Operator Norm and Cauchy Sequence .... Browder, Proposition 8.7 ....

[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 some help in fully understanding the proof of Proposition 8.7 ...Proposition 8.7 and its proof reads as follows:
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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:
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Hope that helps ...

Peter

 

[Opalg]

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.

 

[soloenergy]

Hello Peter,

I can assist you in understanding the proof of Proposition 8.7. First, let's review the definition of a Cauchy sequence in a metric space. A sequence \{x_n\} in a metric space is called a Cauchy sequence if for every positive real number \epsilon, there exists an index N such that for all n,m \geq N, we have d(x_n, x_m) < \epsilon. In simpler terms, this means that the terms of the sequence get closer and closer together as the sequence progresses.

Now, let's look at the proof of Proposition 8.7. In the proof, Browder is showing that the sequence \{S_m\} is a Cauchy sequence in the space \mathscr{L}(\mathbb{R}^n). This means that for any positive real number \epsilon, there exists an index N such that for all n,m \geq N, we have d(S_n, S_m) < \epsilon.

Let's break down the proof step by step. First, Browder defines the sequence \{S_m\} as a sequence of linear maps from \mathbb{R}^n to \mathbb{R}^n. This means that for each m, S_m is a linear map from \mathbb{R}^n to itself.

Next, Browder shows that \{S_m\} is a Cauchy sequence in \mathscr{L}(\mathbb{R}^n). This is done by showing that for any positive real number \epsilon, there exists an index N such that for all n,m \geq N, we have d(S_n, S_m) < \epsilon. This is done by using the definition of a Cauchy sequence in a metric space. Since \{S_m\} is a sequence of linear maps, we can use the metric d(S_n, S_m) = \sup_{\|x\| = 1} \|S_n(x) - S_m(x)\| to show that the terms of the sequence get closer and closer together as the sequence progresses.

Finally, Browder concludes that \{S_m\} is a Cauchy sequence in \mathscr{L}(\mathbb{R}^n) by showing that the sequence is convergent in \mathscr{L}(\mathbb