Regulated functions and the regulated integral

1 Regulated functions and step functions

Let $I=[a,b]$ and let $X$ be a normed space. A function $f:I\to X$ is said to be regulated if for all $t\in [a,b)$ the limit ${lim}_{s\to t^{+}}f(s)$ exists and for all $t\in (a,b]$ the limit ${lim}_{s\to t^{-}}f(s)$ exists. We denote these limits respectively by $f(t^{+})$ and $f(t^{-})$. We define $R(I,X)$ to be the set of regulated functions $I\to X$. It is apparent that $R(I,X)$ is a vector space. One checks that a regulated function is bounded, and that $R(I,X)$ is a normed space with the norm ${\parallel f\parallel }_{\mathrm{\infty }}={sup}_{t\in [a,b]}\parallel f(t)\parallel$.

Theorem 1.

If $I$ is a compact interval in $\mathrm{R}$ and $X$ is a normed algebra, then $R\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}$ is a normed algebra.

Proof.

If $f,g\in R(I,X)$, then $fg\in R(I,X)$ because the limit of a product is equal to a product of limits. For $t\in I$ we have

$$\parallel (fg)(t)\parallel =\parallel f(t)g(t)\parallel \le \parallel f(t)\parallel \parallel g(t)\parallel \le {\parallel f\parallel }_{\mathrm{\infty }}{\parallel g\parallel }_{\mathrm{\infty }},$$

so ${\parallel fg\parallel }_{\mathrm{\infty }}\le {\parallel f\parallel }_{\mathrm{\infty }}{\parallel g\parallel }_{\mathrm{\infty }}$. ∎

A function $f:I\to X$, where $I=[a,b]$, is said to be a step function if there are for which $f$ is constant on each open interval $(s_{i-1},s_i)$. We denote the set of step functions $I\to X$ by $S(I,X)$. It is apparent that $S(I,X)$ is contained in $R(I,X)$ and is a vector subspace, and the following theorem states that if $X$ is a Banach space then $S(I,X)$ is dense in $R(I,X)$.

Theorem 2.

Let $I$ be a compact interval in $\mathrm{R}$, let $X$ be a Banach space, and let $f\mathrm{\in }{X}^I$. $f\mathrm{\in }R\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}$ if and only if for all $ϵ\mathrm{>}\mathrm{0}$ there is some $g\mathrm{\in }S\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}$ such that .

We prove in the following theorem that the set of regulated functions from a compact interval to a Banach space is itself a Banach space.

Theorem 3.

If $I$ is a compact interval in $\mathrm{R}$ and $X$ is a Banach space, then $R\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}$ is a Banach space.

Proof.

Let $f_n\in R(I,X)$ be a Cauchy sequence. For each $t\in I$ we have

$$\parallel f_n(t)-f_m(t)\parallel \le {\parallel f_n-f_m\parallel }_{\mathrm{\infty }},$$

hence $f_n(t)$ is a Cauchy sequence in $X$. As $X$ is a Banach space, this Cauchy sequence converges to some limit, and we define $f(t)$ to be this limit. Thus $f\in X^I$ and ${\parallel f-f_n\parallel }_{\mathrm{\infty }}\to 0$. We have to prove that $f\in R(I,X)$. Let $ϵ>0$. There is some $N$ for which $n\ge N$ implies that ; in particular, . By Theorem 2, there is some $g_N\in S(I,X)$ with . Then,

and by Theorem 2 this implies that $f\in R(I,X)$. ∎

The following lemma shows that the set of points of discontinuity of a regulated function taking values in a Banach space is countable.

Lemma 4.

If $I$ is a compact interval in $\mathrm{R}$, $X$ is a Banach space, and $f\mathrm{\in }R\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}$, then

$$\{t\in I:f\mathit{\text{ is discontinuous at }}t\}$$

is countable.

Proof.

For each $n$ let $g_n\in S(I,X)$ satisfy $\parallel f-g_n\parallel \le \frac{1}{n}$, and let

$$D_n=\{t\in I:g_n\text{ is discontinuous at }t\}.$$

$g_n$ is a step function so $D_n$ is finite, and hence $D={\bigcup }_{n=1}^{\mathrm{\infty }}{D}_n$ is countable. It need not be true that $f$ is discontinuous at each point in $D$, but we shall prove that if $t\in I\setminus D$ then $f$ is continuous at $t$, which will prove the claim.

Suppose that $t\in I\setminus D$, let $ϵ>0$, and take $N>\frac{1}{ϵ}$. As $t\notin D_N$, the step function $g_N$ is continuous at $t$, and hence there is some $\delta >0$ for which implies that . If , then

$\parallel f(s)-f(t)\parallel$ $\le$ $\parallel f(s)-g_N(s)\parallel +\parallel g_N(s)-g_N(t)\parallel +\parallel g_N(t)-f(t)\parallel$

$\le$ $2{\parallel f-g_N\parallel }_{\mathrm{\infty }}+\parallel g_N(s)-g_N(t)\parallel$

${\displaystyle \frac{2}{N}}+ϵ$

$3ϵ,$

showing that $f$ is continuous at $t$. ∎

2 Integrals of step functions

Let $I=[a,b]$ and let $X$ be a normed space. If $f\in S(I,X)$ then there is a subdivision of $[a,b]$ and there are $c_i\in X$ such that $f$ takes the value $c_i$ on the open interval $(s_{i-1},s_i)$. Suppose that there is a subdivision of $[a,b]$ and $d_i\in X$ such that $f$ takes the value $d_i$ on the open interval $(t_{i-1},t_i)$. One checks that

$$\sum _{i=1}^k(s_i-s_{i-1})c_i=\sum _{i=1}^l(t_i-t_{i-1})d_i.$$

We define the integral of $f$ to be the above element of $X$, and denote this element of $X$ by ${\int }_{I}f={\int }_a^{b}f$.

Lemma 5.

If $I$ is a compact interval in $\mathrm{R}$ and $X$ is a normed space, then ${\mathrm{\int }}_I\mathrm{:}S\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}\mathrm{\to }X$ is linear.

Lemma 6.

If $I\mathrm{=}\mathrm{[}a\mathrm{,}b\mathrm{]}$ and $X$ is a normed space, then ${\mathrm{\int }}_I\mathrm{:}S\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}\mathrm{\to }X$ is a bounded linear map with operator norm $b\mathrm{-}a$.

Proof.

If $f\in S(I,X)$, let be a subdivision of $[a,b]$ and let $c_i\in X$ such that $f$ takes the value $c_i$ on the open interval $(s_{i-1},s_i)$. Then,

$$\parallel {\int }_{I}f\parallel \le \sum _{i=1}^k(s_i-s_{i-1})\parallel c_i\parallel \le \sum _{i=1}^k(s_i-s_{i-1}){\parallel f\parallel }_{\mathrm{\infty }}=(b-a){\parallel f\parallel }_{\mathrm{\infty }}.$$

This shows that $\parallel {\int }_I\parallel \le b-a$, and if $f$ is constant, say $f(t)=c\in X$ for all $t\in I$, then ${\int }_{I}f=(b-a)c$ and $\parallel {\int }_{I}f\parallel =(b-a)\parallel c\parallel =(b-a){\parallel f\parallel }_{\mathrm{\infty }}$, showing that $\parallel {\int }_I\parallel =b-a$. ∎

Lemma 7.

If $a\mathrm{\le }b\mathrm{\le }c$, if $X$ is a normed space, and if $g\mathrm{\in }S\mathit{}\mathrm{(}\mathrm{[}a\mathrm{,}c\mathrm{]}\mathrm{,}X\mathrm{)}$, then

$${\int }_a^{c}g={\int }_a^{b}g+{\int }_b^{c}g.$$

3 The regulated integral

Let $I$ be a compact interval in $ℝ$ and let $X$ be a Banach space. Theorem 2 shows that $S(I,X)$ is a dense subspace of $R(I,X)$, and therefore if $T_0\in ℬ(S(I,X),X)$ then there is one and only one $T\in ℬ(R(I,X),X)$ whose restriction to $S(I,X)$ is equal to $T_0$, and this operator satisfies $\parallel T\parallel =\parallel T_0\parallel$. Lemma 6 shows that ${\int }_I:S(I,X)\to X$ is a bounded linear operator, thus there is one and only one bounded linear operator $R(I,X)\to X$ whose restriction to $S(I,X)$ is equal to ${\int }_I$, and we denote this operator $R(I,X)\to X$ also by ${\int }_I$. With $I=[a,b]$, we have $\parallel {\int }_I\parallel =b-a$. We call ${\int }_I:R(I,X)\to X$ the regulated integral.

Lemma 8.

If $a\mathrm{\le }b\mathrm{\le }c$, if $X$ is a Banach space, and if $f\mathrm{\in }R\mathit{}\mathrm{(}\mathrm{[}a\mathrm{,}c\mathrm{]}\mathrm{,}X\mathrm{)}$, then

$${\int }_a^{c}f={\int }_a^{b}f+{\int }_b^{c}f.$$

Proof.

Let $I_1=[a,b]$, $I_2=[b,c]$, $I=[a,c]$, and let $f_1$ and $f_2$ be the restriction of $f$ to $I_1$ and $I_2$ respectively. From the definition of regulated functions, $f_1\in R(I_1,X)$ and $f_2\in R(I_2,X)$. By Theorem 2, for any $ϵ>0$ there is some $g\in S(I,X)$ satisfying . Taking $g_1$ and $g_2$ to be the restriction of $g$ to $I_1$ and $I_2$, we check that $g_1\in S(I_1,X)$ and $g_2\in S(I_2,X)$. Then by Lemma 7,

${\parallel {\displaystyle {\int }_I}f-{\displaystyle {\int }_{I_1}}{f}_1-{\displaystyle {\int }_{I_2}}{f}_2\parallel }_{\mathrm{\infty }}$ $\le$ ${\parallel {\displaystyle {\int }_I}f-{\displaystyle {\int }_I}g\parallel }_{\mathrm{\infty }}+{\parallel {\displaystyle {\int }_I}g-{\displaystyle {\int }_{I_1}}{g}_1-{\displaystyle {\int }_{I_2}}{g}_2\parallel }_{\mathrm{\infty }}$

$+{\parallel {\displaystyle {\int }_{I_1}}{g}_1+{\displaystyle {\int }_{I_2}}{g}_2-{\displaystyle {\int }_{I_1}}{f}_1-{\displaystyle {\int }_{I_2}}{f}_2\parallel }_{\mathrm{\infty }}$

$=$ ${\parallel {\displaystyle {\int }_I}(f-g)\parallel }_{\mathrm{\infty }}+0$

$+{\parallel {\displaystyle {\int }_{I_1}}(g_1-f_1)\parallel }_{\mathrm{\infty }}+{\parallel {\displaystyle {\int }_{I_2}}(g_2-f_2)\parallel }_{\mathrm{\infty }}$

$\le$ $(c-a){\parallel f-g\parallel }_{\mathrm{\infty }}+(b-a){\parallel g_1-f_2\parallel }_{\mathrm{\infty }}$

$+(c-b){\parallel g_2-f_2\parallel }_{\mathrm{\infty }}.$

But ${\parallel g_1-f_1\parallel }_{\mathrm{\infty }}\le {\parallel g-f\parallel }_{\mathrm{\infty }}$ and ${\parallel g_2-f_2\parallel }_{\mathrm{\infty }}\le {\parallel g-f\parallel }_{\mathrm{\infty }}$, hence we obtain

Since $ϵ>0$ was arbitrary, we get

$${\parallel {\int }_{I}f-{\int }_{I_1}{f}_1-{\int }_{I_2}{f}_2\parallel }_{\mathrm{\infty }}=0,$$

so

$${\int }_{I}f={\int }_{I_1}{f}_1+{\int }_{I_2}{f}_2,$$

proving the claim. ∎

We prove that applying a bounded linear map and taking the regulated integral commute.

Lemma 9.

Suppose that $I$ is a compact interval in $\mathrm{R}$ and that $X$ and $Y$ are Banach spaces. If $f\mathrm{\in }R\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}$ and $T\mathrm{\in }\mathrm{B}\mathit{}\mathrm{(}X\mathrm{,}Y\mathrm{)}$, then $T\mathrm{\circ }f\mathrm{\in }R\mathit{}\mathrm{(}I\mathrm{,}Y\mathrm{)}$ and

$${\int }_{I}T\circ f=T{\int }_{I}f.$$

Proof.

Because $T$ is continuous we have $T\circ f\in R(I,Y)$. For $ϵ>0$, there is some $g\in S(I,X)$ satisfying . Write $I=[a,b]$. Because $g$ is a step function, there is a subdivision of $I$ and there are $c_i\in X$ such that $g$ takes the value $c_i$ on the open interval $(s_{i-1},s_i)$. Furthermore, $T\circ g$ takes the value $T{c}_i$ on the open interval $(s_{i-1},s_i)$ so $T\circ g\in S(I,Y)$, and then because $T$ is linear,

$${\int }_{I}T\circ g=\sum _{i=1}^k(s_i-s_{i-1})T{c}_i=T\sum _{i=1}^k(s_i-s_{i-1})c_i=T{\int }_{I}g.$$

Using this,

$\parallel {\displaystyle {\int }_I}T\circ f-T{\displaystyle {\int }_I}f\parallel$ $\le$ $\parallel {\displaystyle {\int }_I}T\circ f-{\displaystyle {\int }_I}T\circ g\parallel +\parallel {\displaystyle {\int }_I}T\circ g-T{\displaystyle {\int }_I}g\parallel$

$+\parallel T{\displaystyle {\int }_I}g-T{\displaystyle {\int }_I}f\parallel$

$=$ $\parallel {\displaystyle {\int }_I}T\circ (f-g)\parallel +\parallel T{\displaystyle {\int }_I}(f-g)\parallel$

$\le$ $(b-a){\parallel T\circ (f-g)\parallel }_{\mathrm{\infty }}+\parallel T\parallel \parallel {\displaystyle {\int }_I}(f-g)\parallel$

$\le$ $(b-a)\parallel T\parallel {\parallel f-g\parallel }_{\mathrm{\infty }}+\parallel T\parallel (b-a){\parallel f-g\parallel }_{\mathrm{\infty }}$

$2(b-a)\parallel T\parallel ϵ.$

As $ϵ>0$ is arbitrary, this means that

$$\parallel {\int }_{I}T\circ f-T{\int }_{I}f\parallel =0,$$

and so

$${\int }_{I}T\circ f=T{\int }_{I}f.$$

∎

4 Left and right derivatives

Suppose that $I$ is an open interval in $ℝ$, $X$ is a normed space, $f\in X^I$, and $t\in I$. We say that $f$ is right-differentiable at $t$ if $\frac{f(t+h)-f(t)}{h}$ has a limit as $h\to 0^{+}$, and that $f$ is left-differentiable at $t$ if $\frac{f(t+h)-f(t)}{h}$ has a limit as $h\to 0^{-}$. We call these limits respectively the right derivative of $f$ at $t$ and the left derivative of $f$ at $t$, denoted respectively by $f_{+}^{\prime }(t)$ and $f_{-}^{\prime }(t)$. For $f$ to be differentiable at $t$ means that $f_{+}^{\prime }(t)$ and $f_{-}^{\prime }(t)$ exist and are equal.

The following is the mean value theorem for functions taking values in a Banach space.

Theorem 10 (Mean value theorem).

Suppose that $I\mathrm{=}\mathrm{[}a\mathrm{,}b\mathrm{]}$, that $X$ is a Banach space, and that $f\mathrm{:}I\mathrm{\to }X$ and $g\mathrm{:}I\mathrm{\to }\mathrm{R}$ are continuous functions. If there is a countable set $D\mathrm{\subset }I$ such that $t\mathrm{\in }I\mathrm{\setminus }D$ implies that $f_{\mathrm{+}}^{\mathrm{\prime }}\mathit{}\mathrm{(}t\mathrm{)}$ and $g_{\mathrm{+}}^{\mathrm{\prime }}\mathit{}\mathrm{(}t\mathrm{)}$ exist and satisfy $\mathrm{\parallel }{f}_{\mathrm{+}}^{\mathrm{\prime }}\mathit{}\mathrm{(}t\mathrm{)}\mathrm{\parallel }\mathrm{\le }{g}_{\mathrm{+}}^{\mathrm{\prime }}\mathit{}\mathrm{(}t\mathrm{)}$, then

$$\parallel f(b)-f(a)\parallel \le g(b)-g(a).$$

Corollary 11.

Suppose that $I\mathrm{=}\mathrm{[}a\mathrm{,}b\mathrm{]}$, that $X$ is a Banach space, and that $f\mathrm{:}I\mathrm{\to }X$ is continuous. If there is a countable set $D\mathrm{\subset }I$ such that $t\mathrm{\in }I\mathrm{\setminus }D$ implies that $f_{\mathrm{+}}^{\mathrm{\prime }}\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}\mathrm{0}$, then $f$ is constant on $I$.

5 Primitives

Let $I=[a,b]$, let $X$ be a normed space, and let $f,g\in X^I$. We say that $g$ is a primitive of $f$ if $g$ is continuous and if there is a countable set $D\subset I$ such that $t\in I\setminus D$ implies that $g$ is differentiable at $t$ and $g^{\prime }(t)=f(t)$.

Lemma 12.

Suppose that $I$ is a compact interval in $\mathrm{R}$, that $X$ is a Banach space, and that $f\mathrm{:}I\mathrm{\to }X$ is a function. If $g_{\mathrm{1}}\mathrm{,}{g}_{\mathrm{2}}\mathrm{:}I\mathrm{\to }X$ are primitives of $f$, then $g_{\mathrm{1}}\mathrm{-}{g}_{\mathrm{2}}$ is constant on $I$.

Proof.

For $i=1,2$, as $g_i$ is a primitive of $f$ there is a countable set $D_i\subset I$ such that $t\in I\setminus D_i$ implies that $g_i$ is differentiable at $t$ and $g_i^{\prime }(t)=f(t)$. Let $D=D_1\cup D_2$, which is a countable set. Both $g_1$ and $g_2$ are continuous so $g=g_1-g_2:I\to X$ is continuous, and if $t\in I\setminus D$ then $g$ is differentiable at $t$ and $g^{\prime }(t)=g_1^{\prime }(t)-g_2^{\prime }(t)=f(t)-f(t)=0$. Then Corollary 11 shows that $g$ is constant on $I$, i.e., that $g_1-g_2$ is constant on $I$.∎

We now give a construction of primitives of regulated functions.

Theorem 13.

If $I\mathrm{=}\mathrm{[}a\mathrm{,}b\mathrm{]}$, $X$ is a Banach space, and $f\mathrm{\in }R\mathit{}\mathrm{(}I\mathrm{,}X\mathrm{)}$, then the map $g\mathrm{:}I\mathrm{\to }X$ defined by $g\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}{\mathrm{\int }}_a^{t}f$ is a primitive of $f$ on $I$.

Proof.

For $t\in [a,b)$ and $ϵ>0$, because $f$ is regulated there is some such that implies that $\parallel f(t+r)-f(t^{+})\parallel \le ϵ$. For and for any , using Lemma 8 we have

$\parallel {\displaystyle {\int }_a^{t+r}}f-{\displaystyle {\int }_a^t}f-{\displaystyle {\int }_t^{t+r}}f(t^{+})\parallel$ $=$ $\parallel {\displaystyle {\int }_t^{t+r}}f-{\displaystyle {\int }_t^{t+r}}f(t^{+})\parallel$

$=$ $\parallel {\displaystyle {\int }_t^{t+\eta }}(f-f(t^{+}))+{\displaystyle {\int }_{t+\eta }^{t+r}}(f-f(t^{+}))\parallel$

$\le$ $\eta \underset{t\le s\le t+\eta }{sup}\parallel f(s)-f(t^{+})\parallel$

$+(r-\eta )\underset{t+\eta \le s\le t+r}{sup}\parallel f(s)-f(t^{+})\parallel$

$\le$ $2{\parallel f\parallel }_{\mathrm{\infty }}\eta +(r-\eta )ϵ.$

This is true for all , so

$$\parallel {\int }_a^{t+r}f-{\int }_a^{t}f-{\int }_t^{t+r}f(t^{+})\parallel \le rϵ,$$

i.e.

$$\parallel \frac{g(t+r)-g(t)}{r}-f(t^{+})\parallel \le ϵ.$$

This shows that

$$g_{+}^{\prime }(t)=f(t^{+}).$$

Similarly,

$$g_{-}^{\prime }(t)=f(t^{-}).$$

Because $f$ is regulated, Lemma 4 shows that there is a countable set $D\subset I$ such that $t\in I\setminus D$ implies that $f$ is continuous at $t$. Therefore, if $t\in I\setminus D$ then $f(t^{+})=f(t^{-})=f(t)$, so $g_{+}^{\prime }(t)=g_{-}^{\prime }(t)$, which means that if $t\in I\setminus D$ then $g$ is differentiable at $t$, with $g^{\prime }(t)=f(t)$. To prove that $g$ is a primitive of $f$ on $I$ it suffices now to show that $g$ is continuous. For $ϵ>0$ and $t\in I$, let $\delta =\frac{ϵ}{{\parallel f\parallel }_{\mathrm{\infty }}}$, and then for we have by Lemma 8 that

showing that $g$ is continuous at $t$, completing the proof. ∎

Suppose that $X$ is a Banach space and that $f:[a,b]\to X$ is a primitive of a regulated function $h:[a,b]\to X$. Because $h$ is regulated, by Theorem 13 the function $g:[a,b]\to X$ defined by $g(t)={\int }_a^{t}f$ is a primitive of $f$ on $[a,b]$. Then applying Lemma 12, there is some $c\in X$ such that $f(t)-g(t)=c$ for all $t\in [a,b]$. But $f(a)-g(a)=f(a)$, so $c=f(a)$. Hence, for all $t\in [a,b]$,

$$f(t)=f(a)+{\int }_a^{t}h.$$

But

$${\int }_a^{t}h={\int }_a^{a+{\eta }_1}h+{\int }_{a+{\eta }_1}^{t-{\eta }_2}h+{\int }_{t-{\eta }_2}^{t}h={\int }_a^{a+{\eta }_1}h+{\int }_{a+{\eta }_1}^{t-{\eta }_2}{f}^{\prime }+{\int }_{t-{\eta }_2}^{t}h$$

and

$$\parallel {\int }_a^{a+{\eta }_1}h\parallel \le {\eta }_1{\parallel h\parallel }_{\mathrm{\infty }},\parallel {\int }_{t-{\eta }_2}^{t}h\parallel \le {\eta }_2{\parallel h\parallel }_{\mathrm{\infty }},$$

hence as ${\eta }_1\to 0^{+}$ and ${\eta }_2\to 0^{+}$,

$${\int }_{a+{\eta }_1}^{t-{\eta }_2}{f}^{\prime }\to {\int }_a^{t}h,$$

and so it makes sense to write

$${\int }_a^t{f}^{\prime }={\int }_a^{t}h,$$

and thus for all $t\in [a,b]$,

$$f(t)=f(a)+{\int }_a^t{f}^{\prime }.$$