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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.

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If it helps, let $y=e^{-1/x}$ to transform the function to $g(y)=y^2\sin(1/y)$ on the interval $[0,1/e]$, which is of bounded variation by essentially what you said using the mean value theorem. –  Alex R. Nov 29 '12 at 2:29
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@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? –  Did Dec 2 '12 at 12:48

2 Answers 2

By taking the derivative we have $f'(x)=2y\sin(1/y)-\cos(1/y)$. Since we are over $[0,1]$, this is bounded by $2+1=3$. So we can assert bounded variation via mean value theorem.

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What is the relationship between $x$ and $y$ in your answer? –  Jonas Meyer Jul 1 '13 at 6:24

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.

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