Show that $\phi$ achieves its maximum on the set of $1-$ Lipschitz functions in $C[0, 1]$ passing through the origin

Question

Let $E = \{u \in C[0,1]: |u(x)-u(y)|\leq |x-y|, x,y, \in [0,1], u(0)=0\}$ and let $\phi: E \to \mathbb{R}$ be define by $\phi(u)=\displaystyle\int_{0}^{1}(u(x)^2-u(x))dx.$ Show that $\phi$ achieves its maximum at some element of $E.$ ($C[0,1]$ is equipped with the sup norm)

Answer 1

It suffices to show that $E$ is compact and $\phi$ is continuous. $\phi$ is continuous since:

Let $\varepsilon >0$ be given, set $\delta = 3\varepsilon$ and let $u, v \in E$ such that $||u-v||_{\infty} <\delta.$ Consider:

$\begin{align}|\phi(u)-\phi(v)|&= \left|\displaystyle \int_{0}^{1}(u(x)^2-u(x)-(v(x)^2-v(x)))dx\right|\\&\leq \displaystyle \int_{0}^{1}\left|(u(x)-v(x))||(u(x)+v(x)+1)\right|dx\\&\leq ||u-v||_\infty\displaystyle \int_{0}^{1}(|u(x)|+|v(x)|+1)dx\\&\leq ||u-v||_\infty\displaystyle \int_{0}^{1}(2|x|+1)dx\\&=3||u-v||_{\infty}< 3\delta= \varepsilon\end{align}$

so $\phi$ is continuous on $E.$ For compactness of $E$, it suffices to show completeness and total boundedness of $E.$

Claim: $E$ is closed.

Let $p$ be a limit point of $E.$ The there exists a sequence $(u_n) \in E$ such that $u_n \to p.$ So $0=u_n(0)\to p(0)$ and $p(0)=0.$ Let $x, y \in [0,1].$ Consider

$\begin{align}|p(x)-p(y)|&= |u_n(x)-p(x)-(u_n(y)-p(y))+u_n(y)-u_n(x)|\\&\leq |u_n(x)-p(x)|+|u_n(y)-p(y)|+|u_n(y)-u_n(x)|\\&\leq|u_n(x)-p(x)|+|u_n(y)-p(y)|+|x-y|\end{align}$

so by taking limits as $n \to \infty$ and using the fact that $u_n \to p$ we get $|p(x)-p(y)|< |x-y|$ for every $x, y \in [0,1].$ Then $p \in E$ and $E$ is closed. Since $E$ is a closed subset of a complete metric space $C[0,1], E$ is complete.

Claim: $E$ is totally bounded

It suffices to show that every sequence in $E$ has a Cauchy subsequence. Let $(u_n)$ be any sequence in $E.$ Note that for all $u \in E$ we have $|u(x)|= |u(x)-u(0)|\leq |x|\leq 1$ so the sequence $(u_n)$ is uniformly bounded and since each $u_n$ is $1-$ Lipschitz, the $(u_{n})$ form an equicontinuous family and hence by the Arzela-Ascoli theorem, it follows that there exists a subsequence $u_{n_k}$ that converges uniformly in $E.$ Since this sequences converges uniformly, it is Cauchy. Hence $E$ is totally bounded.

Hence $E$ being complete and totally bounded must be sequentially compact and hence compact.

Therefore we showed that a continuous function $\phi$ is defined on a compact set $E.$ Hence it achieves its maximum on $E$ and we are done.