This is an old preliminary exam problem: Let $T:C([0,1])\rightarrow C([0,1])$ be defined by $Tf(x)=\int_0^x f(t)dt$. Prove that $T$ is a compact operator, i.e. the image of the unit ball is pre compact.

So, if I understand correctly, I need to show that the image of $B:=\{f\in C([0,1]) | \sup_{x\in[0,1]}|f(x)|<1\}$ has compact closure. Here is my attempt:

Note that $TB=\{f\in C[0,1]| \sup_{x\in[0,1]} \int_0^x f(t)dt<1\}$

My first thought is to rewrite this as $TB=\{F'(x) | \sup_{x\in[0,1]} F(x)<1\}$

I think this is justified because if $F(x)=\int_0^x f(t)dt$ then $F'(x)=f(x)$ (since w.l.o.g. $F(0)=0$) for almost all $x$.

So the next step is to figure out what the closure of $TB$ looks like, but this is where I'm stuck. Any help would be appreciated. Thanks!

Update: I understand, from the answer given by user251257 that $TB$ is uniformly bounded and equicontinuous, so that, by Arzela Ascoli, any sequence in $TB$ has a uniformly convergent subsequence. But I still don't understand why this shows that the closure of $TB$ is compact. In order for $\overline{TB}$ to be compact, don't we need for any sequence in $\overline{TB}$ to have a convergent subsequence? What if we take a sequence in $\overline{TB}\backslash TB$?

  • 1
    $\begingroup$ Hint: Arzela ascoli? $\endgroup$
    – user251257
    Commented Sep 29, 2016 at 18:13
  • $\begingroup$ If I understand your suggestion correctly, your saying that I should use the sequential definition of compactness. So I would have to take a sequence of functions in $\overline{TB}$ and then use Arzela Ascoli to justify there being a uniformly convergent subsequence. However, in order to do this, I still need to know what $\overline{TB}$ looks like, so I'm still stuck at the same point. $\endgroup$
    – user140776
    Commented Sep 29, 2016 at 19:19

1 Answer 1



  1. $TB = \{ g\in C^1[0,1] \mid \|g'\| < 1, g(0) = 0 \}$
  2. $TB$ is uniformly Lipschitz continuous by the Lipschitz constant $1$, in particular it is equicontinuous.
  3. $TB$ is uniformly bounded, as for $g\in TB$ we have $$ |g(x)| \le \int_0^x |g'(t)| dt \le x \le 1. $$
  4. By Arzela Ascoli theorem $TB$ is pre compact.
  5. For $g_k \in \overline{TB}$ there exists a $h_k\in TB$ with $\|g_k-h_k\| < 1/k$. Then, $h_k$ has a convergent subsequence and the corresponding subsequence of $g_k$ has the same limit.
  • $\begingroup$ @user140776: For $f\in B$, $g=Tf$ is obviously $C^1$, $\|g'\|=\|f\| < 1$ by assumption, and $g(0) = \int_0^0 f(t) dt = 0$. Now, for $g\in C^1[0,1]$ with $\|g'\| < 1$ and $g(0) = 0$ we have $g'\in B$ and $g(x) = g(0) + \int_0^x g'(t) dt$ by fundamental theorem of calculus. $\endgroup$
    – user251257
    Commented Sep 29, 2016 at 20:04
  • $\begingroup$ yes, I saw my error. Thank you for clarifying though. $\endgroup$
    – user140776
    Commented Sep 29, 2016 at 20:07
  • $\begingroup$ @ user251257 I see that this gives us a uniformly convergent subsequence for each sequence in $TB$ but what if we take a sequence in $\overline{TB}\backslash TB$ then I don't see how your argument implies the existence of a convergent subsequence in this case. Sorry if this seems like a dumb question. I'm slow. $\endgroup$
    – user140776
    Commented Sep 29, 2016 at 20:57
  • 1
    $\begingroup$ @user140776: See Item 5. $\endgroup$
    – user251257
    Commented Sep 29, 2016 at 21:08
  • $\begingroup$ Thanks so much! I'm not sure I would have figured that out on my own, but it makes total sense! Your awesome! :) $\endgroup$
    – user140776
    Commented Sep 29, 2016 at 21:13

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