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Spring 2010 - Problem 7

convex sets, Hilbert spaces

Let $H$ be a Hilbert space and let $E$ be a closed convex subset of $H$ . Prove that there exists a unique element $x \in E$ such that

\norm{x} = \inf_{y \in E}\,\norm{y}.

Solution.

First observe that we have the parallelogram law:

\norm{x - y}^2 + \norm{x + y}^2 = 2\norm{x}^2 + 2\norm{y}^2

Set $\delta = \inf_{y \in E}\,\norm{y}$ and let $\set{x_n}_n \subseteq E$ be a sequence in $E$ such that $\norm{x_n} \to \delta$ . Since $E$ is convex, $\frac{1}{2}\p{x_n + x_m} \in E$ for any $n, m \geq 1$ . Combined with the triangle inequality, this implies

\delta \leq \norm{\frac{x_n + x_m}{2}} \leq \frac{\norm{x_n} + \norm{x_m}}{2} \xrightarrow{n,m\to\infty} \delta.

Thus, by the parallelogram law,

\begin{gathered} \norm{\frac{x_n - x_m}{2}}^2 + \norm{\frac{x_n + x_m}{2}}^2 = 2\norm{\frac{x_n}{2}}^2 + 2\norm{\frac{x_m}{2}}^2 \\ \implies \norm{x_n - x_m}^2 = \frac{\norm{x_n}^2 + \norm{x_m}^2}{2} - \norm{\frac{x_n + x_m}{2}}^2 \xrightarrow{n,m\to\infty} \frac{\delta^2 + \delta^2}{2} - \delta^2 = 0. \end{gathered}

In other words, $\set{x_n}_n$ is Cauchy, so it converges to some $x \in H$ . Since $E$ was closed, it follows that $x \in E$ , which proves existence. To show uniqueness, suppose $x^* \in E$ is another minimizer. Then by convexity, $\frac{1}{2}\p{x + x^*} \in E$ and so by another application of the parallelogram law,

\begin{gathered} \norm{\frac{x - x^*}{2}}^2 + \norm{\frac{x + x^*}{2}}^2 = 2\norm{\frac{x}{2}}^2 + 2\norm{\frac{x^*}{2}}^2 = \frac{\delta^2}{2} + \frac{\delta^2}{2} = \delta^2 \\ \implies \delta^2 \leq \norm{\frac{x + x^*}{2}}^2 = \delta^2 - \norm{\frac{x - x^*}{2}}^2. \end{gathered}

The first inequality comes from the fact that $\frac{1}{2}\p{x + x^*} \in E$ and the fact that $\delta$ is the minimum norm. Thus, $\norm{x - x^*} = 0 \implies x = x^*$ , which completes the proof.