitrary this proves that k · k and ||| · ||| cannot be equivalent.
3.1. SCHAUDER BASES
67
Definition 3.1.8. (Basic Sequences)
Let X be a Banach space. A sequence (xn ) ⇢ X \{0} is called basic sequence
if it is a basis for span(xn : n 2 N).
If (ej ) and (fj ) are two basic sequences (in possibly two di↵erent Banach
spaces X and Y ). We say that (ej ) and (fj ) are isomorphically equivalent
if the map
T : span(ej : j 2 N) ! span(fj : j 2 N),
n
X
j=1
aj ej 7!
n
X
aj fj ,
j=1
extends to an isomorphism between the Banach spaces between span(ej : j 2 N)
and span(fj : j 2 N).
Note that this is equivalent with saying that there are constants 0 < c 
C so that for any n 2 N and any sequence of scalars ( j )nj=1 it follows that
c
n
X
j=1
j ej 
n
X
j=1
j fj  C
n
X
j ej
.
j=1
Proposition 3.1.9. Let X be Banach space and (xn : n 2 N) ⇢ X \ {0}.
The (xn ) is a basic sequence if and only if there is a constant K 1, so that
for all m < n and all scalars (aj )nj=1 ⇢ K we have
(3.2)
m
X
i=1
ai xi  K
n
X
ai xi .
i=1
In that case the basis constant is the smallest of all K
holds.
1 so that (3.2)
Proof.
Theorem 3.1.6, since K := supn2N kPn k < 1 and
Pn“)” Follows
Pfrom
m
n
Pm
a
x
=
a
i=1 i i
i=1 i xm , if m  n and (ai )i=1 ⇢ K.
“(” Assume that there is a constant K
1 so that for all m < n and all
scalars (aj )nj=1 ⇢ K we have
m
X
i=1
ai xi  K
n
X
ai xi .
i=1
We first note P
that this implies that (xn ) is linear independent. Indeed, if we
assume that nj=1 aj xj = 0, for some choice of n 2 N and (aj )nj=1 ⇢ K, and
not all of the aj are vanishing, we first observe that at least two of a0j s cannot
be equal to 0 (since xj 6= 0, for j 2 N), thus if we let m := min{j : aj 6= 0},
68
CHAPTER 3. BASES IN BANACH SPACES
P
Pn
it follows that m
j=1 aj xj 6= 0, but
j=1 aj xj = 0, which contradicts our
assumption.
It follows therefore that (xn ) is a Hamel basis for (the vector space)
span(xj : j 2 N), which implies that the projections Pn are well defined on
span(xj : j 2 N), and satisfy (a), (b), and (c) of Proposition 3.1.3. Moreover,
it follows from our assumption that
kPm k = sup
m
n X
j=1
aj xj : n 2 N, (aj )nj=1 ⇢ K,
n
X
j=1
o
aj xj  1  K.
Thus, our claim follows from Proposition 3.1.5.
Also note that the proof of “)” implies that the smallest constant so
that 3.2 is at most as big as the basis constant, and the proof of “(” yielded
that it is at least as large as the basis constant.
Exercises
1.
Let (e : 2 ) be a Hamel basis of an infinite dimensional Banach
space X.Show that some of the coordinate functionals associated with
that basis are not continuous.
Hint: pick a sequence ( n ) ⇢ of pairwise di↵erent elements of and
consider
1
X
e n
x=
2 n
.
ke n k
n=1
2.
Show that the Hamel basis of an infinite dimensional Banach space X
must be uncountable.
3.
For n 2 N define in c0
sn =
n
X
j=1
ej = (1, 1, . . . 1, 0, . . .).
| {z }
n times
Prove that (sn ) is a basis for c0 , but that one can reorder (sn ) so that
it is not a basis of c0 .
(sn ) is called the summing basis of c0 .
Hint: Play around with alternating series of (sn ).
3.1. SCHAUDER BASES
69
4.⇤ Let 1 < p < 1 and assume that (xn ) is a weakly null sequence in
`p with inf n2N kxn k > 0. Show that (xn ) has a subsequence which is
isomorphically equivalent to the unit vector basis of `p .
And then deduce from this:
Let T : `p ! `q with 1 < q < p < 1, be a bounded linear operator.
Show that T is compact., meaning that T (B`p ) is relatively compact
in `q .