Prove that the space of function $C^{k,\gamma}(\bar{\Omega})$ is a Banach space

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…)

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 $u\in C^{0, \gamma}(\overline\gamma)$. 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 $uu_m(x) \to u(x)$ (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.

• Ok, thanks. Another thing I forgot to ask. In step 1 of the proof, $\{u_m(x)\}$ is uniformly bounded sequence. Why uniformly? Is it because $\bar{\Omega}$ is compact and so the Cauchy sequence $u_m$ converges uniformly to some $u$ on the compact set $\bar{\Omega}$? – math101 Jul 6 '15 at 4:45
• @math101 : As $\{u_n\}$ is a Cauchy sequence, so the $C^{k, \gamma}$ norm is bounded. This in particular implies that the $C^0$ norm is bounded. Note that $C^0$ bound alone does not imply that $u_n$ converges uniformly to a function (for example, $u_n = x^n$ on $[0,1]$). One need to use Ascoli-Arzela theorem, each require that $u_n$ are equicontinuous. – user99914 Jul 6 '15 at 5:47
• Thanks. Uniform boundedness and equicontinuous are equivalent, right? – math101 Jul 6 '15 at 5:57
• No. Think of that $u_n =x^n$ on $[0,1]$. It's uniformly bounded, but not equicontinuous. @math101 – user99914 Jul 6 '15 at 5:58
• but i found books.google.com.au/… – math101 Jul 6 '15 at 5:58