Let \(E\) be the set of continuous functions \(u:[0,1]\to\mathbb R\) such that \[ |u(x)-u(y)|\le |x-y| \qquad(0\le x,y\le1), \] and \(u(0)=0\). Define \[ \varphi(u)=\int_0^1\bigl(u(x)^2-u(x)\bigr)\,dx. \] Show that \(\varphi\) attains its maximum value at some \(u\in E\).
Solution
Equip \(C([0,1],\RR)\) with the supremum norm \[ \norm{u}_\infty = \max_{0\leq x\leq1}\abs{u(x)}. \]
1. The family \(E\) is uniformly bounded:\[ \norm{u}_\infty\leq1 \]for every \(u\in E\).
For \(u\in E\) and \(x\in[0,1]\), \[ \abs{u(x)} = \abs{u(x)-u(0)} \leq \abs{x-0} \leq1. \] Taking the maximum over \(x\) gives the claim.
2. The family \(E\) is equicontinuous.
Every \(u\in E\) satisfies \[ \abs{u(x)-u(y)}\leq\abs{x-y} \] for all \(x,y\in[0,1]\). Thus for every \(\varepsilon>0\), taking \(\delta=\varepsilon\) gives \[ \abs{x-y}<\delta \quad\Longrightarrow\quad \abs{u(x)-u(y)}<\varepsilon \] simultaneously for all \(u\in E\).
3. The set \(E\) is closed in the supremum norm.
Suppose \(u_j\in E\) and \[ \norm{u_j-u}_\infty\longrightarrow0 \] for some \(u\in C([0,1],\RR)\). Then \[ u(0) = \lim_{j\to\infty}u_j(0) =0. \] For every \(x,y\in[0,1]\), \[ \begin{aligned} \abs{u(x)-u(y)} &= \lim_{j\to\infty} \abs{u_j(x)-u_j(y)}\\ &\leq \abs{x-y}. \end{aligned} \] Hence \(u\in E\).
4. The set \(E\) is compact in the supremum norm.
By steps 1 and 2, \(E\) is uniformly bounded and equicontinuous. The Arzelà–Ascoli theorem therefore says that the closure of \(E\) in \(C([0,1],\RR)\) is compact. Step 3 says that \(E\) is already closed, so \(E\) itself is compact.
5. For all \(u,v\in E\),\[ \abs{\varphi(u)-\varphi(v)} \leq 3\norm{u-v}_\infty. \]
By step 1, \[ \abs{u(x)}\leq1, \qquad \abs{v(x)}\leq1 \] for every \(x\in[0,1]\). Therefore \[ \begin{aligned} \abs{\varphi(u)-\varphi(v)} &\leq \int_0^1 \abs{ u(x)^2-v(x)^2-u(x)+v(x) } \,dx\\ &\leq \int_0^1 \abs{u(x)-v(x)} \bigl(\abs{u(x)+v(x)}+1\bigr) \,dx\\ &\leq 3\norm{u-v}_\infty. \end{aligned} \]
6. The functional\[ \varphi:E\longrightarrow\RR \]is continuous.
Step 5 shows that \(\varphi\) is Lipschitz with constant \(3\).
7. There is \(u_*\in E\) such that\[ \boxed{ \varphi(u_*) = \max_{u\in E}\varphi(u) }. \]
The set \(E\) is nonempty because the zero function belongs to it. By step 4, \(E\) is compact, and by step 6, \(\varphi\) is continuous. The extreme value theorem therefore gives a point \(u_*\in E\) at which \(\varphi\) attains its maximum.
8. Q.E.D.
Step 7 is the required conclusion.