Re: Norm with infinite values


In article <ddtbtb$p68$1@xxxxxxxxxxxxxxxxxxxxxxxxx>, Lee Rudolph
wrote:

> Jeremy Henty <jeremy@xxxxxxxxxxxx> writes:
>
>>These axioms imply that the set of points where the norm is finite
>>is
>
> either empty or
>
>>a subspace, ... So norms, which can take infinite values, are just
>>norms defined on a subspace
>
> with the single exception of the identically-infinite map.

Good call! But there's a slight issue (which I've only just thought
of) to do with definedness: we have ||0|| = ||0.0|| = |0|.||0|| =
0.||0|| . If ||0|| is finite then this implies ||0|| = 0 , but what
does this equation mean if ||0|| is \infty ? It's common to regard
0.\infty as undefined, so if we interpret the equation as saying that
its sides are either both undefined or both defined and equal, then
this rules out ||0|| = \infty . OTOH, if we take it to mean that the
two sides are equal if the RHS is defined then ||0|| = \infty is
allowed. I'm not sure which is the "right" way to take it.

I can also see a case for arguing there is really an extra axiom for a
normed space, namely ||0|| = 0 , which is usually omitted because it
is redundant. The idea is that 0 is a nullary operator and we have
axioms describing the norm of a sum and the norm of a scalar product,
so the norm of 0 deserves its own axiom too (and so does the norm of a
negative, I guess). In that case generalising the "correct"
definition of a normed space to allow infinite norms does not admit a
norm that is everywhere \infty .

This is all just nitpicking of course; however you cut it allowing
norms to take infinity as a value doesn't introduce any new
interesting structures.

Cheers,

Jeremy Henty
.

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