Compact-open and Whitney $C^\infty$-topologies agree on $C^\infty(M, N)$ for compact $M$.

Question

Let $M$ and $N$ be smooth manifolds. There are different topologies we can equip the space $C^\infty (M, N)$ of smooth mappings between them with. Two of them are the compact-open topology and the Whitney $C^k$-topologies for $k \in \mathbb{N} \cup \{\infty\}$ (I'm using this definition based on jet bundles).

I have read claims that when $M$ is compact these agree. Is this true? Could someone provide a proof or a reference?

EDIT: I'm starting to seriously doubt that this is true for $k \geq 1$. I can show that the Whitney topology is finer than the compact-open, but for say $f: \mathbb{S}^1 \to \mathbb{R}$ allowing to change the value by any $\varepsilon$ on a compact set with nonempty interior is sufficient to make the derivative arbitrarily large.

Answer 1

Let $d$ be a metric on $N$. Then the following family is a local basis of the Whitney topology on $C(M,N)$: $$B(f,\varepsilon) := \{g \in C(M,N): d(g(x),f(x)) \lt \varepsilon(x) \text{ for all x} \in M, \varepsilon \in C(M, (0; +\infty))\}$$

(Michor P.W. Manifold of differentiable mappings, lemma 3.3)

Since $M$ is compact, we see that for any continuous function $\varepsilon \colon M \to (0; +\infty)$ there exist $n \in \mathbb{N}$ such that $\frac{1}{n} \lt \varepsilon(x)$ for all $x \in M$ and compact-open topology and topology of uniform convergence coincide. Hence $\{B(f, \frac{1}{n}), n \in \mathbb{N}\} $ is a local basis for both topologies. Therefore compact-open topology and Whitney topology on $C(M,N)$ coincide. Compact-open $C^{k}$-topology and Whitney $C^k$-topology can be defined by embedding $j^k \colon C^{\infty}(M,N) \to C(M, J^k(M,N))$. For details see Michor P.W. Manifold of differentiable mappings, page 33. From previous it follows that they coincide too.