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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)?

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3 Answers 3

The answer by jathd has a right idea, but one step is missing: one should consider points on the line segment from $h$ to $g$ (which belong to $A$ by convexity). Assume $\operatorname{Re}\langle f-g,h-g\rangle<0$. To simplify notation, write $v=h-g$. For all sufficiently small $t>0$ the point $g+tv$ is in $A$. Since $$\|g+tv-f\|^2=\|g-f\|^2+2t\operatorname{Re}\langle g-f,v\rangle+O(t^2)<\|g-f\|^2$$ for small $t$, we have a contradiction.

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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 $$ \|x\|^2 - \|y\|^2 = \mathrm{Re}\langle x+y, x-y\rangle.$$

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I end up with a term involving $2f-h-g$, how should I deal with that? –  newbie Nov 29 '12 at 14:48

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

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

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

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i tried but couldn't get to the expected inequality. –  newbie Nov 29 '12 at 13:32

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