I am trying to prove that $f(x)$ is in BV or in BV$([0,1])$?

Question

I am trying to prove that $f(x)$ is a Bounded Variation function (BV) or in BV$([0,1])$? I have already verified that $f(x)=\tan^{-1}(x)$ is continuous on $[0,1]$ and differentiable on $[0,1]$. Also $f'(x)$ is continuous on $[0,1]$ and therefore continuous on $(0,1)$. $f'(x)$ is bounded. In fact $|f'(x)|\leq 1$ for all $x \in [0,1]$. Basing on these facts, I am concluding that $f$ is in BV$([0,1])$.
My question is:

Is this correct? Is it sufficient? If $f$ is in BV$([0,1])$. How do I proceed to show that $f$ is in BV?
$\arctan x$

Answer 1

Your proof has the right ingredients, but it can be made much simpler: $f(x)=\arctan x$ is monotonically increasing, hence it is of bounded variation. In fact the total variation of $f$ over $[0,1]$ is $$\arctan(1)-\arctan(0)=\frac{\pi}{4}. $$