Multivariable uniform convergence and differentiation There is a well known theorem on the relationship between uniform convergence of univariate functions and differentiation.  Quoting Theorem 7.17 from Rudin 1976, Principles of Mathematical Analysis:
Theorem: Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a, b]$ and such that $\{f_n(x_0)\}$ converges for some point $x_0$ on $[a,b]$. If $\{f_n'\}$ converges uniformly on $[a,b]$, then $\{f_n\}$ converges uniformly on $[a,b]$ to a function $f$, and $f'(x) = \lim_{n\rightarrow\infty} f_n'(x)$.
My question is, does a generalisation of this theorem exist for sequences of functions $f_n: \mathbb{R}^n \rightarrow \mathbb{R}?$ For example, can $f'_n$ be replaced by $\nabla f_n$ and $[a,b]$ be replaced by a compact set?
 A: After some thought, I have this corollary to the theorem quoted in the question.
Corollary:
Let $f_n : \mathcal{X} \rightarrow \mathbb{R}$ be a family of functions on a
    subset $\mathcal{X} \subseteq \mathbb{R}^n$ and suppose $\{f_n(x)\}$ converges pointwise to a 
    function $f$ on a convex set $\mathcal{C} \subseteq \mathcal{X}$.  Further,
    suppose that $\nabla f_n$ converges uniformly on $\mathcal{C}$.  Then
    $\{f_n\}$ converges uniformly on $\mathcal{C}$ and $\nabla f =
 \lim_{n\rightarrow \infty} \nabla f_n$.
Proof:
Fix two points $a, b \in \mathcal{C}$.  By convexity of $\mathcal{C}$, all
    points $g(t)$ on the straight line connecting $a$ and $b$ are in
    $\mathcal{C}$:
    $$
  g(t) = (1-t)a + tb \in \mathcal{C}\; \forall\; t\in[0,1].
 $$
    Then we have $f_n \circ g$ pointwise convergent on $t\in[1,0]$.  Furthermore,
    $(\nabla f_n \cdot (b-a))\circ g$ is uniformly convergent on $t\in[0,1]$.
    Indeed, fixing $\epsilon > 0$, by the Cauchy-Schwarz inequality, we have
    $$
  |(\nabla f_n (x) - \nabla f_m (x))\cdot(b-a)| \leq
  ||b-a||\,||\nabla f_n(x) - \nabla f_m(x)||,
 $$
    and by uniform convergence of $\nabla f_n$, there is some $N(\epsilon)$ such
    that, whenever $n > N$ and $m > N$, the right hand side is less than
    $||b-a||\epsilon$.  It is quick to verify that $(\nabla f_n \cdot (b-a))\circ
 g = \partial f_n \circ g/\partial t$.  It follows from the theorem stated in the question
    that $f_n \circ g(t)$ converges uniformly to $f\circ g(t)$ on $[0, 1]$ and
    $$
  \frac{\partial}{\partial t}f \circ g(t) = \lim_{n\rightarrow \infty} \frac{\partial}{\partial t} f_n \circ
  g(t).
 $$
    As the choice of $a$ and $b$ is arbitrary, this shows that $\nabla f_n\cdot l
 \rightarrow \nabla f \cdot l$ in all directions $l$.  Choosing $l$ to be the
    basis vectors, this shows $\nabla f_n \rightarrow \nabla f$.
