Why is the function $y(x) = \frac{1}{x}$ not a distribution?

Question

Why is this function not a distribution? I am trying to understand how this function is related to the Cauchy Principle Value theorem or how the theorem is able to explain this. A simple explanation is greatly appreciated, thankyou

Answer 1

A function $f :\mathbb{R}^n \to \mathbb{R}$ defines a distribution iff it is locally integrable, and $\frac 1 x$ is not around $0$. Indeed, you will have trouble defining $\int_{-1}^1 \frac 1 x \phi(x) dx$ for tests function for whose support is a neighbourhood of $0$.

A way to go around the problem is by using Cauchy's principal value, because the integrability only fails at one point, and thus we are able to define a distribution (which formaly is just a linear form from the tests functions to $\mathbb{R}$) in some way from such a function.

EDIT : As said before, a distribution is a continuous linear form over the space of test functions $\mathcal{D}$. The non-integrability of $\frac 1 x$ prevent us from using the standard distribution definition from a function $f$, which is $T_f:\phi \mapsto \int_\mathbb{R} f \phi $. Indeed, in this case, such operator is not continuous (you can try to construct a sequence such that $|T_f (\phi_n)| \to \infty$ while $\|\phi_n\|<1$, their support have to be neighbourhood of $0$ of course). However, we can enforce continuity by considering the principal value instead, and this is exactly what is being prooven in the previous link.