Operator norm .... Field, Theorem 9.2.9 ....

In summary: Hi Peter,Yes, your reasoning is correct. The inequality holds because the supremum of a sum is always less than or equal to the sum of the suprema. This is because the supremum is the smallest upper bound, so adding two terms can only increase the value. Therefore, the inequality in Theorem 9.2.9 (3) is a strict one. I hope this helps clarify things for you. Happy studying!In summary, Theorem 9.2.9 (3) states that for a normed vector space of linear maps, the supremum of the sum of the norms of two linear maps is less than or equal to the sum of the suprema of the individual norms. This is a strict
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I am reading Michael Field's book: "Essential Real Analysis" ... ...

I am currently reading Chapter 9: Differential Calculus in \(\displaystyle \mathbb{R}^m\) and am specifically focused on Section 9.2.1 Normed Vector Spaces of Linear Maps ...

I need some help in fully understanding Theorem 9.2.9 (3) ... Theorem 9.2.9 (3) reads as follows:
View attachment 9375
View attachment 9376

In the proof of Theorem 9.2.9 (3) Field asserts the following:

\(\displaystyle \text{sup}_{ \| u \| = 1} ( \| Au \| + \| Bu \| ) \le \text{sup}_{ \| u \| = 1} ( \| Au \| ) +\text{sup}_{ \| u \| = 1} ( \| Bu \| ) \)
Can someone please explain why this is true ... surely the above is a strict equality ...
Help will be much appreciated ...

Peter
 

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Hi Peter,

Peter said:
In the proof of Theorem 9.2.9 (3) Field asserts the following:

\(\displaystyle \text{sup}_{ \| u \| = 1} ( \| Au \| + \| Bu \| ) \le \text{sup}_{ \| u \| = 1} ( \| Au \| ) +\text{sup}_{ \| u \| = 1} ( \| Bu \| ) \)

Can someone please explain why this is true ... surely the above is a strict equality ...

Here is a hint: Use the fact that $\|Au\|\leq \sup_{\|u\|=1}\|Au\|$ and $\|Bu\|\leq \sup_{\|u\|=1}\|Bu\|$ for all $\|u\|=1.$ Can you proceed from here?
 
  • #3
GJA said:
Hi Peter,
Here is a hint: Use the fact that $\|Au\|\leq \sup_{\|u\|=1}\|Au\|$ and $\|Bu\|\leq \sup_{\|u\|=1}\|Bu\|$ for all $\|u\|=1.$ Can you proceed from here?
Hi GJA ...

I think that a way to proceed is as follows:

Since $\|Au\|\leq \sup_{\|u\|=1}\|Au\|$ and $\|Bu\|\leq \sup_{\|u\|=1}\|Bu\|$ for all $\|u\|=1.$

we have ... ...

\(\displaystyle \|Au\| + \|Bu\| \leq \sup_{\|u\|=1}\|Au\| + \sup_{\|u\|=1} \|Bu\| \) for all $\|u\|=1.$ Therefore ...\(\displaystyle \sup_{\|u\|=1} ( \|Au\| + \|Bu\| ) \leq \sup_{\|u\|=1} \|Au\| + \sup_{\|u\|=1}\|Bu\|\)... since \(\displaystyle \sup_{\|u\|=1} ( \|Au\| + \|Bu\| ) = \text{ max }_{ \| u \| = 1 } ( \|Au\| + \|Bu\| )\)
Is that correct?Peter
 
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FAQ: Operator norm .... Field, Theorem 9.2.9 ....

1. What is the operator norm?

The operator norm, also known as the induced norm, is a way of measuring the size or magnitude of a linear operator on a vector space. It is the maximum amplification of any input vector under the operator.

2. How is the operator norm calculated?

The operator norm is calculated by taking the supremum (or maximum) of the ratio of the output norm to the input norm over all non-zero input vectors. In other words, it is the largest possible value that the operator can multiply a vector by without increasing its norm.

3. What is the significance of the operator norm in mathematics?

The operator norm is important in many areas of mathematics, including functional analysis, linear algebra, and differential equations. It allows us to measure the "size" of a linear operator and determine if it is bounded or unbounded. It also plays a crucial role in the study of convergence of numerical methods for solving differential equations.

4. Can the operator norm be negative?

No, the operator norm is always a non-negative value. This is because it is defined as the maximum amplification of a vector, which cannot be negative. However, it is possible for the operator norm to be zero, which would indicate that the operator is the zero operator.

5. How is the operator norm related to Theorem 9.2.9?

Theorem 9.2.9, also known as the Banach-Steinhaus theorem, states that if a sequence of linear operators converges pointwise to a bounded linear operator, then the sequence is uniformly bounded. This theorem is often used to prove the existence of the operator norm, as it guarantees that the supremum in the definition of the operator norm is finite.

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