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}

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