I'm trying to prove $T$ is a compact operator where $T$ is $$T:C[0,1] \to C[0,1] \text{ such that } T u(t)=\int_{0}^{t} u(s) d s.$$

But I "proved" that $T(B)$ is closed in $F$, which is not by this.

Let $u_n\in B$, i.e. $||u_n||\leqslant 1$ and $v_n=Tu_n$. Then $\left|v_{n}(x)-v_{n}(y)\right|=\int_{x}^{y}\left|u_{n}(t)\right| d t \leqslant|x-y|$. By Arzela-Ascoli Theorem $v_n$ has a convergent subsequence, which means $T(B)$ is compact. Therefore, $T(B)$, being a compact subset of a metric space, is closed.

What's wrong with my proof?

  • 1
    $\begingroup$ Depends on whether the limit of the convergent subsequence has to be in $T(B)$ or not. $\endgroup$ – Lord Shark the Unknown Nov 21 '19 at 5:54

Arzela -Ascoli Theorem does not tell you that $T(B)$ is compact. It tells you that $T(B)$ is relatively compact in the sense its closure is compact.


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