Given some function $g:\mathbb{R} \to \mathbb{R}$, it is twice differentiable if $g'(x)$ and $g''(x)$ both exist. But what about something like $f:\mathbb{R}^2 \to \mathbb{R}$. Now I have partials to worry about. What is the definition of being twice differentiable in this case?
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1For a precise definition of higher order differentiability, see A question about derivatives between Euclidean spaces: how should we construct it and interpret its definition?, and regarding the currently accepted answer’s latest edit, a proof (of a very slightly modified statement) can be found here: Does Frechet differentiability of the first order partials imply twice Frechet differentiability? – peek-a-boo Feb 12 '23 at 19:05
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@peek-a-boo: That means my answer, after modification is correct? Sadly the answer was accepted before the call to action, so I couldn't delete it and correct it. So, I had to do it on the fly. – A. P. Feb 12 '23 at 19:10
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It is similar, only now with partial derivatives. To be $f:{\bf R}^2\to {\bf R}$ twice $\color{red}{\text{[edit: continuously]}}$ differentiable, $\color{red}{\text{[edit: it is necessary and sufficient]}}$ that all second-order partial derivatives exist $\color{red}{\text{[edit: and to be continuous]}}$. That is, $f_{xx},f_{yy},f_{xy}$ and $f_{yx}$ they have to exist $\color{red}{\text{[edit: and to be continuous]}}$.
NB: Notice that more generally $f$ is twice differentiable iff $f$ is differentiable and each partial derivatives $f_{x_j}$ is differentiable. Then, the proposition given above it a result from here.
A. P.
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This requires proof though. You need to be more careful. If $f:\mathbb R^n\to \mathbb R^m$ and assuming differentiability, then $f'$ is a linear operator $f':\mathbb{R}^n\to L(\mathbb{R}^n,\mathbb{R}^m)$ so $f''$ becomes a map $f'':\mathbb{R}^n\to L(\mathbb{R}^n, L(\mathbb{R}^n,\mathbb{R}^m)).$ Relating these ideas to the corresponding partial derivatives is not hard, but it is not trivial, either. – Matematleta Feb 12 '23 at 17:18
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This is false. Existence of partial derivatives does not imply differentiability. – Nicholas Todoroff Feb 12 '23 at 18:18
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@NicholasTodoroff It seems that when I wrote, I forgot to write the important continuity hypothesis. I think that is now correct. – A. P. Feb 12 '23 at 18:30
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Indeed, it should works. Since $f_{x_i,x_j}$ there exists and they are continuous, then they are differentiables. Therefore, $f$ is twice differentiable. – A. P. Feb 12 '23 at 18:38
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1I believe the edited version is now too restrictive – it is now equivalent to $f$ beeing twice continuously differentiable. But the OP only asked about differentiability. – junjios Feb 12 '23 at 18:50
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1First line: “To be $f:\Bbb{R}^2\to\Bbb{R}$ twice continuously differentiable it is necessary and sufficient that all second-order partial derivatives exist and are continuous. The first instance of ‘continuous’ is still missing in your answer, and rather than “need”, the more precise statement is ‘necessary and sufficient’. The subsequent edit is fine. – peek-a-boo Feb 12 '23 at 19:12
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Thanks, I was just trying to highlight the differentiability conclusion. I will add that part. @peek-a-boo – A. P. Feb 12 '23 at 19:17