Can a norm take infinite value? For example, $\|\cdot \|_1$?

A definition for norm from Wikipedia says

Given a vector space $V$ over a subfield $F$ of the complex numbers, a norm on $V$ is a function $p: V → \mathbb{R}$ with the following properties:

For all $a ∈ F$ and all $u, v ∈ V$,

• $p(av) = |a| p(v)$, (positive homogeneity or positive scalability).
• $p(u + v) ≤ p(u) + p(v)$ (triangle inequality or subadditivity).
• If $p(v) = 0$ then $v$ is the zero vector (separates points).

Can a norm take value $+\infty$? I think topology and convergence are what I had in mind. If we modify the definition of a norm to allow it take $\infty$, in such a generalized norm space, does it induce a topology, so that we can talk about convergence relative to the generalized norm being equivalent to convergence relative the induced topology?

My question comes from an example: can $\|\cdot \|_1$ be defined on all measurable functions which are allowed to have infinite integrals not just finite integrals?

Thanks!

Solution 1:

The definition of the norm over some linear space $V$ explicitly says that $\|\cdot\|:V\to\Bbb R_+$ which means that there does not exist any $x\in V$ such that $\|x\| = \infty$. This makes $V$ to be a normed space, and the fact that $\|\cdot\|$ has a finite range is important each time you deal with a norm.

However, when we are already working with some linear space, say $V = \mathfrak B([0,1])$ being the space of all Borel-measurable functions with a domain $[0,1]$, we cannot always make exactly this space to be a normed space. What we can do instead is to introduce a norm-like function $\|\cdot\|'$ on $V$ with a range $[0,\infty]$ and define $$ V':=\{x\in B:\|x\|'<\infty\}\tag{1} $$ to be the normed space, which is a linear subspace of $V$. For example, we can say that $$ \|x\|':=\sup_{t\in [0,1]}|x(t)| $$ which is not a norm on $V = \mathfrak B(\Bbb R)$ since for $x(t) = 1_{t>0}\cdot\frac1t$ we have $\|x\|' = \infty$. However, when restricted to $V'$ - the space of all measurable functions whose $\|\cdot\|'$ is bounded, it is a norm.

The very same argument applies to the norm $\|\cdot\|_1$:

1. You pick up a candidate linear space $V$ to introduce a norm over, e.g. a space of measurable functions.

2. You introduce a candidate $\|\cdot\|'$ for a norm, which can take infinite values. Note that in such case $\|\cdot\|'$ is not a norm, and thus $(V,\|\cdot\|')$ is not a normed space.

3. You define $V'\subseteq V$ according to $(1)$ and then show that $(V',\|\cdot\|')$ is a normed space.

Solution 2:

Norms with infinite values are discussed here, where they are called extended norms.

As pointed out in Anguepa's answer, extended norms $||\cdot||$ define extended metrics $d(x,y)=||x-y||$ which define topologies generated by balls $B_x^r=\{y\in V:d(x,y)<r\}$ in the usual way. However, this topology will be disconnected if $||x||=\infty$ for any $x\in V$. In fact, if the scalar field $F$ is $\mathbb{R}$ or $\mathbb{C}$ then the connected components will be precisely the equivalence classes defined by the relation $||x-y||<\infty$, as mentioned in the link above. In particular, any $x\in V$ with $||x||=\infty$ will span a discrete one dimesional subspace, something which never happens in a (finitely) normed real or complex vector space.

Alternatively, one might consider the topology generated by holes $H_x^r=\{y\in V:d(x,y)>r\}$, but this has its own weird properties. For example, the hole topology on $\mathbb{R}$ is $T_1$ but not Hausdorff and even hyperconnected in that there are no disjoint non-empty open subsets.

Generally, when dealing with a vector space without a canonical (finite) norm, one tends to find other ways to define topologies with natural convergence properties. For example, with measurable functions you might consider convergence in measure which, while not defined by a norm, is at least defined by a metric.