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Let $\mathbb{H}$ be a Hilbert space on which $\Phi$ is a bounded linear functional. Define $f : \mathbb{H} \rightarrow \mathbb{R} : x \mapsto ||x||^2 + \Phi (x)$. Prove that each (nonempty) closed convex $K\subset \mathbb{H}$ contains a unique point at which the restriction $f|K$ is minimized.

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up vote 3 down vote accepted


  1. By Riesz representation, you can complete the square and it suffices to consider $f(x) = \|x\|^2$.

  2. Let $x,y$ attain the minimum. Denote $z = (x+y)/2$. By convexity $z$ is also in $K$. Show that $f(z) < f(x)$. This shows uniqueness.

  3. For existence, you should recall that closed, bounded, convex sets in a Hilbert space is weakly compact (Hanh-Banach + Banach-Alaoglu). Then it suffices to show that $f$ is sequentially lower semi-continuous with respect to the weak topology. To show this last step note that if $x_n \to x$ weakly, then $\|x\|^2 = \lim \langle x_n,x\rangle$.

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