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Is there a function $f\colon(-\delta,\delta)\to\Bbb R$ satisfying the folowing conditions(real number $ \delta\gt0$)?

(i) $f$ is differentiable on $(-\delta,\delta)$;

(ii) the second derivative of $f$ exists at $0$, that is $f''(0)$ exists

(iii) there is a sequence $\{x_n\}$, $-\delta\lt x_n\lt \delta$, $\lim\limits_{n\to\infty}x_n=0$, such that $f'$ is not continuous at all $x_n$.


I can't construct such a example

Any help will be appreciated!

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  • $\begingroup$ I thought a function has to be at least continuous to be differentiable? $\endgroup$ – 3x89g2 Jul 20 '15 at 17:19
  • $\begingroup$ @Misakov $f'$ is differentiable at $0$, but$f'$ is not continuous at $x_n$ $\endgroup$ – Clin Jul 20 '15 at 17:25
  • $\begingroup$ @Misakov: $f'$ needs be continuous at $x=0$, nothing more. $\endgroup$ – user251257 Jul 20 '15 at 17:26
  • $\begingroup$ You might be able to use something like Volterra's function $\endgroup$ – Omnomnomnom Jul 20 '15 at 17:38
  • $\begingroup$ Such an example definitely exists. Writing one down might be a bit of a pain... $\endgroup$ – David C. Ullrich Jul 20 '15 at 17:46
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The standard way to get a function that is differentiable everywhere (with bounded derivative) but whose derivative has a discontinuity point is

$$ g(x) = \begin{cases} 0 & x=0 \\ x^2\sin(1/x) & x\ne 0 \end{cases} $$

Now select your $x_n$s and define

$$ f(x) = x^2 \sum_{k=1}^\infty \frac{g(x-x_k)}{2^k} $$

This places a discontinuity of $f'$ at each $x_k$, and the overall factor of $x^2$ squeezes the range of $f'$ around $0$ enough to make sure $f''(0)$ exists.

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  • $\begingroup$ Why does $f''(0)$ exist? I'm not saying it doesn't exist. But, say, $g$ bounded and $f(x) = x^2g(x)$ does not imply that $f''(0)$ exists, so it seems that at least more explanation is needed... $\endgroup$ – David C. Ullrich Jul 20 '15 at 20:17
  • $\begingroup$ Never mind - the point is your $g'$ is bounded... $\endgroup$ – David C. Ullrich Jul 20 '15 at 20:34

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