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!


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 $N$.

  • $\begingroup$ 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

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 $B_2$.

  • $\begingroup$ 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

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