$\psi$ continuous and g(Bounded variation) $\Rightarrow$ $\psi \circ g$ bounded variation?

Question

Suppose I have a function $\psi$ that is continuous on $\mathbb{R}$ and $g$ is a function with a bounded variation on $[0,1]$. Is it true that $\psi \circ g$ has a bounded variation on [0,1]?

If so, why? Thanks

Answer 1

No it is false. The function $\psi(x)=x\sin(1/x)$ with $\psi(0)=0$ is continuous. The identity function $g(x)=x$ has bounded variation on $[0,1]$, but $\psi \circ g=\psi$ does not have bounded variation on $[0,1]$.

To show $\psi$ does not have bounded variation, let $\{x_j\}=\{1/(\pi(k+1/2))\}$ then $$\sum_{k=0}^\infty |x_j \sin(1/x_j)|=\sum_{k=0}^\infty \frac{2}{\pi(k+1/2)}=\infty.$$