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Let $(f_{n})_{n}\subseteq C([0,1])$ be an arbitrary sequence such that $\vert \vert f_{n} \vert \vert_{\infty}\leq 1$ for all $n \in \mathbb N$ and $k \in C([0,1]^{2})$.

Define the bounded linear map $T: (C([0,1]),\vert \vert \cdot \vert \vert_{\infty}) \to (C([0,1]),\vert \vert \cdot \vert \vert_{\infty}), f\mapsto Tf$

where $Tf: [0,1]\to \mathbb K, x\mapsto \int_{0}^{x}k(x,y)f(y)dy$

Show that $(Tf_{n})_{n}$ has a convergent subsequence.

I have been told to use Arzela-Ascoli, but I believe I have found a much easier way (which is most probably incorrect, and thus I need someone to correct my thinking).

Let $(f_{n})_{n}\subseteq C([0,1])$ be an arbitrary sequence, note that:

$\vert\vert Tf_{n}\vert\vert_{\infty}=\sup\limits_{x\in [0,1]}\vert Tf_{n}(x)\vert=\sup\limits_{x\in [0,1]}\vert\int_{0}^{x}k(x,y)f_{n}(y)dy\vert\leq \sup\limits_{x\in [0,1]}\int_{0}^{x}\vert k(x,y)f_{n}(y)\vert dy\leq\vert\vert f_{n}\vert\vert_{\infty}\sup\limits_{x\in [0,1]}\int_{0}^{x}\vert k(x,y)\vert dy\leq \int_{0}^{x}\vert k(x,y)\vert dy<\infty$ for all $n \in \mathbb N$. By Bolzano Weierstrass there must be a subsequence $f_{n_{k}}$ so that $Tf_{n_{k}}$ converges.

Why can I not simply use Bolzano-Weierstrass here?

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  • $\begingroup$ How exactly are you trying to use Bolzano-Weierstrass? $\endgroup$ – carmichael561 Jul 13 at 19:23
  • $\begingroup$ I am using using Bolzano-Weierstrass as I have a sequence $(\sup\limits_{x \in [0,1]}Tf_{n}(x))_{n} \subseteq \mathbb R$ and thus showing it is bounded, means that there must be a convergent subsequence $\endgroup$ – SABOY Jul 13 at 19:25
  • $\begingroup$ @Matematleta for all $x \in [0,1]$ because $k \in C([0,1]^{2})$ $\endgroup$ – SABOY Jul 13 at 19:26
  • $\begingroup$ How does the fact that the $numbers$, $\|Tf_n\|$ imply that there is a convergent subsequence in the sup norm? $\endgroup$ – Matematleta Jul 13 at 19:33
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You cannot simply use the Bolzano-Weirstrass theorem here, as you are using it, because the fact that the sequence of numbers $\|Tf_n\|_{\infty}$ has a convergent subsequence does not, usually, imply that the sequence of functions $\{Tf_n\}_{n=1}^{\infty}$ has a uniformly-convergent subsequence, which is what you need to prove.

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