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Operator Norm and Sequences ... Another 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 some 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





In the above proof by Browder we read the following:


"... ... it follows from Proposition 8.6 that \(\displaystyle S_m \to S\) for some \(\displaystyle S \in \mathscr{L} ( \mathbb{R}^n)\).


In particular, taking \(\displaystyle m = 0\) above, we find


\(\displaystyle \| I - S_p \| \leq \frac{t}{ 1 - t }\)


for every \(\displaystyle p\), and hence \(\displaystyle \| I - S \| \leq t/(1 - t )\) ... ...

... ... ... "



My question is as follows:


Can someone please explain exactly why/how that \(\displaystyle \| I - S_p \| \leq \frac{t}{ 1 - t }\)


for every \(\displaystyle p\) ... implies that \(\displaystyle \| I - S \| \leq t/(1 - t )\) ... ... ?


In other words if some relation is true for every term of a sequence ... why then is it true for the limit of a sequence ... ...




Help will be much appreciated ...

Peter
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,184
Let $\varepsilon>0.$ Since $S_p$ is Cauchy in a complete space, it converges to $S$ in the space. That is, there exists $N>0$ such that for every $n>N,$ we have $\|S_n-S\|<\varepsilon.$ Now then, we have that
\begin{align*}
\|I-S\|&=\|I-S_p+S_p-S\|\\
&\le \|I-S_p\|+\|S_p-S\|\\
&\le \frac{t}{1-t}+\varepsilon.
\end{align*}
Now we are nearly there.

The final step of the argument is something like this. Suppose we have a number $x\ge 0,$ and we let $\delta>0$ be arbitrary. What can we say about $x$ if $x<\delta$ for all $\delta?$ We can actually claim that $x=0.$ For, if $x>0,$ then suppose $\delta=x/2.$ Then $\delta>0,$ but $x=2\delta>\delta,$ contrary to the assumption.

Similarly, for the above argument, because $\varepsilon>0$ is arbitrary, we can conclude that $\|I-S\|\le t/(1-t).$
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Let $\varepsilon>0.$ Since $S_p$ is Cauchy in a complete space, it converges to $S$ in the space. That is, there exists $N>0$ such that for every $n>N,$ we have $\|S_n-S\|<\varepsilon.$ Now then, we have that
\begin{align*}
\|I-S\|&=\|I-S_p+S_p-S\|\\
&\le \|I-S_p\|+\|S_p-S\|\\
&\le \frac{t}{1-t}+\varepsilon.
\end{align*}
Now we are nearly there.

The final step of the argument is something like this. Suppose we have a number $x\ge 0,$ and we let $\delta>0$ be arbitrary. What can we say about $x$ if $x<\delta$ for all $\delta?$ We can actually claim that $x=0.$ For, if $x>0,$ then suppose $\delta=x/2.$ Then $\delta>0,$ but $x=2\delta>\delta,$ contrary to the assumption.

Similarly, for the above argument, because $\varepsilon>0$ is arbitrary, we can conclude that $\|I-S\|\le t/(1-t).$



Thanks Ackbach ...

Your post was most helpful ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,679
This is a particular case of the general rule that weak inequalities are preserved by limits (but strict inequalities may not be). If $(x_n)$ is a sequence with \(\displaystyle \lim_{n\to\infty}x_n = x\), and $x_n\leqslant a$ for all $n$, then $x\leqslant a$. (But if $x_n<a$ for all $n$ then it need not be true that $x<a$. All you can assert is that $x\leqslant a$.)
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
This is a particular case of the general rule that weak inequalities are preserved by limits (but strict inequalities may not be). If $(x_n)$ is a sequence with \(\displaystyle \lim_{n\to\infty}x_n = x\), and $x_n\leqslant a$ for all $n$, then $x\leqslant a$. (But if $x_n<a$ for all $n$ then it need not be true that $x<a$. All you can assert is that $x\leqslant a$.)


Thanks for a most helpful post, Opalg ...


Peter