Distribution $P_{\frac{1}{x^3}}$

Question

How to show that for $\phi\in D(R)$, $<P_{\frac{1}{x^3}},\phi>=v.p.\int_{-\infty}^{\infty}\frac{\phi(x)-x\phi'(0)}{x^3}$ defines a distribution?

It is easy to check that $P_{\frac{1}{x^3}}$ is correctly defined and linear on D(R), but how to show that it continuously depends on $\phi$?

Thanks in advance.

Answer 1

One approach is begin with $f(x)=\log|x|$, which is locally integrable and therefore defines a distribution via $\phi\mapsto \int \phi(x)\log|x|\,dx$.

The third distributional derivative $f'''$ is the distribution $\phi\mapsto -\int \phi'''(x)\log|x|\,dx$. Formally, this can be viewed as $(\log |x|)'''=2x^{-3}$. To make this rigorous, we should integrate by parts (taking due diligence around $0$). To simplify the computations, observe that

• Every function $\mathbb R\to\mathbb R$ is the sum of an odd function and an even function.
• The distribution $\log|x|$ is even in the sense that it annihilates odd test functions.
• The third derivative of an even distribution is odd: it annihilates even test functions.

Therefore, it suffices to work with $-\int \phi'''(x)\log|x|\,dx$ when $\phi$ is odd. Take $\epsilon>0$: $$ \begin{split} -\int_{|x|>\epsilon}\phi'''(x)\log |x|\,dx &= -2\int_{\epsilon}^\infty \phi'''(x)\log x\,dx \\ &= 2(\log \epsilon)\phi''(\epsilon)+2\int_\epsilon^\infty \phi''(x)\frac{dx}{x} \\ &= 2(\log \epsilon)\phi''(\epsilon)- 2\epsilon^{-1}\phi'(\epsilon) +2\int_\epsilon^\infty \phi'(x)\frac{dx}{x^2} \\ &= 2(\log \epsilon)\phi''(\epsilon)- 2\epsilon^{-1}\phi'(\epsilon) -2\epsilon^{-2}\phi(\epsilon) + 4\int_\epsilon^\infty \phi(x)\frac{dx}{x^3} \end{split} $$ Since $\phi$ is odd, $\phi(0)=\phi''(0)=0$. As $\epsilon\to 0$, $$\begin{split} 2(\log \epsilon)\phi''(\epsilon) &= o(1) \\ - 2\epsilon^{-1}\phi'(\epsilon) &= - 2\epsilon^{-1}\phi'(0)+o(1) \\ -2\epsilon^{-2}\phi(\epsilon)&=-2\epsilon^{-1}\phi'(0)+o(1) \\ \end{split}$$

Add these up, express $-4\epsilon^{-1}\phi'(0)$ as $\int_{|x|>\epsilon} \frac{-2\phi'(0)}{x^2}\,dx$, and there you have it: $$ -\int_{|x|>\epsilon}\phi'''(x)\log x\,dx = 2\int_{|x|>\epsilon}\frac{\phi(x)-x\phi'(0)}{x^3}\,dx $$ The limit as $\epsilon\to 0$ is the principal value in question.