# Understanding Proposition 8.7: Operator Norm and Sequences

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In summary, Peter found that 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.
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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:
View attachment 9397
View attachment 9398In 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

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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 $\|I-S\|\le t/(1-t).$

Ackbach said:
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

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 $x\leqslant a$.)

Opalg said:
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

## 1. What is Proposition 8.7 in mathematics?

Proposition 8.7 is a theorem in functional analysis that relates to the operator norm and sequences in a normed vector space. It states that if a sequence of bounded linear operators converges to a bounded linear operator in the operator norm, then the limit operator is also bounded and its norm is less than or equal to the limit of the norms of the sequence.

## 2. What is the operator norm?

The operator norm is a way to measure the size or magnitude of a bounded linear operator in a normed vector space. It is defined as the supremum of the norms of the operator applied to all unit vectors in the space. In other words, it represents the maximum amount by which the operator can scale a vector without increasing its norm.

## 3. How is Proposition 8.7 useful in mathematics?

Proposition 8.7 is useful in mathematics because it provides a way to prove the continuity of certain operations on bounded linear operators, such as addition and multiplication. It also helps in understanding the behavior of sequences of operators and their limits in a normed vector space.

## 4. What is the significance of Proposition 8.7 in functional analysis?

In functional analysis, Proposition 8.7 is significant because it connects the concept of operator norm to the convergence of sequences of operators. This allows for a more thorough understanding of the behavior of operators and their limits, which is important in many areas of mathematics and physics.

## 5. Are there any real-world applications of Proposition 8.7?

Yes, there are many real-world applications of Proposition 8.7. For example, it is used in the study of differential equations, where operators are often used to represent differential operators. It is also important in the analysis of quantum mechanics, where operators are used to represent physical observables. Additionally, Proposition 8.7 has applications in computer science, particularly in the field of signal processing and image recognition.

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