1 Coordinate Functionals

Schauder Bases

Viktor Zeh

101.379 Seminar aus Analysis mit Seminararbeit SS2010

Introduction

Due to the restriction to (finite) linear combinations classical vector space bases are not always suitable for the analysis of infinite dimensional spaces. Therefore, it is natural in some way to consider generalized basis concepts.

Note, that all vector spaces in this paper are spaces over the field F, where F denotes the field of real numbersRor the field of complex numbersC. Whenever reference is made to some topological property, the norm topology is implied.

Definition 0.1 A sequence(x_n)_n∈

Nin a Banach space(X,k.k_X)is a called aSchauder basis of Xif for everyx∈Xthere exists a unique sequence(α_n)_n∈

Nof scalars such thatx=P∞

n=1α_nx_n, i.e. such thatlimN→∞

x−PN

n=1α_nx_n X = 0.

Throughout this paper we make the convention that a basis for a Banach space shall be a Schauder basis, unless explicit reference is made to a vector space basis.

Propostion 0.2 Suppose that(x_n)_n∈Nis a basis for a Banach spaceX. Then(x_n)_n∈Nis linearly independent. In particular, every banach space with a basis is infinite dimensional.

Proof: Suppose that an elementxofXcould be written in two different ways as a finite linear combination of the terms of(xn)_n∈N, i.e. forn, m∈N,(αi)ⁿ_i=1 ∈Fⁿ,(βi)^m_i=1 ∈F^msatisfyingαn6=

0 6= β_m and (α)ⁿ_i=1 6= (β_i)^m_i=1 we had x = Pn

i=1α_ix_i = Pm

i=1β_ix_i. Then we had P∞

i=1α˜_ix_i = P∞

i=1β˜ixi, for ( ˜αi)i∈N := (α1, . . . , αn,0,0, . . .) and ( ˜βi)i∈N := (β1, . . . , βm,0,0, . . .). This is a contradiction to the uniqueness of expansions of vectors in terms of a basis in X.

1 Coordinate Functionals

Our first aim is to show that the continuity of the coordinate functionals - a property that has been required in Schauder’s original defintion - follows from the rest of Definition 0.1.

Definition 1.1 Let X be a Banach space with basis (x_n)_n∈

N. For each m ∈ N the maps x^∗_m :X → F:P∞

n=1αnxn 7→ αm and Pm :X → X :P∞

n=1αnxn 7→ Pm

n=1αnxn are called the m^th coordinate functional and them^th natural projection associated with(xn)_n∈N, respectively.

Remark 1.2 Due to the uniqueness of expansions of each vector in terms of a basis required in definition 0.1 it is instantly verified, that the coordinate functionals actually are linear and the natural projections are projections.

For the sake of convenience we will not work with the original norm of the underlying Banach space, but with the following one.

Lemma 1.3 Let (X,k.k_X) be a Banach space with basis (xn)_n∈N. Then the norm k.k defined by the formula kP∞

n=1αnxnk = sup_m∈NkPm

n=1αnxnk_X is a Banach space norm equivalent to the normk.k_X (i.e. they induce the same topology) satisfying kxk ≥ kxk_X for all x∈X.

Proof: We prove this lemma in four steps. In step one and step two we show, thatk.kactually is a norm and the claimed inequality, respectively. In step three we find a limit in X for a random Cauchy sequence with respect tok.k inX. Finally, in step four we complete the proof by using the open mapping theorem to show the equivalence of the two norms.

Note that in this proof convergence of series in Xisalways meant to be with respect tok.k_X. 1. Since the other two requirements of the definition of a norm follow instantly from the fact thatk.k_X is a norm, we will confine ourselves to showing that the triangle inequality holds for k.k. For this purpose pick two vectors x and y in X having the expansions x=P∞

n=1αnxn and y=P∞

n=1α˜nxn. Then by the linearity of Pm

∞

X

n=1

αnxn+

∞

X

n=1

˜ αnxn

= sup

m∈N

m

X

n=1

αnxn+

m

X

n=1

˜ αnxn

X

≤ sup

m∈N

m

X

n=1

α_nx_n X

+

m

X

n=1

˜ α_nx_n

X

!

≤ sup

m∈N

m

X

n=1

αnxn

X

+ sup

m∈N

m

X

n=1

˜ αnxn

X

=

∞

X

n=1

α_nx_n

+

∞

X

n=1

˜ α_nx_n

= kxk+kyk

2. For a vector xinX with the expansion x=P∞

n=1αnxn we obtain from the continuity of the normk.k_X

kxk=

∞

X

n=1

α_nx_n

= sup

m∈N

m

X

n=1

α_nx_n X

≥ lim

n→∞

m

X

n=1

α_nx_n X

=

∞

X

n=1

α_nx_n X

.

3. We want to show next, that an arbitrary Cauchy sequence(bi)_i∈N= (P∞

n=1βn,ixn)_i∈

Nwith respect to k.k in X converges towards P∞

n=1βnxn in X, for βn := limi→∞βn,i, n ∈ N. In order to see that the sequence (β_n,i)_i∈

N is Cauchy and hence convergent in Ffor each n∈Nletj, k, n be inNand n≥2. Then we have

|β_1,j−β_1,k| kx₁k_X = k(β_1,j−β_1,k)·x₁k_X

≤ sup

m∈N

m

X

l=1

(βl,j−βl,k)·xl

X

=

∞

X

l=1

(βl,j−βl,k)·xl

and

|β_n,j −β_n,k| kx_nk_X = k(β_n,j −β_n,k)·xnk_X

=

n

X

l=1

(β_l,j−β_l,k)·x_l−

n−1

X

l=1

(β_l,j−β_l,k)·x_l X

≤

n

X

l=1

(βl,j−βl,k)·xl

X

+

n−1

X

l=1

(βl,j−βl,k)·xl

X

≤ sup

m∈N

m

X

l=1

(βl,j−βl,k)·xl

X

+ sup

m∈N

m

X

l=1

(βl,j−βl,k)·xl

X

= 2·

∞

X

l=1

(βl,j−βl,k)·xl

. As (P∞

n=1β_n,ix_n)_i∈

N is Cauchy so must be (β_n,i)_i∈

N for each n∈ N. Thus the sequence (βn)_n∈Nis well-defined.

Since (P∞

n=1β_n,ix_n)_i∈

N is Cauchy with respect to k.k we may choose i() ∈ N for each fixed >0such that for i, j, M ∈N, i, j≥i()

3 >

∞

X

n=1

β_n,jx_n−

∞

X

n=1

β_n,ix_n

= sup

m∈N

m

X

n=1

β_n,jx_n−

m

X

n=1

β_n,ix_n X

≥

M

X

n=1

β_n,jx_n−

M

X

n=1

β_n,ix_n X

The inequality holds true for j→ ∞. This way we obtain for alli≥i()

M

X

n=1

βnxn−

M

X

n=1

βn,ixn

X

≤

3. (1)

We will complete the proof of this step by showing, that (P∞

n=1β_n,ix_n)_i∈

N converges towards P∞

n=1βnxn. Therefore, it is necessary to show, that P∞

n=1βnxn exists. Due to the completeness of (X,k.k_X) it is sufficient to proof that P∞

n=1βnxn is Cauchy with respect tok.k_X.

Suppose thatm1, m2 ∈N, m2 ≥m1 >1. Using (1) we have

m2

X

n=m1

β_nx_n−

m2

X

n=m1

β_n,ix_n X

(2)

=

m2

X

n=m1

β_nx_n−

m2

X

n=m1

β_n,ix_n±

m1−1

X

n=1

β_nx_n±

m1−1

X

n=1

β_n,ix_n X

=

m2

X

n=1

β_nx_n−

m2

X

n=1

β_n,ix_n−

m1−1

X

n=1

β_nx_n+

m1−1

X

n=1

β_n,ix_n X

≤

m2

X

n=1

βnxn−

m2

X

n=1

βn,ixn

X

+

m1−1

X

n=1

βnxn−

m1−1

X

n=1

βn,ixn

X

≤ 3 +

3 = 2

3. (3)

As the series P∞

n=1β_n,ix_n is the expansion of a vector inX in terms of (x_n)_n∈

N, it must be convergent and so must be

P_N

n=1β_n,ix_{n X}

N∈N

. Therefore we can choosem()∈N such that

m2

X

n=m1

βn,ixn

X

<

3 (4)

for m2, m1 ∈ N, m2 ≥ m1 > m(). Finally (3) and (4) give us the Cauchy criterion for our seriesP∞

n=1β_nx_n:

m2

X

n=m1

β_nx_n X

=

m2

X

n=m1

β_nx_n±

m2

X

n=m1

β_n,ix_n X

≤

m2

X

n=m1

βnxn−

m2

X

n=m1

βn,ixn

X

+

m2

X

n=m1

βn,ixn

X

≤ 2 3 +

3 = . and the convergence ofP∞

n=1β_nx_n is proven.

Since inequality (1) still holds when taking the supremum over allM ∈Nwe see

∞

X

n=1

β_nx_n−

∞

X

n=1

β_n,ix_n

= sup

M∈N

M

X

n=1

β_nx_n−

M

X

n=1

β_n,ix_n X

≤ 3, which completes the proof of this step.

4. The identity mapI : (X,k.k) →(X,k.k_X) is a bijective, linear and due to the inequality, proofed in step 2, continuous operator. The open mapping theorem¹ ensures, that I⁻¹ : (X,k.k_X)→(X,k.k) is continuous too.

Theorem 1.4 Let(X,k.k_X)be a Banach space with basis(x_n)_n∈

N. Then all natural projections and all coordinate functionals associated with (xn)_n∈N are continuous.

Proof: Fixm inN and a member of X having the expansionP∞

n=1αnxn. Define a sequence ( ˜αn)n∈N by

˜ α_n=

(αn, n≤m 0, else.

Then the unique expansion of P_m(P∞

n=1α_nx_n) in terms of (x_n)_n∈

N is given by P∞

n=1α˜_nx_n. Now the continuity of the natural projection Pm for (xn)_n∈N follows from

Pm

∞

X

n=1

αnxn

!

=

∞

X

n=1

˜ αnxn

= sup

M∈N

M

X

n=1

˜ αnxn

X

= sup

M=1,...,m

M

X

n=1

˜ αnxn

X

= sup

M=1,...,m

M

X

n=1

αnxn

X

≤ sup

M∈N

M

X

n=1

αnxn

X

=

∞

X

n=1

αnxn

.

1see e.g. [1] theorem 12.1 and 12.5

For each m >1 the coordinate functional x^∗_m associated with (x_n)_n∈

N is continuous as it is the composition of continuous maps

∞

X

n=1

αnxn7→(Pm−Pm−1)

∞

X

n=1

αnxn

!

=αmxm7→αm

and so it is form= 1

∞

X

n=1

αnxn7→P1

∞

X

n=1

αnxn

!

=α1x17→α1.

2 Banach’s Basis Problem

In the following proposition we will see, that every Banach Space having a basis is separable.

The question whether the converse is true, i.e. whether every infinite dimensional separable Banach space has a basis, is known as the classicalbasis problem for Banach spaces. It remained open for forty years until Per Enflo found a counterexample in 1973.

Propostion 2.1 Every Banach spaceX with a basis(x_n)_n∈N is separable.

Proof: We want to show, that the countable set A:=

( _m X

n=1

α_nx_n:α₁, . . . , α_n∈Q˜, m∈N )

,

where Q˜ denotes the rational numbers or respectively the complex numbers with rational real and imaginary part, is dense in X. Since an arbitrary element of X having the expansion P∞

n=1α_nx_n in terms of (x_n)_n∈

N can be written as limN→∞PN

n=1α_nx_n, it follows that X = span({x_n:n∈N}). Therefore, it suffices to show thatAis a dense subset of span({x_n:n∈N}).

Fix k ∈ N, x₁, . . . , x_k ∈ {x_n : n∈ N} and α₁, . . . , α_k ∈F. Due to the fact, that Q˜ is dense inF, there is a sequence (αj,i)i∈N∈Q˜ⁿ converging towards αj for each j = 1, . . . , k. From the continuity of the vector space operations it follows, that

i→∞lim

k

X

j=1

αj,ixj =

k

X

j=1

αjxj,

i.e. an arbitrary element of span({x_n:n∈N}) can be written as the limit of a sequence inA, which completes the proof of this proposition.

In fact Enflo showed how to construct a separable infinite dimensional Banach space lacking the approximation property. Therefore, it is of some interest, that every Banach space with a basis has the approximation property.

Definition 2.2 Let X be a Banach space. X has the approximation property if the subset of operators of B(Y, X) having finite rank is dense in K(Y, X) for each Banach space Y, where B(Y, X) and K(Y, X) denote the set of all bounded linear operators and the set of all compact operators from Y intoX, respectively.

The following lemma gives us a sufficient condition for a Banach space to have the approximation property.

Lemma 2.3 Suppose that (X,k.k_X) is Banach space and that for each compact set C ⊂X and each > 0 there exists a bounded linear Operator A_C, from X into X having finite rank such that kA_C,x−xk_X < for eachx in C. Then X has the approximation property.

Proof: LetA be a compact operator from Y into X for an arbitrary Banach space Y. Since A(U) is a compact subset of X, when U denotes the closed unit ball inY, there is a sequence (B_n)_n∈

N of bounded linear operators from X intoX satisfying kB_nx−xk_X < ¹_n for each nin Nand eachx inA U

⊂A U

. For each ninNwe have kB_nA−Ak= sup{kB_nAy−Ayk_X :y∈Y,kyk ≤1}= sup

kB_nx−xk_X :x∈A U ≤ 1 n, which completes the proof of this lemma.

Theorem 2.4 Let (X,k.k_X) be a Banach space with basis (xn)_n∈N. Then X has the approxi- mation property.

Proof: Let C be a compact subset ofX and >0. By the preceding lemma it is sufficient to show that there is anN inN such thatkP_Nx−xk_X < for each x inC.

It follows readily from the uniform boundedness principle² that Π := sup_n∈NkP_nk is finite.

Due to the compactness ofCwe can pick finitely manyy₁, . . . , y_linCsuch thatmin_i=1,...,lkx−y_ik_X ≤

2(1+Π) for eachx inC.

Letx₀ be inC. Then there isj∈Nsuch thatkx₀−y_jk_X ≤ _2(1+C) and, since it follows from definition 0.1 that limn→∞ky_j −Pnyjk_X = 0, there is N ∈Nsuch that ky_j−Pnyjk_X ≤ ₂ for each n≥N. We conclude

kP_nx₀−x₀k_X = kP_nx₀−x₀±y_j ±P_ny_jk_X

≤ ky_j −x0k_X+ kP_nyj−Pnx0k_X

| {z }

≤kP_nkky_j−x₀k_X≤Πky_j−x₀k_X

+ky_j−Pnyjk_X

| {z }

2

≤ (1 + Π)ky_j −x₀k_X+

2 ≤ (1 + Π)

2(1 + Π) + 2 =

2+ 2 =

References

[1] John B. Conway, A course in functional analysis, Springer, New York, 2nd edition, 1990.

[2] Robert E. Megginson,An Introduction to Banach Space Theory, Springer, New York, 1998.

[3] Ivan Singer, Bases in Banach Spaces I, Springer, Berlin, 1970.

2see e.g. [1] theorem 14.1