In Rudin's Functional Analysis, §1.46, he defines the space $C^\infty(\Omega)$, where $\Omega\subseteq \mathbb R^n$ is open. He defines a topology on this space, in part by introducing an increasing sequence of compact sets $K_i$ that covers $\Omega$. He argues that this topology turns $C^\infty(\Omega)$ into a Fréchet space by showing (among other things) that a Cauchy sequence of functions $f_i$ converges. To make the argument work, he appeals to the following fact (where $\alpha$ is a multi-index).
...each $D^\alpha f_i$ converges (uniformly on compact subsets of $\Omega$) to a function $g_\alpha$. In particular, $f_i(x) \to g_0(x)$. It is now evident that $g_0 \in C^\infty (\Omega)$, that $g_\alpha = D^\alpha g_0$, and that $f_i \to g$ in the topology of $C^\infty(\Omega)$.
I am familiar with the more basic result that if $f_n : [a, b]\to \mathbb R$ is a sequence of functions such that $f_n'$ converges uniformly to some $g$ and $f_n(x)$ converges for some $x\in[a, b]$, then $f_n$ converges to some $f$ and $f' = g$. However, I'm confused about how to generalize this to Rudin's case, where we must consider arbitrary partial derivatives and arbitrary compact sets in $\mathbb R^n$ rather than intervals in $\mathbb R$.
Is there a simple way to prove the more general result using the basic one? Or does the general case require a separate proof? What does such a proof look like?
I'm also not sure, when he says "uniformly on compact sets," if he means just those in the sequence $K_i$ or general compact sets. This seems to affect the claim Rudin is making, which adds to my confusion.
Edit. I think I was confused about the claim Rudin was making, and my confusion has been cleared up by the exposition in Trèves's Topological Vector Spaces, Distributions and Kernels, Ch. 10, Example 1.
This is the relevant proof in the book by Rudin (Example 1.46):
Where $\Omega$ is a non-empty subset of $\mathbb{R}^k$ for some $k \in \mathbb{N}$.
To elaborate on
Let $\alpha$ be an arbitrary multi-index. Then for all $N \in \mathbb{N}$ larger than $|\alpha|$ it follows that $(D^\alpha f_i|_{K_N})_{i \in \mathbb{N}}$ converges to a function $g_\alpha|_{K_N}$ in $C (K_N)$, because it is a Banach space and $\| D^\alpha f|_{K_N} \|_{C(K_N)} \leq p_N (f) $ for all $f \in C^\infty (\Omega)$.
Define $g_\alpha : \Omega \to \mathbb{K}$ by $g_{\alpha}(x) = \lim_{n \to \infty} D^\alpha f_n (x)$. Then the above shows that this limit always exists and that $f_i|_{K_N} \to g_\alpha |_{K_N}$ in $C(K_N)$ for all $N \geq |\alpha|$.
Now let $x \in \Omega$ arbitrary. Then there exists a $K_N$ with $N\geq |\alpha|$ so that $x$ is in the interior of $K_N$. Therefore $g_\alpha$ is continuous at $x$, because $g_\alpha|_{K_N}$ is continuous.
It is left to show that $g_\alpha = D^\alpha g_0$.
Let $h = g_0$ and $s = (g_{(1,0, \dots,0)}, g_{(0,1,0,\dots)}, \dots, g_{(0, \dots , 0,1)} )$. It suffices to show that $ h^\prime = s$. Since the desired result follows from this one by induction.
Let $x \in \Omega$ be arbitrary, then as above there exist some $N\in\mathbb{N}$ and $\delta>0$ so that $B_\delta (x) \subset K_N$ (ball with radius $\delta$ centered at $x$). For $y \in B_\delta (x) $ define $u : [0,1]\to B_\delta (x) $ by $u(t) = ty+(1-t)x$. By the fundamental theorem of calculus and the chain rule we have: $$ f_i(y)-f_i(x) = \int_0^1 (f_i\circ u)^\prime (t) dt = \int_0^1 f_i^\prime ( u (t)) (y-x) dt. $$
Therefore (using the above, the Cauchy-Schwartz inequality and $|\int | \leq \int | | $)
$$|f_i (y)-f_i(x) - f_i^\prime (x) (y-x) | \leq \int_0^1 \| f_i^\prime (u (t)) - f_i^\prime (x) \| dt \|y-x\| $$
Furthermore $$ \begin{align*} \| f_i^\prime (u (t)) - f_i^\prime (x) \| &= \| f_i^\prime (u(t) ) - s (u(t)) + s(u(t))- f_i^\prime (x) + s(x)-s(x)\| \\ &\leq \sup_{t \in [0,1]} \| s(u(t))-s(x) \| + 2 \sup_{x \in K_N}\| f^\prime_i (x) - s(x)\| \end{align*} $$ insertion into the above equation and attacking with $\lim_{i\to \infty}$ yields the desired result.