# partial derivatives continuous $\implies$ differentiability in Euclidean space

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 $c$.

• 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 $I$. – littleO Oct 10 '13 at 9:28
• 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. – Daniel Fischer Oct 10 '13 at 9:40
• @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? – Ryan G Oct 10 '13 at 10:41
• 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. – Daniel Fischer Oct 10 '13 at 10:44
• @DanielFischer That's the slyest thing I've ever heard! Thanks! – Ryan G Oct 10 '13 at 10:50

$A$ 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.
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 $\forall \mathbf v=(v_1,\ldots,v_n)\in\mathbf R^n$ $$f'(x_0)\mathbf v=\sum\limits_{i=1}^n\frac{\partial f}{\partial x_i}(x_0)v_i$$
• If $c$ is not an interior point, I think the limit isn't well defined – Student Oct 10 '13 at 9:12