Editing Operator norm (section)

== Properties ==

The operator norm is indeed a norm on the space of all [[bounded operator]]s between <math>V</math> and <math>W</math>. This means
<math display="block">\|A\|_\text{op} \geq 0 \mbox{ and } \|A\|_\text{op} = 0 \mbox{ if and only if } A = 0,</math>
<math display="block">\|aA\|_\text{op} = |a| \|A\|_\text{op} \mbox{ for every scalar } a ,</math>
<math display="block">\|A + B\|_\text{op} \leq \|A\|_\text{op} + \|B\|_\text{op}.</math>

The following inequality is an immediate consequence of the definition:
<math display="block">\|Av\| \leq \|A\|_\text{op} \|v\| \ \mbox{ for every }\ v \in V.</math>

The operator norm is also compatible with the composition, or multiplication, of operators: if <math>V</math>, <math>W</math> and <math>X</math> are three normed spaces over the same base field, and <math>A : V \to W</math> and <math>B : W \to X</math> are two bounded operators, then it is a [[sub-multiplicative norm]], that is: 
<math display="block">\|BA\|_\text{op} \leq \|B\|_\text{op} \|A\|_\text{op}.</math>

For bounded operators on <math>V</math>, this implies that operator multiplication is jointly continuous.

It follows from the definition that if a sequence of operators converges in operator norm, it [[converges uniformly]] on bounded sets.