# $C[0,1]$: the Faber-Schauder basis, the Riesz representation theorem, and the Borel $\sigma$-algebra

Jordan Bell
July 24, 2015

## 1 Introduction

In this note I work out some results about the space of continuous functions $[0,1]\to\mathbb{R}$. 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 $K\to\mathbb{R}$. With the norm

 $\left\|f\right\|=\sup_{t\in K}|f(t)|,\qquad f\in C(K),$

this is a real Banach algebra, with unity $1:K\to\mathbb{R}$ defined by $1(t)=1$ for all $t\in K$.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 $(h_{k})$ in $X$ such that for each $x\in X$ there is a unique sequence of real numbers $(c_{k}(x))$ such that $\sum_{k=1}^{n}c_{k}(x)h_{k}\to x$. A sequence $(h_{k})$ 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 $(h_{k})$ is a Schauder basis, then $c_{k}(0)=0$ for all $k$. Suppose that for some real numbers $a_{1},\ldots,a_{n}$ we have $\sum_{k=1}^{n}a_{k}h_{k}=0$. Then $a_{k}=c_{k}(0)=0$ for $1\leq k\leq n$, which means that the set $\{h_{1},\ldots,h_{n}\}$ 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 $(h_{k})$ is a Schauder basis for a normed space $X$ then

 $\left\{\sum_{k=1}^{n}a_{k}h_{k}:n\geq 1,a_{1},\ldots,a_{n}\in\mathbb{Q}\right\}$

is dense in $X$.

###### Proof.

Let $x\in X$ and let $\epsilon>0$. There is some $n$ for which

 $\left\|\sum_{k=1}^{n}c_{k}(x)h_{k}-x\right\|\leq\epsilon.$

For each $1\leq k\leq n$, let $a_{k}\in\mathbb{Q}$ satisfy $|a_{k}-c_{k}(x)|\leq\frac{\epsilon}{n\left\|h_{k}\right\|}$. Then

 $\displaystyle\left\|\sum_{k=1}^{n}a_{k}h_{k}-x\right\|$ $\displaystyle\leq\left\|\sum_{k=1}^{n}(a_{k}-c_{k}(x))h_{k}\right\|+\epsilon$ $\displaystyle\leq\sum_{k=1}^{n}|a_{k}-c_{k}(x)|\left\|h_{k}\right\|+\epsilon$ $\displaystyle\leq 2\epsilon,$

which proves the claim. ∎

For each $k$ we define $h_{k}^{*}:X\to\mathbb{R}$ by $h_{k}^{*}(x)=c_{k}(x)$. For $x,y\in X$, $\sum_{k=1}^{n}h_{k}^{*}(x)h_{k}\to x$ and $\sum_{k=1}^{n}h_{k}^{*}(y)h_{k}\to y$, so $\sum_{k=1}^{n}(h_{k}^{*}(x)+h_{k}^{*}(y))h_{k}\to x+y$ and therefore $h_{k}^{*}(x+y)=h_{k}^{*}(x)+h_{k}^{*}(y)$, and for $\alpha\in\mathbb{R}$, $\sum_{k=1}^{n}\alpha h_{k}^{*}(x)h_{k}\to\alpha x$ and therefore $h_{k}^{*}(\alpha x)=\alpha h_{k}^{*}(x)$. This shows that $h_{k}^{*}:X\to\mathbb{R}$ is linear. It is apparent that $h_{k}^{*}(h_{j})=\delta_{j,k}$.

We define $P_{n}:X\to X$ by

 $P_{n}x=\sum_{k=1}^{n}h_{k}^{*}(x)h_{k},$

and it is immediate that $P_{n}$ is linear and that for each $x\in X$, $P_{n}x\to x$ as $n\to\infty$. It is also immediate that $P_{n}$ is a finite-rank operator: $P_{n}(X)$ is a finite-dimensional linear subspace of $X$. For $m,n\geq 1$,

 $\displaystyle P_{n}P_{m}x$ $\displaystyle=P_{n}\sum_{k=1}^{m}h_{k}^{*}(x)h_{k}$ $\displaystyle=\sum_{k=1}^{m}h_{k}^{*}(x)P_{n}h_{k}$ $\displaystyle=\sum_{k=1}^{m}h_{k}^{*}(x)\sum_{j=1}^{n}h_{j}^{*}(h_{k})h_{j}$ $\displaystyle=\sum_{k=1}^{m}h_{k}^{*}(x)\sum_{j=1}^{n}\delta_{j,k}h_{j}$ $\displaystyle=P_{\min(m,n)}x,$

showing that

 $P_{n}P_{m}=P_{\min(m,n)}.$

In particular, $P_{n}^{2}=P_{n}$, namely, $P_{n}$ is a projection operator. We prove that when $X$ is a Banach space, $P_{n}$ 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 $(h_{n})$ is a Schauder basis for $X$, then each $P_{n}$ is continuous, and $\sup_{n\geq 1}\left\|P_{n}\right\|<\infty$. Furthermore, each $h_{k}^{*}$ is continuous.

###### Proof.

For $x\in X$, $P_{n}x\to x$, so $\left\|P_{n}x\right\|\to\left\|x\right\|$, which implies that $\sup_{n\geq 1}\left\|P_{n}x\right\|<\infty$. It thus makes sense to define

 $p(x)=\sup_{n\geq 1}\left\|P_{n}x\right\|.$

It is immediate that $p(\alpha x)=|\alpha|p(x)$ and that $p(x+y)\leq p(x)+p(y)$. If $p(x)=0$, then $P_{n}x=0$ for all $n$, which implies that $x=0$. Therefore $p$ is a norm on $X$.

For $n\geq 1$ and $x\in X$,

 $\left\|P_{n}x\right\|\leq p(x),$

showing that $P_{n}:(X,p)\to(X,\left\|\cdot\right\|)$ is a bounded linear operator with operator norm $\leq 1$. Suppose that $(x_{k})$ is a Cauchy sequence in the norm $p$. Then we have for each $n$ that $P_{n}x_{k}$ is a Cauchy sequence in the norm $\left\|\cdot\right\|$, and because $(X,\left\|\cdot\right\|)$ is a Banach space, there is some $y_{n}\in X$ such that $P_{n}x_{k}\to y_{n}$ in the norm $\left\|\cdot\right\|$ as $k\to\infty$. For $\epsilon>0$ there is some $k_{\epsilon}$ such that when $j,k\geq k_{\epsilon}$, $p(x_{j}-x_{k})\leq\epsilon$, and thus for $k\geq k_{\epsilon}$ and for any $n$,

 $\left\|P_{n}x_{k}-y_{n}\right\|=\lim_{j\to\infty}\left\|P_{n}x_{k}-P_{n}x_{j}% \right\|\leq\limsup_{j\to\infty}p(x_{k}-x_{j})\leq\epsilon.$ (1)

Now, for any $m,n$ and any $k$,

 $\left\|y_{n}-y_{m}\right\|\leq\left\|y_{n}-P_{n}x_{k}\right\|+\left\|P_{n}x_{k% }-x_{k}\right\|+\left\|P_{m}x_{k}-x_{k}\right\|+\left\|P_{m}x_{k}-y_{m}\right\|,$

so using the above, for $k\geq k_{\epsilon}$,

 $\left\|y_{n}-y_{m}\right\|\leq 2\epsilon+\left\|P_{n}x_{k}-x_{k}\right\|+\left% \|P_{m}x_{k}-x_{k}\right\|.$

Because $P_{n}x_{k}\to x_{k}$ as $n\to\infty$, there is some $n_{\epsilon}$ such that $\left\|P_{n}x_{k}-x_{k}\right\|\leq\epsilon$ when $n\geq n_{\epsilon}$, and in this case for $n,m\geq n_{\epsilon}$,

 $\left\|y_{n}-y_{m}\right\|\leq 4\epsilon,$

which shows that $y_{n}$ is a Cauchy sequence in the norm $\left\|\cdot\right\|$, and because $(X,\left\|\cdot\right\|)$ is a Banach space there is some $y\in X$ such that $y_{n}\to y$ in the norm $\left\|\cdot\right\|$.

For any $n,m$, the restriction of the linear map $P_{n}:X\to X$ to the finite-dimensional linear subspace $P_{m}(X)$ is continuous in the norm $\left\|\cdot\right\|$, thus

 $\displaystyle P_{n}y_{m}$ $\displaystyle=P_{n}\left(\lim_{k\to\infty}P_{m}x_{k}\right)$ $\displaystyle=\lim_{k\to\infty}P_{n}P_{m}x_{k}$ $\displaystyle=\lim_{k\to\infty}P_{\min(m,n)}x_{k}$ $\displaystyle=y_{\min(m,n)}.$

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

 $y_{n}=\sum_{k=1}^{n}h_{k}^{*}(y_{k})h_{k},$

and because $y_{n}\to y$ in the norm $\left\|\cdot\right\|$ as $n\to\infty$ and $(h_{k})$ is a Schauder basis, this implies that $h_{k}^{*}(y)=h_{k}^{*}(y_{k})$ for all $k$, and thus $P_{n}y=y_{n}$. Therefore

 $p(x_{k}-y)=\sup_{n\geq 1}\left\|P_{n}(x_{k}-y)\right\|=\sup_{n\geq 1}\left\|P_% {n}x_{k}-y_{n}\right\|.$

For $\epsilon>0$ and $k\geq k_{\epsilon}$, by (1) we have $\left\|P_{n}x_{k}-y_{n}\right\|\leq\epsilon$ and therefore $p(x_{k}-y)\leq\epsilon$, which shows that $x_{k}\to y$ in the norm $p$ as $k\to\infty$. Thus, $(X,p)$ is a Banach space.

The identity map $\mathrm{id}_{X}:X\to X$ is a linear isomorphism, and for $x\in X$,

 $\left\|\mathrm{id}_{X}x\right\|=\left\|x\right\|=\lim_{n\to\infty}\left\|P_{n}% x\right\|\leq p(x),$

showing that $\mathrm{id}_{X}$ is continuous $(X,p)\to(X,\left\|\cdot\right\|)$. Because $(X,p)$ and $(X,\left\|\cdot\right\|)$ are Banach spaces and $\mathrm{id}_{X}$ 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

 $\left\|x\right\|=\left\|\mathrm{id}_{X}\right\|\geq ap(x),\qquad x\in X.$

Then

 $\left\|P_{n}x\right\|\leq p(x)\leq\frac{1}{a}\left\|x\right\|,$

which means that $P_{n}:(X,\left\|\cdot\right\|)\to(X,\left\|\cdot\right\|)$ is a bounded linear operator with operator norm $\leq\frac{1}{a}$. ∎

Each $P_{n}$ is a bounded linear operator on $X$, and for each $x\in X$, $P_{n}x\to x$, which means that $P_{n}\to\mathrm{id}_{X}$ in the strong operator topology. The fact that the functions $h_{k}^{*}:X\to\mathbb{R}$ are linear and continuous means that they belong to the dual space $X^{*}$.

In Theorem 2, if $\sup_{n\geq 1}\left\|P_{n}\right\|=1$, we call the Schauder basis $(x_{k})$ 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 $(x_{k})$ is a sequence of nonzero elements of $X$. There is some $K$ such that for any sequence of real numbers $(a_{k})$ and any $n,

 $\left\|\sum_{k=1}^{n}a_{k}x_{k}\right\|\leq K\left\|\sum_{k=1}^{N}a_{k}x_{k}\right\|$ (2)

if and only if $(x_{k})$ is a basic sequence.

###### Proof.

It is apparent that $K\geq 1$. Let $S$ be the linear span of $(x_{k})$, and let $S_{n}$ be the linear span of $\{x_{1},\ldots,x_{n}\}$. For $n,

 $|a_{n}|\left\|x_{n}\right\|\leq\left\|\sum_{k=1}^{n-1}a_{k}x_{k}\right\|+\left% \|\sum_{k=1}^{n}a_{k}x_{k}\right\|\leq 2K\left\|\sum_{k=1}^{N}a_{k}x_{k}\right\|.$ (3)

Thus if $\sum_{k=1}^{N}a_{k}x_{k}=0$ then using the above with $a_{N+1}=0$, for $1\leq n\leq N$ we get $a_{n}=0$, showing that $(x_{k})$ is linearly independent. Because $(x_{k})$ is linearly independent, it makes sense to define a linear map $Q_{n}:S\to S_{n}$ by

 $Q_{n}x_{k}=\begin{cases}x_{k}&k\leq n\\ 0&k>n.\end{cases}$

For $x=\sum_{k=1}^{N}a_{k}x_{k}\in S$, if $n then by (2), $\left\|Q_{n}x\right\|\leq K\left\|x\right\|$, and if $n\geq N$ then $Q_{n}x=x$. Thus for each $n$, $Q_{n}:S\to S_{n}$ is a bounded linear operator with operator norm $\leq K$, and because $X$ is a Banach space and $\overline{S}_{n}=S_{n}$, there is a unique bounded linear operator $P_{n}:\overline{S}\to S_{n}$ whose restriction to $S$ is equal to $Q_{n}$, which satisfies $\left\|P_{n}\right\|=\left\|Q_{n}\right\|\leq K$.88 8 Gert K. Pedersen, Analysis Now, revised printing, p. 47, Proposition 2.1.11. For $s\in S$, $Q_{n}Q_{m}s=Q_{\min(m,n)}s$, and for $x\in\overline{S}$, there is a sequence $s_{k}\in S$ that tends to $x$, thus

 $P_{n}P_{m}x=\lim_{k\to\infty}P_{n}P_{m}s_{k}=\lim_{k\to\infty}P_{\min(m,n)}s_{% k}=P_{\min(m,n)}x.$ (4)

For $x\in\overline{S}$ and $\epsilon>0$, there is some $s\in S$ with $\left\|s-x\right\|\leq\epsilon$. Then there are some $a_{1},\ldots,a_{m}\in\mathbb{R}$ for which $s=\sum_{i=1}^{m}a_{i}x_{i}$, and for $n>m$,

 $\displaystyle\left\|x-P_{n}x\right\|$ $\displaystyle\leq\left\|x-s\right\|+\left\|s-P_{n}s\right\|+\left\|P_{n}s-P_{n% }x\right\|$ $\displaystyle=\left\|x-s\right\|+\left\|P_{n}s-P_{n}x\right\|$ $\displaystyle\leq\epsilon+K\left\|s-x\right\|$ $\displaystyle\leq(K+1)\epsilon,$

showing that $P_{n}x\to x$. It follows from this and (4) that for each $x\in\overline{S}$ there is a sequence of real numbers $(c_{k})$ such that $\sum_{k=1}^{n}c_{k}x_{k}\to x$. If $(b_{k})$ is another such sequence, let $a_{k}=c_{k}-b_{k}$, and then we obtain from (3) that $a_{k}=0$ for each $k$. Therefore, $(c_{k})$ is the unique sequence of real numbers such that $\sum_{k=1}^{n}c_{k}x_{k}\to x$, which establishes that $(x_{k})$ is a Schauder basis for $\overline{S}$. ∎

## 4 Haar system and Faber-Schauder system

For $k\geq 0$, for $1\leq i\leq 2^{k}$, and for $n=2^{k}+i$, write

 $\Delta_{n}=\Delta_{k}^{i}=\left(\frac{i-1}{2^{k}},\frac{i}{2^{k}}\right),$

and we write $\Delta_{1}=\Delta_{0}^{0}=(0,1)$.99 9 We are partly following the presentation in B. S. Kashin and A. A. Saakyan, Orthogonal Series, p. 61, Chapter III. Thus

 $\Delta_{1}=\Delta_{0}^{0}=(0,1),\;\Delta_{2}=\Delta_{0}^{1}=(0,1),\;\Delta_{3}% =\Delta_{1}^{1}=\Big{(}0,\frac{1}{2}\Big{)},\;\Delta_{4}=\Delta_{1}^{2}=\Big{(% }\frac{1}{2},1\Big{)}.$

If $(a,b)\subset[0,1]$, let $(a,b)^{-}=\left(a,\frac{a+b}{2}\right)$ and $(a,b)^{+}=\left(\frac{a+b}{2},b\right)$. Thus

 $\Delta_{n}^{-}=(\Delta_{k}^{i})^{-}=\left(\frac{i-1}{2^{k}},\frac{2i-1}{2^{k+1% }}\right)=\left(\frac{2i-2}{2^{k+1}},\frac{2i-1}{2^{k+1}}\right)=\Delta_{k+1}^% {2i-1}$

and

 $\Delta_{n}^{+}=(\Delta_{k}^{i})^{+}=\left(\frac{2i-1}{2^{k+1}},\frac{i}{2^{k}}% \right)=\left(\frac{2i-1}{2^{k+1}},\frac{2i}{2^{k+1}}\right)=\Delta_{k+1}^{2i}.$
###### Lemma 4.

If $n\geq m$ then either $\Delta_{n}\subset\Delta_{m}$ or $\Delta_{n}\cap\Delta_{m}=\emptyset$.

###### Proof.

Let $n=2^{k}+i$ and $m=2^{l}+j$, with $k\geq l$. Then

 $\Delta_{l}^{j}=\left(\frac{j-1}{2^{l}},\frac{j}{2^{l}}\right)=\left(\frac{2^{k% -l}(j-1)}{2^{k}},\frac{2^{k-l}j}{2^{k}}\right).$

There are three cases: (i) $i\leq 2^{k-l}(j-1)$, (ii) $2^{k-l}(j-1), (iii) $i>2^{k-l}j$. In the first case, $\Delta_{k}^{i}\cap\Delta_{l}^{j}=\emptyset$, i.e. $\Delta_{n}\cap\Delta_{m}=\emptyset$. In the second case, $i-1\geq 2^{k-l}(j-1)$ and $i\leq 2^{k-l}$, so $\Delta_{k}^{i}\subset\Delta_{l}^{j}$, i.e. $\Delta_{n}\subset\Delta_{m}$. In the third case, $i-1\geq 2^{k-l}j$ so $\Delta_{k}^{i}\cap\Delta_{l}^{j}=\emptyset$, i.e. $\Delta_{n}\cap\Delta_{m}=\emptyset$. ∎

We define $\chi_{1}=1$, and for $k\geq 0$, $1\leq i\leq 2^{k}$, and $n=2^{k}+i$, we define

 $\chi_{n}(t)=\begin{cases}1&t\in\Delta_{n}^{-}=\left(\frac{2i-2}{2^{k+1}},\frac% {2i-1}{2^{k+1}}\right)\\ -1&t\in\Delta_{n}^{+}=\left(\frac{2i-1}{2^{k+1}},\frac{2i}{2^{k+1}}\right)\\ 0&\textrm{otherwise}.\end{cases}$

For example,

 $\chi_{2}(t)=\begin{cases}1&t\in\left(0,\frac{1}{2}\right)\\ -1&t\in\left(\frac{1}{2},1\right)\\ 0&\textrm{otherwise}\end{cases}$

and

 $\chi_{3}(t)=\begin{cases}1&t\in\left(0,\frac{1}{4}\right)\\ -1&t\in\left(\frac{1}{4},\frac{1}{2}\right)\\ 0&\textrm{otherwise}\end{cases}$

and

 $\chi_{4}(t)=\begin{cases}1&t\in\left(\frac{1}{2},\frac{3}{4}\right)\\ -1&t\in\left(\frac{3}{4},1\right)\\ 0&\textrm{otherwise}.\end{cases}$

We call $(\chi_{n})$ the Haar system. It is a fact that $(\chi_{n})$ is a monotone Schauder basis for the Banach space $L^{1}[0,1]$ with the norm $\left\|f\right\|_{L^{1}}=\int_{0}^{1}|f(t)|dt$.1010 10 Joram Lindenstrauss and Lior Tzafriri, Classical Banach Spaces I and II, p. 3.

Now we define $\phi_{1}=1$ and for $n>1$ we define $\phi_{n}:[0,1]\to\mathbb{R}$ by

 $\phi_{n}(t)=\int_{0}^{t}\chi_{n-1}(u)du.$

Each $\phi_{n}$ belongs to $C[0,1]$, and we call $(\phi_{n})$ the Faber-Schauder system. For example,

 $\phi_{2}(t)=\int_{0}^{t}\chi_{1}(u)du=t,$

and

 $\phi_{3}(t)=\begin{cases}\int_{0}^{t}1du&t\in\left[0,\frac{1}{2}\right]\\ \int_{0}^{1/2}1du+\int_{1/2}^{t}-1du&t\in\left[\frac{1}{2},1\right]\end{cases}% =\begin{cases}t&t\in\left[0,\frac{1}{2}\right]\\ -t+1&t\in\left[\frac{1}{2},1\right]\end{cases}$

and

 $\phi_{4}(t)=\begin{cases}\int_{0}^{t}1du&t\in\left[0,\frac{1}{4}\right]\\ \int_{0}^{1/4}1du+\int_{1/4}^{t}-1du&t\in\left[\frac{1}{4},\frac{1}{2}\right]% \\ 0&t\in\left[\frac{1}{2},1\right]\end{cases}=\begin{cases}t&t\in\left[0,\frac{1% }{4}\right]\\ -t+\frac{1}{2}&t\in\left[\frac{1}{4},\frac{1}{2}\right]\\ 0&t\in\left[\frac{1}{2},1\right]\end{cases}$

and

 $\phi_{5}(t)=\begin{cases}0&t\in\left[0,\frac{1}{2}\right]\\ \int_{1/2}^{t}1du&t\in\left[\frac{1}{2},\frac{3}{4}\right]\\ \int_{1/2}^{3/4}1du+\int_{3/4}^{t}-1du&t\in\left[\frac{3}{4},1\right]\end{% cases}=\begin{cases}0&t\in\left[0,\frac{1}{2}\right]\\ t-\frac{1}{2}&t\in\left[\frac{1}{2},\frac{3}{4}\right]\\ -t+1&t\in\left[\frac{3}{4},1\right].\end{cases}$

Generally for $k\geq 0$, for $1\leq i\leq 2^{k}$, and for $n=2^{k}+i$,

 $\phi_{n+1}(t)=\begin{cases}t-\frac{i-1}{2^{k}}&t\in\Delta_{n}^{-}=\left(\frac{% i-1}{2^{k}},\frac{2i-1}{2^{k+1}}\right)\\ -t+\frac{i}{2^{k}}&t\in\Delta_{n}^{+}=\left(\frac{2i-1}{2^{k+1}},\frac{i}{2^{k% }}\right)\\ 0&\textrm{otherwise}.\end{cases}$

We remark that

 $\left\|\phi_{n+1}\right\|=\frac{2i-1}{2^{k+1}}-\frac{i-1}{2^{k}}=2^{-k-1}.$

## 5 Riesz representation theorem

Let $(X,\mathfrak{M})$ be a measurable space. A signed measure is a function $\mu:\mathfrak{M}\to[-\infty,\infty]$ such that (i) $\mu(\emptyset)=0$, (ii) $\mu$ assumes at most one of the values $-\infty,\infty$, and (iii) if $(E_{j})$ is a sequence of disjoint elements of $\mathfrak{M}$ then $\mu\left(\bigcup E_{j}\right)=\sum\mu(E_{j})$. A finite signed measure is a signed measure whose image is contained in $\mathbb{R}$. We denote by $\mathrm{ca}(\mathfrak{M})$ the collection of all finite signed measures on $\mathfrak{M}$. For $\mu,\lambda\in\mathrm{ca}(\mathfrak{M})$ and for $c\in\mathbb{R}$, define

 $(\mu+\lambda)(E)=\mu(E)+\lambda(E),\qquad(c\mu)(E)=c\mu(E)$

for $E\in\mathfrak{M}$, and we check that with addition and scalar multiplication thus defined $\mathrm{ca}(\mathfrak{M})$ is a real vector space. A positive measure is a signed measure whose imaged is contained in $[0,\infty]$. For $\mu,\lambda\in\mathrm{ca}(\mathfrak{M})$, we write

 $\mu\geq\lambda$

if $\mu-\lambda$ is a positive measure; in any case $\mu-\lambda\in\mathrm{ca}(\mathfrak{M})$. We check that $\leq$ is a partial order on $\mathrm{ca}(\mathfrak{M})$ with which $\mathrm{ca}(\mathfrak{M})$ is an ordered vector space. Finally, a probability measure is a positive measure satisfying $\mu(X)=1$, which in particular belongs to $\mathrm{ca}(\mathfrak{M})$.

For $E\in\mathfrak{M}$, a partition of $E$ is a countable subset $\{E_{i}\}$ of $\mathfrak{M}$ whose members are pairwise disjoint. For $\mu\in\mathrm{ca}(\mathfrak{M})$ and $E\in\mathfrak{M}$ we define

 $|\mu|(E)=\sup\left\{\sum_{i=1}^{\infty}|\mu(E_{i})|:\textrm{\{E_{i}\} is a % partition of E}\right\}.$

It is immediate that $|\mu(E)|\leq|\mu|(E)$. It is proved that $|\mu|$ is a finite positive measure on $\mathfrak{M}$, called the total variation measure of $\mu$.1111 11 Walter Rudin, Real and Complex Analysis, third ed., pp. 117–118, Theorem 6.2 and Theorem 6.4. For $\mu\in\mathrm{ca}(\mathfrak{M})$, define

 $\mu^{+}=\frac{1}{2}(|\mu|+\mu),\qquad\mu^{-}=\frac{1}{2}(|\mu|-\mu).$

Because $|\mu|$ is a finite positive measure and $|\mu(E)|\leq|\mu(E)|$, $\mu^{+}$ and $\mu^{-}$ are finite positive measures, called the positive and negative variations of $\mu$. Then $\mu=\mu^{+}-\mu^{-}$, called the Jordan decomposition of $\mu$.

For $\mu\in\mathrm{ca}(\mathfrak{M})$, define

 $\left\|\mu\right\|=|\mu|(X)=\mu^{+}(X)+\mu^{-}(X).$

One checks that $\left\|\cdot\right\|$ is a norm on $\mathrm{ca}(\mathfrak{M})$.

For a compact Hausdorff space $K$ and for $f,g\in C(K)$, write $g\geq f$ when $(g-f)(t)\geq 0$ for all $t\in K$. A positive linear functional is a linear map $\phi:C(K)\to\mathbb{R}$ such that $\phi(f)\geq 0$ when $f\geq 0$. In this case, because $\left\|f\right\|\cdot 1+f\geq 0$,

 $\left\|f\right\|\phi(1)+\phi(f)=\phi(\left\|f\right\|\cdot 1+f)\geq 0$

i.e. $-\phi(f)\leq\phi(1)\left\|f\right\|$, and because $\left\|f\right\|\cdot 1-f\geq 0$,

 $\left\|f\right\|\phi(1)-\phi(f)=\phi(\left\|f\right\|\cdot 1-f)\geq 0,$

i.e. $\phi(f)\leq\phi(1)\left\|f\right\|$, showing because $\left\|1\right\|=1$ that the operator norm of $\phi$ is $\left\|\phi\right\|=\phi(1)$, and in particular that $\phi\in C(K)^{*}$.

For normed spaces $(V,\left\|\cdot\right\|_{V})$ and $(W,\left\|\cdot\right\|_{W})$, an isometric isomorphism from $V$ to $W$ is a linear isomorphism $T:V\to W$ satisfying $\left\|Tv\right\|_{W}=\left\|v\right\|_{V}$ for all $v\in V$. 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 $\Lambda:\mathrm{ca}(\mathscr{B}_{K})\to C(K)^{*}$ by

 $\Lambda(\mu)(f)=\int_{K}fd\mu,\qquad\mu\in\mathrm{ca}(\mathscr{B}_{K}),\quad f% \in C(K).$

$\Lambda$ is an isometric isomorphism, and is order preserving: if $\mu\geq 0$ then $\Lambda(\mu)\in C(K)^{*}$ is a positive linear functional, and thus if $\mu\geq\lambda$ then $\Lambda(\mu)\geq\Lambda(\lambda)$.

## 6 The Borel $\sigma$-algebra of $C[0,1]$

Let $I=[0,1]$, with the relative topology inherited from $\mathbb{R}$, with which $I$ is a compact metric space. For $t_{1},\ldots,t_{n}\in I$, we define $\pi_{t_{1},\ldots,t_{n}}:C(I)\to\mathbb{R}^{n}$ by

 $\pi_{t_{1},\ldots,t_{n}}(x)=(x(t_{1}),\ldots,x(t_{n})),\qquad x\in C(I),$

which is continuous.

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

###### Theorem 6.

Let $\mathscr{A}$ be the $\sigma$-algebra generated by the family $\{\pi_{t}:t\in[0,1]\}$. Then $\mathscr{B}_{C[0,1]}=\mathscr{A}$.

###### Proof.

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: $B(x,r)=\bigcup\overline{B(x,r-1/n)}$. Therefore each open subset of $C(I)$ is equal to the union of countably many closed balls, and to prove that $\mathscr{B}_{C(I)}\subset\mathscr{A}$ it suffices to prove that all closed balls belong to $\mathscr{A}$. To this end, let $q_{1},q_{2},\ldots$ be an enumeration of $[0,1]\cap\mathbb{Q}$, let $x\in C(I)$, and let $r>0$. Suppose that $|y(q_{n})-x(q_{n})|\leq r$ for all $n$, and take $t\in[0,1]$. Then there is a subsequence $q_{a_{n}}$ of $q_{n}$ that tends to $t$, and because $y-x:[0,1]\to\mathbb{R}$ is continuous, $|y(q_{a_{n}})-x(q_{a_{n}})|\to|y(t)-x(t)|$, and then because for each $n$ we have $|y(q_{a_{n}})-x(q_{a_{n}})|\leq r$ it follows that $|y(t)-x(t)|\leq r$. This establishes

 $\{y\in C(I):\left\|y-x\right\|\leq r\}=\bigcap_{n=1}^{\infty}\{y\in C(I):|y(q_% {n})-x(q_{n})|\leq r\}.$

But

 $\{y\in C(I):|y(q_{n})-x(q_{n})|\leq r\}=\pi_{q_{n}}^{-1}([\pi_{q_{n}}(x)-r,\pi% _{q_{n}}(x)+r]),$

which belongs to $\mathscr{A}$. Thus $\overline{B(x,r)}$ is a countable intersection of elements of $\mathscr{A}$ and so belongs to $\mathscr{A}$, which shows that $\mathscr{B}_{C(I)}\subset\mathscr{A}$.

On the other hand, for each $t\in I$ the map $\pi_{t}:C(I)\to\mathbb{R}$ is continuous and hence is measurable $\mathscr{B}_{C(I)}\to\mathscr{B}_{\mathbb{R}}$.1414 14 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 140, Corollary 4.26. Therefore $\mathscr{A}\subset\mathscr{B}_{C(I)}$. ∎

## 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 $\epsilon>0$ there are finitely many points $x_{1},\ldots,x_{n}\in M$ such that for any point $x\in M$ there is some $i$ for which $d(x,x_{i})<\epsilon$. 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 $\overline{A}$ 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 $\overline{A}$ is itself a complete metric space (because $(M,d)$ is a complete metric space), and because $\overline{A}$ is complete and totally bounded, by the Heine-Borel theorem it is compact.

If $X$ is a topological space and $\mathscr{F}$ is a subset of $C(X)$, we say that $\mathscr{F}$ is equicontinuous at $x\in X$ if for each $\epsilon>0$ there is a neighborhood $U_{x,\epsilon}$ of $x$ such that $|f(x)-f(y)|<\epsilon$ for all $f\in\mathscr{F}$ and for all $y\in U_{x,\epsilon}$, and we call $\mathscr{F}$ equicontinuous if it is equicontinuous at each $x\in X$. We call $\mathscr{F}$ pointwise bounded if for each $x\in X$, $\{f(x):f\in\mathscr{F}\}$ is a bounded subset of $\mathbb{R}$. The Arzelà-Ascoli theorem states that for a compact Hausdorff space $X$ and for $\mathscr{F}\subset C(X)$, $\mathscr{F}$ is equicontinuous and pointwise bounded if and only if $\mathscr{F}$ 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 $\overline{\mathscr{F}}$ is totally bounded and is itself a complete metric space, and hence is a compact metric space. That is, for a compact Hausdorff space $X$, $\mathscr{F}$ is equicontinuous and pointwise bounded if and only if $\mathscr{F}$ is relatively compact in $C(X)$.

For $x\in C[0,1]$ and $\delta>0$, define

 $\omega_{x}(\delta)=\sup_{s,t\in I,|s-t|\leq\delta}|x(s)-x(t)|.$

For $x\in C[0,1]$ because $x:I\to\mathbb{R}$ is continuous and $I$ is compact, $x$ is uniformly continuous on $I$. Thus for $\epsilon>0$ there is some $\delta_{\epsilon}>0$ such that when $|s-t|\leq\delta_{\epsilon}$, $|x(s)-x(t)|\leq\epsilon$, hence if $\delta\leq\delta_{\epsilon}$ then $\omega_{x}(\delta)\leq\epsilon$, i.e. $\omega_{x}(\delta)\to 0$ as $\delta\to 0$. We give sufficient and necessary conditions for a subset of $C[0,1]$ to be relatively compact.

###### Lemma 7.

Let $A\subset C[0,1]$. $A$ is relatively compact if and only if

 $\sup_{x\in A}|x(0)|<\infty$ (5)

and

 $\lim_{\delta\downarrow 0}\sup_{x\in A}\omega_{x}(\delta)=0.$ (6)
###### Proof.

For $t\in I$ and for $r>0$, let

 $B_{r}(t)=\{s\in I:|s-t|

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 $\epsilon>0$. Because $A$ is equicontinuous, for each $t\in I$ there is some $\delta_{t}>0$ such that for all $x\in A$ and for all $s\in B_{\delta_{t}}(t)$ we have $|x(s)-x(t)|<\epsilon/2$. Then because $I$ is compact, there are $t_{1},\ldots,t_{n}\in I$ such that $I=\bigcup_{i=1}^{n}B_{\delta_{t_{i}}}(t_{i})$. Let $\delta_{\epsilon}>0$ be the Lebesgue number for this open cover: for each $t\in I$ there is some $i$ for which $B_{\delta_{\epsilon}}(t)\subset B_{\delta_{t_{i}}}(t_{i})$. For $0<\delta\leq\delta_{\epsilon}$, for $x\in A$, and for $|s-t|\leq\delta/2$, there is some $i$ for which $B_{\delta}(t)\subset B_{\delta_{t_{i}}}(t_{i})$, so $s,t\in B_{\delta_{t_{i}}}(t_{i})$, which means that $|x(s)-x(t_{i})|<\epsilon/2$ and $|x(t)-x(t_{i})|<\epsilon/2$, and thus $|x(s)-x(t)|\leq|x(s)-x(t_{i})|+|x(t_{i})-x(t)|<\epsilon$. Therefore $\omega_{\delta/2}(x)\leq\epsilon$, and this is true for all $x\in A$ which proves (6).

Suppose now that (5) and (6) are true. By (6), there is some $m\geq 1$ such that $\sup_{x\in A}\omega_{x}(1/m)\leq 1$, and therefore for $x\in A$, $\left\|x\right\|\leq|x(0)|+m$. With (5) this yields $\sup_{x\in A}\left\|x\right\|<\infty$, whence $A$ is pointwise bounded. Let $t\in I$ and let $\epsilon>0$. By (6) there is some $\delta>0$ such that $\sup_{x\in A}\omega_{x}(\delta)<\epsilon$. Thus for $x\in A$ and for $|s-t|\leq\delta$, $|x(s)-x(t)|<\epsilon$, which shows that $A$ is equicontinuous at $t$. This is true for all $t\in I$, 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 $f:[0,1]\to\mathbb{R}$ be a function. We say that $f$ is differentiable at $t\in[0,1]$ if there is some $f^{\prime}(t)\in\mathbb{R}$ 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.

 $\lim_{s\to t,s\in[0,1]\setminus\{t\}}\frac{f(s)-f(t)}{s-t}=f^{\prime}(t).$

If $f$ is differentiable at each $t\in[0,1]$, we say that $f$ is differentiable on $[0,1]$ and define $f^{\prime}:[0,1]\to\mathbb{R}$ by $t\mapsto f^{\prime}(t)$. We define $C^{1}[0,1]$ to be the collection of those $f:[0,1]\to\mathbb{R}$ such that $f$ is differentiable on $[0,1]$ and $f^{\prime}:[0,1]\to\mathbb{R}$ is continuous. $C^{1}[0,1]$ is contained in $C[0,1]$, and it turns out that $C^{1}[0,1]$ 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 $[0,1]\to\mathbb{R}$, 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.