For $\Omega \subset \mathbb{R}^n$ be an open set, let $\{K_n\}_{n=1}^\infty$ be compact subsets of $\Omega$ such that $K_n\subset \mathrm{int}\left(K_{n+1}\right)$ for all $n$ and $\Omega = \bigcup_{n=1}^\infty K_n$, look at the seminorms \begin{equation*} p_{m}(f) = \sup_{x\in K_m} \sup_{|\alpha|\leq m} \left|\left(\frac{\partial}{\partial x}\right)^\alpha f(x)\right| \end{equation*} where $\alpha= (\alpha_1,\ldots, \alpha_n)$ is multi-index. We have

  • A set $E\subset C^\infty(\Omega)$ is bounded if and only if for any $m\in \mathbb{N}$, there exists a constant $C_{m}$ such that $p_{m}(f) \leq C_m$ for all $f\in E$.
  • A sequence of function $f_n$ in $C^\infty(\Omega)$ converges to $f\in C^\infty(\Omega)$ if $p_{m}(f_n-f) \longrightarrow 0$ as $n\longrightarrow \infty$ for all $(m)\in \mathbb{N}$.

  • $C^\infty(\Omega)$ is a complete metric space with the metric can be defined as \begin{equation*} d(f,g) = \sum_{m=1}^\infty 2^{-m} \frac{p_m(f-g)}{1+p_m(f-g)} \end{equation*} is an translation-invariant metric.

Question: How to prove that it has the Heine-Borel property? I.e any closed, bounded subset is compact? It is suggest that we can prove it by using Arzela-Ascoli theorem but I cannot see how we can do it.

  • $\begingroup$ You are right to apply Arzela-Ascoli here to show that every bounded set in $C^k(\Omega)$ is relatively compact in $C^{k-1}(\Omega)$. A full proof is too long to post it here, see e.g. Trèves, Topological Vector Spaces ..., Theorem 14.4. $\endgroup$ – Vobo Feb 25 '17 at 22:16
  • $\begingroup$ @ Vobo: Thank you for reference! $\endgroup$ – Sean Feb 26 '17 at 16:51

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