# Which class of functions does 1/x belong to?

• I
For historical reasons the hyperbola always was considered to be one of the «classical» curves. The function, obviously, does not belong to C0. Apparently, is does not fit L2 or any other Lp? What is the smallest class?

member 587159
It depends.
$$(x \mapsto 1/x) \in C^0(\mathbb{R}\setminus \{0\})$$
In fact even
$$(x \mapsto 1/x) \in C^\infty(\mathbb{R}\setminus \{0\})$$

So I guess you will have to be a little more precise what your question is.

bhobba
It depends.
$$(x \mapsto 1/x) \in C^0(\mathbb{R}\setminus \{0\})$$
In fact even
$$(x \mapsto 1/x) \in C^\infty(\mathbb{R}\setminus \{0\})$$

So I guess you will have to be a little more precise what your question is.
Yes, you are right. I was thinking about interval [-1,1].

member 587159
Yes, you are right. I was thinking about interval [-1,1].

So, how do you define ##x \mapsto 1/x## in ##0## then? You give it an arbitrary value?

Infrared
Gold Member
It is an element of ##L^p## when ##0<p<1## but these spaces aren't as nice; e.g. ##||f||_p=\left(\int |f|^p\right)^{1/p}## doesn't define a norm.

bhobba, WWGD, SVN and 1 other person
So, how do you define x↦1/xx \mapsto 1/x in 00 then? You give it an arbitrary value?

I am not sure I understand your point. The analytic functions form a small and restrictive class of functions. It can be broadened by dropping some requirements imposed on class members. It gives us this sequence (incomplete, I guess, but it illustrates the basic idea):
##C^\omega \subset C^\infty \subset C^0 \subset L^p \in##... well, I do not know what comes next, possibly, distributions.

So my question is not how to define ##1/x## in ##0##, but which class does ##1/x## belongs to, precisely because of its irregular behaviour in ##0##. Well, again the interval is ##[-1,1]##.

It is an element of ##L^p## when ##0<p<1## but these spaces aren't as nice; e.g. ##||f||_p=\left(\int |f|^p\right)^{1/p}## doesn't define a norm.
Thank you!

mathwonk
Homework Helper
2020 Award
my first response would be to call it a rational function.

lavinia
my first response would be to call it a rational function.
Well, of course, but I meant its classification from the viewpoint of functional analysis.

mathman
It is not ##L_p## for ##p\ge 1##. For ##p\lt 1##, it won't be a Banach space.

Svein
That is why I prefer complex analysis. The function $\frac{1}{z}$ is ubiquitous in complex integration and is the basis of residue theory etc.

bhobba and lavinia
mathwonk