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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!

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You can define some norm-like functions with a range $[0,\infty]$ but every normed space consists exactly of those vectors whose norms are finite. The finiteness of norm of elements of a normed space is crucial for many properties of such space. –  Ilya Jan 29 '13 at 16:14
    
@xavierm02: No. –  Ilya Jan 29 '13 at 16:17
    
@Ilya: A normed space is a vector space with a norm. If a norm can take $\infty$ value, why "every normed space consists exactly of those vectors whose norms are finite"? /// If a norm can take $\infty$ value, do the three axioms still hold? –  Ethan Jan 29 '13 at 16:19
    
I didn't say the norm can take value $\infty$, I rather said that you can start with a norm-like function which takes this value. I better put an answer. –  Ilya Jan 29 '13 at 16:20
    
@Ilya: If a norm cannot take $\infty$, why? –  Ethan Jan 29 '13 at 16:21

4 Answers 4

up vote 4 down vote accepted

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.

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If we generalize the definition of a norm to allow it take $\infty$ without changing other aspects in the definition, $V$ will be a generalized norm space? Similarly, if we allow $\| \cdot \|_1$ to take $\infty$, will the set of all measruable functions which may have infinite or finite integral become a generalized norm space? 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? –  Ethan Jan 29 '13 at 16:31
    
@Ethan: I am not familiar with generalized norm, but that would be a possible way to define it. Since norm is an instance of metric, consult the discussion of the generalizations here. There is also a point on metric that take infinite values, but it does not seem to be formal. –  Ilya Jan 29 '13 at 16:37
    
Thanks for the link, where I found a sentence "Every extended metric can be transformed to a finite metric such that the metric spaces are equivalent as far as notions of topology (such as continuity or convergence) are concerned". Does it mean that the extended metric induce a topology in the same way as a metric induces a topology? The topology induced by an extended metric is the same as the topology induced by its equivalent metric? –  Ethan Jan 29 '13 at 16:51
    
An important class of such extended norms arises not on $\mathfrak{B}[0,1]$ but on the space $L^0$ of equivalence classes of measurable functions modulo null functions. See Banach function spaces. –  Martin Jan 29 '13 at 16:57
    
@Ethan: as I mentioned, the statement there is not precise there. I.e. if $d(x,y) = \infty$ what is $\frac{d(x,y)}{1+d(x,y)}$? This construction mostly works when you would like to make a equivalent (topologically) bounded metric space out of a given unbounded one. –  Ilya Jan 29 '13 at 17:03

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).

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$\infty \notin \mathbb R$, therefore $\|\cdot\|$ is not a map into $\mathbb R$.

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I mean can we generalize the definition? –  Ethan Jan 29 '13 at 16:13

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 $\|x\|_\infty = \infty$.

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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? –  Ethan Jan 29 '13 at 16:35

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