Show an element is in the orthogonal complement of a Hilbert space?

Question

Let $M$ be a linear subspace of a Hilbert space $H$. I want to show that if $$ ||x-y|| \ge ||x||, \quad \text{for all} \ y \in M, $$ then $x\in M^\perp$.

Let $x\in H$ and $y\in M$ such that $||x-y||^2 \ge ||x||^2$. This is equivalent to $$ ||x||^2 - 2(x,y) + ||y||^2 - ||x||^2 \ge 0. $$ That is $$ - 2(x,y) + ||y||^2 \ge 0. $$ As this inequality this has to hold for all $y\in M$ we have that $(x,y)$ must equal to zero. Is this correct or have I missed some details?

Is $M$ closed subspace of a $H$.? – Nosrati Aug 29 '17 at 10:34 — Nosrati, Aug 29 '17 at 10:34
@MyGlasses Just a linear subspace. – ManUtdBloke Aug 29 '17 at 10:46 — ManUtdBloke, Aug 29 '17 at 10:46

score 1 · Accepted Answer · answered Aug 29 '17 at 10:45

1

You have to better motivate the last step. A possible way to do it is the following. Suppose that $(x,y)\neq0$. If necessary, take $-y$ instead of $y$ so that $(x,y)>0$. Now take $$\epsilon < \frac{(x,y)}{\|y\|^2}$$ and set $y':=\epsilon y\in M$ to get $$\|y'\|^2 = \epsilon^2\|y\|^2<\epsilon(x,y)=(x,y')\ ,$$ which is a contradiction.

answered Aug 29 '17 at 10:45

Daniel Robert-Nicoud

29,795

I don't really see what the contradiction is, and how this relates to my last step? I have $-2(x,y) + ||y||^2 > 0$, I don't see how you are using that in your approach? – ManUtdBloke Sep 01 '17 at 05:25
@eurocoder I constructed a $y'\in M$ contradicting the assumption... – Daniel Robert-Nicoud Sep 01 '17 at 11:20

Show an element is in the orthogonal complement of a Hilbert space?

1 Answers1