Orthogonal projections on a complete convex set

Question

Let assume it is already known that:

If $H$ is an inner product space and $\varnothing \neq A \subset H$ is a complete convex subset, then there exists a unique vector $P_A f:=g\in A$ with $\|f-g\| = d(f,A): = \inf\{\,\|f-h\|\, : h\in A\}.$

Then I want to prove the following statements are equivalent:

i) $g = P_A f$

ii) $g\in A$ and $\operatorname{Re}\langle f-g,h-g \rangle \leq0, \ \forall f\in A .$

(ii) to (i): know $$\|f-h\|^2 = \|f-g+g-h\|^2 = \|f-g\|^2+\|g-h\|^2+2\operatorname{Re}\langle f-g,g-h\rangle,$$ so $$\|f-g\|^2\leq\|f-h\|^2, \ \forall h\in A.$$

How to prove that (i) implies (ii)?

Answer 1

Assume that $\mathrm{Re}\langle f-g, h-g\rangle > 0$ for some vector $h\in A$ (I'm assuming you meant $\forall h\in A$ in your (ii)). Then you show that $h$ is closer to $f$ than $g$ is, for instance using the polarization identity $$ \|x\|^2 - \|y\|^2 = \mathrm{Re}\langle x+y, x-y\rangle.$$

Answer 2

Hint: $(i)\Rightarrow (ii),$ Let $g = P_A f$, that is

$$ \|f-g\| = d(f,A): = \inf\{\,(f,h)\,|\, h\in A\} \implies ||f-g||\leq ||f-h||\quad \forall h\in A $$

$$ \implies ||f-g||^2\leq ||f-h||^2 \implies <f-g,f-g>\, \leq \,<f-h,f-h>\dots\,.$$