C[0,1]: the Faber-Schauder basis, the Riesz representation theorem, and the Borel σ-algebra

Jordan Bell
July 24, 2015

1 Introduction

In this note I work out some results about the space of continuous functions [0,1]. In some cases these are instances of general results about continuous functions on compact Hausdorff spaces or compact metrizable spaces.

2 C(K)

Let K be a compact topological space (not necessarily Hausdorff) and let C(K) be the collection of continuous functions K. With the norm


this is a real Banach algebra, with unity 1:K defined by 1(t)=1 for all tK.11 1 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 124, Lemma 3.97. Generally, if X is a compact metrizable space and Y is a separable metrizable space then C(X,Y) is separable.22 2 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 125, Lemma 3.99. Thus, when K is a compact metrizable space, C(K) is a separable unital Banach algebra.

3 Schauder bases

If X is a real normed space, a Schauder basis for X is a sequence (hk) in X such that for each xX there is a unique sequence of real numbers (ck(x)) such that k=1nck(x)hkx. A sequence (hk) in X is called a basic sequence if it is a Schauder basis for the closure of its linear span.33 3 Because Schauder bases are not as familiar objects as orthonormal bases (Hilbert space bases) or vector space bases (Hamel bases), it is worth explicitly checking their properties to make ourselves familiar with how they work.

If (hk) is a Schauder basis, then ck(0)=0 for all k. Suppose that for some real numbers a1,,an we have k=1nakhk=0. Then ak=ck(0)=0 for 1kn, which means that the set {h1,,hn} is linearly independent. Therefore a Schauder basis is linearly independent. It is immediate that the linear span of a Schauder basis is a dense linear subspace of X. The following shows that a normed space with a Schauder basis is separable.44 4 In particular, a Banach space with a Schauder basis is separable. There is a celebrated counterexample found by Per Enflo of a separable Banach space for which there does not exist a Schauder basis.

Lemma 1.

If (hk) is a Schauder basis for a normed space X then


is dense in X.


Let xX and let ϵ>0. There is some n for which


For each 1kn, let ak satisfy |ak-ck(x)|ϵnhk. Then

k=1nakhk-x k=1n(ak-ck(x))hk+ϵ

which proves the claim. ∎

For each k we define hk*:X by hk*(x)=ck(x). For x,yX, k=1nhk*(x)hkx and k=1nhk*(y)hky, so k=1n(hk*(x)+hk*(y))hkx+y and therefore hk*(x+y)=hk*(x)+hk*(y), and for α, k=1nαhk*(x)hkαx and therefore hk*(αx)=αhk*(x). This shows that hk*:X is linear. It is apparent that hk*(hj)=δj,k.

We define Pn:XX by


and it is immediate that Pn is linear and that for each xX, Pnxx as n. It is also immediate that Pn is a finite-rank operator: Pn(X) is a finite-dimensional linear subspace of X. For m,n1,

PnPmx =Pnk=1mhk*(x)hk

showing that


In particular, Pn2=Pn, namely, Pn is a projection operator. We prove that when X is a Banach space, Pn is continuous.55 5 N. L. Carothers, A Short Course in Banach Space Theory, p. 26, Theorem 3.1. We indicate explicitly in the proof when we use that X is a Banach space rather than merely a normed space.

Theorem 2.

If X is a Banach space and (hn) is a Schauder basis for X, then each Pn is continuous, and supn1Pn<. Furthermore, each hk* is continuous.


For xX, Pnxx, so Pnxx, which implies that supn1Pnx<. It thus makes sense to define


It is immediate that p(αx)=|α|p(x) and that p(x+y)p(x)+p(y). If p(x)=0, then Pnx=0 for all n, which implies that x=0. Therefore p is a norm on X.

For n1 and xX,


showing that Pn:(X,p)(X,) is a bounded linear operator with operator norm 1. Suppose that (xk) is a Cauchy sequence in the norm p. Then we have for each n that Pnxk is a Cauchy sequence in the norm , and because (X,) is a Banach space, there is some ynX such that Pnxkyn in the norm as k. For ϵ>0 there is some kϵ such that when j,kkϵ, p(xj-xk)ϵ, and thus for kkϵ and for any n,

Pnxk-yn=limjPnxk-Pnxjlim supjp(xk-xj)ϵ. (1)

Now, for any m,n and any k,


so using the above, for kkϵ,


Because Pnxkxk as n, there is some nϵ such that Pnxk-xkϵ when nnϵ, and in this case for n,mnϵ,


which shows that yn is a Cauchy sequence in the norm , and because (X,) is a Banach space there is some yX such that yny in the norm .

For any n,m, the restriction of the linear map Pn:XX to the finite-dimensional linear subspace Pm(X) is continuous in the norm , thus

Pnym =Pn(limkPmxk)

Using this, we find by induction that for each n,


and because yny in the norm as n and (hk) is a Schauder basis, this implies that hk*(y)=hk*(yk) for all k, and thus Pny=yn. Therefore


For ϵ>0 and kkϵ, by (1) we have Pnxk-ynϵ and therefore p(xk-y)ϵ, which shows that xky in the norm p as k. Thus, (X,p) is a Banach space.

The identity map idX:XX is a linear isomorphism, and for xX,


showing that idX is continuous (X,p)(X,). Because (X,p) and (X,) are Banach spaces and idX is a continuous linear isomorphism, by the open mapping theorem66 6 Walter Rudin, Functional Analysis, second ed., p. 49, Corollary 2.12c. there is some a>0 such that




which means that Pn:(X,)(X,) is a bounded linear operator with operator norm 1a. ∎

Each Pn is a bounded linear operator on X, and for each xX, Pnxx, which means that PnidX in the strong operator topology. The fact that the functions hk*:X are linear and continuous means that they belong to the dual space X*.

In Theorem 2, if supn1Pn=1, we call the Schauder basis (xk) monotone.

The following theorem gives sufficient and necessary conditions under which a sequence in a Banach space X is a basic sequence.77 7 Joseph Diestel, Sequences and Series in Banach Spaces, p. 36, Chapter V, Theorem 1. Thus if a sequence satisfies this condition and its linear span is dense in X space, then it is a Schauder basis.

Theorem 3.

Suppose that X is a Banach space and that (xk) is a sequence of nonzero elements of X. There is some K such that for any sequence of real numbers (ak) and any n<N,

k=1nakxkKk=1Nakxk (2)

if and only if (xk) is a basic sequence.


It is apparent that K1. Let S be the linear span of (xk), and let Sn be the linear span of {x1,,xn}. For n<N,

|an|xnk=1n-1akxk+k=1nakxk2Kk=1Nakxk. (3)

Thus if k=1Nakxk=0 then using the above with aN+1=0, for 1nN we get an=0, showing that (xk) is linearly independent. Because (xk) is linearly independent, it makes sense to define a linear map Qn:SSn by


For x=k=1NakxkS, if n<N then by (2), QnxKx, and if nN then Qnx=x. Thus for each n, Qn:SSn is a bounded linear operator with operator norm K, and because X is a Banach space and S¯n=Sn, there is a unique bounded linear operator Pn:S¯Sn whose restriction to S is equal to Qn, which satisfies Pn=QnK.88 8 Gert K. Pedersen, Analysis Now, revised printing, p. 47, Proposition 2.1.11. For sS, QnQms=Qmin(m,n)s, and for xS¯, there is a sequence skS that tends to x, thus

PnPmx=limkPnPmsk=limkPmin(m,n)sk=Pmin(m,n)x. (4)

For xS¯ and ϵ>0, there is some sS with s-xϵ. Then there are some a1,,am for which s=i=1maixi, and for n>m,

x-Pnx x-s+s-Pns+Pns-Pnx

showing that Pnxx. It follows from this and (4) that for each xS¯ there is a sequence of real numbers (ck) such that k=1nckxkx. If (bk) is another such sequence, let ak=ck-bk, and then we obtain from (3) that ak=0 for each k. Therefore, (ck) is the unique sequence of real numbers such that k=1nckxkx, which establishes that (xk) is a Schauder basis for S¯. ∎

4 Haar system and Faber-Schauder system

For k0, for 1i2k, and for n=2k+i, write


and we write Δ1=Δ00=(0,1).99 9 We are partly following the presentation in B. S. Kashin and A. A. Saakyan, Orthogonal Series, p. 61, Chapter III. Thus


If (a,b)[0,1], let (a,b)-=(a,a+b2) and (a,b)+=(a+b2,b). Thus



Lemma 4.

If nm then either ΔnΔm or ΔnΔm=.


Let n=2k+i and m=2l+j, with kl. Then


There are three cases: (i) i2k-l(j-1), (ii) 2k-l(j-1)<i2k-lj, (iii) i>2k-lj. In the first case, ΔkiΔlj=, i.e. ΔnΔm=. In the second case, i-12k-l(j-1) and i2k-l, so ΔkiΔlj, i.e. ΔnΔm. In the third case, i-12k-lj so ΔkiΔlj=, i.e. ΔnΔm=. ∎

We define χ1=1, and for k0, 1i2k, and n=2k+i, we define


For example,






We call (χn) the Haar system. It is a fact that (χn) is a monotone Schauder basis for the Banach space L1[0,1] with the norm fL1=01|f(t)|𝑑t.1010 10 Joram Lindenstrauss and Lior Tzafriri, Classical Banach Spaces I and II, p. 3.

Now we define ϕ1=1 and for n>1 we define ϕn:[0,1] by


Each ϕn belongs to C[0,1], and we call (ϕn) the Faber-Schauder system. For example,








Generally for k0, for 1i2k, and for n=2k+i,


We remark that


5 Riesz representation theorem

Let (X,𝔐) be a measurable space. A signed measure is a function μ:𝔐[-,] such that (i) μ()=0, (ii) μ assumes at most one of the values -,, and (iii) if (Ej) is a sequence of disjoint elements of 𝔐 then μ(Ej)=μ(Ej). A finite signed measure is a signed measure whose image is contained in . We denote by ca(𝔐) the collection of all finite signed measures on 𝔐. For μ,λca(𝔐) and for c, define


for E𝔐, and we check that with addition and scalar multiplication thus defined ca(𝔐) is a real vector space. A positive measure is a signed measure whose imaged is contained in [0,]. For μ,λca(𝔐), we write


if μ-λ is a positive measure; in any case μ-λca(𝔐). We check that is a partial order on ca(𝔐) with which ca(𝔐) is an ordered vector space. Finally, a probability measure is a positive measure satisfying μ(X)=1, which in particular belongs to ca(𝔐).

For E𝔐, a partition of E is a countable subset {Ei} of 𝔐 whose members are pairwise disjoint. For μca(𝔐) and E𝔐 we define

|μ|(E)=sup{i=1|μ(Ei)|:{Ei} is a partition of E}.

It is immediate that |μ(E)||μ|(E). It is proved that |μ| is a finite positive measure on 𝔐, called the total variation measure of μ.1111 11 Walter Rudin, Real and Complex Analysis, third ed., pp. 117–118, Theorem 6.2 and Theorem 6.4. For μca(𝔐), define


Because |μ| is a finite positive measure and |μ(E)||μ(E)|, μ+ and μ- are finite positive measures, called the positive and negative variations of μ. Then μ=μ+-μ-, called the Jordan decomposition of μ.

For μca(𝔐), define


One checks that is a norm on ca(𝔐).

For a compact Hausdorff space K and for f,gC(K), write gf when (g-f)(t)0 for all tK. A positive linear functional is a linear map ϕ:C(K) such that ϕ(f)0 when f0. In this case, because f1+f0,


i.e. -ϕ(f)ϕ(1)f, and because f1-f0,


i.e. ϕ(f)ϕ(1)f, showing because 1=1 that the operator norm of ϕ is ϕ=ϕ(1), and in particular that ϕC(K)*.

For normed spaces (V,V) and (W,W), an isometric isomorphism from V to W is a linear isomorphism T:VW satisfying TvW=vV for all vV. The simplest version of the Riesz representation theorem is for compact metrizable spaces, for which we do not need to speak about regular Borel measures or continuous functions vanishing at infinity.1212 12 Walter Rudin, Real and Complex Analysis, third ed., p. 130, Theorem 6.19.

Theorem 5 (Riesz representation theorem for compact metrizable spaces).

Let K be a compact metrizable space and define Λ:ca(K)C(K)* by


Λ is an isometric isomorphism, and is order preserving: if μ0 then Λ(μ)C(K)* is a positive linear functional, and thus if μλ then Λ(μ)Λ(λ).

6 The Borel σ-algebra of C[0,1]

Let I=[0,1], with the relative topology inherited from , with which I is a compact metric space. For t1,,tnI, we define πt1,,tn:C(I)n by


which is continuous.

For a set X and a collection of functions ft:X, the coarsest σ-algebra on X such that each ft is measurable X, where has the Borel σ-algebra, is called the σ-algebra generated by {ft:tI}, and is denoted by σ({ft:tI}). We show that the Borel σ-algebra of C(I) is equal to the σ-algebra generated by the family of projection maps C(I).1313 13 K. R. Parthasarathy, Probability Measures on Metric Spaces, p. 212, Theorem 2.1.

Theorem 6.

Let 𝒜 be the σ-algebra generated by the family {πt:t[0,1]}. Then C[0,1]=𝒜.


Because C(I) is a separable metric space it is second-countable, and so if U is an open subset of C(I) then U is equal to the union of countably many open balls. Each open ball is equal to the union of countably many closed balls: B(x,r)=B(x,r-1/n)¯. Therefore each open subset of C(I) is equal to the union of countably many closed balls, and to prove that C(I)𝒜 it suffices to prove that all closed balls belong to 𝒜. To this end, let q1,q2, be an enumeration of [0,1], let xC(I), and let r>0. Suppose that |y(qn)-x(qn)|r for all n, and take t[0,1]. Then there is a subsequence qan of qn that tends to t, and because y-x:[0,1] is continuous, |y(qan)-x(qan)||y(t)-x(t)|, and then because for each n we have |y(qan)-x(qan)|r it follows that |y(t)-x(t)|r. This establishes




which belongs to 𝒜. Thus B(x,r)¯ is a countable intersection of elements of 𝒜 and so belongs to 𝒜, which shows that C(I)𝒜.

On the other hand, for each tI the map πt:C(I) is continuous and hence is measurable C(I).1414 14 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 140, Corollary 4.26. Therefore 𝒜C(I). ∎

7 Relatively compact subsets of C[0,1]

If (M,d) is a metric space, a subset A of M is called totally bounded if for each ϵ>0 there are finitely many points x1,,xnM such that for any point xM there is some i for which d(x,xi)<ϵ. It is immediate that a compact metric space is totally bounded, and the Heine-Borel theorem states that a metric space is compact if and only if it is complete and totally bounded.1515 15 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 86, Theorem 3.28. On the other hand, one checks that if (M,d) is a metric space and A is a totally bounded subset of M, then the closure A¯ is a totally bounded subset of M. Thus, if A is a totally bounded subset of a complete metric space (M,d), then the closure A¯ is itself a complete metric space (because (M,d) is a complete metric space), and because A¯ is complete and totally bounded, by the Heine-Borel theorem it is compact.

If X is a topological space and is a subset of C(X), we say that is equicontinuous at xX if for each ϵ>0 there is a neighborhood Ux,ϵ of x such that |f(x)-f(y)|<ϵ for all f and for all yUx,ϵ, and we call equicontinuous if it is equicontinuous at each xX. We call pointwise bounded if for each xX, {f(x):f} is a bounded subset of . The Arzelà-Ascoli theorem states that for a compact Hausdorff space X and for C(X), is equicontinuous and pointwise bounded if and only if is totally bounded.1616 16 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second ed., p. 137, Theorem 4.43. Then the closure ¯ is totally bounded and is itself a complete metric space, and hence is a compact metric space. That is, for a compact Hausdorff space X, is equicontinuous and pointwise bounded if and only if is relatively compact in C(X).

For xC[0,1] and δ>0, define


For xC[0,1] because x:I is continuous and I is compact, x is uniformly continuous on I. Thus for ϵ>0 there is some δϵ>0 such that when |s-t|δϵ, |x(s)-x(t)|ϵ, hence if δδϵ then ωx(δ)ϵ, i.e. ωx(δ)0 as δ0. We give sufficient and necessary conditions for a subset of C[0,1] to be relatively compact.

Lemma 7.

Let AC[0,1]. A is relatively compact if and only if

supxA|x(0)|< (5)


limδ0supxAωx(δ)=0. (6)

For tI and for r>0, let


which is an open neighborhood of t. The Arzelà-Ascoli theorem tells us that A is relatively compact if and only if A is equicontinuous and pointwise bounded. If A is relatively compact, then being pointwise bounded yields (5). Let ϵ>0. Because A is equicontinuous, for each tI there is some δt>0 such that for all xA and for all sBδt(t) we have |x(s)-x(t)|<ϵ/2. Then because I is compact, there are t1,,tnI such that I=i=1nBδti(ti). Let δϵ>0 be the Lebesgue number for this open cover: for each tI there is some i for which Bδϵ(t)Bδti(ti). For 0<δδϵ, for xA, and for |s-t|δ/2, there is some i for which Bδ(t)Bδti(ti), so s,tBδti(ti), which means that |x(s)-x(ti)|<ϵ/2 and |x(t)-x(ti)|<ϵ/2, and thus |x(s)-x(t)||x(s)-x(ti)|+|x(ti)-x(t)|<ϵ. Therefore ωδ/2(x)ϵ, and this is true for all xA which proves (6).

Suppose now that (5) and (6) are true. By (6), there is some m1 such that supxAωx(1/m)1, and therefore for xA, x|x(0)|+m. With (5) this yields supxAx<, whence A is pointwise bounded. Let tI and let ϵ>0. By (6) there is some δ>0 such that supxAωx(δ)<ϵ. Thus for xA and for |s-t|δ, |x(s)-x(t)|<ϵ, which shows that A is equicontinuous at t. This is true for all tI, so A is equicontinuous and by the Arzelà-Ascoli theorem we get that A is relatively compact in C[0,1]. ∎

8 Continuously differentiable functions

Let f:[0,1] be a function. We say that f is differentiable at t[0,1] if there is some f(t) such that1717 17 cf. Nicolas Bourbaki, Elements of Mathematics. Functions of a Real Variable: Elementary Theory, p. 3, Chapter I, §1, no. 1, Definition 1.


If f is differentiable at each t[0,1], we say that f is differentiable on [0,1] and define f:[0,1] by tf(t). We define C1[0,1] to be the collection of those f:[0,1] such that f is differentiable on [0,1] and f:[0,1] is continuous. C1[0,1] is contained in C[0,1], and it turns out that C1[0,1] is a Borel set in C[0,1].1818 18 Alexander S. Kechris, Classical Descriptive Set Theory, p. 70, §11.B, Example 2. On the other hand, the collection differentiable functions [0,1], which is a subset of C[0,1], is not a Borel set in C[0,1].1919 19 S. M. Srivastava, A Course on Borel Sets, p. 139, Proposition 4.2.7.