The Banach algebra of functions of bounded variation and the pointwise Helly selection theorem

Jordan Bell
January 22, 2015

1 $BV[a,b]$

Let $a. For $f:[a,b]\to\mathbb{R}$, we define11 1 In this note we speak about functions that take values in $\mathbb{R}$, because this makes it simpler to talk about monotone functions. Once the machinery is established we can then apply it to the real and imaginary parts of a function that takes values in $\mathbb{C}$.

 $\left\|f\right\|_{\infty}=\sup_{t\in[a,b]}|f(t)|,$

and if $\left\|f\right\|_{\infty}<\infty$ we say that $f$ is bounded. We define $B[a,b]$ to be the set of bounded functions $[a,b]\to\mathbb{R}$, which with the norm $\left\|\cdot\right\|_{\infty}$ is a Banach algebra.

A partition of $[a,b]$ is a set $P=\{t_{0},\ldots,t_{n}\}$ such that $a=t_{0}<\cdots. For example, $P=\{a,b\}$ is a partition of $[a,b]$. If $Q$ is a partition of $[a,b]$ and $P\subset Q$, we say that $Q$ is a refinement of $P$. For $f:[a,b]\to\mathbb{R}$, we define

 $V(f,P)=\sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|.$

It is straightforward to show using the triangle inequality that if $Q$ is a refinement of $P$ then

 $V(f,P)\leq V(f,Q).$

In particular, any partition $P$ is a refinement of $\{a,b\}$, so

 $|f(b)-f(a)|\leq V(f,P).$

The total variation of $f:[a,b]\to\mathbb{R}$ is

 $V_{a}^{b}f=\sup\{V(f,P):\textrm{P is a partition of [a,b]}\},$

and if $V_{a}^{b}f<\infty$ we say that $f$ is of bounded variation. We denote by $BV[a,b]$ the set of functions $[a,b]\to\mathbb{R}$ of bounded variation. For a function $f\in BV[a,b]$, we define $v:[a,b]\to\mathbb{R}$ by $v(x)=V_{a}^{x}f$ for $x\in[a,b]$, called the variation of $f$.

If $f:[a,b]\to\mathbb{R}$ is monotone, it is straightforward to check that $V_{a}^{b}f=|f(b)-f(a)|$, hence that $f$ is of bounded variation.

We first show that $BV[a,b]\subset B[a,b]$.

Lemma 1.

If $f:[a,b]\to\mathbb{R}$ is of bounded variation, then

 $\left\|f\right\|_{\infty}\leq|f(a)|+V_{a}^{b}f.$
Proof.

Let $x\in[a,b]$ If $x=a$ the result is immediate. If $x=b$, then

 $|f(b)|\leq|f(a)|+|f(b)-f(a)|\leq|f(a)|+V_{a}^{b}f.$

Otherwise, $P=\{a,x,b\}$ is a partition of $[a,b]$ and

 $|f(x)-f(a)|\leq V(f,P)\leq V_{a}^{b}f.$

The total variation of functions has several properties. The following lemma and that fact that functions of bounded variation are bounded imply that $BV[a,b]$ is an algebra.22 2 N. L. Carothers, Real Analysis, p. 204, Lemma 13.3.

Lemma 2.

If $f,g\in BV[a,b]$ and $c\in\mathbb{R}$, then the following statements are true.

1. 1.

$V_{a}^{b}f=0$ if and only if $f$ is constant.

2. 2.

$V_{a}^{b}(cf)=|c|V_{a}^{b}(f)$.

3. 3.

$V_{a}^{b}(f+g)\leq V_{a}^{b}f+V_{a}^{b}g$.

4. 4.

$V_{a}^{b}(fg)\leq\left\|f\right\|_{\infty}V_{a}^{b}g+\left\|g\right\|_{\infty}% V_{a}^{b}f$.

5. 5.

$V_{a}^{b}|f|\leq V_{a}^{b}f$.

6. 6.

$V_{a}^{b}f=V_{a}^{x}f+V_{x}^{b}$ for $a\leq x\leq b$.

Lemma 3.

If $f:[a,b]\to\mathbb{R}$ is differentiable on $(a,b)$ and $\left\|f^{\prime}\right\|_{\infty}<\infty$, then

 $V_{a}^{b}f\leq\left\|f^{\prime}\right\|_{\infty}(b-a).$
Proof.

Suppose that $P=\{a=t_{0}<\cdots is a partition of $[a,b]$. By the mean value theorem, for each $j=1,\ldots,n$ there is some $x_{j}\in(t_{j-1},t_{j})$ at which

 $f^{\prime}(x_{j})=\frac{f(t_{j})-f(t_{j-1})}{t_{j}-t_{j-1}}.$

Then

 $\displaystyle V(f,P)$ $\displaystyle=\sum_{j=1}^{n}|f(t_{j})-f(t_{j-1})|$ $\displaystyle=\sum_{j=1}^{n}(t_{j}-t_{j-1})|f^{\prime}(x_{j})|$ $\displaystyle\leq\left\|f^{\prime}\right\|_{\infty}\sum_{j=1}^{n}(t_{j}-t_{j-1})$ $\displaystyle=\left\|f^{\prime}\right\|_{\infty}(b-a).$

Lemma 4.

If $f\in C^{1}[a,b]$, then

 $V_{a}^{b}f\leq\int_{a}^{b}|f^{\prime}(t)|dt.$
Proof.

Let $P=\{t_{0},\ldots,t_{n}\}$ be a partition of $[a,b]$. Then, by the fundamental theorem of calculus,

 $\displaystyle V(f,P)$ $\displaystyle=\sum_{j=1}^{n}|f(t_{j})-f(t_{j-1})|$ $\displaystyle\leq\sum_{j=1}^{n}\left|\int_{t_{j-1}}^{t_{j}}f^{\prime}(t)\right|$ $\displaystyle\leq\sum_{j=1}^{n}\int_{t_{j-1}}^{t_{j}}|f^{\prime}(t)|dt$ $\displaystyle=\int_{a}^{b}|f^{\prime}(t)|dt.$

Therefore

 $V_{a}^{b}f=\sup_{P}V(f,P)\leq\int_{a}^{b}|f^{\prime}(t)|dt.$

Lemma 5.

If $f:[a,b]\to\mathbb{R}$ is a polynomial, then

 $V_{a}^{b}f=\int_{a}^{b}|f^{\prime}(t)|dt.$
Proof.

Because $f$ is a polynomial, $f$ is also, so $f^{\prime}$ is piecewise monotone, say $f^{\prime}=c_{j}|f^{\prime}|$ on $(t_{j-1},t_{j})$ for $j=1,\ldots,n$, for some $c_{j}\in\{+1,-1\}$ and $a=t_{0}<\cdots. Then

 $\int_{t_{j-1}}^{t_{j}}|f^{\prime}(t)|dt=c_{j}\int_{t_{j-1}}^{t_{j}}f^{\prime}(% t)dt=c_{j}(f(t_{j})-f(t_{j-1})),$

giving, because $t_{0}<\cdots is a partition of $[a,b]$,

 $\displaystyle\int_{a}^{b}|f^{\prime}(t)|dt$ $\displaystyle=\sum_{j=1}^{n}\int_{t_{j-1}}^{t_{j}}|f^{\prime}(t)|dt$ $\displaystyle=\sum_{j=1}^{n}c_{j}(f(t_{j})-f(t_{j-1}))$ $\displaystyle\leq\sum_{j=1}^{n}|f(t_{j})-f(t_{j-1})|$ $\displaystyle\leq V_{a}^{b}f.$

Lemma 6.

If $f_{m}$ is a sequence of functions $[a,b]\to\mathbb{R}$ that converges pointwise to some $f:[a,b]\to\mathbb{R}$ and $P$ is some partition of $[a,b]$, then

 $V(f_{m},P)\to V(f,P).$

If $f_{m}$ is a sequence in $BV[a,b]$ that converges pointwise to some $f:[a,b]\to\mathbb{R}$, then

 $V_{a}^{b}f\leq\liminf_{m\to\infty}V_{a}^{b}f_{m}.$
Proof.

Say $P=\{t_{0},\ldots,t_{n}\}$. Then, because taking the limit of convergent sequences is linear,

 $\displaystyle\lim_{m\to\infty}V(f_{m},P)$ $\displaystyle=\lim_{m\to\infty}\sum_{j=1}^{n}|f_{m}(t_{j})-f_{m}(t_{j-1})|$ $\displaystyle=\sum_{j=1}^{n}\lim_{m\to\infty}|f_{m}(t_{j})-f_{m}(t_{j-1})|$ $\displaystyle=\sum_{j=1}^{n}|f(t_{j})-f(t_{j-1})|$ $\displaystyle=V(f,P).$

Let $P=\{t_{0},\ldots,t_{n}\}$ be a partition of $[a,b]$. Then

 $\displaystyle V(f,P)$ $\displaystyle=\sum_{j=1}^{n}|f(t_{j})-f(t_{j-1})|$ $\displaystyle=\sum_{j=1}^{n}\lim_{m\to\infty}|f_{m}(t_{j})-f_{m}(t_{j-1})|$ $\displaystyle=\lim_{m\to\infty}V(f_{m},P)$ $\displaystyle\leq\liminf_{m\to\infty}V_{a}^{b}f_{m}.$

This is true for any partition $P$ of $[a,b]$, which yields

 $V_{a}^{b}f\leq\liminf_{m\to\infty}V_{a}^{b}f_{m}.$

We now prove that $BV[a,b]$ is a Banach space.33 3 N. L. Carothers, Real Analysis, p. 206, Theorem 13.4.

Theorem 7.

With the norm

 $\left\|f\right\|_{BV}=|f(a)|+V_{a}^{b}f.$

$BV[a,b]$ is a Banach space.

Proof.

Using Lemma 2, it is straightforward to check that $BV[a,b]$ is a normed linear space. Suppose that $f_{m}$ is a Cauchy sequence in $BV[a,b]$. By Lemma 1 it follows that $f_{m}$ is a Cauchy sequence in $B[a,b]$, and thus converges in $B[a,b]$ to some $f\in B[a,b]$.

Let $P$ be a partition of $[a,b]$ and let $\epsilon>0$. Because $f_{n}$ is a Cauchy sequence in $BV[a,b]$, there is some $N$ such that if $n,m\geq N$ then $\left\|f_{m}-f_{n}\right\|_{BV}<\epsilon$. For $n\geq N$, Lemma 6 yields

 $\displaystyle\left\|f-f_{n}\right\|_{BV}$ $\displaystyle\leq|f(a)-f_{n}(a)|+V(f-f_{n},P)$ $\displaystyle=\lim_{m\to\infty}\left(|f_{m}(a)-f_{n}(a)|+V(f_{m}-f_{n},P)\right)$ $\displaystyle\leq\sup_{m\geq N}\left(|f_{m}(a)-f_{n}(a)|+V(f_{m}-f_{n},P)\right)$ $\displaystyle=\sup_{m\geq N}\left\|f_{m}-f_{n}\right\|_{BV}$ $\displaystyle\leq\epsilon.$

Because $f-f_{N}\in BV[a,b]$ and $f_{N}\in BV[a,b]$ and $BV[a,b]$ is an algebra, $f=(f-f_{N})+f_{N}\in BV[a,b]$. That is, the Cauchy sequence $f_{n}$ converges in $BV[a,b]$ to $f\in BV[a,b]$, showing that $BV[a,b]$ is a complete metric space and thus a Banach space. ∎

The following theorem shows that a function of bounded of variation can be written as the difference of nondecreasing functions.44 4 N. L. Carothers, Real Analysis, p. 207, Theorem 13.5.

Theorem 8.

Let $f\in BV[a,b]$ and let $v$ be the variation of $f$. Then $v-f$ and $v$ are nondecreasing.

Proof.

If $x,y\in[a,b]$, $x, then, using Lemma 2,

 $\displaystyle v(y)-v(x)$ $\displaystyle=V_{a}^{y}f-V_{a}^{x}f$ $\displaystyle=V_{x}^{y}f$ $\displaystyle\geq|f(y)-f(x)|$ $\displaystyle\geq f(y)-f(x).$

That is, $v(y)-f(y)\geq v(x)-f(x)$, showing that $v-f$ is nondecreasing, and because $f$ is nondecreasing we have $f(y)-f(x)\geq 0$ and so $v(y)-v(x)\geq 0$. ∎

The following theorem tells us that a function of bounded variation is right or left continuous at a point if and only if its variation is respectively right or left continuous at the point.55 5 N. L. Carothers, Real Analysis, p. 207, Theorem 13.9.

Theorem 9.

Let $f\in BV[a,b]$ and let $v$ be the variation of $f$. For $x\in[a,b]$, $f$ is right (respectively left) continuous at $x$ if and only if $v$ is right (respectively left) continuous at $x$.

Proof.

Assume that $v$ is right continuous at $x$. If $\epsilon>0$, there is some $\delta>0$ such that $x\leq y implies that $v(y)-v(x)=|v(y)-v(x)|<\epsilon$. If $x\leq y, then

 $|f(y)-f(x)|\leq v(y)-v(x)<\epsilon,$

showing that $f$ is right continuous at $x$.

Assume that $f$ is right continuous at $x$, with $a\leq x. Let $\epsilon>0$. There is some $\delta>0$ such that $x\leq y implies that $|f(y)-f(x)|<\frac{\epsilon}{2}$. Because $V_{x}^{b}f$ is a supremum over partitions of $[x,b]$, there is some partition $P=\{t_{0},t_{1},\ldots,t_{n}\}$ of $[x,b]$ such that $V_{x}^{b}f-\frac{\epsilon}{2}\leq V(f,P)$. Let $x\leq y<\min\{\delta,t_{1}-x\}$. Then $Q=\{t_{0},y,t_{1},\ldots,t_{n}\}$ is a refinement of $P$, so

 $\displaystyle V_{x}^{b}f-\frac{\epsilon}{2}$ $\displaystyle\leq V(f,P)$ $\displaystyle\leq V(f,Q)$ $\displaystyle=|f(y)-f(t_{0})|+V(f,\{y,t_{1},\ldots,t_{n}\})$ $\displaystyle<\frac{\epsilon}{2}+V_{y}^{b}f.$

Hence

 $\epsilon>V_{x}^{b}f-V_{y}^{b}f=V_{x}^{y}f=v(y)-v(x)=|v(y)-v(x)|,$

showing that $v$ is right continuous at $x$. ∎

For $f\in BV[a,b]$ and for $v$ the variation of $f$, we define the positive variation of $f$ as

 $p(x)=\frac{v(x)+f(x)-f(a)}{2},\qquad x\in[a,b],$

and the negative variation of $f$ as

 $n(x)=\frac{v(x)-f(x)+f(a)}{2},\qquad x\in[a,b].$

We can write the variation as $v=p+n$. We now establish properties of the positive and negative variations.66 6 N. L. Carothers, Real Analysis, p. 209, Proposition 13.11.

Theorem 10.

Let $f\in BV[a,b]$, let $v$ be its variation, let $p$ be its positive variation, and let $n$ be its negative variation. Then $0\leq p\leq v$ and $0\leq n\leq v$, and $p$ and $n$ are nondecreasing.

Proof.

For $x\in[a,b]$, $v(x)=V_{a}^{x}f\geq|f(x)-f(a)|$. Because $v(x)\geq-(f(x)-f(a))$, we have $p(x)\geq 0$, and because $v(x)\geq f(x)-f(a)$ we have $n(x)\geq 0$. And then $v=p+n$ implies that $p\leq v$ and $n\leq v$.

For $x,

 $\displaystyle p(y)-p(x)$ $\displaystyle=\frac{v(y)+f(y)-v(x)-f(x)}{2}$ $\displaystyle=\frac{1}{2}\left(V_{x}^{y}f+(f(y)-f(x))\right)$ $\displaystyle\geq\frac{1}{2}\left(|f(y)-f(x)|+(f(y)-f(x))\right)$ $\displaystyle\geq 0$

and

 $\displaystyle n(y)-n(x)$ $\displaystyle=\frac{v(y)-f(y)-v(x)+f(x)}{2}$ $\displaystyle=\frac{1}{2}\left(V_{x}^{y}f-(f(y)-f(x))\right)$ $\displaystyle\geq\frac{1}{2}\left(|f(y)-f(x)|-(f(y)-f(x))\right)$ $\displaystyle\geq 0.$

We now prove that $BV[a,b]$ is a Banach algebra.77 7 N. L. Carothers, Real Analysis, p. 209, Proposition 13.12.

Theorem 11.

$BV[a,b]$ is a Banach algebra.

Proof.

For $f_{1},f_{2}\in BV[a,b]$, let $v_{1},v_{2}$, $p_{1},p_{2}$, $n_{1},n_{2}$ be their variations, positive variations, and negative variations, respectively. Then

 $\displaystyle f_{1}f_{2}$ $\displaystyle=(f_{1}(a)+p_{1}-n_{1})(f_{2}(a)+p_{2}-n_{2})$ $\displaystyle=f_{1}(a)f_{2}(a)+p_{1}p_{2}+n_{1}n_{2}-n_{1}p_{2}-n_{2}p_{1}$ $\displaystyle+f_{1}(a)p_{2}+f_{2}(a)p_{1}-f_{1}(a)n_{2}-f_{2}(a)n_{1}.$

Using this and the fact that if $f$ is nondecreasing then $V_{a}^{b}f=f(b)-f(a)$,

 $\displaystyle\left\|f_{1}f_{2}\right\|_{BV}$ $\displaystyle=|f_{1}(a)||f_{2}(a)|+V_{a}^{b}(f_{1}f_{2})$ $\displaystyle\leq|f_{1}(a)||f_{2}(a)|+V_{a}^{b}(p_{1}p_{2})+V_{a}^{b}(n_{1}n_{% 2})+V_{a}^{b}(n_{1}p_{2})+V_{a}^{b}(n_{2}p_{1})$ $\displaystyle+|f_{1}(a)|V_{a}^{b}p_{2}+|f_{2}(a)|V_{a}^{b}p_{1}+|f_{1}(a)|V_{a% }^{b}n_{2}+|f_{2}(a)|V_{a}^{b}n_{1}$ $\displaystyle=|f_{1}(a)||f_{2}(a)|+p_{1}(b)p_{2}(b)+n_{1}(b)n_{2}(b)+n_{1}(b)p% _{2}(b)+n_{2}(b)p_{1}(b)$ $\displaystyle+|f_{1}(a)|p_{2}(b)+|f_{2}(a)|p_{1}(b)+|f_{1}(a)|n_{2}(b)+|f_{2}(% a)|n_{1}(b)$ $\displaystyle=(|f_{1}(a)|+p_{1}(b)+n_{1}(b))(|f_{2}(a)|+p_{2}(b)+n_{2}(b))$ $\displaystyle=(|f_{1}(a)+v_{1}(b))(|f_{2}(a)|+v_{2}(b))$ $\displaystyle=\left\|f_{1}\right\|_{BV}\left\|f_{2}\right\|_{BV},$

which shows that $BV[a,b]$ is a normed algebra. And $BV[a,b]$ is a Banach space, so $BV[a,b]$ is a Banach algebra. ∎

Theorem 12.

If $f\in C^{1}[a,b]$, then

 $V_{a}^{b}f=\int_{a}^{b}|f^{\prime}(t)|dt.$

Let $(f^{\prime})^{+}$ and $(f^{\prime})^{-}$ be the positive and negative parts of $f^{\prime}$ and let $p$ and $n$ be the positive and negative variations of $f$. Then, for $x\in[a,b]$,

 $p(x)=\int_{a}^{x}(f^{\prime})^{+}(t)dt,\qquad n(x)=\int_{a}^{x}(f^{\prime})^{-% }(t)dt.$
Proof.

Lemma 4 states that $V_{a}^{b}\leq\int_{a}^{b}|f^{\prime}(t)|dt$. Because $f^{\prime}$ is continuous it is Riemann integrable, hence for any $\epsilon>0$ there is some partition $P=\{t_{0},\ldots,t_{n}\}$ of $[a,b]$ such that if $x_{j}\in[t_{j-1},t_{j}]$ for $j=1,\ldots,n$ then

 $\left|\int_{a}^{b}|f^{\prime}(t)|dt-\sum_{j=1}^{n}|f^{\prime}(x_{j})|(t_{j}-t_% {j-1})\right|<\epsilon.$

By the mean value theorem, for each $j=1,\ldots,n$ there is some $x_{j}\in(t_{j-1},t_{j})$ such that $f^{\prime}(x_{j})=\frac{f(t_{j})-f(t_{j-1})}{t_{j}-t_{j-1}}$. Then

 $V(f,P)=\sum_{j=1}^{n}|f(t_{j})-f(t_{j-1})|=\sum_{j=1}^{n}|f^{\prime}(x_{j})|(t% _{j}-t_{j-1}),$

so

 $\left|\int_{a}^{b}|f^{\prime}(t)|dt-V(f,P)\right|<\epsilon,$

and thus

 $\int_{a}^{b}|f^{\prime}(t)|dt

This is true for all $\epsilon>0$, therefore

 $\int_{a}^{b}|f^{\prime}(t)|dt\leq V_{a}^{b}f,$

which is what we wanted to show.

Write

 $g(t)=(f^{\prime})^{+}(t)=\max\{f^{\prime}(t),0\},\qquad h(t)=(f^{\prime})^{-}(% t)=-\min\{f^{\prime}(t),0\}.$

These satisfy $g+h=|f^{\prime}|$ and $g-h=f^{\prime}$. Using the fundamental theorem of calculus,

 $\displaystyle p(x)$ $\displaystyle=\frac{1}{2}\left(v(x)+f(x)-f(a)\right)$ $\displaystyle=\frac{1}{2}\left(V_{a}^{x}f+\int_{a}^{x}f^{\prime}(t)dt\right)$ $\displaystyle=\frac{1}{2}\left(\int_{a}^{b}|f^{\prime}(t)|dt+\int_{a}^{b}f^{% \prime}(t)dt\right)$ $\displaystyle=\int_{a}^{b}g(t)dt$

and

 $\displaystyle n(x)$ $\displaystyle=\frac{1}{2}\left(v(x)-f(x)+f(a)\right)$ $\displaystyle=\frac{1}{2}\left(V_{a}^{x}f-\int_{a}^{x}f^{\prime}(t)dt\right)$ $\displaystyle=\frac{1}{2}\left(\int_{a}^{x}|f^{\prime}(t)|dt-\int_{a}^{x}f^{% \prime}(t)dt\right)$ $\displaystyle=\int_{a}^{x}h(t)dt.$

2 Helly’s selection theorem

We will use the following lemmas in the proof of the Helly selection theorem.88 8 N. L. Carothers, Real Analysis, p. 210, Theorem 13.13; p. 211, Lemma 13.14; p. 211, Lemma 13.15.

Lemma 13.

Suppose that $X$ is a set, that $f_{n}:X\to\mathbb{R}$ is a sequence of functions, and that there is some $K$ such that $\left\|f_{n}\right\|_{\infty}\leq K$ for all $n$. If $D$ is a countable subset of $X$, then there is a subsequence of $f_{n}$ that converges pointwise on $D$ to some $\phi:D\to\mathbb{R}$, which satisfies $\left\|\phi\right\|_{\infty}\leq K$.

Proof.

Say $D=\{x_{k}:k\geq 1\}$. Write $f_{n}^{0}=f_{n}$. The sequence of real numbers $f_{n}^{0}(x_{1})$ satisfies $f_{n}^{0}(x_{1})\in[-K,K]$ for all $n$, and since the set $[-K,K]$ is compact there is a subsequence $f^{1}_{n}(x_{1})$ of $f_{n}^{0}(x_{1})$ that converges, say to $\phi(x_{1})\in[-K,K]$. Suppose that $f_{n}^{m}(x_{m})$ is a subsequence of $f_{n}^{m-1}(x_{m})$ that converges to $\phi(x_{m})\in[-K,K]$. Then the sequence of real numbers $f_{n}^{m}(x_{m+1})$ satisfies $f_{n}^{m}(x_{m+1})\in[-K,K]$ for all $n$, and so there is a subsequence $f_{n}^{m+1}(x_{m+1})$ of $f_{n}^{m}(x_{m+1})$ that converges, say to $\phi(x_{m+1})\in[-K,K]$. Let $k\geq 1$. Then one checks that $f_{n}^{n}(x_{k})\to\phi(x_{k})$ as $n\to\infty$, namely, $f_{n}^{n}$ is a subsequence of $f_{n}$ that converges pointwise on $D$ to $\phi$, and for each $k$ we have $\phi(x_{k})\in[-K,K]$. ∎

Lemma 14.

Let $D\subset[a,b]$ with $a\in D$ and $b=\sup D$. If $\phi:D\to\mathbb{R}$ is nondecreasing, then $\Phi:[a,b]\to\mathbb{R}$ defined by

 $\Phi(x)=\sup\{\phi(t):t\in[a,x]\cap D\}$

is nondecreasing and the restriction of $\Phi$ to $D$ is equal to $\phi$.

Lemma 15.

If $f_{n}:[a,b]\to\mathbb{R}$ is a sequence of nondecreasing functions and there is some $K$ such that $\left\|f_{n}\right\|_{\infty}\leq K$ for all $n$, then there is a nondecreasing function $f:[a,b]\to\mathbb{R}$, satisfying $\left\|f\right\|_{\infty}\leq K$, and a subsequence of $f_{n}$ that converges pointwise to $f$.

Proof.

Let $D=(\mathbb{Q}\cap[a,b])\cup\{a\}$. By Lemma 13, there is a function $\phi:D\to\mathbb{R}$ and a subsequence $f_{a_{n}}$ of $f_{n}$ that converges pointwise on $D$ to $\phi$, and $\left\|\phi\right\|_{\infty}\leq K$. Because each $f_{n}$ is nondecreasing, if $x,y\in D$ and $x then

 $\phi(x)=\lim_{n\to\infty}f_{a_{n}}(x)\leq\lim_{n\to\infty}f_{a_{n}}(y)=\phi(y),$

namely, $\phi$ is nondecreasing. $D$ is a dense subset of $[a,b]$ and $a\in D$, so applying Lemma 14, there is a nondecreasing function $\Phi:[a,b]\to\mathbb{R}$ such that for $x\in D$,

 $\Phi(x)=\phi(x)=\lim_{n\to\infty}f_{a_{n}}(x).$

Suppose that $\Phi$ is continuous at $x\in[a,b]$ and let $\epsilon>0$. Using the fact that $\Phi$ is continuous at $x$, there are $p,q\in\mathbb{Q}\cap[a,b]$ such that $p and $\Phi(q)-\Phi(p)=|\Phi(q)-\Phi(p)|<\frac{\epsilon}{2}$. Because $p,q\in D$, $f_{a_{n}}(p)\to\Phi(p)$ and $f_{a_{n}}(q)\to\Phi(q)$, so there is some $N$ such that $n\geq N$ implies that both $|f_{a_{n}}(p)-\Phi(p)|<\frac{\epsilon}{2}$ and $|f_{a_{n}}(q)-\Phi(q)|<\frac{\epsilon}{2}$. Then for $n\geq N$, because each function $f_{a_{n}}$ is nondecreasing,

 $\displaystyle f_{a_{n}}(x)$ $\displaystyle\geq f_{a_{n}}(p)$ $\displaystyle\geq\Phi(p)-\frac{\epsilon}{2}$ $\displaystyle\geq\Phi(q)-\epsilon$ $\displaystyle\geq\Phi(x)-\epsilon.$

Likewise, for $n\geq N$,

 $\displaystyle f_{a_{n}}(x)$ $\displaystyle\leq f_{a_{n}}(q)$ $\displaystyle\leq\Phi(q)+\frac{\epsilon}{2}$ $\displaystyle<\Phi(p)+\epsilon$ $\displaystyle\leq\Phi(x)+\epsilon.$

This shows that if $\Phi$ is continuous at $x\in[a,b]$ then $f_{a_{n}}(x)\to\Phi(x)$.

Let $D(\Phi)$ be the collection of those $x\in[a,b]$ such that $\Phi$ is not continuous at $x$. Because $\Phi$ is monotone, $D(\Phi)$ is countable. So we have established that if $x\in[a,b]\setminus D(\Phi)$ then $f_{a_{n}}(x)\to\Phi(x)$. Because $f_{a_{n}}:[a,b]\to\mathbb{R}$ satisfies $\left\|f_{a_{n}}\right\|_{\infty}\leq K$ and $D(\Phi)$ is countable, Lemma 13 tells us that there is a function $F:D\to\mathbb{R}$ and a subsequence $f_{b_{n}}$ of $f_{a_{n}}$ such that $f_{b_{n}}$ converges pointwise on $D$ to $F$, and $\left\|F\right\|_{\infty}\leq K$. We define $f:[a,b]\to\mathbb{R}$ by $f(x)=\Phi(x)$ for $x\not\in D(\Phi)$ and $f(x)=F(x)$ for $x\in D(\Phi)$. $\left\|f\right\|_{\infty}\leq K$. For $x\not\in D(\Phi)$, $f_{a_{n}}(x)$ converges to $\Phi(x)=f(x)$, and $f_{b_{n}}(x)$ is a subsequence of $f_{a_{n}}(x)$ so $f_{b_{n}}(x)$ converges to $f(x)$. For $x\in D(\Phi)$, $f_{b_{n}}(x)$ converges to $F(x)=f(x)$. Therefore, for any $x\in[a,b]$ we have that $f_{b_{n}}(x)\to f(x)$, namely, $f_{b_{n}}$ converges pointwise to $f$. Because each function $f_{b_{n}}$ is nondecreasing, it follows that $f$ is nondecreasing. ∎

Finally we prove the pointwise Helly selection theorem.99 9 N. L. Carothers, Real Analysis, p. 212, Theorem 13.16.

Theorem 16.

Let $f_{n}$ be a sequence in $BV[a,b]$ and suppose there is some $K$ with $\left\|f_{n}\right\|_{BV}\leq K$ for all $n$. There is some subsequence of $f_{n}$ that converges pointwise to some $f\in BV[a,b]$, satisfying $\left\|f\right\|_{BV}\leq K$.

Proof.

Let $v_{n}$ be the variation of $f_{n}$. This satisfies, for any $n$,

 $\left\|v_{n}\right\|_{\infty}=V_{a}^{b}f_{n}\leq K$

and

 $\left\|v_{n}-f_{n}\right\|_{\infty}\leq\left\|v_{n}\right\|_{\infty}+\left\|f_% {n}\right\|_{\infty}\leq K+\left\|f_{n}\right\|_{BV}\leq 2K.$

Theorem 8 tells us that $v_{n}-f_{n}$ and $v_{n}$ are nondecreasing, so we can apply Lemma 15 to get that there is a nondecreasing function $g:[a,b]\to\mathbb{R}$ and a subsequence $v_{a_{n}}-f_{a_{n}}$ of $v_{n}-f_{n}$ that converges pointwise to $g$. Then we use Lemma 15 again to get that there is a nondecreasing function $h:[a,b]\to\mathbb{R}$ and a subsequence $v_{b_{n}}$ of $v_{a_{n}}$ that converges pointwise to $h$. Because $g$ and $h$ are pointwise limits of nondecreasing functions, they are each nondecreasing and so belong to $BV[a,b]$. We define $f=h-g\in BV[a,b]$. For $x\in[a,b]$,

 $\displaystyle\lim_{n\to\infty}f_{b_{n}}(x)$ $\displaystyle=\lim_{n\to\infty}v_{b_{n}}(x)-\lim_{n\to\infty}(v_{b_{n}}(x)-f_{% b_{n}}(x))$ $\displaystyle=h(x)-g(x)$ $\displaystyle=f(x),$

namely the subsequence $f_{b_{n}}$ of $f_{n}$ converges pointwise to $f$. By Lemma 6, because $f_{b_{n}}$ is a sequence in $BV[a,b]$ that converges pointwise to $f$ we have

 $\displaystyle\left\|f\right\|_{BV}$ $\displaystyle=|f(a)|+V_{a}^{b}f$ $\displaystyle\leq|f(a)|+\liminf_{n\to\infty}V_{a}^{b}f_{b_{n}}$ $\displaystyle=\liminf_{n\to\infty}\left(|f_{b_{n}}(a)|+V_{a}^{b}f_{b_{n}}\right)$ $\displaystyle=\liminf_{n\to\infty}\left\|f_{b_{n}}\right\|_{BV}$ $\displaystyle\leq K,$

completing the proof. ∎