Why does $C^\infty(\Omega)$ have the Heine-Borel property?

Question

$ \let\uto\rightrightarrows \let\ii\infty \let\W\Omega \let\a\alpha \let\b\beta \let\e\varepsilon \let\d\delta \let\sbe\subseteq $

I'm struggling to understand the examples at the end of the first chapter of Rudin's Functional Analysis. I will focus the post on $C^\ii(\W)$ as hopefully getting a grasp of $C^\ii(\W)$ will clarify the spaces $C(\W)$ and $\mathcal{D}_K$.

An original post with all my questions on $C^\ii(\W)$ was closed due to lack of focus, so on the current post I will focus on proving $C^\ii(\W)$ has the Heine-Borel property. Below I try to to fill in the gaps in the proof given. I then write my doubts on the proofs and include pictures of the relevant book excerpts.

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Theorem: $C^\ii(\W)$ has the Heine-Borel property.

Proof: let $E\sbe C^\ii(\W)$ be closed and bounded. We wish to show $E$ is compact.

   a) There are numbers $M_N < \ii$ such that $$p_N(f) \le M_N \iff \sup_{\substack{x\in K_N\\ |\a|\le N}} \{|D^\a f(x)|\} \le M_N.$$

   b) Fix $\b$ and $N$ with $|\b|\le N-1$. The family $\mathcal{F}_{\b,N}$ of functions $$\{ D^\b f : f\in E\}$$ is equicontinuous on $K_{N-1}$ i.e. for any $\e>0$ there is a $\d>0$ such that $$d(x,y)< \d \implies |D^\b f(x) - D^\b f(y)| < \e$$ for any $x,y\in K_{N-1}$ and $D^\b f\in\mathcal{F}_{\b,N}$.

   c) Since $\mathcal{F}_{\b,N}$ is pointwise bounded and equicontinuous on $K_{N-1}$, Ascoli's Theorem (and its corollary) implies that $\mathcal{F}_{\b,N}$ is compact and that every sequence in $\mathcal{F}_{\b,N}$ contains a uniformly convergent subsequence. Therefore $E$ is compact.

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My questions:

Firstly: are my outlines of Rudin's proof correct? If at any point I'm deviating from the argument in the book, please let me know.

   b) Why is $\mathcal{F}_{\b,N}$ equicontinuous on $K_{N-1}$?

   c) While I understand the rest of the paragraph, I do not see why the last sentence "Therefore $E$ is compact" follows.

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The main passage in question (regarding $C^\ii(\W)$:

The main results used:

Answer 1

(b) This follows from the mean value theorem. Since $K_{N-1}$ is contained in the interior of $K_{N}$ you can cover it by finitely many open balls $B(x_1,r_1), \ldots, B(x_m,r_m)$ which are contained in $K_{N}$. Since each of these balls are convex, for every $\xi,\eta\in B(x_i,r_i)$ you have $$\partial^\beta(\xi)-\partial^\beta(\eta) = \langle\nabla(\partial^{\beta})(t\xi+(1-t)\eta), \xi-\eta\rangle$$ for some $t\in[0,1]$. Since $|\nabla(\partial^{\beta})(x)|\leq C \max_{|\alpha|=N}\sup_{x\in K_N} |\partial^\alpha(x)|$ you obtain $$|\partial^\beta(\xi)-\partial^\beta(\eta)| \leq C_{\beta,i} |\xi-\eta|.$$ Now taking the maximum of these constants gives you what you need.

(c) Pick a sequence in $(f_i)_i$ in $E$ and note that by compactness of the $\mathcal{F}_{\beta,N}$ for every $N$ you can find a subsequence $(g_{k}^1)$ such that $\partial^\beta g_{k}^{1}$ converges uniformly on $K_{1}$ for $|\beta|\leq 1$. Now repeat this process for $(g_{k}^1)$ to obtain a subsequence $(g_{i}^2)$ with the property that $\partial^\beta g_k^2$ converges uniformly on $K_2$ for $|\beta|\leq 2$ and so on. Now $(g_k^k)_{k\in\mathbb{N}}$ is a subsequence of $(f_i)_i$ with the property that $\partial^\beta f_k$ converges uniformly on all compact subset of $\Omega$ for all $\beta$.

Answer 2

In higher dimensions one usually has to make do with a "Mean-Value Inequality", which is obtained by using the 1-variable theorem. For example, if $g\in \mathcal C^1(\Omega)$ and $y \in B(x,r)\subseteq \Omega$ then if we write $y=x+u$ and consider the path $\gamma\colon [0,1]\to\Omega$, $\gamma(t)=x+tu$ for $t \in [0,1]$ we may apply the $1$-variable mean-value theorem to $t\mapsto g(\gamma(t))$. Using the chain rule, we see that $$ \begin{split} g(y)-g(x) &= dg_{\gamma(\xi)}(\gamma'(\xi))= dg_{\gamma(\xi)}(u) \end{split} $$ for some $\xi \in (0,1)$. Here we write $dg_x$ for the total derivative of $g$ at $x \in \Omega$. But $dg_{\gamma(\xi)}(u) = \sum_{i=1}^n D^i g(\gamma(\xi)).u_i$ and so $$ \begin{split} |g(y)-g(x)| &= |dg_{\gamma(\xi)}(u)| = |\sum_{i=1}^n D^ig(\gamma(\xi))u_i| \\ &\leq \left(\sum_{i=1}^n D^ig(\gamma(\xi))^2\right)^{1/2}.\|u\|_2 \\ &\leq \sup_{\substack{1\leq i \leq n\\x \in \gamma([0,1])}} |D^i g(x)|.\|y-x\|_2 \end{split} $$ where in the last line we use the definition of $u$, that is $y=x+u$.

Applying this to $g = D^{\beta}(f)$ with $|\beta|\leq N-1$, noting that $D^iD^{\beta}f = D^{\alpha}f$ where $\alpha= (i,\beta)$ and hence $|\alpha|= |\beta|+1 \leq N$ we have $|D^iD^\beta f| = |D^{\alpha}f|\leq M_N$ on $K_N$, and thus we obtain $$ \|D^{\beta}f(y)-D^{\beta}f(x)\|\leq M_N.\|y-x\|=M_N.\|y-x\|,\quad \forall f \in E $$ on $K_{N-1}\subset K_N$.

Thus we bound $|D^\beta f(y)-D^{\beta} f(x)|$ by $M_N\|y-x\|$ on $K_{N-1}$ and hence we can ensure the equicontinuity condition for $E_N = \{D^{\beta}(f): f\in E, |\beta|\leq N-1\}$. Thus $E_N$ satisfies the conditions of Ascoli's theorem, and hence is totally bounded -- in other words, given any $\epsilon>0$, there is a finite collection of functions $\mathcal G \subseteq E_N$ such that every $f \in E_N$ is within $\epsilon$ of some $ g \in\mathcal G$.

Now a complete and totally bounded space is compact (in a totally bounded space, every sequence has a Cauchy subsequence using a version of Cantor's diagonal argument, and since the space is complete, that sequence converges), so the $E_N$s are compact.

Thus given a sequence $(f_i)$ in $E$, we can therefore find a subsequence $(f_{n_i^1})$ converging uniformly on $K_1$. Then consider that subsequence on $K_2$. Using the compactness of $E_2$ we may find a subsequence of it $(f_{n_i^2})$ such that $(D^\alpha f_{n^2_i})$ converges uniformly on $K_2$ for $|\alpha|\leq 1$ (c.f. the paragraph after (5) in the page of Rudin you quote). Continuing in this way we get an infinite nested sequence of subsequences, and then by using Cantor's diagonal trick one obtains a subsequence $(f_{m_i})$ (which is eventually a subsequence of any given $(f_{n_i^k})$) with the property that $(D^{\alpha}f_{m_i})$ converges uniformly on compact subsets of $\Omega$ for all $\alpha$, and hence $E$ is (sequentially) compact.