My professor mentioned that for $V=C([0,1])$, equipped with the norm $||*||_{\infty}$, the function $T:V \rightarrow V$ given as $T(f(t))=\int_0^t f(s) ds$ is a "compact operator". I'm not too familiar with compact operators so I'm attempting to prove this as practice, but I'm having some trouble and was hoping someone could help.
My solution so far:
Since C([0,1]) with the infinity norm is a Banach Space it is sufficient to show that $T$ maps a bounded subset of $V$ to a totally bounded subset. Thus, I'm considering the image of the unit ball,$U$, of $V$ under $T$ and trying to prove that that is totally bounded. From uperbounding the integral as $T(f) \leq \int_0^t |f(s)|ds \leq \int_0^t1ds$ I've determined that $T[U] \subset U$ and so the image is bounded. I don't think that helps me too much though. From what I've read online it seems that I somehow need to incorporate equicontinuity, which I believe is given since I'm looking at a subset of continuous functions (is that correct?). My friend suggested incorporating it into my proof via the Arzelà–Ascoli Theorem. However, I don't see how this is possible since $U$ is not a compact set since the Banach space is infinite dimensional. Can anyone offer any further advice or tips? -Thank you!
Best Answer
If $\{f_n\}$ is any sequence in the unit ball $U$ of $V$ and $g_n=T(f_n)$ then $|g_n(x)-g_n(y)| =\int_x^{y} |f_n(t)| \, dt \leq |x-y|$. By Arzela-Ascoli Theorem $\{g_n\}$ has a convergent subsequence. Thus every sequence in $T(U)$ has a convergent subesequence which implies $T$ is compact.