Please give an example (if it exists) for a function which is differentiable everywhere but not absolutely continuous.
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Let $f(x) = \sin(x^2)$. This is clearly $C^\infty$ but it's not even uniformly continuous. Let $\varepsilon = \frac{1}{2}$. Let $x = \sqrt{n\pi}$ and $y = \sqrt{n\pi + \frac{\pi}{2}}$. Then (exercise) we can make $x$ and $y$ arbitrarily close by making $n$ large, but $f(x) = 0$ and $f(y) = \pm 1$.
Julien Clancy
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Can we find such a function defined on closed and bounded interval – Sushil Dec 07 '14 at 05:39
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Since every Lipschitz function is absolutely continuous, such a function must have unbounded derivative, so it can't be even $C^1$. I can't think of one off of the top of my head, ask it as a separate question? – Julien Clancy Dec 07 '14 at 20:31
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Consider $f(x)=x^2\sin{\frac{1}{x^2}}$ on $[0,1]$. It is differentiable, and, were it absolutely continuous, its derivative would be integrable.
As a consequence, $x \longmapsto \frac{-2}{x}\cos(x^{-2})$ would be integrable over $[0,1]$, thus $x \longmapsto \frac{\cos{x^2}}{x}$ would be integrable over $[1,\infty)$, therefore $u \longmapsto \frac{\cos{u}}{u}$ would be integrable over $(1,\infty)$.
So the function is differentiable yet not absolutely integrable.
Aphelli
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Where did you get $\frac{-2}{x}\sin(x^{-2})$? Don't you have $f'(x)=-\frac{2 \cos \left({x^{-2}}\right)}{x}+2 x \sin ({x^{-2}})$? – Jara Dec 11 '19 at 18:56
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