Definition of convergence in $C^\infty(\Omega)$

Question

I am not convinced or may be don't understand, the way they define convergence and then topology as a consequence of convergence.

$\Omega$ is open subset of $\Bbb R^n.$Define standard topology on $\mathcal C^\infty(\Omega)$ through the following notion of convergence:

Which says, that a sequence $h^k $, $h^k \in \mathcal C^\infty(\Omega)$ converges to $h$ in $\mathcal C^\infty(\Omega)$ if $h^k$ converges to h uniformly with all partial derivatives of any order on any compact subset of $\Omega$.

I am wondering, why did they define convergence in such a way and how does it give birth to a topology?

[In above definition they are talking about the convergence of the sequence of derivatives of a function $h$ i.e. ($h^1, h^2....\to h)$. Now, if we want to discuss the convergence of any random sequence of functions (I mean, a sequence which is not made of derivatives of a particular function), how will we consider that sequence and the convergence....?]

Then writer says:

If $K$ is a compact subset of $\Omega$, we use the standard norm i.e $||\mathcal u||_{C^k(K)}= \sum _{|\alpha|\le k} sup_{x\in K}|D^\alpha\mathcal u(x)|$

Here, I've a basic question - why does he call it a standard norm and why does he consider only this specific norm...??

I will appreciate, if you can give me few minutes out of your precious time. Thank you so much.

Answer 1

Note: I am pretty sure this is done very well in Rudin's Functional Analysis. But I don't have it with me so I can't point to a precise page.

In a metric space $(X,d)$, the topology is characterized by convergent sequences and their limits. Indeed, a set $S$ is closed in $X$ if whenever $(s_n)$ in $S$ converges to $x\in X$, we have $x\in S$. One says that an $X$ is a sequential space. Every first countable space is sequential. Metric spaces in particular.

So you need a metric on $C^\infty(\Omega)$ for which convergent sequences obey the description above.

One way to do that is to use the family of seminorms $\|\cdot \|_{C^k(K)}$ where $K$ is compact in $\Omega$ (they have positive homogeneity and triangular inequality, but not the separation).

Actually, you need only countably many of them, and that's why the topology is metrizable. Write $\Omega=\bigcup_{n\geq 1}K_n$ with $K_n$ a nondecreasing sequence of compact sets (this is called a compact exhaustion). You can define, for instance, the metric $$ d(u,v):=\sum_{n\geq 1}\frac{\min \left(1,\|u-v\|_{C^n(K_n)}\right)}{2^n}. $$ Then you have to check that this is indeed a distance (that's easy) on $C^\infty(\Omega)$, and that (more tedious) $d(u_n,u)\longrightarrow 0$ if and only if $u_n$ converges to $u$ uniformly on every compact set, together with its derivatives.

Answer 2

Julien has already given the construction for a topology. Let me give you some more motivation:

I am wondering, why did they define convergence in such a way and how does it give birth to a topology?

This is a very natural convergence: If you have a sequence of differentiable functions $h_n$ converging uniformly on compact sets to a function $h$ and the derivatives converge uniformly too, then you know from classical analysis that $h$ is differentiable and $h_n'$ converges to $h'$ uniformly on compact sets. And there are well-know examples that all these assumptions are necessary for the limit function $h$ being differentiable. So essentially this type of convergence guarantees that the limit function is again $C^\infty$.

Here, I've a basic question - why does he call it a standard norm and why does he consider only this specific norm...??

This is rather sloppy from the author: $\|u\|_{C^k(K)}$ is a norm on $C^k(K)$ but it is only a seminorm on $C^\infty$. And the family of seminorms $\{ \|\cdot\|_{C^k(K)}: K\subset \Omega \text{ compact }, k\in\mathbb{N}\}$ defines in a natural way a locally convex topology described here, in fact the coarsest locally convex topology such that all seminorms are continuous. And "luckily", this topology coincides with the topology described in the construction of julien. This is why these (semi-)norms are called standard.