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It is known that 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?

  • You probably mean $x \in (0,1]$ for $f(x) = 1/x$. – Mark Fantini Jul 03 '14 at 05:27
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    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. –  Jul 03 '14 at 05:33

2 Answers2

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

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

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

user71352
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  • You are right. I made a mistake in calculating. Thank you. – user71352 Jul 03 '14 at 05:18
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    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! –  Jul 03 '14 at 06:27
  • Thank you for catching my mistake. I shall update. – user71352 Jul 03 '14 at 06:43
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Sticking with the example $f(x) = x^2\sin (1/x^2),$ which is everywhere differentiable, I think it's easier to deal with its total variation directly from the definition. Let $a_n = 1/\sqrt {\pi/2+2n\pi},\, b_n = 1/\sqrt {2n\pi}.$ Then

$$\sum_{n=1}^{\infty}|f(b_n)-f(a_n)| = \sum_{n=1}^{\infty}|f(a_n)|= \sum_{n=1}^{\infty}\frac{1}{\pi/2+2n\pi}.$$

Apply the limit comparsion test, using the terms $1/n$ for comparison, to see this series diverges. Since the total variation of $f$ on $[0,1]$ is at least the $\sup$ of all partial sums of this infinite series, we see $f$ has unbounded total variation on $[0,1].$

zhw.
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