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It is well-known that if $f$ is twice differentiable at $a$, then

$$ f''(a) = \lim_{h\to 0} \frac{f(a+2h)-2f(a+h) + f(a)}{h^2}. $$

See e.g. this question or this question.

On the other hand, the RHS limit may exist without $f$ being twice differentiable. For instance, if $f(x) = x^3 \sin(1/x)$ for $x\neq 0$ and $f(0)=0$, then at $a=0$ the RHS is $2h(4\sin(1/2h) - \sin(1/h))$, which has limit $0$; but this function $f$ is not twice differentiable at $0$.

My question is, under what additional hypotheses can we conclude from the existence of the RHS limit that $f$ is twice differentiable?

We probably need to assume that $f$ is once differentiable on a neighborhood of $a$, but that is not sufficient since the counterexample above is in fact smooth at all points $x\neq 0$. For the same reason, it is not sufficient to assume that the above limit exists for all $a$ in some neighborhood.

Note that this is not the same as this question, which is asking for a limit definition of the second derivative that doesn't require extra hypotheses; here I'm asking what extra hypotheses can be added to this particular definition to make it work.

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Mike Shulman
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  • Without looking into it, I'd assume that we need the limit to exist and be continuous (w.r.t. $a$) in a neighbourhood around the relevant point. – Arthur May 13 '14 at 16:59
  • @Arthur that was my first guess at a stronger condition that eliminates the particular counterexample, but I don't have any other reason to believe that it would suffice. – Mike Shulman May 13 '14 at 21:15
  • Do you really mean a forward difference in the numerator ? – Tony Piccolo May 14 '14 at 11:04
  • @TonyPiccolo what do you mean? – Mike Shulman May 15 '14 at 16:41
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    Why don't you consider the 2nd order symmetric derivative ? – Tony Piccolo May 15 '14 at 21:57
  • @TonyPiccolo My best guess is that you're thinking of replacing the numerator with $f(x+h)-2f(x)+f(x-h)$. That has the problem that it equals zero whenever f is an odd function around x. – Mike Shulman May 16 '14 at 04:54
  • Can you explain why that is a problem ? – Tony Piccolo May 16 '14 at 23:34
  • @TonyPiccolo Many odd functions are not twice differentiable. – Mike Shulman May 17 '14 at 23:51
  • The same thing happens for the example in your question ... – Tony Piccolo May 18 '14 at 05:36
  • @TonyPiccolo I really don't understand what you're getting at. I'm asking a question. You seem to be suggesting that I should have asked a different question. Well, this is the question that I'm asking. If you want to know about symmetric second differences, you can ask another question. – Mike Shulman May 19 '14 at 14:32
  • I cannot understand why, after producing the limit of a difference quotient, you suggest two links that speak of another. I thought it was clear. – Tony Piccolo May 19 '14 at 15:35
  • @TonyPiccolo Oh, is that all? It's basically the same proof in both cases, and one of the links mentions both. In fact, I suspect that the answer to this question, if there is one, will work in both cases too. I just chose the version I did because it works as-is in slightly more cases. – Mike Shulman May 19 '14 at 19:45
  • I agree and find the question very interesting. – Tony Piccolo May 19 '14 at 22:10
  • The function $f(x)=x^3\sin(1/x)$ is twice differentiable at $x=0$. But it is not twice continuously differentiable at $x=0$. – Mark Viola Jul 27 '22 at 16:59

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