Hi this is problem 1 in Chapter 5 Evans PDE.
I have trouble showing the second part, i.e. completeness of $C^{k,\gamma}$. I know there is some previous posts but I did not quite get the answers. And I found a version that makes me feel better…
enter link description here
However I am still lost…simply I do not get the logical steps in tackling the problem.
well, my understanding is that, to show any arbitrary space $M$ is complete. One needs to show
1) Every Cauchy sequence in $M$ has a limit. (Hence, I think this is step 1 in the link)
2) The limit is also within the same space $M$. (Step 3)
Could anyone explain the flow in the proof of Theorem 2.1.2 p.12 ? (especially, Im lost what is step 2 for…)
Best Answer
Note that in step one, it is given that for any multi-index $\alpha$ with $\|\alpha\| \le k$, $D^\alpha u_n$ converges uniformly to some function $u_\alpha$. In particular, $u_n \to u $ uniformly.
You do not know, a priori, how all these $u_\alpha$'s are related. In particular, you do not know $u_\alpha = D^\alpha u$. (Actually, from step $1$ alone you do not know even if $u$ is differentiable. So step 2 is basically showing that $u \in C^k(\overline\Omega)$ .
Step $3$ shows furthermore that $u \in C^{k, \gamma}(\overline\Omega)$, while step 4 shows that you do have $u_n \to u$ in $C^{k, \gamma}$ norm.
Remark I will write a proof for the case $k=0$, illustrating the general procedure.
Recall that the $C^{0, \gamma}$ norm is given by
$$\|u\| = \sup_{x\in \overline\Omega}|u(x)| + \sup_{x, y\in \overline\Omega,x\neq y} \frac{|u(x) - u(y)|}{|x-y|^\gamma}.$$
Now we show that $C^{0, \gamma}(\overline\Omega)$ is complete. Let $\{u_n\}$ be a Cauchy sequence in $C^{0, \gamma}$. Then there is $C>0$ so that $\|u_n\| \le C$ (Do you know why?).
Claim $1$: The family $\{u_n\}$ are equicontinuous.
To see this let $\epsilon >0$. Let $\delta$ be small so that $C \delta^\gamma <\epsilon$. Then for all $n$ and $x, y\in \overline\Omega$ and $|x-y|<\delta$, we have
$$|u_n(x) - u_n(y)| \le C|x-y|^\gamma < C\delta^\gamma <\epsilon$$
Thus the claim is shown.
Now as $\{u_n\}$ are also uniformly bounded (that is, $|u_n(x)|\le C$ for all $n$ and all $x\in \Omega$), by Arzelà–Ascoli_theorem there is $u\in C^0(\overline\Omega)$ so that $u_n \to u$ in $C^0$-norm.
Now we show that $u\in C^{0, \gamma}(\overline\Omega)$. Fix $x, y\in \overline\Omega$, $x\neq y$. Then as $u_n(x) \to u(x)$, we have
$$C\ge \frac{|u_n(x) - u_n(y)|}{|x-y|^\gamma} \to \frac{|u(x) - u(y)|}{|x-y|^\gamma}.$$
As $x, y$ are arbitrary, we have
$$\sup_{x, y\in \overline\Omega,x\neq y}\frac{|u(x)-u(y)|}{|x-y|^\gamma} \le C .$$
Together with the fact that $u \in C^0$, we have $u \in C^{0, \gamma}(\overline\Omega)$.
Lastly, we show that $u_n \to u$ in $C^{0, \gamma}(\overline \Omega)$. That is, $\|u_n-u\|\to 0$ as $n\to \infty$ (where $\|\cdot\|$ is the $C^{0, \gamma}$-norm).
Note that we have already that $u_n\to u$. Now for any $x, y,n$ fixed, as $u_m(x) \to u(x)$ (similar for $y$),
$$\begin{split} \frac{|u_n(x) - u(x) - (u_n(y) - u(y))|}{|x-y|^\gamma} &\le \limsup_{m\to \infty} \frac{|u_n(x) - u_m(x) - (u_n(y) - u_m(y))|}{|x-y|^\gamma}\\ &\le \limsup_{m\to \infty} \|u_m -u_n\|. \end{split}$$
As $\{u_n\}$ is a Cauchy sequence, the right hand side converges to zero (independent of $x, y$). Thus
$$\|u_n - u\| \to 0$$
as $n\to \infty$. Thus $u_n \to u$ in $C^{0, \gamma}$ norm and so the space is complete.