Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $${\cal F}=\left\{ f:\left[0,1\right]\to\mathbb{R} : \left|f\left(x\right)-f\left(y\right)\right|\le\left|x-y\right|\mbox{ and }{\displaystyle \int_{0}^{1}f\left(x\right)dx=1}\right\}.$$ Show that ${\cal F}$ is a compact subset of $C\left(\left[0,1\right]\right)$.

When I am trying to show a set is compact, I usually resort to the every open cover has a finite subcover definition. But if this case, we are dealing with functions. So I am having a difficulty "visualizing" what's going on. Any help or solutions would be appreciated.

Edit: I should mention that we are working with respect to the sup norm.

share|cite|improve this question
What topology does $C([0,1])$ have? The uniform one? If yes, then work with sequences instead of covers. Try showing that any sequence has a uniformly convergent subsequence. – T. Eskin Jun 3 '12 at 5:09
Have you seen the Arzelà–Ascoli theorem? – Jonas Meyer Jun 3 '12 at 5:17
Yes we have seen it. Never thought of using it though!! – Galois Jun 3 '12 at 5:18
Perhaps it's not particularly usual for this question, but since you mentioned that you don't know how to visualize $C[0,1]$ with sup-norm, you might have a look here. – Martin Sleziak Jun 3 '12 at 5:29
up vote 10 down vote accepted

According to Arzelà–Ascoli theorem you only have to show that $\mathcal F$ is

  • equicontinuous, i.e. $(\forall x\in[0,1])(\forall \varepsilon>0)(\exists \delta>0)(\forall f\in\mathcal F) (\forall y) |y-x|<\delta \Rightarrow |f(y)-f(x)|<\varepsilon$;
  • pointwise bounded;
  • closed in $C[0,1]$.

Both equicontinuity and pointwise boundedness follow from the Lipschitz condition $|f(x)-f(y)|\le |x-y|$.

To show pointwise boundedness you can notice that $|f(x)-f(0)|\le |x|=x$, which means $$f(0)-x \le f(x) \le f(0)+x.$$ If you apply integral $\int_0^1$ to the left inequality, you get $f(0)\le\int_0^1 (x+f(x))\,\mathrm{d}x=\frac32$. Now the right inequality implies $f(x)\le \frac52$ for each $x$. (Thanks to Nate Eldredge, who pointed in his comment, that this was missing in my original answer.)

To show that it is closed in sup-norm, you only have to show that if $f_n$ converges to $f$ uniformly and $f_n\in\mathcal F$, then the limit is in $\mathcal F$. We know that integral behaves well w.r.t. uniform convergence, see this questions. Proof of the fact that the condition $(\forall x,y\in [0,1])|f(x)-f(y)|\le |x-y|$ is preserved by uniform convergence is more-or-less standard. (In fact, in this part we only use pointwise convergence.)

share|cite|improve this answer
Pointwise boundedness doesn't follow from Lipschitz alone. The condition on the integrals will have to be used as well. – Nate Eldredge Jun 3 '12 at 5:35
Thanks Nate, I hope the new version is correct. – Martin Sleziak Jun 3 '12 at 5:43
@Galois: It would be a worthwhile and probably enlightening exercise not to use the Arzelà-Ascoli theorem and do the proof of (sequential) compactness of $\mathcal{F}$ by hand. If you're getting stuck, look at the proof of the Arzelà-Ascoli theorem and adapt it to the present situation. – t.b. Jun 3 '12 at 12:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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