# Show set of functions compact and convex

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?

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

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.

• 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 $x_1,x_2$? – sequence Oct 24 '17 at 9:24
• @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? – José Carlos Santos Oct 24 '17 at 9:37
• Was too tired yesterday, thanks for clarifying. – sequence Oct 24 '17 at 13:12
• @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 $\{v\in C([0,1])\,|\,|v(x_2)-v(x_1)|\leqslant\frac M2|x_2-x_1|\}$ ($x_1,x_2\in[0,1]$). – José Carlos Santos Oct 24 '17 at 21:50
• @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 $G^{-1}$. – José Carlos Santos Oct 26 '17 at 3:45