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Operator Norm ... differences between Browder and Field ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
I am reader 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 differences between Andrew Browder and Michael Field (Essential Real Analysis) concerning the "operator norm" for linear transformations ...

The relevant notes form Browder read as follows:



Browder - Remarks on Norm of an LT ... Section 8.1, Page 179 ... .png




In the above text from Browder we read the following:


" ... ... A perhaps more natural way to define the distance between linear transformations is by using the so-called "operator norm" defined by

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert \le 1 \}
\)

... ... ... ... ... "



Now the above definition, differs (apparently anyway) from the definition of the operator norm by Michael Field in his book: "Essential Real Analysis" ... Field writes the following:


Field - Operator Norm ... Section 9.2.1 ... Page 355 ... .png


Thus Field's version of the operator norm (if we write it in Browder's notation is as follows:

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert = 1 \}
\)



My question is as follows:

Are Browder's and Field's definition essentially the same ... if so how are they equivalent ... ... ?

Maybe in Browder's definition the supremum is actually reached when \(\displaystyle \lvert v \rvert = 1\) ... ...



Help will be appreciated ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,696
" ... ... A perhaps more natural way to define the distance between linear transformations is by using the so-called "operator norm" defined by

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert \le 1 \}
\)

... ... ... ... ... "


Thus Field's version of the operator norm (if we write it in Browder's notation is as follows:

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert = 1 \}
\)

My question is as follows:

Are Browder's and Field's definition essentially the same ... if so how are they equivalent ... ... ?

Maybe in Browder's definition the supremum is actually reached when \(\displaystyle \lvert v \rvert = 1\) ... ...
The two definitions are equivalent. If $v \in \mathbb{R}^n$ is nonzero and $|v|<1$, let $w = \frac v{|v|}$. Then $|w|=1$, and $$|Tv| = |T(|v|w)| = |\,|v|Tw| = |v|\,|Tw| < |Tw|.$$ So $|Tv|$ for any $v$ "inside" the unit ball is less than $|Tw|$ for the corresponding vector on the "surface" of the unit ball. It follows that to find $\sup\{|Tv|:|v|\leqslant1\}$ it is sufficient to take the supremum over $v$ with $|v|=1$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
The two definitions are equivalent. If $v \in \mathbb{R}^n$ is nonzero and $|v|<1$, let $w = \frac v{|v|}$. Then $|w|=1$, and $$|Tv| = |T(|v|w)| = |\,|v|Tw| = |v|\,|Tw| < |Tw|.$$ So $|Tv|$ for any $v$ "inside" the unit ball is less than $|Tw|$ for the corresponding vector on the "surface" of the unit ball. It follows that to find $\sup\{|Tv|:|v|\leqslant1\}$ it is sufficient to take the supremum over $v$ with $|v|=1$.



Thanks Opalg ..

I appreciate your help ...

Peter