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