# Operator Norm and Sequences ... Another Question ... Browder, Proposition 8.7 ...

#### Peter

##### Well-known member
MHB Site Helper
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:

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
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
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 ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
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
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