# The compactness of an operator from $C[0,1]$ to $C[0,1]$

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?

• Depends on whether the limit of the convergent subsequence has to be in $T(B)$ or not. – 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.