Let $f:[a,b] \to \mathbb R$ a function of class $C^1$ on the interval $[a,b]$. Prove that:
i) $f$ is a function of bounded variation.
ii) The equality $V_a^b f= \int_a^b |f'(x)|dx$ holds.
My attempt at a solution.
For point i) what I did was:
By hypothesis, $f'$ is a continuous function on $[a,b]$, this means $f'$ attains a maximum and a minimum, call them $M$ and $m$ respectively and call $A=max\{|M|,|m|\}$. Then, $|f'(x)|\leq A$ for every $x \in [a,b]$. Let $\pi$ be a partition of $[a,b]$ with $\pi=\{x_0,x_1,…,x_n\}$ and $\sum(\pi)=\sum_{k=1}^n |f(x_k)-f(x_{k-1})|$. By the mean value theorem, for each subinterval $[x_{k-1},x_k]$, there exists $c_k \in [x_{k-1},x_k]$ such that $f(x_k)-f(x_{k-1})=f'(c)(x_k-x_{k-1}) \implies |f(x_k)-f(x_{k-1})|=|f'(c)|(x_k-x_{k-1})\leq A(x_k-x_{k-1})$.
From here, it follows that given a partition $\pi$ of $[a,b]$, $\sum_{k=1}^n |f(x_k)-f(x_{k-1})| \leq \sum_{k=1}^n A(x_k-x_{k-1})=A\sum_{k=1}^n x_k-x_{k-1}=A(b-a)$. This proves $f$ is of bounded variation.
I have problems with part ii), I must prove $V_a^b(f)\leq \int_a^b |f'(x)|dx$ and $\int_a^b |f'(x)|dx\leq V_a^b(f)$ but I don't know how to do this.
Best Answer
Hint:
From the definition,
$$V_a^b(f) = \sup_P \big( \sum_{i=1}^{k_p} |f(x_i) - f(x_{i-1})|\big)\ ,$$
thus you can find a sequence of partition $P_n$ such that $||P_n|| \to 0$ and
$$V_a^b(f) = \lim_{n\to \infty} \big( \sum_{i=1}^{k_{p_n}} |f(x_i) - f(x_{i-1})|\big)\ \ \ \ (How?)$$
On the other hand, what is
$$\lim_{n\to \infty} \sum_{i=1}^{k_{p_n}} |f'(c_i)| (x_i -x_{i-1})\ \ ?$$
Why the limit exists?