Suppose $f$ is continuous on $[a,b]$, and its derivative $f'$ exists and is continuous on $(a,b]$ . Suppose $\int_a^b |f'|<\infty$. Show that $TV(f) \leq \int_a^b |f'|$.
When $f$ is continuously differentiable, this property holds trivially, and we could even prove the equality holds. However, in this case, $f'(a)$ does not exist, and I wonder how things would change. Many thanks!
Hint: Let $\{x_i: 0\leq i \leq n\}$ be a partition of $[a,b]$. Then $f(x_1)-f(a+r)=\int_{a+r}^{x_1} f'(t)dt$ for every $r \in (0,x_1-a)$. By continuity of $f$ and integrability of $f'$ this implies that $f(x_1)-f(a)=\int_a^{x_1} f'(t)dt$. Can you finish?