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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.

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6  
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. –  Thomas E. Jun 3 '12 at 5:09
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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
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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

1 Answer 1

up vote 9 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.)

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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
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@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

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