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If i have a differentiable function $f$ . I know in general $f'$ isn't always continuous. My question is if we have $f''(0)$ exists, can i say that $f'$ is continuous on some open interval contain 0?

I know that if $f'(0)$ exist then we can't say $f$ is local continuous on 0. But since the derivatives need satisfy more properties , i can't say if this is true or false. I also looked up some descriptions of the discontinued set of derivatives but i still couldn't find a counterexample...

zhw.
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Idele
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    Let $g=f'$. Now the question reads "if $g'(0)$ exists, can we say that $g$ is continuous on some open interval containing $0$?" It's a bit more concrete. – Arthur Aug 25 '17 at 20:33
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    @Arthur yes, but now $g$ satisfies more properties than a usual function do since it's some function's derivative.(such as intermediate value property and the set of the continuous points is a dense G delta set ). – Idele Aug 25 '17 at 20:42
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    @Arthur I don't think those two questions are really the same. Consider $$g(x) = \begin{cases} x^2,& x\in \mathbb{Q} \ -x^2,& x\notin \mathbb{Q}\end{cases}$$ Then, $g$ is differentiable at $0$ and continuous nowhere else, so $g$ is a counterexample of $g'(0)$ existing implying $g$ being continuous in a neighborhood of $0$, but $g$ is not the derivative of a function, so it does not function as a counterexample to the question that was asked. – Michael L. Aug 25 '17 at 20:43
  • Yes, you're right. It was an overly simple simplification. – Arthur Aug 25 '17 at 20:43
  • This seems relevant. – Michael L. Aug 25 '17 at 20:47
  • @MichaelLee oh! I see, take a differential function f such that D,the discontinued points set of $f'$, is positive lebesgue measure (for example f be the Volterra function ),take a limit point $a$ of D. Then take function g such that g'=(x-a)^2f(x) . Then g is a counterexample. – Idele Aug 25 '17 at 21:21

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Counterexample: On $(-2,2)$ define $f(x) = x^2\sin(1/x), x\ne 0,$ $f(0)=0.$ Then $f$ is differentiable on $(-2,2),$ with $f'$ bounded there, and $f'$ is continuous on $(-2,2)$ except at $0.$

For $x\in (-1,1),$ set

$$g(x)= \sum_{n=1}^{\infty}\frac{f(x-1/n)}{2^n}.$$

Check with the usual theorems that $g$ is differentiable in $(-1,1),$ with $g'$ being discontinuous at each $1/n, n = 1,2,\dots.$

Now define

$$h(x) = x^2g(x)$$

for $x\in (-1,1).$ Then $h$ is differentiable on $(-1,1),$ with $h'(x) = 2xg(x)+x^2g'(x).$ Because $g'$ is a bounded function, $h''(0)$ exists. And since $g'$ is discontinuous at each $1/n,$ so is $h'.$ Thus there is no neighborhood of $0$ where $h'$ is continuous.

zhw.
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