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

[Math Amateur]

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

 

[Ackbach]

I'm a little confused. Isn't that what the proof in the book is supposed to do?

 

[Math Amateur]

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

 

[Opalg]

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 $\|I-L^{-1}\| \leqslant \dfrac {\|I-L\|}{1- \|I-L\|}$.

 

[Math Amateur]

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