Prove that every closed convex subset of a Hilbert space has a unique element of minimal norm.
Solution
1. Choose a minimizing sequence.
Let \(C\) be a nonempty closed convex subset of a Hilbert space \(H\), and set \[ d:=\inf_{x\in C}\|x\|. \] Choose \(x_n\in C\) so that \[ \|x_n\|^2\longrightarrow d^2. \]
2. Prove that the minimizing sequence is Cauchy.
By convexity, \[ \frac{x_n+x_m}{2}\in C, \] so \[ \left\|\frac{x_n+x_m}{2}\right\|\ge d. \] The parallelogram identity gives \[ \|x_n-x_m\|^2 =2\|x_n\|^2+2\|x_m\|^2-\|x_n+x_m\|^2. \] Since \[ \|x_n+x_m\|^2 =4\left\|\frac{x_n+x_m}{2}\right\|^2 \ge4d^2, \] we obtain \[ \|x_n-x_m\|^2 \le2\|x_n\|^2+2\|x_m\|^2-4d^2. \] The right-hand side tends to \(0\) as \(m,n\to\infty\). Thus \((x_n)\) is Cauchy.
3. Obtain a minimum-norm point.
Since \(H\) is complete, \(x_n\to x\) for some \(x\in H\). Since \(C\) is closed, \(x\in C\). Continuity of the norm gives \[ \|x\|=\lim_n\|x_n\|=d. \] Thus \(x\) has minimum norm in \(C\).
4. Prove uniqueness.
Suppose \(x,y\in C\) both satisfy \[ \|x\|=\|y\|=d. \] Again \((x+y)/2\in C\), so its norm is at least \(d\). The parallelogram identity yields \[ \begin{aligned} \|x-y\|^2 &=2\|x\|^2+2\|y\|^2-\|x+y\|^2\\ &\le4d^2-4d^2=0. \end{aligned} \] Hence \(x=y\). Therefore every nonempty closed convex subset of a Hilbert space has a unique element of minimum norm.