0
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Couldn't this same argument be applied to $\{f_n\} \subset U$ by noting that as a sequence of continuous functions it is equicontinuous and uniformly bounded? Thus Arzela-Ascoli Theorem also says that $\{f_n\}$ has a convergent subsequence, meaning that $U$ is compact, which is not the case for an infinite dimensional Banach space? $\endgroup$ – P. Reinecke Oct 10 '18 at 6:37
  • 1
    $\begingroup$ @G.Richardson No. All you know about $\{f_n\}$ is that it is bounded. Why is it equicontinuous? For example, if $f_n(x)=x^{n}$ then $\{f_n\}$ is not equicontinuous and it does not have a uniformly convergent subsequence. The sequence $\{Tf_n\}$ has the property that the derivatives are bounded and that helps us to apply Arzela-Ascoli. $\endgroup$ – Kabo Murphy Oct 10 '18 at 6:41
  • $\begingroup$ I think I'm misunderstanding what equicontinuity is then. I thought a collection of subsets was equicontinuous if each element of the family was continuous. However, I just looked at the definition again and I think I understand now. Equicontinuous is a stronger condition than even saying that every function in the family is uniformly continuous. Equicontinuity is essentially saying that for any $\epsilon >0$ there's some "ultimate" $\delta >0$ for which every function in the family is uniformly continuous. Is that more accurate understanding of equicontinuous, or am I still missing something? $\endgroup$ – P. Reinecke Oct 10 '18 at 6:56
  • $\begingroup$ You have got the point. Equicontinuity means there must be one $\delta$ corresponding a given $\epsilon$ that works for all members of the family. $\endgroup$ – Kabo Murphy Oct 10 '18 at 7:15
  • $\begingroup$ Ahh! Ok. I think I understand now. Because we only know that every element of $\{f_n\}$ is continuous, we can't apply Arzela_Ascoli to it. However, because we can manipulate $\{g_n\}$ to show that for any $\epsilon >0$ picking $\delta = \epsilon$ makes all elements of the sequence continuous, and because all the elements of the sequence are uniformly bounded, we can apply Arzela-Ascoli to it, which gives us a definition of compactness for $f[U]$. Thank you so much! I think I need to review Arzela-Ascoli a bit more to fully get it, but this has been a great help! Thank you again. $\endgroup$ – P. Reinecke Oct 10 '18 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.