Understanding Andrew Browder's Prop 8.7: Operator Norm and Sequences

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Discussion Overview

The discussion revolves around understanding Proposition 8.7 from Andrew Browder's "Mathematical Analysis: An Introduction," specifically focusing on the proof related to operator norms and sequences. Participants are seeking clarification on the formal proof of a particular assertion within the proposition.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Peter expresses confusion regarding the proof of the assertion $$ \| I - L^{ -1 } \| \leq \| I - L \| ( 1 - \| I - L \|)^{-1} $$ and requests a formal demonstration.
  • Ackbach questions whether the proof in the book is intended to address this assertion, suggesting it may be considered obvious by Browder.
  • Peter notes that he finds the assertion unproven in the text and struggles to formulate a proof, despite recognizing a potential series expansion related to the inequality.
  • Another participant points out that Browder has established an inequality $$ \|I-S\| \leqslant \dfrac t{1-t}$$ during the proof, where $S=L^{-1}$ and $t = \|I-L\|$, leading to the conclusion that $$ \|I-L^{-1}\| \leqslant \dfrac {\|I-L\|}{1- \|I-L\|}$$.
  • Peter acknowledges this clarification and expresses appreciation for the assistance provided by Opalg.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the clarity of the proof or the completeness of Browder's presentation. There is a shared sense of confusion regarding the proof's rigor and the assertion's demonstration.

Contextual Notes

Participants highlight the potential assumption that the proof is trivial, which may not hold for all readers. There is also mention of a series expansion that could relate to the proof, but its application remains unclear.

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I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.1 Linear Algebra ...

I need yet further help in fully understanding the proof of Proposition 8.7 ...Proposition 8.7 and its proof reads as follows:
View attachment 9399
View attachment 9400My question is as follows:Can someone please demonstrate, formally and rigorously, the last assertion of the above proposition ... ... That is, can someone please demonstrate, formally and rigorously, that ... ... $$ \| I - L^{ -1 } \| \leq \| I - L \| ( 1 - \| I - L \|)^{-1} $$
Help will be much appreciated ... ...

Peter
 

Attachments

  • Browder - 1 - Proposition 8.7 ... PART 1 ....png
    Browder - 1 - Proposition 8.7 ... PART 1 ....png
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  • Browder - 2 - Proposition 8.7 ... PART 2 ... ....png
    Browder - 2 - Proposition 8.7 ... PART 2 ... ....png
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Last edited:
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I'm a little confused. Isn't that what the proof in the book is supposed to do?
 
Ackbach said:
I'm a little confused. Isn't that what the proof in the book is supposed to do?
Hi Ackbach ... ...

Hmmm ... can only say I agree with you ...

But Browder leaves the assertion unproven ...

I can only assume that Browder thinks the proof is obvious and trivial ... but I am having problems formulating a proof .. so I hope that someone can help ...

PeterEDIT: I note in passing that in relation to the assertion $$ \| I - L^{ -1 } \| \leq \| I - L \| ( 1 - \| I - L \|)^{-1} $$we have that $$ \| I - L \| ( 1 - \| I - L \|)^{-1} = \frac{ \| I - L \| }{ ( 1 - \| I - L \|) }$$$$= \| I - L \| + \| I - L \|^2 + \| I - L \|^3 + \ ... \ ... \ ... $$
But I cannot see how to use this in the proof ... but it might be helpful :) ... Peter
 
Last edited:
Browder has shown during the proof of Proposition 8.7 that $\|I-S\| \leqslant \dfrac t{1-t}$. But $S=L^{-1}$ and $t = \|T\| = \|I-L\|$. So that inequality becomes $\|I-L^{-1}\| \leqslant \dfrac {\|I-L\|}{1- \|I-L\|}$.
 
Opalg said:
Browder has shown during the proof of Proposition 8.7 that $\|I-S\| \leqslant \dfrac t{1-t}$. But $S=L^{-1}$ and $t = \|T\| = \|I-L\|$. So that inequality becomes $\|I-L^{-1}\| \leqslant \dfrac {\|I-L\|}{1- \|I-L\|}$.
... Hmmm ... I should have seen that .

Appreciate the help, Opalg ...

Peter
 

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