# 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
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
Homework Helper
Re post #9, Let me remark on the role of 1/z from the viewpoint of functional analysis as I understand it, and in relation to the insightful remarks of Svein. To me functional analysis is the analysis of linear operators on Banach space, of which spaces the particular examples L^p are of basic interest. You seem to have meant to ask which of these particular spaces, if any, the function 1/z belongs to, and you have received answers of that nature.

Beyond a study of examples of specific Banach spaces, functional analysis deals more deeply with the behavior of, and structure of, linear operators on these spaces, e.g. such as integral operators on function spaces. In that study it is fundamental to study the construction of new operators out of old, and the decomposition of arbitrary operators into combinations of standard ones.

Just as in finite dimensions it is crucial to examine the result of operators defined by applying polynomials to linear operators, so in infinite dimensions does it become important to apply more general functions, in particular holomorphic and rational functions to given operators. A fundamental question is whether a given operator has an inverse, say an operator of form (T-cI), and the inverse of this operator may be sought by substituting T into the rational function 1/(z-c). The theory of complex path integration, i.e. complex residues, is crucial here.

Some comments on the evaluation of rational functions at linear operators can be found in the fundamental reference Functional Analysis, by Riesz-Nagy, pages 422, 431. I hope this may be of interest.

lavinia