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operator norm (Definition)

Definition

Let $ A\colon {\mathsf V}\to{\mathsf W}$ be a linear map between normed vector spaces $ {\mathsf V}$ and $ {\mathsf W}$ . To each such map (operator) $A$ we can assign a non-negative number $\|A\|_{\op}$ defined by

$\displaystyle \Vert A\Vert _{{\rm op}} := \mathop{\sup_{{\mathbf v}\in{\mathsf ... ..._{{\mathbf v}\ne{\bf0}} \frac{\Vert A{\mathbf v}\Vert}{\Vert{\mathbf v}\Vert}, $
where the supremum $\|A\|_{\op}$ could be finite or infinite. Equivalently, the above definition can be written as

$\displaystyle \Vert A\Vert _{\rm op}:= \mathop{\sup_{{\mathbf v}\in{\mathsf V}}... ...thbf v}\in{\mathsf V}}}_{0<\Vert{\mathbf v}\Vert\le1} \Vert A{\mathbf v}\Vert. $
By convention, if $ {\mathsf V}$ is the zero vector space, any operator from $ {\mathsf V}$ to $ {\mathsf W}$ must be the zero operator and is assigned zero norm.

$\|A\|_{\op}$ is called the the operator norm (or the induced norm) of $A$ , for reasons that will be clear in the next section.

Operator norm is in fact a norm

Definition - If $\|A\|_{\op}$ is finite, we say that $A$ is a bounded. Otherwise, we say that $A$ is unbounded.

It turns out that, for bounded operators, $\|\cdot\|_{\op}$ satisfies all the properties of a norm (hence the name operator norm). The proof follows immediately from the definition:

Positivity:
Since $\|A\v\|\ge 0$ , by definition $\|A\|_\op \ge 0$ . Also, $\|A\v\| = 0$ identically only if $A=0$ . Hence $\|A\|_\op = 0$ only if $A = 0$ .
Absolute homogeneity:
Since $\|\lambda A\v\|=|\lambda| \|A\v\|$ , by definition $\|\lambda A\|_\op = |\lambda| \|A\|_\op$ .
Triangle inequality:
Since $\|(A+B)\v\|=\|A\v+B\v\|\le \|A\v\| + \|B\v\|$ , by definition $\|A+B\|_\op \le \|A\|_\op + \|B\|_\op$ .

The set $ L({\mathsf V},{\mathsf W})$ of bounded linear maps from $ {\mathsf V}$ to $ {\mathsf W}$ forms a vector space and $\|\cdot\|_{\op}$ defines a norm in it.

Example

Suppose that $ {\mathsf V}=({\mathbb{R}}^n,\Vert\cdot\Vert _p)$ and $ {\mathsf W}=({\mathbb{R}}^n,\Vert\cdot\Vert _p)$ , where $\|\cdot\|_p$ is the vector p-norm. Then the operator norm $\|\cdot\|_\op = \|\cdot\|_p$ is the matrix p-norm.




"operator norm" is owned by asteroid. [ full author list (3) | owner history (2) ]
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See Also: vector norm, operator topologies, $C^*$-algebra homomorphisms are continuous, $C^*$-algebra

Other names:  induced norm
Also defines:  bounded linear map, unbounded linear map, bounded operator, unbounded operator

Attachments:
examples of bounded and unbounded operators (Example) by matte
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Cross-references: matrix p-norm, vector p-norm, vector space, proof, properties, clear, norm, zero operator, zero vector space, infinite, finite, supremum, number, operator, map, normed vector spaces, linear map
There are 53 references to this entry.

This is version 12 of operator norm, born on 2002-06-03, modified 2008-09-17.
Object id is 3018, canonical name is OperatorNorm.
Accessed 22270 times total.

Classification:
AMS MSC46A32 (Functional analysis :: Topological linear spaces and related structures :: Spaces of linear operators; topological tensor products; approximation properties)
 47L25 (Operator theory :: Linear spaces and algebras of operators :: Operator spaces )
 47A30 (Operator theory :: General theory of linear operators :: Norms )

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help: a proof about uniformly partiallly differentiaable by nusjinxin on 2008-09-04 14:50:35
I faced a problem:
f:U-->R is uniformly partiallly differentiaable at a , i.e.,
for any epsilon, there exists a delta so that:
|[f(a+tu)-f(a)]/t-Dfu(a)|<=epsilon
for all u belongs to Rn and ||u||=1 and all t with 0<|t|<delta.

show that f is diffentiable at a

*Dfu(a) is the patial dirivative of f in the direction of u.

I got a proof but I am not confident in it because this kind of basic concept problem need to be very accurate..but I am quite confused about some symbol and concept .
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