1
$\begingroup$

Let $M>0$, $S=\{v\in C[0,1]:|v(x_2)-v(x_1)|\le \frac{M}2|x_2-x_1|, \forall x_1, x_2\in[0,1], v(0)=0\}$. Show that $S$ is a compact convex set in $(C[0,1], \|\cdot\|_\infty$).

I'm not sure how to approach this problem. For compactness, I think we can consider a sequence of functions $\{v_s\}_s\subset S$ and prove that it has a convergent subsequence. Since $[0,1]$ is compact and $v_s$ is continuous, $v_s([0,1])$ is compact for all $s$. Thus $\{v_s\}_s$ is a bounded sequence, so it has a convergent subsequence.

Now how does one show that $\{v_s\}_s$ converges in $S$ under $\|\cdot\|_\infty$? How do the Lipschitz condition and $v(0)=0$ help?

EDIT: It appears that we can apply the Arzela-Ascoli theorem here, but in order to do so we need to somehow prove that $S$ contains uniformly bounded and equicontinuous functions. How do we do so?

That's where I'm stuck. Would appreciate some help.

For convexity, would it be correct to argue that if $u, v\in S$ and $x\in [0,1],\alpha\in\mathbb{R}$ then $\alpha u(x) + (1-\alpha)v(x)=\alpha(u(x)-v(x))+v(x)=\alpha w(x) + v(x) \in S$, since $S$ is a vector space?

$\endgroup$
  • $\begingroup$ It's just some positive real number. I've just edited. $\endgroup$ – sequence Oct 24 '17 at 8:55
1
$\begingroup$

The set $S$ is relatively compact, by the Arzelà–Ascoli theorem. We can apply it here because, if $f\in S$:

  • if $x\in[0,1]$, then $\bigl|f(x)\bigr|=\bigl|f(x)-f(0)\bigr|\leqslant\frac M2|x-0|\leqslant\frac M2$;
  • if $x,y\in[0,1]$, if $\varepsilon>0$ and if you take $\delta=\frac{2\varepsilon}M$, then$$|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<\varepsilon.$$

The set $S$ is also closed. Therefore, it is compact.

If $u,v\in S$ and if $\alpha\in[0,1]$, then $\alpha u+(1-\alpha)v\in S$, because:

  • $\bigl(\alpha u+(1-\alpha)v\bigr)(0)=\alpha u(0)+(1-\alpha)v(0)=0$;
  • if $x_1,x_2\in[0,1]$, then\begin{multline}\bigl|\bigl(\alpha u+(1-\alpha)v\bigr)(x_2)-\bigl(\alpha u+(1-\alpha)v\bigr)(x_1)\bigr|=\\=\bigl|\alpha\bigl(u(x_2)-u(x_1)\bigr)+(1-\alpha)\bigl(v(x_2)-v(x_1)\bigr)\bigr|\leqslant\\\leqslant\alpha\bigl|u(x_2)-u(x_1)\bigr|+(1-\alpha)\bigl|v(x_2)-v(x_1)\bigr|\leqslant\\\leqslant\frac M2|x_2-x_1|.\end{multline}

Therefore, $S$ is convex.

$\endgroup$
  • $\begingroup$ Can you please clarify the following: what makes us able to apply the Arzela-Ascoli theorem in this case? That is, how do we know that the conditions are satisfied? Unfortunately, I don't see how they are satisfied. Also, can you please clarify how convexity follows from $\bigl(\alpha u+(1-\alpha)v\bigr)(0)=\alpha u(0)+(1-\alpha)v(0)=0$ and the inequality with $x_1,x_2$? $\endgroup$ – sequence Oct 24 '17 at 9:24
  • 1
    $\begingroup$ @sequence I've edited my question in order to explain why can we apply the Arzelà-Ascoli theorem. Concerning the convexity part, what does “convex” mean to you? $\endgroup$ – José Carlos Santos Oct 24 '17 at 9:37
  • $\begingroup$ Was too tired yesterday, thanks for clarifying. $\endgroup$ – sequence Oct 24 '17 at 13:12
  • 1
    $\begingroup$ @sequence It's much more simple than that. It is the intersection of a family of closed sets. The set $\{v\in C([0,1])\,|\,v(0)=0\}$ and the sets of the type $\{v\in C([0,1])\,|\,|v(x_2)-v(x_1)|\leqslant\frac M2|x_2-x_1|\}$ ($x_1,x_2\in[0,1]$). $\endgroup$ – José Carlos Santos Oct 24 '17 at 21:50
  • 1
    $\begingroup$ @sequence $F^{-1}\{0\})$ is the inverse image of $\{0\}$ with respect to the map $F$. Since $\{0\}$ is closed, $F^{-1}\{0\})$ is also closed. It's similar for $G^{-1}$. $\endgroup$ – José Carlos Santos Oct 26 '17 at 3:45

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.