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This is the statement of Theorem 2.8 from Spivak's Calculus on Manifolds. I'd like feedback on if this looks fine as far as a generalization to his proof goes:

Theorem: If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, then $Df(a)$ exists if all $D_jf^i(x)$ exist in an open set containing $a$ and if each function $D_jf^i$ is continuous at $a$.

Proof: Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ and suppose that all $D_jf^i(x)$ exist in an open set containing $a=(a^1,...,a^n)$ and that each function $D_jf^i$ is continuous at $a$. Then, for each $j$ such that $1 \leq j \leq n$, by the mean value theorem, we can find $b^j$ satisfying $a^j<b^j<a^j+h^j$, so that, $$\lim_{h \to 0} \frac{|f(a+h)-f(a)-(\sum_{j=1}^n D_jf^1(a)(h^j),...,\sum_{j=1}^n D_jf^m(a)(h^j))|}{|h|}=$$ $$\lim_{h \to 0} \frac{|f(a^1+h^1,a^2...,a^n)-f(a)+...+f(a+h)-f(a^1+h^1,...,a^{n-1}+h^{n-1},a^n)-(...)|}{|h|}=$$ $$\lim_{h \to 0} \frac{|D_1f(b^1,a^2...,a^n)(h^1)+...+D_nf(a^1+h^1,...,a^{n-1}+h^{n-1},b^n)(h^n)-(...)|}{|h|}=$$ $$\lim_{h \to 0} \frac{| (\sum_{j=1}^n [D_jf^1(c_j)-D_jf^1(a)](h^j),...,\sum_{j=1}^n [D_jf^m(c_j)-D_jf^m(a)](h^j))|}{|h|} \leq$$ $$\lim_{h \to 0} |(\sum_{j=1}^n |D_jf^1(c_j)-D_jf^1(a)|\frac{|h^j|}{|h|},...,\sum_{j=1}^n |D_jf^m(c_j)-D_jf^m(a)|\frac{|h^j|}{|h|})| \leq$$ $$\lim_{h \to 0} |(\sum_{j=1}^n |D_jf^1(c_j)-D_jf^1(a)|(1),...,\sum_{j=1}^n |D_jf^m(c_j)-D_jf^m(a)|(1))|=0,$$ where $h=(h^1,...,h^m)$, each $c_j$ is defined suitably in terms of $a^j$'s, $b^j$'s and $h^j$'s, and the last equality holds by the continuity hypothesis. Therefore $Df(a)$ exists.

Thanks in advance.

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    What are the $c_j$? You have to establish that the limit exists first. – copper.hat Nov 05 '14 at 18:10
  • Oh sorry, I had to edit the thing and throw away a couple of lines because the original thing I did was way too big to fit here, so I copied the thing from the 4th line to the 1st without thinking. What do you mean by establishing that the limit exists first? Can I not try to evaluate a limit in terms of f, a and the partials? –  Nov 05 '14 at 18:22
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    Maybe you could consider to change the title of the question. I suggest Proof that continuous partial derivatives imply differenciability. – mfl Nov 05 '14 at 18:24
  • To show that the derivative exists you need to show the limit exists. You can't start off with $\lim_{h \to 0} | \text{something}|$. Also, you can't just replace ${ |h_j| \over |h| }$ by one. You need to show that for any $\epsilon>0$ there is a $\delta>0$ such that if $|h|<\delta$, then $|f(a+h)-f(a) -Ah| \le \epsilon |h|$. You have part of the right idea above, but your last step is way off. – copper.hat Nov 05 '14 at 18:31
  • Are you saying I should drop all the $\lim$'s from the inequality? I'm not sure if I'm missing something, but it seems to me the author of the book just straights away right down everything in terms of limits in his inequality (but his proof is done for a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, so I'm not sure I'm not missing something that doesn't work here). (Edit: I'm not using $Df(a)$ in the limit, I'm using the partials, which exist by hypothesis). –  Nov 05 '14 at 18:42
  • I'm not sure what to say and my typing skills are pedestrian. I have given a proof below to show how I would approach the proof. The main point is that I establish that the limit exists, and explicitly show where the continuity is used. – copper.hat Nov 05 '14 at 19:09

1 Answers1

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To simplify life, use the $\|\cdot\|_1$ norm. To reduce clutter, let $\phi_k(h) = (h_1,...,h_k,0,...0)$. Let $\phi_0(h) = 0$, and note that $\|\phi_k(h)-\phi_j(h)\|_1 \le \|h\|_1$.

First suppose $m=1$.

Let $\epsilon>0$.

By continuity, we can choose $\delta>0$ such that $|D_kf(a+h)-D_kf(a)| < \epsilon$ for all $k$ and $\|h\| < \delta$.

Let $A=(D_1f(a),...,D_n f(a))$ and suppose $\|h\| < \delta$, then we have \begin{eqnarray} |f(a+h)-f(a)-Ah| &=& |\sum_{k=1}^n (f(a+\phi_k(h))-f(a+\phi_{k-1}(h))-D_kf(a)h_k)| \\ &\le& \sum_{k=1}^n |f(a+\phi_k(h))-f(a+\phi_{k-1}(h))-D_kf(a)h_k| \end{eqnarray} By the mean value theorem, there are $c_k \in [a+\phi_{k-1}(h), a+\phi_k(h)]$ (that is, each $c_k$ lies on the line segment) such that $f(a+\phi_k(h))-f(a+\phi_{k-1}(h)) = D_k f(c_k) h_k$. Note that $\|c_k -a\|_1 \le \|h\|_1$. Continuing: \begin{eqnarray} |f(a+h)-f(a)-Ah| &\le& \sum_{k=1}^n |D_k f(c_k) h_k-D_kf(a)h_k| \\ &=& \sum_{k=1}^n |D_k f(c_k) -D_kf(a)||h_k| \\ &<& \epsilon \sum_{k=1}^n |h_k| \\ &=& \epsilon \|h\|_1 \end{eqnarray} Since $\epsilon>0$ was arbitrary, this shows that $f$ is differentiable at $a$ and $Df(a)h = Ah$.

It is straightforward to show that if $f_1,...f_m$ are differentiable, then so is $f(x) = (f_1(x),...,f_n(x))$, and $Df(x)h = (Df_1(x)h, ..., D f_m(x)h)$.

copper.hat
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  • Thanks, this makes sense, but I'm still wondering if and how what I did up there is incorrect. I mean the point was really to see if I could do the thing directly for a $f$ with values in $\mathbb{R}^m$ instead of $\mathbb{R}$. –  Nov 05 '14 at 19:42
  • You can, you would need to use a a slightly different form of the mean value theorem. Pick the $\delta>0$ such that $\sup_{|h| < \delta} | D_k f(a+h)|_\infty < \epsilon$. Its just messier. – copper.hat Nov 05 '14 at 19:52
  • Right, but is it fine if I don't specify the norm in the space of transformations, and use the standard norm for the Euclidean space (since as far as I understand, that's what this particular book does)? And is it a problem if I start out writing a limit which I don't know exists instead of writing out things in terms of epsilon-deltas but showing it's squeezed between 0 and 0 (since as far as I understand, this is what this particular book does in this proof)? Or is there something that I'm simply missing? –  Nov 05 '14 at 20:00
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    Well, since this is finite dimensions you can use whatever norm you like, but I think the $l_1$ norm is simpler for the above estimates. I think it is a problem (or at best very sloppy) to use $\lim$ when the main purpose is to show it exists. Space requirements in Spivak's book may force a more 'relaxed' approach. – copper.hat Nov 05 '14 at 20:13
  • Specifically can't you aviod it with something like this: https://math.stackexchange.com/questions/84524/question-on-differentiability-at-a-point?noredirect=1&lq=1 – Vivaan Daga Sep 15 '23 at 21:54
  • @VivaanDaga I am not sure what avoiding the mean value theorem accomplishes. If you use the other approach you need to establish the $O(t^2)$ bound. – copper.hat Sep 16 '23 at 05:06
  • So the issue with that argument is it is not "uniform"? – Vivaan Daga Sep 16 '23 at 07:20
  • I don't that answer showed it goes to zero uniformly in $(x_1,y_1)$, only that for a small enough $y_{1}$ you can find an $x_{1}$ such that the quotient is small, which is not what we want to prove, no? – Vivaan Daga Sep 16 '23 at 08:14
  • @VivaanDaga If you have a question then ask it on MSE. The comments are not really a venue for exploring. Plus, I think you are mistaken, the other answer must use some form of estimation to get a uniform $O(t^2)$ bound, and the usual routes are the mean value theorem (which is fairly basic) and integration (warranted here by continuity, but stronger than the mean value theorem). – copper.hat Sep 16 '23 at 20:01