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

1 $BV[a,b]$

Let . For $f:[a,b]\to ℝ$, we define

$${\parallel f\parallel }_{\mathrm{\infty }}=\underset{t\in [a,b]}{sup}|f(t)|,$$

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

A partition of $\mathrm{[}a\mathrm{,}b\mathrm{]}$ is a set $P=\{t_0,\mathrm{\dots },t_n\}$ such that . 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 $f:[a,b]\to ℝ$, 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

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

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

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

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

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

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

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

$${\parallel f\parallel }_{\mathrm{\infty }}\le |f(a)|+V_a^{b}f.$$

Proof.

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

$$|f(b)|\le |f(a)|+|f(b)-f(a)|\le |f(a)|+V_a^{b}f.$$

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

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

∎

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.

Lemma 2.

If $f\mathrm{,}g\mathrm{\in }B\mathit{}V\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$ and $c\mathrm{\in }\mathrm{R}$, then the following statements are true.

1. $V_a^{b}f=0$ if and only if $f$ is constant.

2. $V_a^b(cf)=|c|V_a^b(f)$.

3. $V_a^b(f+g)\le V_a^{b}f+V_a^{b}g$.

4. $V_a^b(fg)\le {\parallel f\parallel }_{\mathrm{\infty }}{V}_a^{b}g+{\parallel g\parallel }_{\mathrm{\infty }}{V}_a^{b}f$.

5. $V_a^b|f|\le V_a^{b}f$.

6. $V_a^{b}f=V_a^{x}f+V_x^b$ for $a\le x\le b$.

Lemma 3.

If $f\mathrm{:}\mathrm{[}a\mathrm{,}b\mathrm{]}\mathrm{\to }\mathrm{R}$ is differentiable on $\mathrm{(}a\mathrm{,}b\mathrm{)}$ and , then

$$V_a^{b}f\le {\parallel f^{\prime }\parallel }_{\mathrm{\infty }}(b-a).$$

Proof.

Suppose that is a partition of $[a,b]$. By the mean value theorem, for each $j=1,\mathrm{\dots },n$ 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

$V(f,P)$ $={\displaystyle \sum _{j=1}^n}|f(t_j)-f(t_{j-1})|$

$={\displaystyle \sum _{j=1}^n}(t_j-t_{j-1})|f^{\prime }(x_j)|$

$\le {\parallel f^{\prime }\parallel }_{\mathrm{\infty }}{\displaystyle \sum _{j=1}^n}(t_j-t_{j-1})$

$={\parallel f^{\prime }\parallel }_{\mathrm{\infty }}(b-a).$

∎

Lemma 4.

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

$$V_a^{b}f\le {\int }_a^b|f^{\prime }(t)|𝑑t.$$

Proof.

Let $P=\{t_0,\mathrm{\dots },t_n\}$ be a partition of $[a,b]$. Then, by the fundamental theorem of calculus,

$V(f,P)$ $={\displaystyle \sum _{j=1}^n}|f(t_j)-f(t_{j-1})|$

$\le {\displaystyle \sum _{j=1}^n}\left|{\displaystyle {\int }_{t_{j-1}}^{t_j}}{f}^{\prime }(t)\right|$

$\le {\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{t_{j-1}}^{t_j}}|f^{\prime }(t)|𝑑t$

$={\displaystyle {\int }_a^b}|f^{\prime }(t)|𝑑t.$

Therefore

$$V_a^{b}f=\underset{P}{sup}V(f,P)\le {\int }_a^b|f^{\prime }(t)|𝑑t.$$

∎

Lemma 5.

If $f\mathrm{:}\mathrm{[}a\mathrm{,}b\mathrm{]}\mathrm{\to }\mathrm{R}$ 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 $j=1,\mathrm{\dots },n$, for some $c_j\in \{+1,-1\}$ and . 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 is a partition of $[a,b]$,

${\displaystyle {\int }_a^b}|f^{\prime }(t)|𝑑t$ $={\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{t_{j-1}}^{t_j}}|f^{\prime }(t)|𝑑t$

$={\displaystyle \sum _{j=1}^n}{c}_j(f(t_j)-f(t_{j-1}))$

$\le {\displaystyle \sum _{j=1}^n}|f(t_j)-f(t_{j-1})|$

$\le V_a^{b}f.$

∎

Lemma 6.

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

$$V(f_m,P)\to V(f,P).$$

If $f_m$ is a sequence in $B\mathit{}V\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$ that converges pointwise to some $f\mathrm{:}\mathrm{[}a\mathrm{,}b\mathrm{]}\mathrm{\to }\mathrm{R}$, then

$$V_a^{b}f\le \underset{m\to \mathrm{\infty }}{lim\; inf}{V}_a^b{f}_m.$$

Proof.

Say $P=\{t_0,\mathrm{\dots },t_n\}$. Then, because taking the limit of convergent sequences is linear,

$\underset{m\to \mathrm{\infty }}{lim}V(f_m,P)$ $=\underset{m\to \mathrm{\infty }}{lim}{\displaystyle \sum _{j=1}^n}|f_m(t_j)-f_m(t_{j-1})|$

$={\displaystyle \sum _{j=1}^n}\underset{m\to \mathrm{\infty }}{lim}|f_m(t_j)-f_m(t_{j-1})|$

$={\displaystyle \sum _{j=1}^n}|f(t_j)-f(t_{j-1})|$

$=V(f,P).$

Let $P=\{t_0,\mathrm{\dots },t_n\}$ be a partition of $[a,b]$. Then

$V(f,P)$ $={\displaystyle \sum _{j=1}^n}|f(t_j)-f(t_{j-1})|$

$={\displaystyle \sum _{j=1}^n}\underset{m\to \mathrm{\infty }}{lim}|f_m(t_j)-f_m(t_{j-1})|$

$=\underset{m\to \mathrm{\infty }}{lim}V(f_m,P)$

$\le \underset{m\to \mathrm{\infty }}{lim\; inf}{V}_a^b{f}_m.$

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

$$V_a^{b}f\le \underset{m\to \mathrm{\infty }}{lim\; inf}{V}_a^b{f}_m.$$

∎

We now prove that $BV[a,b]$ is a Banach space.

Theorem 7.

With the norm

$${\parallel f\parallel }_{BV}=|f(a)|+V_a^{b}f.$$

$BV[a,b]$ 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 $ϵ>0$. Because $f_n$ is a Cauchy sequence in $BV[a,b]$, there is some $N$ such that if $n,m\ge N$ then . For $n\ge N$, Lemma 6 yields

${\parallel f-f_n\parallel }_{BV}$ $\le |f(a)-f_n(a)|+V(f-f_n,P)$

$=\underset{m\to \mathrm{\infty }}{lim}\left(|f_m(a)-f_n(a)|+V(f_m-f_n,P)\right)$

$\le \underset{m\ge N}{sup}\left(|f_m(a)-f_n(a)|+V(f_m-f_n,P)\right)$

$=\underset{m\ge N}{sup}{\parallel f_m-f_n\parallel }_{BV}$

$\le ϵ.$

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.

Theorem 8.

Let $f\mathrm{\in }B\mathit{}V\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$ and let $v$ be the variation of $f$. Then $v\mathrm{-}f$ and $v$ are nondecreasing.

Proof.

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

$v(y)-v(x)$ $=V_a^{y}f-V_a^{x}f$

$=V_x^{y}f$

$\ge |f(y)-f(x)|$

$\ge f(y)-f(x).$

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

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.

Theorem 9.

Let $f\mathrm{\in }B\mathit{}V\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$ and let $v$ be the variation of $f$. For $x\mathrm{\in }\mathrm{[}a\mathrm{,}b\mathrm{]}$, $f$ 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 $ϵ>0$, there is some $\delta >0$ such that implies that . If , then

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

Assume that $f$ is right continuous at $x$, with . Let $ϵ>0$. There is some $\delta >0$ such that implies that . Because $V_x^{b}f$ is a supremum over partitions of $[x,b]$, there is some partition $P=\{t_0,t_1,\mathrm{\dots },t_n\}$ of $[x,b]$ such that $V_x^{b}f-\frac{ϵ}{2}\le V(f,P)$. Let . Then $Q=\{t_0,y,t_1,\mathrm{\dots },t_n\}$ is a refinement of $P$, so

$V_x^{b}f-{\displaystyle \frac{ϵ}{2}}$ $\le V(f,P)$

$\le V(f,Q)$

$=|f(y)-f(t_0)|+V(f,\{y,t_1,\mathrm{\dots },t_n\})$

Hence

$$ϵ>V_x^{b}f-V_y^{b}f=V_x^{y}f=v(y)-v(x)=|v(y)-v(x)|,$$

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

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

and the negative variation of $f$ as

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

We can write the variation as $v=p+n$. We now establish properties of the positive and negative variations.

Theorem 10.

Let $f\mathrm{\in }B\mathit{}V\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$, let $v$ be its variation, let $p$ be its positive variation, and let $n$ be its negative variation. Then $\mathrm{0}\mathrm{\le }p\mathrm{\le }v$ and $\mathrm{0}\mathrm{\le }n\mathrm{\le }v$, and $p$ and $n$ are nondecreasing.

Proof.

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

For ,

$p(y)-p(x)$ $={\displaystyle \frac{v(y)+f(y)-v(x)-f(x)}{2}}$

$={\displaystyle \frac{1}{2}}\left(V_x^{y}f+(f(y)-f(x))\right)$

$\ge {\displaystyle \frac{1}{2}}\left(|f(y)-f(x)|+(f(y)-f(x))\right)$

$\ge 0$

and

$n(y)-n(x)$ $={\displaystyle \frac{v(y)-f(y)-v(x)+f(x)}{2}}$

$={\displaystyle \frac{1}{2}}\left(V_x^{y}f-(f(y)-f(x))\right)$

$\ge {\displaystyle \frac{1}{2}}\left(|f(y)-f(x)|-(f(y)-f(x))\right)$

$\ge 0.$

∎

We now prove that $BV[a,b]$ is a Banach algebra.

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

$f_1{f}_2$ $=(f_1(a)+p_1-n_1)(f_2(a)+p_2-n_2)$

$=f_1(a)f_2(a)+p_1{p}_2+n_1{n}_2-n_1{p}_2-n_2{p}_1$

$+f_1(a)p_2+f_2(a)p_1-f_1(a)n_2-f_2(a)n_1.$

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

${\parallel f_1{f}_2\parallel }_{BV}$ $=|f_1(a)||f_2(a)|+V_a^b(f_1{f}_2)$

$\le |f_1(a)||f_2(a)|+V_a^b(p_1{p}_2)+V_a^b(n_1{n}_2)+V_a^b(n_1{p}_2)+V_a^b(n_2{p}_1)$

$+|f_1(a)|V_a^b{p}_2+|f_2(a)|V_a^b{p}_1+|f_1(a)|V_a^b{n}_2+|f_2(a)|V_a^b{n}_1$

$=|f_1(a)||f_2(a)|+p_1(b)p_2(b)+n_1(b)n_2(b)+n_1(b)p_2(b)+n_2(b)p_1(b)$

$+|f_1(a)|p_2(b)+|f_2(a)|p_1(b)+|f_1(a)|n_2(b)+|f_2(a)|n_1(b)$

$=(|f_1(a)|+p_1(b)+n_1(b))(|f_2(a)|+p_2(b)+n_2(b))$

$=(|f_1(a)+v_1(b))(|f_2(a)|+v_2(b))$

$={\parallel f_1\parallel }_{BV}{\parallel f_2\parallel }_{BV},$

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

$$V_a^{b}f={\int }_a^b|f^{\prime }(t)|𝑑t.$$

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

$$p(x)={\int }_a^x{(f^{\prime })}^{+}(t)𝑑t,n(x)={\int }_a^x{(f^{\prime })}^{-}(t)𝑑t.$$

Proof.

Lemma 4 states that $V_a^b\le {\int }_a^b|f^{\prime }(t)|𝑑t$. Because $f^{\prime }$ is continuous it is Riemann integrable, hence for any $ϵ>0$ there is some partition $P=\{t_0,\mathrm{\dots },t_n\}$ of $[a,b]$ such that if $x_j\in [t_{j-1},t_j]$ for $j=1,\mathrm{\dots },n$ then

By the mean value theorem, for each $j=1,\mathrm{\dots },n$ 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

and thus

This is true for all $ϵ>0$, therefore

$${\int }_a^b|f^{\prime }(t)|𝑑t\le V_a^{b}f,$$

which is what we wanted to show.

Write

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

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

$p(x)$ $={\displaystyle \frac{1}{2}}\left(v(x)+f(x)-f(a)\right)$

$={\displaystyle \frac{1}{2}}\left(V_a^{x}f+{\displaystyle {\int }_a^x}{f}^{\prime }(t)𝑑t\right)$

$={\displaystyle \frac{1}{2}}\left({\displaystyle {\int }_a^b}|f^{\prime }(t)|𝑑t+{\displaystyle {\int }_a^b}{f}^{\prime }(t)𝑑t\right)$

$={\displaystyle {\int }_a^b}g(t)𝑑t$

and

$n(x)$ $={\displaystyle \frac{1}{2}}\left(v(x)-f(x)+f(a)\right)$

$={\displaystyle \frac{1}{2}}\left(V_a^{x}f-{\displaystyle {\int }_a^x}{f}^{\prime }(t)𝑑t\right)$

$={\displaystyle \frac{1}{2}}\left({\displaystyle {\int }_a^x}|f^{\prime }(t)|𝑑t-{\displaystyle {\int }_a^x}{f}^{\prime }(t)𝑑t\right)$

$={\displaystyle {\int }_a^x}h(t)𝑑t.$

∎

2 Helly’s selection theorem

We will use the following lemmas in the proof of the Helly selection theorem.

Lemma 13.

Suppose that $X$ is a set, that $f_n\mathrm{:}X\mathrm{\to }\mathrm{R}$ is a sequence of functions, and that there is some $K$ such that ${\mathrm{\parallel }{f}_n\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\le }K$ 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 $\varphi \mathrm{:}D\mathrm{\to }\mathrm{R}$, which satisfies ${\mathrm{\parallel }\varphi \mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\le }K$.

Proof.

Say $D=\{x_k:k\ge 1\}$. 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 $f_n^1(x_1)$ of $f_n^0(x_1)$ that converges, say to $\varphi (x_1)\in [-K,K]$. Suppose that $f_n^m(x_m)$ is a subsequence of $f_n^{m-1}(x_m)$ that converges to $\varphi (x_m)\in [-K,K]$. 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 $\varphi (x_{m+1})\in [-K,K]$. Let $k\ge 1$. Then one checks that $f_n^n(x_k)\to \varphi (x_k)$ as $n\to \mathrm{\infty }$, namely, $f_n^n$ is a subsequence of $f_n$ that converges pointwise on $D$ to $\varphi$, and for each $k$ we have $\varphi (x_k)\in [-K,K]$. ∎

Lemma 14.

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

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

is nondecreasing and the restriction of $\mathrm{\Phi }$ to $D$ is equal to $\varphi$.

Lemma 15.

If $f_n\mathrm{:}\mathrm{[}a\mathrm{,}b\mathrm{]}\mathrm{\to }\mathrm{R}$ is a sequence of nondecreasing functions and there is some $K$ such that ${\mathrm{\parallel }{f}_n\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\le }K$ for all $n$, then there is a nondecreasing function $f\mathrm{:}\mathrm{[}a\mathrm{,}b\mathrm{]}\mathrm{\to }\mathrm{R}$, satisfying ${\mathrm{\parallel }f\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\le }K$, 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 $\varphi :D\to ℝ$ and a subsequence $f_{a_n}$ of $f_n$ that converges pointwise on $D$ to $\varphi$, and ${\parallel \varphi \parallel }_{\mathrm{\infty }}\le K$. Because each $f_n$ is nondecreasing, if $x,y\in D$ and then

$$\varphi (x)=\underset{n\to \mathrm{\infty }}{lim}{f}_{a_n}(x)\le \underset{n\to \mathrm{\infty }}{lim}{f}_{a_n}(y)=\varphi (y),$$

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

$$\mathrm{\Phi }(x)=\varphi (x)=\underset{n\to \mathrm{\infty }}{lim}{f}_{a_n}(x).$$

Suppose that $\mathrm{\Phi }$ is continuous at $x\in [a,b]$ and let $ϵ>0$. Using the fact that $\mathrm{\Phi }$ is continuous at $x$, there are $p,q\in \mathbb{Q}\cap [a,b]$ such that and . Because $p,q\in D$, $f_{a_n}(p)\to \mathrm{\Phi }(p)$ and $f_{a_n}(q)\to \mathrm{\Phi }(q)$, so there is some $N$ such that $n\ge N$ implies that both and . Then for $n\ge N$, because each function $f_{a_n}$ is nondecreasing,

$f_{a_n}(x)$ $\ge f_{a_n}(p)$

$\ge \mathrm{\Phi }(p)-{\displaystyle \frac{ϵ}{2}}$

$\ge \mathrm{\Phi }(q)-ϵ$

$\ge \mathrm{\Phi }(x)-ϵ.$

Likewise, for $n\ge N$,

$f_{a_n}(x)$ $\le f_{a_n}(q)$

$\le \mathrm{\Phi }(q)+{\displaystyle \frac{ϵ}{2}}$

$\le \mathrm{\Phi }(x)+ϵ.$

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

Let $D(\mathrm{\Phi })$ be the collection of those $x\in [a,b]$ such that $\mathrm{\Phi }$ is not continuous at $x$. Because $\mathrm{\Phi }$ is monotone, $D(\mathrm{\Phi })$ is countable. So we have established that if $x\in [a,b]\setminus D(\mathrm{\Phi })$ then $f_{a_n}(x)\to \mathrm{\Phi }(x)$. Because $f_{a_n}:[a,b]\to ℝ$ satisfies ${\parallel f_{a_n}\parallel }_{\mathrm{\infty }}\le K$ and $D(\mathrm{\Phi })$ is countable, Lemma 13 tells us that there is a function $F:D\to ℝ$ and a subsequence $f_{b_n}$ of $f_{a_n}$ such that $f_{b_n}$ converges pointwise on $D$ to $F$, and ${\parallel F\parallel }_{\mathrm{\infty }}\le K$. We define $f:[a,b]\to ℝ$ by $f(x)=\mathrm{\Phi }(x)$ for $x\notin D(\mathrm{\Phi })$ and $f(x)=F(x)$ for $x\in D(\mathrm{\Phi })$. ${\parallel f\parallel }_{\mathrm{\infty }}\le K$. For $x\notin D(\mathrm{\Phi })$, $f_{a_n}(x)$ converges to $\mathrm{\Phi }(x)=f(x)$, and $f_{b_n}(x)$ is a subsequence of $f_{a_n}(x)$ so $f_{b_n}(x)$ converges to $f(x)$. For $x\in D(\mathrm{\Phi })$, $f_{b_n}(x)$ 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.

Theorem 16.

Let $f_n$ be a sequence in $B\mathit{}V\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$ and suppose there is some $K$ with ${\mathrm{\parallel }{f}_n\mathrm{\parallel }}_{B\mathit{}V}\mathrm{\le }K$ for all $n$. There is some subsequence of $f_n$ that converges pointwise to some $f\mathrm{\in }B\mathit{}V\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$, satisfying ${\mathrm{\parallel }f\mathrm{\parallel }}_{B\mathit{}V}\mathrm{\le }K$.

Proof.

Let $v_n$ be the variation of $f_n$. This satisfies, for any $n$,

$${\parallel v_n\parallel }_{\mathrm{\infty }}=V_a^b{f}_n\le K$$

and

$${\parallel v_n-f_n\parallel }_{\mathrm{\infty }}\le {\parallel v_n\parallel }_{\mathrm{\infty }}+{\parallel f_n\parallel }_{\mathrm{\infty }}\le K+{\parallel f_n\parallel }_{BV}\le 2K.$$

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 $g:[a,b]\to ℝ$ 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 $h:[a,b]\to ℝ$ 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 $x\in [a,b]$,

$\underset{n\to \mathrm{\infty }}{lim}{f}_{b_n}(x)$ $=\underset{n\to \mathrm{\infty }}{lim}{v}_{b_n}(x)-\underset{n\to \mathrm{\infty }}{lim}(v_{b_n}(x)-f_{b_n}(x))$

$=h(x)-g(x)$

$=f(x),$

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

${\parallel f\parallel }_{BV}$ $=|f(a)|+V_a^{b}f$

$\le |f(a)|+\underset{n\to \mathrm{\infty }}{lim\; inf}{V}_a^b{f}_{b_n}$

$=\underset{n\to \mathrm{\infty }}{lim\; inf}\left(|f_{b_n}(a)|+V_a^b{f}_{b_n}\right)$

$=\underset{n\to \mathrm{\infty }}{lim\; inf}{\parallel f_{b_n}\parallel }_{BV}$

$\le K,$

completing the proof. ∎