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Let $k: [0,1] \times [0,1] \to \mathbb{R}$ be continuous (with respect to Euclidean metrics, in both spaces), and consider the map $K: (C[0,1],\mathbb{R}) \to (C[0,1],\mathbb{R})$, $(Kf)(x)=\int\limits_{0}^{1}k(x,y)f(y)dy$. Show that any sequence $f_{n}$ in $(C[0,1],\mathbb{R})$ with $\Vert f_{n} \Vert \leq 1$ (the sup norm) has a subsequence $f_{n_{j}}$ with $Kf_{n_{j}}$ uniformly convergent (Using Ascoli Arzela).

Attempt: Let $S=\lbrace Kf_{n}, n \in \mathbb{N}\rbrace$. $k$ is continuous on a compact set hence uniformly continuous. Let $M=|\sup k|+1$. Let $\epsilon>0$. $\exists \delta >0$ such that $\forall a_{1},b_{1},a_{2},b_{2} \in [0,1]$, $\Vert (a_{1},a_{2})-(b_{1},b_{2}) \Vert < \delta \Rightarrow |k(a_{1},a_{2})-k(b_{1},b_{2})|<\frac{\epsilon}{2M}$. Hence $\forall x_{1},x_{2} \in [0,1]$,if $|x_{1}-x_{2}|<\delta, |Kf(x_{1})-Kf(x_{2})|=|\int\limits_{0}^{1} [k(x_{1},y)-k(x_{2},y)]f(y)dy|\leq \int\limits_{0}^{1} \frac{\epsilon}{2} dy < \epsilon$. Hence $S$ is equicontinuous. Also, $S$ is equibounded, say by M. Hence $S$ has compact closure. It remains for me to prove that $S$ is closed so that the result holds, any hint?

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  • $\begingroup$ why should $S$ be closed? And why do you need to show that? You have already show that $Kf_n$ has a convergent subsequence. The limit need not to be in $S$. $\endgroup$ – user251257 Sep 13 '15 at 20:42
  • $\begingroup$ Oh ok! What a misundestanding ...Thanks for your comment! $\endgroup$ – mich95 Sep 13 '15 at 21:07
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You already proved that $\{Kf_n\}$ is equicontinuous and equibounded. Recall that these are precisely the conditions which should be met in order to conclude the existence of a uniformly convergent subsequence.

I think that you confused the Ascoli-Arzela theorem with the following corollary of it:

If $X$ is a compact and $S \subset C(X)$, then $S$ is compact $\iff$ $S$ is equibounded, equicontinuous and closed.

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