# Orthogonal projection to closed, convex subset in a Hilbert space

I don't understand one step in the proof of the following lemma (Projektionssatz):

Let $X$ a Hilbert space with scalar product $(\cdot)_X$ and let $A\subset X$ be convex and closed. Then there is a unique map $P:X\rightarrow A$ that satisfies: $\|x-P(x)\| = \inf_{y\in A} \|x- y\|$. This is equivalent to the following statement:

(1) For all $a\in A$ and fixed $x\in X$, $\mbox{Re}\bigl( x-P(x), a-P(x) \bigr)_X \le 0$.

I don't understand the following step in the proof that (1) implies the properties of $P$:

Let $a\in A$. Then

$\|x-P(x)\|^2 + 2\mbox{Re}\bigl( x-P(x), P(x)-a \bigr)_X + \|P(x)-a\|^2 \ge \|x-P(x)\|^2$.

I don't understand the "$\ge$". How do we get rid of the term $\|P(x)-a\|$ on the left hand side?

Thank you very much!

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## 1 Answer

It is non-negative! Hence, just drop it.

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And what about second terms in lhs? –  Norbert Jul 9 '12 at 9:47
Oh my god - I'm sooo stupid. Sorry to bother you with such trivialities... Thanks! –  Sh4pe Jul 9 '12 at 9:55
@Norbert: Non-negative by assumption. –  Dirk Jul 9 '12 at 9:59
Well, I was inattentive –  Norbert Jul 9 '12 at 11:10