# 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

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}

## Author: mathtuition88

http://mathtuition88.com

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