$
\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 completeness of $C^\ii(\W)$. 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.
Theorem: $C^\ii(\W)$ is complete, hence a Fréchet space.
Proof: let $\{f_i\}$ be a Cauchy sequence in $C^\ii(\W)$ so that (for a fixed $N$) whenever $i$ and $j$ are large enough
$$f_i-f_j\in V_N \implies \sup_{\substack{x\in K_N\\ |\a|\le n}}|D^\a f_i(x) – D^\a f_j(x)| < \frac{1}{n}.$$
a) In $K_nN$ we have
$$\sup_{x\in K_N} |D^\a f_i(x) – D^\a f_j(x)|\to 0$$
as $i,j\to \ii$.
b) $\{D^\a f_i\}_{i\in\mathbb{N}}$ converges uniformly (in $K_N$) to a function $k^\a_N:K_N\to \mathbb{C}$ i.e.
$$\lim_{i\to\ii} \sup_{x\in K_n} |k^\a_N(x) – D^\a f_i(x)| = 0$$
where, for $n<m$, the functions $k_n$ and $k_m$ agree on $K_n$.
c) Define $g_\a:\W\to\mathbb{C}:x\mapsto k^\a_N(x)$ where, for each $x\in \W$, we take $N$ sufficiently large for $k_N(x)$ to be defined ($\W$ being open is essential for such $N$ to exist).
d) $D^\a f_i$ converges uniformly to $g_\a$ (in $\W$):
$$\lim_{i\to\ii}\sup_{x\in\W}|g_\a(x) – D^\a f_i(x)| = \lim_{i\to\ii}\sup_n\sup_{x\in K_n}|g_\a(x)-D^\a f_i(x)| = 0.$$
In particular $f_i\uto g_0$.
e) $g_\a\in C^\ii(\W)$ for any $\a$, as the uniform limit of smooth functions is smooth.
f) Uniform convergence also gives
$$D^\a\lim_{i\to\ii} f_i = \lim_{i\to\ii}D^\a f_i \implies D^\a g_0 = g_\a.$$
g) A small computation shows
$$\lim_{i\to\ii}p_n(g_0,f_i) = \lim_{i\to\ii}\sup_{\substack{x\in K_n\\ |\a|\le n}} |D^\a g_0(x)-D^\a f_i(x)| = 0
\implies
\lim_{i\to\ii}d(g_0,f_i) = \lim_{i\to\ii} \sup_n\frac{2^{-n}p_n(g_0,f_i)}{1+p_n(g_0,f_i)} = 0$$
and thus $f_i\to g_0$ in the established topology, so that $C^\ii(\W)$ is complete.
My many 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 does such a function $k^\a_N$ exist?
d) Why is
$$\lim_{i\to\ii}\sup_n\sup_{x\in K_n}|g_\a(x)-D^\a f_i(x)|$$
equal to zero? Letting
$$c_{i,n} := \sup_{x\in K_n}|g_\a(x)-D^\a f_i(x)|$$
then (2.2) gives $\sup_n \lim_{i\to\ii} c_{i,n} = 0,$
but that does not imply $\lim_{i\to\ii}\sup_n c_{i,n} = 0$ (which is what we want to prove) as the counterexample $c_{i,n} = \frac{n}{i+n}$ shows.
g) A similar issue arises, as while it is true that
$$\lim_{i\to\ii} p_n(g_0,f_i) = 0 \implies \sup _n\lim_{i\to\ii} \frac{2^{-n}p_n(g_0,f_i)}{1+p_n(g_0,f_i)} = 0$$
I fail to see why
$$\lim_{i\to\ii} \sup_n\frac{2^{-n}p_n(g_0,f_i)}{1+p_n(g_0,f_i)} = 0.$$
The main passage in question (regarding $C^\ii(\W)$:
The main results used:
Best Answer
To me your outline seems to be almost correct. The only issue I see is that $D^\alpha f_i \to g_\alpha$ only converges uniformly on every compact subset of $\Omega$ and not on $\Omega$ itself.
Let me address the remaining issues.
(b) Your function $k_n^\alpha$ is just the pointwise limit of the functions $D^\alpha f_i$. Since the sequence $(D^\alpha f_i)_{i=0}^{\infty}$ satisfies $\sup_{x\in K_n} |D^\alpha f_i(x) - D^\alpha f_j(x)| < \frac{1}{n}$ for $i,j\geq N$, they in particular satisfy $|D^\alpha f_i(x) - D^\alpha f_j(x)| < \frac{1}{n}$ for fixed $x\in K_n$ and $i,j\geq N$. So $(D^\alpha f_i(x))_{i=0}^{\infty}$ is a Cauchy sequence in $\mathbb{C}$ and hence has a limit, which you denote by $k_n^\alpha(x)$. Then for every $\varepsilon>0$ and every $x\in K_n$ you have $k\geq N$ such that $$|k_n^\alpha(x)-D^\alpha f_i(x)| \leq |k_n^\alpha(x)-D^\alpha_k f_k| + |D^\alpha f_k-D^\alpha f_i| < \varepsilon + \varepsilon = 2\varepsilon,$$ i.e. $D^\alpha f_i \to k_n^\alpha$ uniformly on $K_n$
(d) It isn't. You only have for $$\sup_{x\in K_n} |D^\alpha f_i(x)-g_\alpha(x)| \to 0$$ for every $n\in\mathbb{N}$. But this is what you need: uniform convergence of every derivative on every compact subset of $\Omega$.
(g) Let us use the notation $p_n(f) = \sup_{x\in K_n} \max_{|\alpha|\leq n} |D^\alpha f(x)|$. Here you may observe that $$ \frac{p_n(f-g)}{1+p_n(f-g)} \leq 1$$ for every $f,g$ (since $\frac{x}{1+x}\leq 1$ for every non-negative number $x$). Now given $\varepsilon>0$ pick an $N$ such that $2^{-n}\leq \varepsilon$ for all $n\geq N$. Hence $$\sup_{n\geq N} \frac{2^{-n}p_n(f_i-g_0)}{1+p_n(f_i-g_0)} < \varepsilon$$ and you are left with $$\max_{n < N} \frac{p_n(f_i-g_0)}{1+p_n(f_i-g_0)}$$ The previous steps show that for every $n$ there is an $N_n$ such that $p_n(f_i-g_0)< \varepsilon$ for all $i\geq N_n$. Pick $K:=\max_{n<N}N_m$ and hence $p_n(f_i-g_0)<\varepsilon$ for $i\geq K$ and $n< N$. This now gives $$\sup_{n}\frac{2^{-n}p_n(f_i-g_0)}{1+p_n(f_i-g_0)} = \max\left\{ \max_{n< N} \frac{2^{-n}p_n(f_i-g_0)}{1+p_n(f_i-g_0)}, \sup_{n\geq N} \frac{2^{-n}p_n(f_i-g_0)}{1+p_n(f_i-g_0)} \right\} < \varepsilon$$ which finishes the proof.