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!
 A: 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.
A: While it's not a norm per se, one can consider valuations into very large ordered fields which extend the real numbers.
For example one can consider a "hyperreal-norm" into a hyperreal field. There are other difficulties which may arise from this (for example the completeness of the underlying field is no longer usable). However it is possible to define something like that, in which case a vector whose norm is a hyperreal which is larger than all standard real numbers can be considered as having an infinite norm.
Do note that this "norm" need not be compatible with the vector space structure, and a lot of modifications may be required. If you wish to seriously consider defining such structure it might be very wise to learn first about valuations, possibly about uniform topology, and try to define this step by step.
Remember that sometimes things are not well-known because they have no actual use for most people, but it can still be a very interesting exercise to sit and come up with a whole new set of definitions (even if you later find out that this set of definition is known).
A: 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 $\|\cdot\|_1$:


*

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

*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.

*You define $V'\subseteq V$ according to $(1)$ and then show that $(V',\|\cdot\|')$ is a normed space.
A: It is not entirely clear what you are asking.
Let $C((0,1))$ be the space of continuous functions on $(0,1)$ with the norm $\|x\|_\infty = \sup_{t \in (0,1)} |x(t)|$, and let $X$ be the continuous functions on $(0,1)$ with the norm $\|x\|_1 = \int_0^1 |x(t)| dt$. If we let $x(t) = \frac{1}{\sqrt{t}}$, then $x \in X$, but $x \notin C((0,1))$. However, we have $\|x\|_\infty = \infty$.
A: Okay let me try this. I guess when you write about generating a topology you wish it to be done in the same way as when having a usual norm. Let me talk more generally about a metric generalised to reach infinite values. 
So the thing here is to check that the set of all balls form a base for a topology. Obviously we only have to think about balls of finite radious because if such a thing as a ball of infinite radious exist they must be the whole space. Now this is going to be true because it is true with the usual finite metric. Still, let's do it. 
You can see that all you need to prove is that given a finite number of balls and a point in the intersection of all of them then there is a ball containing the point that is still contained in the intesection. We can simplify the case to only having two balls: $B_1=B(x,r)$ and $B_2=B(y,s)$, with $z\in B_1\cap B_2$. 
Now let $\delta:= \min\{r-d(x,z), s-d(y,z)\}$ and prove by means of the triangular inequality that the ball $B(z,\delta)$ still lies in the intersection of $B_1$ and $B_2$.
So we've proved (you might need to think a bit to see this) that the set of arbitrary unions of balls is closed not only under arbitrary union but also under finite intersection (and obviously contains the whole space) so the set of all balls is base for a topology. 
Now in this space all balls containing a given point form a base of neighbourhoods for said point. In particular a sequence of balls centered at a point whose radious tends to zero form a basis. It follows that convergence in this top. space is related to convergence in our generalised metric in the same way convergence in a usual metric space is related to convergence in the metric. 
Hope this was the answer you were looking for. Long story short: yes we can think of a topology generated as usual and maintain the classical notion of convergence in a metric space.
A: $\infty \notin \mathbb R$, therefore $\|\cdot\|$ is not a map into $\mathbb R$.
