Property about orthogonal projections

Question

Let $H$ be a Hilbert space and $P_1,P_2$ orthogonal projections on the closed spaces $M_1$ and $M_2$.

I want to show that $M_1\subset M_2$ implies $\|P_1x\|\leq \|P_2x\|$

Some observations:

$M_2^\perp \subset M_1^\perp$, $P_1P_2=P_2P_1$

can someone give me a tip?

score 2 · Answer 1 · answered Feb 07 '15 at 16:31

The answer posted by Pp.. uses techniques worth knowning about, but there's a simpler way in this particular instance. The point $P_1(x)$ is the point that minimizes $y\mapsto \|y-x\|$ subject to the constraint that $y\in M_1$. If $M_2 \supseteq M_1$, then minimizing subject to the condition that $y\in M_2$ is minimizing with respect to a weaker condition; hence you get at least as small a minimum. The idea is that there may or may not be some point in $y\in M_2$ for which $\|y-x\|$ is smaller than it is for all $y\in M_1$. If there is none, then the minimum is what it was when one was restricted to $M_1$; otherwise it's smaller. It can't get bigger.

score 1 · Accepted Answer · edited Feb 07 '15 at 16:11

1

Get an orthonomal basis $\{v_k\}_{k\in I}$ of $M_1$, extend it to an orthonormal basis $\{v_k\}_{k\in J}$, with $I\subset J$, of $M_2$, and express the projections in those bases.

You get $$P_1(x)=\sum_{k\in I}(x,v_k)v_k$$ and $$P_2(x)=\sum_{k\in J}(x,v_k)v_k$$ Compute their norms now.

$$\begin{align}\|P_1(x)\|^2&=\sum_{k\in I}|(x,v_k)|^2\\&\leq\sum_{k\in J}|(x,v_k)|^2=\|P_2(x)\|^2\end{align}$$

edited Feb 07 '15 at 16:11

Michael Hardy

1

answered Feb 07 '15 at 15:08

Pp..

5,973

Question: The orthonormal basis must not be countable, right? – Epsilondelta Feb 07 '15 at 15:19
And another question: Shouldnt there be $P_1(x)=\sum_{k=1}^{n}(P_1(x),v_k)v_k$ instead of $P_1(x)=\sum_{k=1}^{n}(x,v_k)v_k$ ? – Epsilondelta Feb 07 '15 at 15:27
@Epsilondelta It is just $(x,v_k)$. The bases can be uncountable, yes. The number of non-zero terms will be countable, though. – Pp.. Feb 07 '15 at 15:35
I changed $\displaystyle||P_1(x)||$ to $\displaystyle|P_1(x)|$. That is standard usage, as you'll see if you consult any reasonable manual. – Michael Hardy Feb 07 '15 at 16:13
I think this question admits a simpler answer. I'll post one. – Michael Hardy Feb 07 '15 at 16:13
@MichaelHardy Sure. While you are at it, fix this answer too. – Pp.. Feb 07 '15 at 16:26
Thanks for the answers:) If possible could you look at this too? http://math.stackexchange.com/questions/1137872/a-property-about-orthogonal-projections – Epsilondelta Feb 07 '15 at 16:38

Property about orthogonal projections

2 Answers2