Does $f(x)=\frac{1}{x^{2}}$ define a distribution on $\mathbb{R}$?

Question

I wonder whether the following function defines a distribution on $\mathbb{R}$ or not: $\;\;f(x)=\frac{1}{x^{2}}?$

I tried to prove $f\notin\mathbb{L}_{loc}^{1}$ and this is clear as $f$ is not continuous at $0$.

Is my idea correct?

Answer 1

In fact, this sort of problem lies on how we define $f$ as a distribution. As you already know, we often just say $f$ is a distribution if the functional $$F_f(\varphi)=\int f\varphi$$ is a continuous linear functional on $C_0^\infty(\mathbb{R})$, but this is an (clear) abuse of notation. What $f\in\mathcal{D}^*$ means in fact, corresponding linear functional $F_f$ of $f$ is distribution.

But what do you even mean when $F_f$ is not well-defined in above sense (in our case $f(x)=1/x^2$)? Then the answer depends on how you define the corresponding functional of $f$. Maybe you can interpret the corresponding functional as Cauchy principal value, say $$F_f(\varphi)=p.v.(f)(\varphi):=\lim_{\epsilon\rightarrow 0}\int_{\mathbb{R}\backslash[-\epsilon,\epsilon]}\frac{\varphi(x)}{x^2}\,dx.$$ Then this $F_f$ is clearly not continuous since it is linear, however, there exists $\varphi_n$ converging to $0$ in $\mathcal{D}$, but $F_f(\varphi_n)\rightarrow +\infty$.

Or you may consider $F_f$ is a (non-finite) measure and still it is not a distribution by the same reasoning.

But first you need to understand the topology on $C_0^\infty(\mathbb{R})=\mathcal{D}$. Do you know what it means $\varphi_n$ converges to $\varphi$ in $\mathcal{D}$?