Arzela-Ascoli variant for $C^1$ functions Let $K$ be a compact subset of a smooth manifold, and for simplicity let $d(\cdot,\cdot)$ be any metric on $K$. Let $C^1(K)$ be the set of $C^1$ real-valued functions on $K$ equipped with the norm $$\|f\|_{C^1}:= \max \{\|f\|_\infty, \|df\|_\infty\},$$

Definition: Let $\mathcal{F}$ be a subset of $C^1(K)$. $\mathcal{F}$ is bounded if $$\exists M > 0: \forall f \in \mathcal{F}: \|f\|_{C^1} < M.$$ $\mathcal{F}$ is equicontinuous if for every $x \in K$ and $\epsilon > 0$ there exists $\delta > 0$ such that $$d(x,y) < \delta \implies \max\{|f(x)-f(y)|, \|df(x)-df(y)\|\} < \epsilon.$$
  $\mathcal{F}$ is closed if it is a closed subset of $C^1(K)$.

I believe the following variant of the Arzela-Ascoli theorem is true:

Theorem: A subset $\mathcal{F}$ of $C^1(K)$ is compact (in the topology induced by the $C^1$ norm) if and only if $\mathcal{F}$ is closed, bounded, and equicontinuous.

From reading the proof of the Arzela-Ascoli theorem, it seems to me that one could prove this variant with straightforward modifications. Can anyone please provide a reference containing this generalization (or point out that I am mistaken)? 
 A: You need to be a bit careful with definitions. What does $f \in C^1(K)$ mean? If you mean that $f$ extends to a $C^1$ function on $M$ with derivative $df$, this is false. Let $M = \mathbb R$ and let $K = \{0\} \cup \{1/n: n \in \mathbb N\}$, and let $f_n (1/k) = \sqrt k$ for $k \leq n$, $0$ otherwise, and let $df_n = 0$. A less pertinent, but still worth considering, point is that a smooth manifold does not come equipped with a natural norm on the cotangent space, so an appropriate definition of $\|df\|$ takes a little consideration (although you get a natural norm once you have chosen a metric).
Nonetheless, there is something to what you have written here. Mainly you need to add some Taylor remainder type term. Let's first consider $K \subset \mathbb R^n$ compact, $f_n: K \to \mathbb R$, and $G_n : K \to \mathbb R^n$ such that $f_n$ extends to a $C^1$ function on $\mathbb R^n$ with gradient $G_n$ on $K$. Does there exist $f, G$ such that $f_n \to f$, $G_n \to G$, and $f$ extends to a $C^1$ function such that $G = \nabla f$ on $K$? The answer is yes if the $f_n$ are uniformly bounded and uniformly equicontinuous, and the $G_n$ are, and 
$$\sup\limits_n \sup\limits_{x,y \in K, x \neq y} {f_n(y) - f_n(x) + \langle G_n(x), y-x \rangle \over |x-y|^2} < \infty.$$
It follows from the usual Arzela-Ascoli theorem that $f_n$ and $G_n$ converge to some functions $f$ and $G$, and it is easy to check that the pair $(f,G)$ satisfy the above estimate. Any pair (called a $1$-jet) that satisfies this estimate extends to a $C^{1,1}$ function on $\mathbb R^n$ by Whitney's extension theorem. A good reference for further discussion and a proof is Stein's "Singular Integrals and Differentiability Properties of Functions" as well as this expository paper by Bierstone.
On a manifold, you can reduce to the $\mathbb R^n$ case by composing with coordinate functions. You can then extend the limit functions on each coordinate patch, and patch them together with a partition of unity. 
