# Differentiation is continuous

The setting in a functional analysis class:

1. Let $\Omega=\mathbb{R}^m$.
2. Let $C(\Omega)=\{f:\Omega\rightarrow\mathbb{C}\mid f\text{ is continuous}\}$.
3. Let $(K_n)_{n=1}^\infty$ be a sequence of compact subsets of $\Omega$ such that $K_n \subset int(K_{n+1})$ for each $n$.
4. Define seminorms on $C(\Omega)$: $p_n(f)=\sup\{|f(x)| \mid x \in K_n\}$.
5. Define a topology given by the basis $V_n=\{f\in C(\Omega)\mid p_n(f)<\frac{1}{n}\}$.
6. We also get an invariant metric compatible with this topology, given by $d(f,g)=\sum_{i=1}^\infty 2^{-n} \frac{p_n(f-g)}{1+p_n(f-g)}$.
7. We've seen in class that $C(\Omega)$ is complete with regard to this metric (and actually, any other invariant metric compatible with this topology).
8. Define $C^\infty(\Omega)=\{f:\Omega\rightarrow\mathbb{C}\mid D^\alpha f\in C(\Omega) \text{ for any multi-index$\alpha$}\}$.
9. ($D^\alpha f=\frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\dots\frac{\partial^{\alpha_1}}{\partial x_m^{\alpha_m}}f$)
10. In other words, $C^{\alpha}(\Omega)$ is the set of infinitely differentiable functions of $\Omega$.

Prove that for any mutli-index $\alpha$, the mapping $D^\alpha:C^\infty(\Omega)\rightarrow C^\infty(\Omega)$ is continuous.

My attempt:

With so many definitions, I'm a little confused. I tried to go by (6) and bound $d(D^\alpha f,D^\alpha g)$ by means of $d(f,g)$ but was stuck.

Note: This should be the easiest exercise in this homework sheet. If you think you can give me some useful tools to handle this subject, this would be much appreciated too.

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This doesn't make sense; $D^\alpha$ is not even defined on $C(\Omega)$. Presumably you want to show that it is continuous as a mapping $C^\infty(\Omega)\to C(\Omega)$, but to do this you first have to define a topology on $C^\infty(\Omega)$, and once you have done this properly, the solution should be straightforward. –  Florian Nov 27 '12 at 13:10
@Florian: My typo. It should be $C^\infty \rightarrow C^\infty$. I should really check what the topology on $C^\infty$ is. –  Gils Nov 27 '12 at 13:27
A couple of points. First, did you want the $K_n$s to exhaust $\Omega$ in the sense that $\bigcup_n K_n = \Omega$? Second, you have a really weird topology. For example, the $0$ function is in every basic open set, hence is in every open set! I think you meant to define a topology solely through your distance function. Last, just for definitenes, I assume you want to use the subspace topology on $C^\infty (\Omega)$, right? –  Jason DeVito Nov 27 '12 at 13:30
The "right" topology on $C^\infty(\Omega)$ is the topology of uniform convergence of all derivatives on compact sets; of course, it is possible to specify this topology using a countable family of seminorms in a way similar to what you describe. In any event, if you believe this then the proof amounts to saying that if all derivatives of $f_n$ converge uniformly on compact sets to the derivatives of $f$ then all derivatives of $D^\alpha f_n$ converge uniformly on compact sets to the derivatives of $D^\alpha f$. –  Paul Siegel Nov 27 '12 at 14:36