# Is there a function $f''(0)$ exists, $f'$ is not continuous on $(-\delta,\delta)$

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!

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

$$g(x) = \begin{cases} 0 & x=0 \\ x^2\sin(1/x) & x\ne 0 \end{cases}$$
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.
• 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... – David C. Ullrich Jul 20 '15 at 20:17
• Never mind - the point is your $g'$ is bounded... – David C. Ullrich Jul 20 '15 at 20:34