Sequence of projection is Cauchy

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Homework Help Overview

The discussion revolves around proving that the sequence of orthogonal projections {P_i(e)} in a pre-Hilbert space V is Cauchy for any vector e in V. Participants are exploring the properties of orthogonal sequences of complete subspaces and their implications on convergence.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to show that the norm of the difference between projections approaches zero as indices go to infinity. They are questioning the validity of their expressions and exploring the convergence of the series involving projections.

Discussion Status

Some participants have offered suggestions to consider the completion of the space and the convergence of the series of projections. There is an ongoing examination of the relationships between the projections and the original vector, with no clear consensus yet on the correct approach or resolution.

Contextual Notes

There is a mention of the need to treat V as a Hilbert space and the implications of completeness in the context of the problem. Participants are also discussing the validity of certain formulas related to the projections.

yifli
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Homework Statement


Let {M_i} be an orthogonal sequence of complete subspaces of a pre-Hilbert space V, and let P_i be the orthogonal projection on M_i. Prove that {P_i(e)} is Cauchy for any e in V


2. The attempt at a solution
I'm trying to prove as n and m goes infinity, [tex]\left\|P_n(e)-P_m(e)\right\|^2\rightarrow 0[/tex]

Here is what I've got so far:
[tex]\left\|P_n(e)-P_m(e)\right\|^2=\left\|P_n(e)\right\|^2+\left\|P_m(e)\right\|^2[/tex] because P_n(e) is orthogonal to P_m(e), how to proceed?
 
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Hi yifli! :smile:

First, we can take V to be a Hilbert space. Indeed, try to prove this by taking the completion of V.

Second, try to do something with [itex]\sum_{i=1}^{+\infty}{P_i(e)}[/itex]?
 
micromass said:
Second, try to do something with [itex]\sum_{i=1}^{+\infty}{P_i(e)}[/itex]?

I did try that:
[tex]P_n(e)=e-\sum_{i=1,i \neq n}^{+\infty}{P_i(e)}[/tex].

substituting the above formula into [tex]\left\|P_n(e)-P_m(e)\right\|[/tex] gives me the same thing

What did I miss?
 
yifli said:
I did try that:
[tex]P_n(e)=e-\sum_{i=1,i \neq n}^{+\infty}{P_i(e)}[/tex].

Firstly, This formula doesn't hold.
Secondly, my point was that [itex]\sum_{i=1}^{+\infty}{P_i(e)}[/itex] converges. Thus...
 

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