Functional analysis, projection operators

[Fredrik]

Homework Statement

I want to understand the proof of proposition 7.1 in Conway. The theorem says that if \{P_i|i\in I\} is a family of projection operators, and P_i is orthogonal to P_j when i\neq j, then for any x in a Hilbert space H,

\sum_{i\in I}P_ix=Px

where P is the projection operator for the closed subspace that's spanned by the members of the P_i(H).

Homework Equations

Suppose that P_M and P_N are projection operators for closed subspaces M and N respectively. Then

(a) P_M+P_N is a projection operator for M\oplus N if and only if M and N are orthogonal.
(b) P_MP_N is a projection operator for M\cap N if and only if [P_M,P_N]=0.
(c) P_M-P_N is a projection operator for M\cap N^\perp if and only if N\subset M.

The Attempt at a Solution

For each finite F\subset I, define

S_F=\sum_{i\in F}P_ix

The map F\mapsto S_F is a net, and I need to show that it converges to Px. So given \varepsilon>0, I want to find a finite F_0\subset I such that

F\geq F_0\Rightarrow \|S_F-Px\|<\varepsilon

I think I've shown that the norm on the right is equal to

\|P_{M_{I-F}}x\|

where M_{I-F} is the closed subspace spanned by the vectors in the P_i(H) with i\in I-F. But then what? I still don't see how to pick an F that makes the above as small as I want.

 

[micromass]

Allright, I think I finally found it. First, let us fix some notation: let M_i=P_i(H) and M=P(H). Let's focus on this question first:

For \epsilon>0, find a finite set F_0\subseteq I such that

\|Px-\sum_{i\in F_0}{P_ix}\|<\epsilon

We know that Px\in M, and we know that M=\overline{span\{M_i~\vert~i\in F_0\}}. Thus there exist a finite set F₀ and vectors m_i\in M_i for i\in F_0, such that \|Px-\sum_{i\in F_0}{m_i}\|<\epsilon.

Now, do the following:

\|Px-\sum_{i\in F_0}{P_ix}\|\leq \|Px-\sum_{i\in F_0}{m_i}\|+\|\sum_{i\in F_0}{m_i}-\sum_{i\in F_0}{P_ix}\|

I'll let you continue from here... It is of course obvious that you should try to find an upper bound to \|m_i-P_ix\|...

 

[Fredrik]

We know that Px\in M, and we know that M=\overline{span\{M_i~\vert~i\in F_0\}}. Thus there exist a finite set F₀ and vectors m_i\in M_i for i\in F_0, such that \|Px-\sum_{i\in F_0}{m_i}\|<\epsilon.

Aha. (I assume the first \in F_0 should be \in I). Px is in the closure of the set of linear combinations of members of the M_i, and that means that every open ball around Px contains one of those linear combinations. Very clever. (Nothing like this occurred to me when I was thinking about this problem).

\|Px-\sum_{i\in F_0}{P_ix}\|\leq \|Px-\sum_{i\in F_0}{m_i}\|+\|\sum_{i\in F_0}{m_i}-\sum_{i\in F_0}{P_ix}\|

I'll let you continue from here... It is of course obvious that you should try to find an upper bound to \|m_i-P_ix\|...

I actually have to go to bed, but I'll continue tomorrow. Thank you very much.

 

[micromass]

Aha. (I assume the first \in F_0 should be \in I)

Yes, of course. I reread my post 10 times to check for typo's, and they're still one left. I suck at typing correct things...

 

[Fredrik]

I'm so dumb I wasn't able to see how to use the other thing you hinted, but I found a way to solve the problem (using your first hint). Perhaps not the most elegant solution. I decided to express Px as a sum of vectors that belong to the M_i plus a "remainder" that's orthogonal to the other terms.

\varepsilon^2>\Big\|\sum_{i\in F_0}m_i-Px\Big\|^2=\Big\|\sum_{i\in F_0}(m_i-P_i x)+(Px)^\perp_{F_0}\Big\|^2=\sum_{i\in F_0}\|m_i-P_ix\|^2+\|(Px)^\perp_{F_0}\|^2\geq \|(Px)^\perp_{F_0}\|^2

Then I realized that what I called (Px)^\perp_{F_0} here is what I called P_{M_{I-F}}x in my first post. So

\Big\|\sum_{i\in F_0}P_ix-Px\big\|=\|P_{M_{I-F}}x\|=\|(Px)^\perp_{F_0}\|<\varepsilon

And then the rest wasn't too hard. If F\geq F_0, then I-F\subset I-F_0 and

\Big\|\sum_{i\in F}P_ix-Px\big\|=\|P_{M_{I-F}}x\|=\|P_{M_{I-F}}P_{M_{I-F_0}}x\|\leq\|P_{M_{I-F_0}}x\|<\varepsilon

Thanks again for the help. If your solution differs significantly from mine, I would be interested in seeing it.

 

[micromass]

Yes, this was basically what I had in mind. My solution is just a bit longer and less elegant