operator norm
Definition
Let $A:\U0001d5b5\to \U0001d5b6$ be a linear map between normed vector spaces^{} $\U0001d5b5$ and $\U0001d5b6$. To each such map (operator) $A$ we can assign a nonnegative number ${\parallel A\parallel}_{\mathrm{op}}$ defined by
$${\parallel A\parallel}_{\mathrm{op}}:=\underset{\mathbf{v}\in \U0001d5b5}{sup}\frac{\parallel A\mathbf{v}\parallel}{\parallel \mathbf{v}\parallel},$$ 
where the supremum ${\parallel A\parallel}_{\mathrm{op}}$ could be finite or infinite. Equivalently, the above definition can be written as
$$ 
By convention, if $\U0001d5b5$ is the zero vector space, any operator from $\U0001d5b5$ to $\U0001d5b6$ must be the zero operator and is assigned zero norm.
${\parallel A\parallel}_{\mathrm{op}}$ is called the the operator norm^{} (or the induced norm) of $A$, for reasons that will be clear in the next .
Operator norm is in fact a norm
Definition  If ${\parallel A\parallel}_{\mathrm{op}}$ is finite, we say that $A$ is a . Otherwise, we say that $A$ is .
It turns out that, for bounded operators^{}, $\parallel \cdot {\parallel}_{\mathrm{op}}$ satisfies all the properties of a norm (hence the name operator norm). The proof follows immediately from the definition:
 Positivity:

Since $\parallel A\mathbf{v}\parallel \ge 0$, by definition ${\parallel A\parallel}_{\mathrm{op}}\ge 0$. Also, $\parallel A\mathbf{v}\parallel =0$ identically only if $A=0$. Hence ${\parallel A\parallel}_{\mathrm{op}}=0$ only if $A=0$.
 Absolute homogeneity:

Since $\parallel \lambda A\mathbf{v}\parallel =\lambda \parallel A\mathbf{v}\parallel $, by definition ${\parallel \lambda A\parallel}_{\mathrm{op}}=\lambda {\parallel A\parallel}_{\mathrm{op}}$.
 Triangle inequality:

Since $\parallel (A+B)\mathbf{v}\parallel =\parallel A\mathbf{v}+B\mathbf{v}\parallel \le \parallel A\mathbf{v}\parallel +\parallel B\mathbf{v}\parallel $, by definition ${\parallel A+B\parallel}_{\mathrm{op}}\le {\parallel A\parallel}_{\mathrm{op}}+{\parallel B\parallel}_{\mathrm{op}}$.
The set $L(\U0001d5b5,\U0001d5b6)$ of bounded linear maps from $\U0001d5b5$ to $\U0001d5b6$ forms a vector space^{} and $\parallel \cdot {\parallel}_{\mathrm{op}}$ defines a norm in it.
Example
Suppose that $\U0001d5b5=({\mathbb{R}}^{n},\parallel \cdot {\parallel}_{p})$ and $\U0001d5b6=({\mathbb{R}}^{n},\parallel \cdot {\parallel}_{p})$, where $\parallel \cdot {\parallel}_{p}$ is the vector pnorm. Then the operator norm $\parallel \cdot {\parallel}_{\mathrm{op}}=\parallel \cdot {\parallel}_{p}$ is the matrix pnorm^{}.
Title  operator norm 
Canonical name  OperatorNorm 
Date of creation  20130322 12:43:20 
Last modified on  20130322 12:43:20 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  15 
Author  asteroid (17536) 
Entry type  Definition 
Classification  msc 47L25 
Classification  msc 46A32 
Classification  msc 47A30 
Synonym  induced norm 
Related topic  VectorNorm 
Related topic  OperatorTopologies 
Related topic  HomomorphismsOfCAlgebrasAreContinuous 
Related topic  CAlgebra 
Defines  bounded linear map 
Defines  unbounded linear map 
Defines  bounded operator 
Defines  unbounded operator 