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

In summary, the conversation discusses the differences between Andrew Browder and Michael Field's definitions of the "operator norm" for linear transformations. Browder defines it as the supremum of the absolute values of the transformation applied to vectors with norm less than or equal to 1, while Field defines it as the supremum over vectors with norm equal to 1. The conversation concludes that these definitions are essentially the same, as the supremum over vectors with norm less than 1 can be achieved by taking the supremum over vectors with norm equal to 1.
  • #1
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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:
View attachment 9367

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:View attachment 9368Thus 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
 

Attachments

  • Browder - Remarks on Norm of an LT ... Section 8.1, Page 179 ... .png
    Browder - Remarks on Norm of an LT ... Section 8.1, Page 179 ... .png
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  • Field - Operator Norm ... Section 9.2.1 ... Page 355 ... .png
    Field - Operator Norm ... Section 9.2.1 ... Page 355 ... .png
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  • #2
Peter said:
" ... ... 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 $|v|=1$.
 
  • #3
Opalg said:
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
 

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

1. What is the operator norm?

The operator norm is a mathematical concept used to measure the size or magnitude of a linear operator on a vector space. It is defined as the maximum value that the operator can take on a unit vector in the vector space.

2. How is the operator norm calculated?

The operator norm is calculated by taking the supremum (or least upper bound) of the operator's norm on all unit vectors in the vector space. In other words, it is the largest value that the operator can take on any vector with a magnitude of 1.

3. What are the differences between Browder and Field operator norms?

The Browder operator norm is a type of operator norm that is used in the study of Banach spaces, while the Field operator norm is used in the study of Hilbert spaces. The main difference between the two is in the definition of the norm, with the Browder norm using the supremum of the operator on all unit vectors and the Field norm using the square root of the supremum of the operator's adjoint on all unit vectors.

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

The operator norm is an important concept in functional analysis and linear algebra. It is used to study the properties of linear operators, such as boundedness and invertibility, and to prove theorems in these areas. It is also used in applications such as numerical analysis and quantum mechanics.

5. How is the operator norm related to the spectral radius?

The spectral radius of a linear operator is defined as the maximum absolute value of its eigenvalues. The operator norm is closely related to the spectral radius, as it is always less than or equal to the spectral radius. In fact, the operator norm is equal to the spectral radius for certain types of operators, such as self-adjoint operators on a Hilbert space.

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