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I'm reading a note about higher derivative:

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and a remark

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The remark said that "The proof below does NOT assume continuity of the second derivative."

From my textbook and Wikipedia, I got that the necessary condition for the second derivative to be symmetric is that it is continuous, contradicting the above remark.

Could you please explain elaborate on this discrepancy? Thank you for your help!

Here is the proof in the note:

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Akira
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  • I suspect the difference is Gateaux vs Frechet differentiability. The Wikipedia article just assumes the second partials exist, which is essentially Gateaux differentiability. In this case, you need continuity to get symmetry, as you correctly noted. The notes you are reading assume the function is twice Frechet differentiable, which means essentially the 3rd order Taylor expansion holds. This is a stronger condition than merely existence of partial derivatives. – Jeff Nov 09 '19 at 20:35
  • Hi @Jeff, you meant that in the note, $f \in \mathcal C^2$? – Akira Nov 09 '19 at 20:38
  • It suffices that $D^2f$ (in the Fréchet sense) exists at $x$, it need not exist at any other point. – Daniel Fischer Nov 09 '19 at 20:41
  • Hi @DanielFischer, did you meant it suffices that $D^2 f$ is continuous at $x$? – Akira Nov 09 '19 at 20:43
  • No. The second derivative of $f$ need not exist at any other point. Wherever $D^2 f$ exists (as a Fréchet derivative) it is a symmetric bilinear map (when we identify $L(V,L(V,W))$ with $L^2(V,W)$). – Daniel Fischer Nov 09 '19 at 20:45
  • Hi @DanielFischer, I got where the misunderstanding arises. In the case the domain of $f$ is $\mathbb R^n$, $D^2 f$ is symmetric bilinear map as long as it exists. But in general (the domain of $f$ is not necessarily $\mathbb R^n$), then the continuity of $f$ is required. Could you please confirm if my understanding is correct? – Akira Nov 09 '19 at 20:53
  • No. I mean what Daniel said. The note assume Frechet differentiability at $x$, which is stronger than Gateaux. – Jeff Nov 09 '19 at 21:06
  • As Jeff said, it's the difference between the Gateuax derivative and the Fréchet derivative that is at play here. The domain of $f$ isn't important, if $f$ is defined on an open subset of a Banach space and it is twice Fréchet differentiable at $x$, the second derivative is still symmetric. – Daniel Fischer Nov 09 '19 at 21:07
  • Hi @DanielFischer, I have posted my concern about this issue here. Could you please have a look at it? – Akira Nov 09 '19 at 22:44
  • Hi @DanielFischer, the proof in the note mentions that $f$ is differentiable. I could not get why $D^2 f(a)$ exists implies $D f(x)$ exists for all $x$ in some neighborhood of $a$. I think there maybe the case that $D^2 f(a)$ exists, but there exists $x_n$ in arbitrary neighborhood of $a$ such that $D f(x)$ does not exist. As such, $D^2 f(a)$ exists but $f$ maybe not differentiable on any whole neighborhood of $a$. Hence, we can not apply mean value theorem. – Akira Nov 10 '19 at 09:34
  • Hi @DanielFischer, I just posted my attempt on this theorem here. If you don't mind, please have a look at it. Thank you so much! – Akira Nov 10 '19 at 12:08

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