# Operator Norm and Cauchy Sequence ... 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 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:

"... ... Thus, $$\displaystyle \{ S_m \}$$ is a Cauchy sequence in $$\displaystyle \mathscr{L} ( \mathbb{R}^n )$$... ...

My question is as follows:

Can someone please demonstrate formally and rigorously that $$\displaystyle \{ S_m \}$$ is a Cauchy sequence in $$\displaystyle \mathscr{L} ( \mathbb{R}^n )$$... ...

Help will be much appreciated ...

Peter

===============================================================================

Note: Browder defines a Cauchy Sequence in a metric space as follows:

Hope that helps ...

Peter

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
Can someone please demonstrate formally and rigorously that $$\displaystyle \{ S_m \}$$ is a Cauchy sequence in $$\displaystyle \mathscr{L} ( \mathbb{R}^n )$$... ...
From Browder's definition, to prove that a sequence $(x_n)$ is Cauchy you have to show that given $\varepsilon>0$ there exists $n_0$ such that $\rho(x_m,x_n)<\varepsilon$ whenever $m$ and $n$ are greater than $n_0$. By taking $m$ to be the smaller of those two numbers, you can write $n=m+p$, where $p>0$. So you need to find $n_0$ such that $\rho(x_m,x_{m+p})<\varepsilon$ whenever $m\geqslant n_0$ and $p\geqslant1$.

In this example, the sequence $(x_n)$ becomes $(S_m)$ and the metric is given by $\rho(S_m,S_n) = \|S_m-s_n\|$. So we want to show that given $\varepsilon>0$ there exists $n_0$ such that $\|S_m - S_{m+p}\| < \varepsilon$ whenever $m\geqslant n_0$ and $p\geqslant1$. But Browder shows that $\|S_m - S_{m+p}\| < \frac{t^m}{1-t}$. Since $t<1$, the sequence $\left(\frac{t^m}{1-t}\right)$ converges to $0$. Therefore, given $\varepsilon>0$ there exists $n_0$ such that $\frac{t^m}{1-t} < \varepsilon$ whenever $m\geqslant n_0$, from which the rquired result immediately follows.