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Operator Norm and Sequences ... Yet a Further Question ... Browder, Proposition 8.7 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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:



Browder - 1 - Proposition 8.7 ... PART 1 ....png
Browder - 2 - Proposition 8.7 ... PART 2 ... ....png





My 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 ... ...


\(\displaystyle \| I - L^{ -1 } \| \leq \| I - L \| ( 1 - \| I - L \|)^{-1} \)



Help will be much appreciated ... ...

Peter
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,184
I'm a little confused. Isn't that what the proof in the book is supposed to do?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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 ...

Peter





EDIT: I note in passing that in relation to the assertion


\(\displaystyle \| I - L^{ -1 } \| \leq \| I - L \| ( 1 - \| I - L \|)^{-1} \)


we have that


\(\displaystyle \| I - L \| ( 1 - \| I - L \|)^{-1} = \frac{ \| I - L \| }{ ( 1 - \| I - L \|) }\)


\(\displaystyle = \| I - L \| + \| I - L \|^2 + \| I - L \|^3 + \ ... \ ... \ ... \)



But I cannot see how to use this in the proof ... but it might be helpful :) ...


Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
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 $\|I-L^{-1}\| \leqslant \dfrac {\|I-L\|}{1- \|I-L\|}$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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