1
$\begingroup$

I'm studying by myself PDEs without having done Functional Analysis and I'm trying do the following exercise of the book "Partial Differential Equations I" by Michael E. Taylor on Appendix $A$ - Section $6$ (page $592$):

  1. Prove the following result, also known as part of Ascoli's theorem. If $X$ is a compact metric space, $B_j$ are Banach spaces, and $K: B_1 \longrightarrow B_2$ is a compact operator, then $\kappa f(x) = K(f(x))$ defines a compact map $\kappa: \mathcal{C}^{\alpha}(X,B_1) \longrightarrow C(X,B_2)$, for any $\alpha > 0$.

$\mathcal{C}^{\alpha}$ denotes a space with $\alpha$-Holder continuity (this notation was introduced on the final of the page $317$).

I would like to know if my attempt it's correct until where I wrote and, if it is correct, how I can ensure the limit below lives in $C(X,B_2)$.

$\textbf{My attempt:}$

Firstly, we observe that $\kappa$ is well-defined, otherwise, $K$ wouldn't well defined, which would be an absurd since $K$ is a map.

Now, given a bounded sequence $(f_n)$ in $\mathcal{C}^{\alpha}(X,B_1)$ and fixing $x \in X$, we have that $(f_n(x))$ is a bounded sequence in $B_1$. By compactness of the map $K$, there is a subsequence $(f_{n_l}(x))$ in $B_1$ such that $K(f_{n_l}(x)) \rightarrow K(f(x)) \in B_2$, i.e., given $\varepsilon > 0$, there is $N(x) \in \mathbb{N}$ such that

$$l > N(x) \Longrightarrow ||K(f_{n_l}(x)) - K(f(x))||_{B_2} < \frac{\varepsilon }{3}\ (*)$$

The subsequence $((K \circ f_{n_l}))$ is in $C(X,B_2)$ since $K$ and $f_{n_l}$ are continuous functions. Since the limit is unique, $f$ is well-defined.

By continuity of $K \circ f_{n_l}$ in every $x \in X$, there is $\delta_x > 0$ such that

$$||y - x||_X < \delta_x \Longrightarrow ||K (f_{n_l})(y) - K (f_{n_l})(x)||_{B_2} < \frac{\varepsilon}{3} \ (**)$$

W.l.o.g., we assume $\frac{1}{N(x)} < \delta_x$. By compactness of $X$, there is a finite subcover of $X$ by open balls $B(x_i,\frac{1}{N(x_i)}) \subset X$, $i = 1, \cdots, j$. Thus, arguing as before, we find $(K \circ f_{n_l}(x))$ such that $(*)$ holds for $N := \max_\limits{ i \in \{ 1,\cdots,j \} } N(x_i)$. Denoting by $\delta := \frac{1}{N}$ and using $(*)$ and $(**)$, we have that

$||K(f(y)) - K(f(x))||_{B_2} \leq ||K(f(y)) - K(f_{n_l}(y))||_{B_2} + ||K(f_{n_l}(y)) - K(f_{n_l}(x))||_{B_2} + ||K(f_{n_l}(x)) - K(f(x))||_{B_2}$

$\begin{eqnarray*} ||K(f(y)) - K(f(x))||_{B_2} &\leq& ||K(f(y)) - K(f_{n_l}(y))||_{B_2} + ||K(f_{n_l}(y)) - K(f_{n_l}(x))||_{B_2} + ||K(f_{n_l}(x)) - K(f(x))||_{B_2}\\ &\leq& \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon, \end{eqnarray*}$

whenever $l > N$ and $||y - x||_X < \delta$, hence $(K \circ f) \in C(X,B_2)$. $\square$

I'm stuck here, because I can't prove that $(K \circ f) \in C(X,B_2)$. I know that $(K \circ f_{n_l}) \in C(X,B_2)$ for each $l \in \mathbb{N}^*$, $(K \circ f_{n_l})(x) \rightarrow (K \circ f)(x)$, $(K \circ f_{n_l})$ is bounded I know the uniform limit of continuous functions is continuous, but I can't see how I can prove the pointwise convergence is, indeed, uniform. The only thing that I thought about this is use Ascoli's theorem because I know that $(K \circ f_{n_l})$ is bounded, but I can't see how the family $\{ (K \circ f_{n_l}) \in C(X,B_2) \ ; \ l \in \mathbb{N}^* \}$ is equicontinuous.

I thought about the MaoWao's comment and I think I'm close to solving the question, but I don't sure if $(*)$ by the $N$ that I chose. If this is not the right $N$ that I need to take, someone can help me with a hint about how I can choose the $N$ without depending on $x$?

Thanks in advance!

$\textbf{EDIT:}$

The version of Ascoli's theorem that I'm using:

$\textbf{Ascoli-Arzela's theorem:}$ let be $E$ a set of continuous maps $f: K \longrightarrow N$, where $K$ is compact. $E \subset C(K,N)$ is relatively compact if, and only if, the following holds:

1) $E$ is equicontinuous;

2) For each $x \in K$, $E(x) = \{ f(x) \ ; \ f \in E \}$ is relatively compact in $N$.

$\endgroup$
  • $\begingroup$ The problem with your attempt is that the subsequence you choose depends on the point $x$. It is not immediate why there should be a subsequence working for all points simultaneously. Also, could you mention which version of Ascoli's theorem (if any) you already know. $\endgroup$ – MaoWao Mar 6 at 23:48
  • $\begingroup$ @MaoWao, I included the version of Ascoli's theorem that I have in mind. $\endgroup$ – George Mar 7 at 1:08
  • $\begingroup$ In your version of Ascoli-Arzela, what is $N$? $\endgroup$ – Nate Eldredge Mar 7 at 21:41
  • $\begingroup$ @NateEldredge, $N$ is a metric space, this version of AScoli-Arzela is for metric spaces $\endgroup$ – George Mar 7 at 22:11
2
$\begingroup$

This will follow pretty directly from the version of Ascoli-Arzela that you already know.

Let $D$ be the unit ball of $C^\alpha(X, B_1)$. Your goal is to show that $E = \kappa D$ is relatively compact, i.e. that the set $E = \{\kappa f : f \in D\}$ is relatively compact in $C(X, B_2)$ (here $N = B_2$). You want to verify the hypotheses of Ascoli-Arzela.

For (1), use the Hölder continuity. You should be able to show that for any $f \in D$ and any $x,y \in X$, we have $\|(\kappa f)(x) - (\kappa f)(y)\|_{B_2} = \|K(f(x)) - K(f(y))\|_{B_2} \le \|K\| d(x,y)^\alpha$, where $\|K\|$ is the operator norm of $K$ and $d$ is the metric on $X$. Equicontinuity should then follow easily.

For (2), fix $x \in X$ and note that for any $f \in D$, we have $\|f(x)\|_{B_1} \le 1$. Now use the fact that $K$ is a compact operator to conclude that $\{(\kappa f)(x) : f \in D\}$ is relatively compact in $B_2$.

$\endgroup$
  • $\begingroup$ I understood your hints, but I don't sure if I understand why you are considering $D$ as the unit ball of $\mathcal{C}^{\alpha}(X,B_1)$. The idea is consider $D$ as a bounded set, which is contained in a closed ball $\overline{B}(0, A) \subset \mathcal{C}^{\alpha}(X,B_1)$ for some $A > 0$. Since $f \in D$, we must have $||f||_{C(X,B_2)} \leq A$, then $||\frac{1}{A} f||_{C(X,B_2)} \leq 1$. Thus, we can suppose w.l.o.g. that $D = \overline{B}(0,1)$, is this the reason why you assume $D$ the unit ball or the reason is more subtle? $\endgroup$ – George Mar 7 at 23:42
  • $\begingroup$ Sorry if this seems a stupid question, but, as I said in the OP, I never did Functional Analysis, I just want to be sure if I understood the argument. $\endgroup$ – George Mar 7 at 23:43
  • $\begingroup$ For me, the definition of "$\kappa : Y_1 \to Y_2$ is a compact operator" is "the image of the unit ball of $Y_1$ under $\kappa$ is a relatively compact set in $Y_2$". What's your definition? $\endgroup$ – Nate Eldredge Mar 7 at 23:44
  • $\begingroup$ The same of Appendix $D.5$ of Evans' book: "let $X$ and $Y$ be real Banach spaces. A bounded linear operator $$K: X \longrightarrow Y$$ is called compact provided for each bounded sequence $\{ u_k \}_{k=1}^{\infty} \subset X$, the sequence $\{ Ku_k \}_{k=1}^{\infty}$ is precompact in $Y$; that is, there exists a subsequence $\{ u_{k_j} \}_{j=1}^{\infty}$ such that $\{ Ku_{k_j} \}_{j=1}^{\infty}$ converges in $Y$." $\endgroup$ – George Mar 8 at 0:02
  • $\begingroup$ Okay. Then the rescaling argument you gave above can be used to show that my definition is equivalent to Evans'. $\endgroup$ – Nate Eldredge Mar 8 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.