- Let K be a closed convex subset of a Hilbert space H. Show that for each \(x \in H\) , there is a unique \(y \in K\) such that
\[ | | x - y | | = i n f _ { z \in K } | | x - z | | \]
Solution
Let \(d=\inf_{z\in K}\norm{x-z}\). For \(u,v\in K\) the parallelogram law and \(\frac{u+v}2\in K\) give \[\norm{u-v}^2=2\norm{u-x}^2+2\norm{v-x}^2-4\norm{\tfrac{u+v}2-x}^2\le2\norm{u-x}^2+2\norm{v-x}^2-4d^2.\tag{$*$}\]
1. Some \(y\in K\) has \(\norm{x-y}=d\).
Choose \(z_n\in K\) with \(\norm{x-z_n}\to d\). By \((*)\), \(\norm{z_n-z_m}^2\to2d^2+2d^2-4d^2=0\), so \((z_n)\) is Cauchy and converges to some \(y\in H\), since \(H\) is complete. Then \(y\in K\) because \(K\) is closed, and \(\norm{x-y}=d\) by continuity of the norm.
2. Two such points are equal.
If \(\norm{x-y_1}=\norm{x-y_2}=d\) with \(y_1,y_2\in K\), then \((*)\) gives \(\norm{y_1-y_2}^2\le0\).
3. Q.E.D.
Steps 1 and 2 give existence and uniqueness.