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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
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:



Field - 1 - Theorem 9.2.9 ... ... PART 1 ... .png
Field - 2 - Theorem 9.2.9 ... ... PART 2 ... .png




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
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
271
Hi Peter ,

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?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
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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