Prove that every nonempty closed convex subset of a Hilbert space has a unique element of minimal norm.
Nonemptiness is necessary: the empty set is closed and convex but has no 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 \[ d:=\inf_{x\in C}\|x\|. \] Nonemptiness gives \(0\leq d<\infty\). By the defining property of the infimum, choose \(x_n\in C\) with \(d\leq\|x_n\|<d+1/n\). Then \[ \|x_n\|^2\longrightarrow d^2. \]
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 \[ \left\|\frac{x_n-x_m}{2}\right\|^2 =\frac12\|x_n\|^2+\frac12\|x_m\|^2 -\left\|\frac{x_n+x_m}{2}\right\|^2. \] Therefore \[ \frac14\|x_n-x_m\|^2 \le \frac12\|x_n\|^2+\frac12\|x_m\|^2-d^2. \] The right-hand side tends to \(0\) as \(m,n\to\infty\). Hence \((x_n)\) is Cauchy.
3. Obtain existence of a minimum-norm point.
Since \(H\) is complete, \(x_n\to x\) for some \(x\in H\). Because \(C\) is closed, \[ x\in C. \] Continuity of the norm gives \[ \|x\|=d. \] Thus \(x\) has minimal norm in \(C\).
4. Prove uniqueness.
Suppose \(x,y\in C\) both satisfy \[ \|x\|=\|y\|=d. \] Again \((x+y)/2\in C\), hence \[ \left\|\frac{x+y}{2}\right\|\ge d. \] The parallelogram identity gives \[ \frac14\|x-y\|^2 =\frac12d^2+\frac12d^2 -\left\|\frac{x+y}{2}\right\|^2 \le0. \] Therefore \(x=y\).
Hence every nonempty closed convex subset of a Hilbert space has a unique element of minimal norm.