Let $$f(0)=0,\;\;f(x)=e^{-2/x}\sin\left(e^{1/x}\right),$$ is $f$ bounded variation on $[0,1]$?
Here is my thinking:
Since $f$ is differentiable on $(0,1]$ and continuous on $[0,1]$
If $f^\prime$ is bounded, we can use mean value theorem to prove it.
Let $$f(0)=0,\;\;f(x)=e^{-2/x}\sin\left(e^{1/x}\right),$$ is $f$ bounded variation on $[0,1]$?
Here is my thinking:
Since $f$ is differentiable on $(0,1]$ and continuous on $[0,1]$
If $f^\prime$ is bounded, we can use mean value theorem to prove it.
That is a good approach.
Because $\lim\limits_{x\to 0+} f'(x)$ exists, $f$ is in fact in $C^1[0,1]$, hence absolutely continuous by the Fundamental Theorem of Calculus. Absolute continuity implies bounded variation.