extended norm
It is sometimes convenient to allow the norm to take extended real numbers as values. This way, one can accomdate elements of infinite^{} norm in one’s vector space^{}. The formal definition is the same except that one must take care in stating the second condition to avoid the indeterminate form $0\cdot \mathrm{\infty}$.
Definition: Given a real or complex vector space $V$, an extended norm is a map $\parallel \cdot \parallel \to \overline{\mathbb{R}}$ which staisfies the follwing three defining properties:

1.
Positive definiteness: $\parallel v\parallel >0$ unless $v=0$, in which case $\parallel v\parallel =0$.

2.
Homogeneity: $\parallel \lambda v\parallel =\lambda \parallel v\parallel $ for all nonzero scalars $\lambda $ and all vectors $v$.

3.
Triangle inequality^{}: $\parallel u+v\parallel \le \parallel u\parallel +\parallel v\parallel $ for all $u,v\in V$.
Example Let ${C}^{0}$ be the space of continuous functions^{} on the real line. Then the function $\parallel \cdot \parallel $ defined as
$$\parallel f\parallel =\underset{x}{sup}f(x)$$ 
is an extended norm. (We define the supremum^{} of an unbounded set as $\mathrm{\infty}$.) The reason it is not a norm in the strict sense is that there exist continuous functions which are unbounded. Let us check that it satisfies the defining properties:

1.
Because of the absolute value^{} in the definition, it is obvious that $\parallel f\parallel \ge 0$ for all $f$. Furthermore, if $\parallel f\parallel =0,$ then ${sup}_{x}f(x)=0$, so $f(x)\le 0$ for all $x$, which implies that $f=0$.

2.
If $f$ is bounded^{}, then
$$\parallel f\parallel =\underset{x}{sup}\lambda f(x)=\underset{x}{sup}\lambda f(x)=\lambda \underset{x}{sup}f(x)=\lambda \parallel \parallel f\parallel $$ If $f$ is unbounded and $\lambda \ne 0$, then $\lambda f$ is unbounded as well. By the convention that infinity^{} times a finite number is infinity, property (2) holds in this case as well.

3.
If both $f$ and $g$ are bounded, then
$$\parallel f+g\parallel =\underset{x}{sup}f(x)+g(x)\le \underset{x}{sup}(f(x)+g(x))\le \underset{x}{sup}f(x)+\underset{x}{sup}g(x)=\parallel f\parallel +\parallel g\parallel $$ On the other hand, if either $f$ or $g$ is unbounded, then the right hand side of (3) is infinity by the convention that anything other minus infinity added to plus infinity is still infinity and infinity is bigger than anything which could appear on the left.
An extended norm partitions^{} a vector space into equivalence classes^{} modulo the relation^{} $$. The equivalence class of zero is a vector space consisting of all elemets of finite norm. Restricted to this equivalence class the extended norm reduces to an ordinary norm. The other equivalence classes are translates^{} of the equivalence class of zero. Furthermore, when we use the distance function of the norm to define a topology^{} on our vector space, these equivalence classes are exactly the connected components^{} of the topology.
Title  extended norm 

Canonical name  ExtendedNorm 
Date of creation  20130322 14:59:49 
Last modified on  20130322 14:59:49 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  6 
Author  rspuzio (6075) 
Entry type  Definition 
Classification  msc 46B99 