Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question

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.

share|cite|improve this answer

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

share|cite|improve this answer
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

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

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

share|cite|improve this answer
i tried but couldn't get to the expected inequality. – newbie Nov 29 '12 at 13:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.