3
$\begingroup$

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.

  • 1
    @Alex Is it so obvious that if $g$ has bounded variation then $f$ has bounded variation? I mean, more obvious than the direct approach the OP is advocating?2012-12-02

1 Answers 1

1

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.