Functional analysis, projection operators

[Fredrik]

Homework Statement

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

[tex]\sum_{i\in I}P_ix=Px[/tex]

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 [itex]P_M[/itex] and [itex]P_N[/itex] are projection operators for closed subspaces M and N respectively. Then

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

The Attempt at a Solution

For each finite [itex]F\subset I[/itex], define

[tex]S_F=\sum_{i\in F}P_ix[/tex]

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

[tex]F\geq F_0\Rightarrow \|S_F-Px\|<\varepsilon[/tex]

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

[tex]\|P_{M_{I-F}}x\|[/tex]

where [itex]M_{I-F}[/itex] is the closed subspace spanned by the vectors in the [itex]P_i(H)[/itex] with [itex]i\in I-F[/itex]. 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 [tex]M_i=P_i(H)[/tex] and [tex]M=P(H)[/tex]. Let's focus on this question first:

For [tex]\epsilon>0[/tex], find a finite set [tex]F_0\subseteq I[/tex] such that

[tex]\|Px-\sum_{i\in F_0}{P_ix}\|<\epsilon[/tex]

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

Now, do the following:

[tex]\|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}\|[/tex]

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

 

[Fredrik]

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

Aha. (I assume the first [itex]\in F_0[/itex] should be [itex]\in I[/itex]). 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).

[tex]\|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}\|[/tex]

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

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

 

[micromass]

Aha. (I assume the first [itex]\in F_0[/itex] should be [itex]\in I[/itex])

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.

[tex]\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[/tex]

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

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

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

[tex]\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[/tex]

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