Let $\Omega \subset R^n$ be open, and let us equip $\mathcal E = C^\infty$ with the locally convex topology induced by the following family of semi-norms:

For $K \subset \Omega$ and $\alpha, \beta \in \mathbb N _ 0 ^n$, let $|f|_{K,\alpha,\beta} = sup_{x \in K} | \partial_x^{\alpha} f(x) |$.

It is easy to see that compact sets are closed and bounded within that topology. But how does the opposite direction work?

Thank you!


This is far too long to be a comment, so it's an answer instead.

I don't think anyone usually gives $C^\infty(\Omega)$ another name. Now I presume you've already shown the space to be barreled, so we just need $X \subset C^\infty(\Omega)$ compact iff $X$ is closed and bounded.

First note that your sufficient family of seminorms is too large. Because $\Omega$ is $\sigma$-compact, we can thin out the family by taking a countable family of compact sets $K_j$ that exhaust $\Omega$, i.e. $K_j \subsetneq K_{j+1}$ and $\bigcup_j K_j = \Omega$, so $$ \Vert f \Vert_{i,j} = \max_{|\alpha|\leq i} \; \sup_{x \in K_j} |\partial^\alpha f(x)|$$

is a countable family of seminorms that generate the same topology. From these we can define a metric on $C^\infty(\Omega)$ as follows

$$ d(f,g) = \sum_{i=0}^\infty\; \sup_{j \in \mathbb{N}} \frac{\Vert f-g \Vert_{i,j}}{1 + \Vert f-g \Vert_{i,j}} 2^{-i}.$$

It is easy to verify that this metric generates the same topology as the family of seminorms and that the metric is complete and translation-invariant. Thus $C^\infty(\Omega)$ is a Fréchet space and we know that the compact subsets of complete metric spaces are precisely those that are closed and totally bounded.

So the strategy is clear. First show that $X$ bounded w.r.t. the seminorms implies $X$ bounded w.r.t. the metric. Then show that $X$ bounded w.r.t. the metric implies that $X$ is totally bounded.

  • 1
    $\begingroup$ About giving $C^\infty$ another name: it can happen in complex differential geometry when you consider the sheaf of smooth $k$ or $(p,q)$ forms on a manifold, sometimes it's denoted $\mathcal E^k$ or $\mathcal E^{(p,q)}$. Then $C^\infty$ is equal to $\mathcal E^0$ (or the global sections thereof). $\endgroup$ – Gunnar Þór Magnússon Nov 21 '10 at 17:14
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    $\begingroup$ @kahen The metric you introduced is always bounded, so this won't lead anywhere. The proof that $\mathcal{E}$ is a Montel space relies heavily on Ascoli-Arzela theorem. You can find a proof for instance in the book "Topological spaces, distributions and kernels" by Trèves. $\endgroup$ – Ahriman Jan 31 '13 at 13:31

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