Auguste Toulmouche (1829-1890) was a French painter.
\begin{equation}{\label{map13}}\tag{1}\mbox{}\end{equation}
Proposition \ref{map13}. Let \(f\) be an increasing real-valued function defined on the closed interval \([a,b]\), and let \(\{x_{0},x_{1},\cdots ,x_{n+1}\}\) be \(n+1\) points in \([a,b]\) satisfying \(a=x_{0}<x_{1}<x_{2}<\cdots <x_{n}=b\). Then, we have the following inequality
\[\sum_{k=1}^{n-1}\left [f(x_{k}+)-f(x_{k}-)\right ]\leq f(b)-f(a).\]
Proof. For \(y_{k}\in (x_{k},x_{k+1})\) and \(1\leq k\leq n-1\), since \(f\) is increasing, we have
\[f(x_{k}+)=\lim_{x\rightarrow x_{k}+}f(x)\leq f(y_{k})\]
and
\[f(y_{k-1})\leq\lim_{x\rightarrow x_{k}-}f(x)=f(x_{k}-),\]
which implies
\[f(x_{k}+)-f(x_{k}-)\leq f(y_{k})-f(y_{k-1}).\]
By adding these inequalities, we obtain
\[\sum_{k=1}^{n-1}\left [f(x_{k}+)-f(x_{k}-)\right ]\leq f(y_{n-1})-f(y_{0})\leq f(b)-f(a).\]
This completes the proof. \(\blacksquare\)
The difference \(f(x_{k}+)-f(x_{k}-)\) is the jump of \(f\) at \(x_{k}\). The above proposition says that, for every finite collection of points \(x_{k}\) in \((a,b)\), the sum of the jumps at these points is always bounded by \(f(b)-f(a)\).
Proposition. Suppose that the real-valued function \(f\) is monotonic on the closed interval \([a,b]\). Then, the set of discontinuities of \(f\) is countable.
Proof. We assume that \(f\) is increasing on \([a,b]\). For \(m>0\), let \(S_{m}\) be the set of points in \((a,b)\) at which the jump of \(f\) exceeds \(1/m\). For \(\{x_{1},x_{2},\cdots ,x_{n}\}\subset S_{m}\), we have
\[f(x_{k}+)-f(x_{k}-)\geq\frac{1}{m}\]
for \(k=1,\cdots,n-1\). Proposition \ref{map13} says
\[\frac{n-1}{m}\leq\sum_{k=1}^{n-1}\left [f(x_{k}+)-f(x_{k}-)\right ]\leq f(b)-f(a)<+\infty .\]
This means that \(S_{m}\) cannot be an uncountable infinite set. Since the set of discontinuities of \(f\) in \((a,b)\) is a subset of \(\bigcup_{m=1}^{\infty}S_{m}\), we obtain the desired result. If \(f\) is decreasing, the argument can be applied to \(-f\). This completes the proof. \(\blacksquare\)
Definition. Let \({\cal P}=\{x_{0},x_{1},\cdots ,x_{n}\}\) be a partition of \([a,b]\). Given a real-valued function \(f\) defined on the closed interval \([a,b]\), we write
\[\Delta f_{k}=f(x_{k})-f(x_{k-1})\mbox{ for }k=1,2,\cdots ,n.\]
The function \(f\) is said to be of bounded variation on \([a,b]\) when there exists a positive number \(M\) satisfying
\[\sum_{k=1}^{n}\left |\Delta f_{k}\right |\leq M\]
for all partitions of \([a,b]\). \(\sharp\)
Example. Given a real-valued function
\[f(x)=\left\{\begin{array}{ll}
x\cdot\cos\left (\frac{\pi}{2x}\right ) & \mbox{if \(x\neq 0\)}\\
0 & \mbox{if \(x=0\),}
\end{array}\right .\]
it is clear to see that \(f\) is continuous on \([0,1]\). We are going to claim that it is not of bounded variation on \([0,1]\). We consider the following partition
\[{\cal P}=\left\{0,\frac{1}{2n},\frac{1}{2n-1},\cdots ,\frac{1}{3},\frac{1}{2},1\right\}.\]
Then, we have
\begin{align*} \sum_{k=1}^{2n}\left |\Delta f_{k}\right | & =\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n-2}+
\frac{1}{2n-2}+\cdots +\frac{1}{2}+\frac{1}{2}\\ & =1+\frac{1}{2}+\cdots+\frac{1}{n},\end{align*}
which diverges to \(+\infty\). Therefore \(f\) is not of bounded variation on \([0,1]\). \(\sharp\)
\begin{equation}{\label{map16}}\tag{2}\mbox{}\end{equation}
Proposition \ref{map16}. Suppose that the real-valued function \(f\) defined on the closed interval \([a,b]\) is monotonic. Then \(f\) is of bounded variation on \([a,b]\).
Proof. We assume that \(f\) is increasing. For every partition \({\cal P}\) of \([a,b]\), we have \(\Delta f_{k}\geq 0\), which says
\begin{align*} \sum_{k=1}^{n}\left |\Delta f_{k}\right | & =\sum_{k=1}^{n}\Delta f_{k}\\ & =\sum_{k=1}^{n}\left [f(x_{k})-f(x_{k-1})\right ]\\ & =f(b)-f(a).\end{align*}
This completes the proof. \(\blacksquare\)
Proposition. Let \(f\) be a continuous real-valued function defined on the closed interval \([a,b]\). Suppose that \(|f'(x)|\leq A\) for all \(x\in (a,b)\), where \(A\) is a constant. Then \(f\) is of bounded variation on \([a,b]\).
Proof. Using the mean-value theorem, we have
\[\Delta f_{k}=f(x_{k})-f(x_{k-1})=f'(c_{k})(x_{k}-x_{k-1})\]
for some \(c_{k}\in (x_{k-1},x_{k})\), which implies
\begin{align*} \sum_{k=1}^{n}\left |\Delta f_{k}\right | & =\sum_{k=1}^{n}\left |f'(c_{k})\right |(x_{k}-x_{k-1})
\\ & \leq A\cdot\sum_{k=1}^{n}\Delta x_{k}=A\cdot (b-a).\end{align*}
This completes the proof. \(\blacksquare\)
Proposition. Let \(f\) be a real-valued function defined on the closed interval \([a,b]\). Suppose that \(f\) is of bounded variation on \([a,b]\). Then \(f\) is bounded on \([a,b]\). More precisely, suppose that \(f\) is of bounded variation on \([a,b]\) with
\[\sum_{k}\left |\Delta f_{k}\right |\leq M\]
for all partitions of \([a,b]\). Then \(f\) is bounded on \([a,b]\) with
\[\left |f(x)\right |\leq\left |f(a)\right |+M\]
for all \(x\in [a,b]\).
Proof. For \(x\in (a,b)\), using the particular partition \({\cal P}=\{a,x,b\}\), we have
\[\left |f(x)-f(a)\right |+\left |f(b)-f(x)\right |\leq M,\]
which implies \(|f(x)-f(a)|\leq M\). Therefore, we obtain
\[\left ||f(x)|-|f(a)|\right |\leq |f(x)-f(a)|\leq M,\]
which implies \(|f(x)|\leq |f(a)|+M\). This completes the proof. \(\blacksquare\)
Definition. Let \(f\) be a real-valued function defined on the closed interval \([a,b]\), and let \({\cal P}\) be a partition of \([a,b]\). We write
\[V({\cal P},[a,b])=\sum_{k}|\Delta f_{k}|\]
to denote the sum corresponding to the partition \({\cal P}\) of \([a,b]\). The following supremum
\[V_{f}(a,b)=\sup_{{\cal P}\in\mathfrak{P}[a,b]}V({\cal P},[a,b])\]
is called the total variation of \(f\) on \([a,b]\). \(\sharp\)
Suppose that \(f\) is of bounded variation on \([a,b]\). Then, we immediately have \(0\leq V_{f}(a,b)<+\infty\). We also see that \(V_{f}(a,b)=0\) if and only if \(f\) is constant on \([a,b]\).
\begin{equation}{\label{map17}}\tag{3}\mbox{}\end{equation}
Proposition \ref{map17}. Suppose that the real-valued functions \(f\) and \(g\) are of bounded variation on \([a,b]\). Then, we have the following properties.
(i) The sum \(f+g\), the difference \(f-g\) and the product \(fg\) are also of bounded variation on \([a,b]\). Moreover, we also have
\[V_{f\pm g}(a,b)\leq V_{f}(a,b)+V_{g}(a,b)\] and
\[V_{fg}(a,b)\leq A\cdot V_{f}(a,b)+B\cdot V_{g}(a,b),\]
where
\[A=\sup_{x\in [a,b]}|g(x)|\mbox{ and }B=\sup_{x\in [a,b]}|f(x)|.\]
(ii) Suppose that there exists a positive number \(m\) satisfying \(0<m\leq |f(x)|\) for all \(x\in [a,b]\). Then \(1/f\) is of bounded variation on \([a,b]\) satisfying
\[V_{1/f}(a,b)\leq\frac{1}{m^{2}}\cdot V_{f}(a,b).\]
Proof. To prove part (i), let \(h(x)=f(x)g(x)\). For every partition \({\cal P}\) of \([a,b]\), we have
\begin{align*}
\left |\Delta h_{k}\right | & =\left |f(x_{k})g(x_{k})-f(x_{k-1})g(x_{k-1})\right |\\
& \leq\left |f(x_{k})g(x_{k})-f(x_{k-1})g(x_{k})\right |+\left |f(x_{k-1})g(x_{k})-f(x_{k-1})g(x_{k-1})\right |\\
& \leq\left |g(x_{k})\right |\left |f(x_{k})-f(x_{k-1})\right |+\left |f(x_{k-1})\right |\left |g(x_{k})-g(x_{k-1})\right |\\
& \leq A\cdot\left |\Delta f_{k}\right |+B\cdot\left |\Delta g_{k}\right |.
\end{align*}
Therefore, we obtain
\begin{equation}{\label{maeq457}}\tag{4}
\sum_{k=1}^{n}\left |\Delta h_{k}\right |\leq A\cdot\sum_{k=1}^{n}\left |\Delta f_{k}\right |+B\cdot\sum_{k=1}^{n}\left |\Delta g_{k}\right |.
\end{equation}
Since \(f\) and \(g\) are of bounded variation on \([a,b]\), there exists \(M_{1}\) and \(M_{2}\) satisfying \(\sum_{k=1}^{n}\left |\Delta f_{k}\right |\leq M_{1}\) and \(\sum_{k=1}^{n}\left |\Delta g_{k}\right |\leq M_{2}\), which imply
\[\sum_{k=1}^{n}\left |\Delta h_{k}\right |\leq A\cdot M_{1}+B\cdot M_{2}\equiv M.\]
This show that \(h\) is of bounded variation on \([a,b]\). By taking supremum on both sides of (\ref{maeq457}), we obtain
\[\sup_{{\cal P}\in\mathfrak{P}[a,b]}\sum_{k=1}^{n}\left |\Delta h_{k}\right |\leq
A\cdot\sup_{{\cal P}\in\mathfrak{P}[a,b]}\sum_{k=1}^{n}\left |\Delta f_{k}\right |
+B\cdot\sup_{{\cal P}\in\mathfrak{P}[a,b]}\sum_{k=1}^{n}\left |\Delta g_{k}\right |,\]
which says
\[V_{h}(a,b)\leq A\cdot V_{f}(a,b)+B\cdot V_{g}(a,b).\]
The proofs for the sum and difference are left as exercises.
To prove part (ii), let \(g=1/f\), we have
\begin{align*} \left |\Delta g_{k}\right | & =\left |\frac{1}{f(x_{k})}-\frac{1}{f(x_{k-1})}\right |\\ & =\left |\frac{\Delta f_{k}}{f(x_{k})f(x_{k-1})}\right |
\leq\frac{|\Delta f_{k}|}{m^{2}}.\end{align*}
By taking the supremum on both sides, we can obtain the desired results, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{ma100}}\tag{5}\mbox{}\end{equation}
Proposition \ref{ma100}. Suppose that the real-valued function \(f\) is of bounded variation on \([a,b]\). Given any \(c\in (a,b)\), the function \(f\) is also of bounded variation on \([a,c]\) and \([c,b]\). Moreover, we have
\[V_{f}(a,b)=V_{f}(a,c)+V_{f}(c,b).\]
Proof. We first show that \(f\) is of bounded variation on \([a,c]\) and \([c,b]\). Let \({\cal P}_{1}\) and \({\cal P}_{2}\) be two partitions of \([a,c]\) and \([c,b]\), respectively. Then \({\cal P}_{0}={\cal P}_{1}\cup {\cal P}_{2}\) is a partition of \([a,b]\). Therefore, we have
\begin{equation}{\label{maeq14}}\tag{6}
V({\cal P}_{1},[a,c])+V({\cal P}_{2},[c,b])=V({\cal P}_{0},[a,b])\leq V_{f}(a,b),
\end{equation}
which says \(V({\cal P}_{1},[a,c])\leq V_{f}(a,b)\) and \(V({\cal P}_{2},[c,b])\leq V_{f}(a,b)\). Since \({\cal P}_{1}\) and \({\cal P}_{2}\) can be any partitions of \([a,c]\) and \([c,b]\), respectively, it follows that \(f\) is of bounded variation on \([a,c]\) and on \([c,b]\).
From (\ref{maeq14}), by taking supremum on both sides, we also have
\[V_{f}(a,c)+V_{f}(c,b)\leq V_{f}(a,b).\]
To obtain the reverse inequality, let \({\cal P}\) be a partition of \([a,b]\), and let \({\cal P}_{0}={\cal P}\cup\{c\}\) be another partition of \([a,b]\) by adjoining the point \(c\in (a,b)\). If \(c\in [x_{k-1},x_{k}]\), we have
\[\left |f(x_{k})-f(x_{k-1})\right |\leq\left |f(x_{k})-f(c)\right |+\left |f(c)-f(x_{k-1})\right |,\]
which says \(V({\cal P},[a,b])\leq V({\cal P}_{0},[a,b])\). The points of \({\cal P}_{0}\) in \([a,c]\) determine a partition \({\cal P}_{1}\) of \([a,c]\), and the points in \([c,b]\) determine a partition \({\cal P}_{2}\) of \([c,b]\). Therefore, we obtain
\begin{align*} V({\cal P},[a,b]) & \leq V({\cal P}_{0},[a,b])\\ & =V({\cal P}_{1},[a,b])+V({\cal P}_{2},[a,b])\\ & \leq V_{f}(a,c)+V_{f}(c,b).\end{align*}
By taking supremum on both sides, we obtain
\[V_{f}(a,b)\leq V_{f}(a,c)+V_{f}(c,b).\]
This completes the proof. \(\blacksquare\)
\begin{equation}{\label{map15}}\tag{7}\mbox{}\end{equation}
Proposition \ref{map15}. Suppose that the real-valued function \(f\) is of bounded variation on \([a,b]\). Let \(V\) be a real-valued function defined on \([a,b]\) by
\begin{equation}{\label{maeq19}}\tag{8}
V(x)=\left\{\begin{array}{ll}
V_{f}(a,x) & \mbox{if \(a<x\leq b\)}\\
0 & \mbox{if \(x=a\)}.
\end{array}\right .
\end{equation}
Then, we have the following properties.
(i) \(V\) is an increasing function on \([a,b]\).
(ii) \(V-f\) is an increasing function on \([a,b]\).
Proof. To prove part (i), for \(a<x<y<b\), using Proposition \ref{ma100}, we have
\begin{align*} V(y) & =V_{f}(a,y)\\ & =V_{f}(a,x)+V_{f}(x,y)\\ & =V(x)+V_{f}(x,y),\end{align*}
which implies
\begin{equation}{\label{ma102}}\tag{9}
V(y)-V(x)=V_{f}(x,y)\geq 0.
\end{equation}
To prove part (ii), considering a particular partition \({\cal P}=\{x,y\}\), we have
\begin{equation}{\label{ma101}}\tag{10}
f(y)-f(x)=V({\cal P},[x,y])\leq V_{f}(x,y).
\end{equation}
For \(x\in [a,b]\), we define \(D(x)=V(x)-f(x)\). Therefore, for \(a\leq x<y\leq b\), using (\ref{ma101}), we have
\begin{align*} D(y)-D(x) & =V(y)-V(x)-\left [f(y)-f(x)\right ]\\ & =V_{f}(x,y)-\left [f(y)-f(x)\right ]\geq 0.\end{align*}
This completes the proof. \(\blacksquare\)
\begin{equation}{\label{mat18}}\tag{11}\mbox{}\end{equation}
Theorem \ref{mat18}. Let \(f\) be a real-valued function defined on \([a,b]\). Then \(f\) is of bounded variation on \([a,b]\) if and only if \(f\) can be expressed as the difference of two increasing functions.
Proof. Suppose that \(f\) is of bounded variation on \([a,b]\). We can write \(f=V-D\), where \(V\) is the function defined in (\ref{maeq19}) and \(D=V-f\). Both \(V\) and \(D\) are increasing functions on \([a,b]\) by referring to Proposition \ref{map15}. The converse follows immediately from Propositions \ref{map16} and \ref{map17}. This completes the proof. \(\blacksquare\)
The representation of a function of bounded variation as a difference of two increasing functions is not unique. If \(f=f_{1}-f_{2}\), where \(f_{1}\) and \(f_{2}\) are increasing, we also have \(f=(f_{1}+g)-(f_{2}+g)\), where \(g\) is an arbitrary increasing function. Therefore, we obtain a new representation of \(f\). If \(g\) is strictly increasing, then the functions \(f_{1}+g\) and \(f_{2}+g\) are also strictly increasing. Therefore, Theorem \ref{mat18} also holds true when increasing functions are replaced by strictly increasing functions.
\begin{equation}{\label{map20}}\tag{12}\mbox{}\end{equation}
Proposition \ref{map20}. Suppose that the real-valued function \(f\) is of bounded variation on \([a,b]\). We consider the real-valued function \(V\) defined in (\ref{maeq19}). Then \(f\) is continuous at \(c\) if and only if \(V\) is continuous at \(c\).
Proof. Since \(V\) is monotonic, the righthand limit \(V(x+)\) and lefthand limit \(V(x-)\) exist for each \(x\in (a,b)\). Since \(f\) can be expressed as the difference of increasing functions, it also says that \(f(x+)\) and \(f(x-)\) exists for each \(x\in (a,b)\). For \(a<x<y\leq b\), using (\ref{ma101}) and (\ref{ma102}), we have
\begin{align*}
0 & \leq |f(y)-f(x)|\leq V_{f}(x,y)\\
& =V(y)-V(x),
\end{align*}
which implies
\begin{align*} 0 & \leq |f(x+)-f(x)|=\left |\lim_{y\rightarrow x+}f(y)-f(x)\right |\\ & \leq \lim_{y\rightarrow x+}V(y)-V(x)=V(x+)-V(x).\end{align*}
Similarly, we can obtain
\[0\leq |f(x)-f(x-)|\leq V(x)-V(x-).\]
These inequalities say that the continuity of \(V\) at \(c\) implies the continuity of \(f\) at \(c\).
To prove the converse, suppose that \(f\) is continuous at \(c\). Given \(\epsilon >0\), there exists \(\delta >0\) such that
\begin{equation}{\label{ma103}}\tag{13}
0<|x-c|<\delta\mbox{ implies }|f(x)-f(c)|<\frac{\epsilon}{2}.
\end{equation}
For this same \(\epsilon\), using the concept of supremum, there exists a partition \({\cal P}\) of \([c,b]\) satisfying
\[x_{0}=c<x_{1}<x_{2}<\cdots <x_{n}=b\]
and
\begin{equation}{\label{maeq21}}\tag{14}
V_{f}(c,b)-\frac{\epsilon}{2}<\sum_{k=1}^{n}\left |\Delta f_{k}\right |.
\end{equation}
Adding more points to the partition \({\cal P}\) can increase the sum \(\sum_{k}|\Delta f_{k}|\). Therefore, we can assume that \(0<x_{1}-x_{0}<\delta\), which means
\begin{align*} \left |\Delta f_{1}\right | & =|f(x_{1})-f(x_{0})|\\ & =|f(x_{1})-f(c)|<\frac{\epsilon}{2}\end{align*}
by using (\ref{ma103}). Since \(\{x_{1},x_{2},\cdots ,x_{n}\}\) is a partition of \([x_{1},b]\), using (\ref{maeq21}), we have
\begin{align*} V_{f}(c,b)-\frac{\epsilon}{2} & <\frac{\epsilon}{2}+\sum_{k=2}^{n}\left |\Delta f_{k}\right |\\ & \leq\frac{\epsilon}{2}+V_{f}(x_{1},b),\end{align*}
which implies
\[V_{f}(c,b)-V_{f}(x_{1},b)<\epsilon .\]
We also have
\begin{align*} 0 & \leq V(x_{1})-V(c)=V_{f}(a,x_{1})-V_{f}(a,c)\\ & =V_{f}(c,x_{1})=V_{f}(c,b)-V_{f}(x_{1},b)<\epsilon .\end{align*}
This shows that \(0<x_{1}-c=x_{1}-x_{0}<\delta\) implies \(0\leq V(x_{1})-V(c)<\epsilon\), i.e., \(V(c+)=V(c)\). We can similarly prove \(V(c-)=V(c)\). Therefore \(V\) is continuous at \(c\in (a,b)\). We can also show the continuities at the endpoints \(a\) and \(b\) by following the above argument with slight modification. This completes the proof. \(\blacksquare\)
Theorem. Suppose that the real-valued function \(f\) is continuous on \([a,b]\). Then \(f\) is of bounded variation on \([a,b]\) if and only if \(f\) can be expressed as the difference of two increasing continuous functions on \([a,b]\). The result still holds true when increasing functions are replaced by strictly increasing functions.
Proof. Suppose that the continuous function \(f\) is of bounded variation on \([a,b]\). We can write \(f=V-D\), where \(V\) is the function defined in (\ref{maeq19}) and \(D=V-f\). Both \(V\) and \(D\) are continuous and increasing functions on \([a,b]\) by referring to Propositions \ref{map15} and \ref{map20}. The converse follows immediately from Propositions \ref{map16} and \ref{map17}. This completes the proof. \(\blacksquare\)
