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A function with bounded variation is differentiable almost everywhere.

There are also functions with unbounded variation that are differentiable almost everywhere (e.g. take $f:[0,1]\rightarrow\mathbb{R}$ with $f(0)=0$ and $f(x)=1/x$ otherwise).

But does there exist a function on $[0,1]$ with unbounded variation that is everywhere differentiable?

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You probably mean $x \in (0,1]$ for $f(x) = 1/x$. – Mark Fantini Jul 3 '14 at 5:27
I defined $f(0)=0$ specifically so I could include 0 in the domain, but I did label the domain sloppily and I can see where I have miscommunicated. Will edit now to clarify this. – user105475 Jul 3 '14 at 5:33
up vote 4 down vote accepted

Consider $f:[0,1]\to\mathbb{R}$ by $f(x)=x^{2}\sin(\frac{1}{x^{2}})$ for $x\neq0$ and $f(0)=0$.

$\int_{0}^{1}\lvert f'(x)\rvert dx=\int_{0}^{1}\lvert2x\sin(\frac{1}{x^{2}})-\frac{2}{x}\cos(\frac{1}{x^{2}})\rvert\ge\int_{0}^{1}\frac{2}{x}\lvert\cos(\frac{1}{x^{2}})\rvert dx-\int_{0}^{1}2x\vert\sin(\frac{1}{x^{2}})\rvert$

$\ge\int_{0}^{1}\frac{2}{x}\lvert\cos(\frac{1}{x^{2}})\rvert dx-1$.

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You are right. I made a mistake in calculating. Thank you. – user71352 Jul 3 '14 at 5:18
I think that $f'(x)=2x\sin(\frac{1}{x^2})-\frac{2}{x}\cos(\frac{1}{x^2})$, but the rest of your argument still holds even if this is changed. A nice example, thank you! – user105475 Jul 3 '14 at 6:27
Thank you for catching my mistake. I shall update. – user71352 Jul 3 '14 at 6:43

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