Exercise 7.5.9

Question

Given a function $f$ on $[a, b]$, define the total variation of $f$ to be
$$
V f=\sup \left\{\sum_{k=1}^{n}\left|f\left(x_{k}\right)-f\left(x_{k-1}\right)\right|\right\},
$$
where the supremum is taken over all partitions $P$ of $[a, b]$.
\begin{enumerate}[label=(\alph*)] 
\item If $f$ is continuously differentiable ($f^{\prime}$ exists as a continuous function), use the Fundamental Theorem of Calculus to show $V f \leq \int_{a}^{b}\left|f^{\prime}\right|$.
\item Use the Mean Value Theorem to establish the reverse inequality and conclude that $V f=\int_{a}^{b}\left|f^{\prime}\right|$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item For any subinterval $[c,d] \subseteq [a,b]$,
$$\int^d_c \left|f'(x)\right| \geq \left|\int^d_c f'(x)\right| = \left|f(d) - f(c)\right|$$
Applying this to each subinterval of any partition $P$ leaves us with
$$\sum_{k=1}^n \left|f(x_k)-f(x_{k-1})\right| \leq \int^b_a \left|f'\right|$$
and hence $Vf \leq \int^b_a |f'|$.

\item Let $\epsilon > 0$. Since $|f'|$ is uniformly continuous over the closed interval $[a,b]$, we can create a partition $P$ with $n$ elements so that each subinterval has length less than $\delta$, where $|x-y| < \delta \implies \left| \left|f'(x)\right| - \left|f'(y)\right| \right| < \frac{\epsilon}{b-a}$.

Within a representative subinterval $[x_{k-1}, x_k]$ of $P$, by the Mean Value Theorem $\exists \hat{x}_k \in [x_{k-1}, x_k]$ satisfying
$$f'(\hat{x}_k) = \frac{f(x_k) - f(x_{k-1})}{x_k - x_{k-1}} $$
Letting $M_k$ be the supremum of $|f'|$ over $[x_{k-1}, x_k]$, note that
$$
    \begin{aligned}
        U(\left|f'\right|, P) - \sum^n_{k=1} \left|f(x_k) - f(x_{k-1})\right|
&= \sum^n_{k=1} \left(M_k - \left|\frac{f(x_k) - f(x_{k-1})}{x_k - x_{k-1}}\right|\right) (x_k - x_{k-1}) \
&= \sum^n_{k=1} \left(M_k - \left|f'(\hat{x}_k)\right| \right) (x_k - x_{k-1}) \
&\leq \frac{\epsilon}{b-a} \sum^n_{k-1} x_k-x_{k-1} = \epsilon
    \end{aligned}
$$
Since $U(|f'|, P) \geq \int^b_a |f'|$, we have $V f \geq \int^b_a |f'| - \epsilon$, or simply $V f \geq \int^b_a |f'|$.

It's worth noting that we can't use argument with $L(|f'|, P)$ for the $\leq$ inequality because of the $\sup$ in $V f$.

\end{enumerate}

Exercise 7.5.9

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Exercise 7.5.9

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