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I am given this theorem:

If $f \in C^1(A,\mathbb R^m)$, i.e. every partial derivative of $f$ is continuous on $A$, and $A$ is open in $\mathbb R^n$, then $f$ is differentiable on $A$.

Is the following stronger assertion also true?

If every partial derivative of $f:A\to\mathbb R^m$ is continuous at $c\in A$, and $A$ is open in $\mathbb R^n$, then $f$ is differentiable at $c$.

Also, is the requirement that $A$ be an open set really necessary in the above statements?

EDIT: After doing a little more research, here is the strongest version of the theorem I've come up with (there is an even stronger version, but it isn't quite as neat or succinct):

If every partial derivative of $f:A\subset\mathbb R^n\to\mathbb R^m$ exists, and is continuous, at $c\in $int$(A)$, then $f$ is differentiable at $c$.

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    $\begingroup$ In multivariable calculus, if $x$ is not in the interior of the domain of $f$, then a matrix $f'(x)$ may not be uniquely determined by the requirement that $f(x + h) = f(x) + f'(x) h + o(h)$ as $h \to x$. So the definition of the statement "$f$ is differentiable at $x$" requires that $x$ is in the interior of the domain of $f$. This isn't a problem in single variable calculus where (according to baby Rudin at least) a function whose domain is a closed interval $I$ can be differentiable at an endpoint of $I$. $\endgroup$ – littleO Oct 10 '13 at 9:28
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    $\begingroup$ The stronger assertion is true, existence of the partial derivatives in a neighbourhood plus continuity of the partial derivatives in one point implies differentiability in that point. $\endgroup$ – Daniel Fischer Oct 10 '13 at 9:40
  • $\begingroup$ @DanielFischer If the partial is continuous at c, and if c is not an isolated point, then is it automatic that the partial exists near (i.e. in a neighbourhood of) c? $\endgroup$ – Ryan G Oct 10 '13 at 10:41
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    $\begingroup$ That depends on your definitions. If a function is defined only on a subspace $D\subset A$, you could silently call it continuous at $x$ if it is continuous at $x$ as a function $D \to Y$. To explicitly forbid that, I included existence in my statement. $\endgroup$ – Daniel Fischer Oct 10 '13 at 10:44
  • $\begingroup$ @DanielFischer That's the slyest thing I've ever heard! Thanks! $\endgroup$ – Ryan G Oct 10 '13 at 10:50
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$A$ needn't be an open set in the "stronger" assertion, but $c$ should be an interior point of $A$ for the definition of the derivative to work correctly.

A general version is (see Tao's Analysis II for a proof)

Let $E\subseteq \mathbf R^n$, $f:E\to\mathbf R^m$, $F\subseteq E$ and $x_0$ an interior point of $F$. If all partial derivatives of $f$ exist on $F$ and are continuous at $x_0$, then $f$ is differentiable at $x_0$ with $\forall \mathbf v=(v_1,\ldots,v_n)\in\mathbf R^n$ $$f'(x_0)\mathbf v=\sum\limits_{i=1}^n\frac{\partial f}{\partial x_i}(x_0)v_i$$

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  • $\begingroup$ What do you mean by "work correctly"? $\endgroup$ – dfeuer Oct 10 '13 at 9:09
  • $\begingroup$ If $c$ is not an interior point, I think the limit isn't well defined $\endgroup$ – Student Oct 10 '13 at 9:12
  • $\begingroup$ The stronger assertion is stronger in that it demands only continuity of the partial derivatives at a single point, and not continuity on a neighbourhood of the point. $\endgroup$ – Daniel Fischer Oct 10 '13 at 9:39
  • $\begingroup$ O alright, I wasn't thinking right $\endgroup$ – Student Oct 10 '13 at 9:42
  • $\begingroup$ Thanks, check out my latest edit, incorporating your input. $\endgroup$ – Ryan G Oct 10 '13 at 9:44

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