Composition of Continuously Differentiable Function and Function of Bounded Variation

Assume $\phi$ is a continuously differentiable function on $\mathbb{R}$ and $f$ is a function of bounded variation on $[0,1]$ . Then $\phi(f)$ is also a function of bounded variation on $[0,1]$ .

Proof:

$\displaystyle V_a^b(\phi(f))=\sup_{P\in\mathcal{P}}\sum_{i=0}^{N_P-1}|\phi(f(x_{i+1}))-\phi(f(x_i))|$ where $\displaystyle \mathcal{P}=\{P|P:a=x_0<x_1<\dots<x_{N_P}=b\ \text{is a partition of}\ [a,b]\}.$

By Mean Value Theorem, $\displaystyle |\phi(f(x_{i+1}))-\phi(f(x_i))|=|f(x_{i+1})-f(x_i)||\phi'(c)|$ for some $c\in(x_i, x_{i+1})$ .

Since $\phi'$ is continuous, it is bounded on $[0,1]$ , say $|\phi'(x)|\leq K$ for all $x\in[0,1]$ . Thus
$\begin{aligned} V_a^b(\phi(f))&=\sup_{P\in\mathcal{P}}\sum_{i=0}^{N_P-1}|\phi(f(x_{i+1}))-\phi(f(x_i))|\\ &\leq K\sup_{P\in\mathcal{P}}\sum_{i=0}^{N_P-1}|f(x_{i+1})-f(x_i)|\\ &=KV_a^b(f)\\ &<\infty. \end{aligned}$