# $\operatorname{p.\tmspace-{.1667em}v.}(\frac{1}{x})$ is a distribution of first order

(Following [1, 2].) Let $u\in\mathcal{D}(U)$. Then $\operatorname{supp}u\subset[-k,k]$ for some $k>0$. For any $\varepsilon>0$, $u(x)/x$ is Lebesgue integrable in $|x|\in[\varepsilon,k]$. Thus, by a change of variable, we have

 $\displaystyle\operatorname{p.\!v.}(\frac{1}{x})(u)$ $\displaystyle=$ $\displaystyle\lim_{\varepsilon\to 0+}\int_{[\varepsilon,k]}\frac{u(x)-u(-x)}{x% }dx.$

Now it is clear that the integrand is continuous for all $x\in\mathbb{R}\setminus\{0\}$. What is more, the integrand approaches $2u^{\prime}(0)$ for $x\to 0$, so the integrand has a removable discontinuity at $x=0$. That is, by assigning the value $2u^{\prime}(0)$ to the integrand at $x=0$, the integrand becomes continuous in $[0,k]$. This means that the integrand is Lebesgue measurable on $[0,k]$. Then, by defining $f_{n}(x)=\chi_{[1/n,k]}\big{(}u(x)-u(-x)\big{)}/x$ (where $\chi$ is the characteristic function), and applying the Lebesgue dominated convergence theorem, we have

 $\displaystyle\operatorname{p.\!v.}(\frac{1}{x})(u)$ $\displaystyle=$ $\displaystyle\int_{[0,k]}\frac{u(x)-u(-x)}{x}dx.$

It follows that $\operatorname{p.\!v.}(\frac{1}{x})(u)$ is finite, i.e., $\operatorname{p.\!v.}(\frac{1}{x})$ takes values in $\mathbb{C}$. Since $\mathcal{D}(U)$ is a vector space, if follows easily from the above expression that $\operatorname{p.\!v.}(\frac{1}{x})$ is linear.

To prove that $\operatorname{p.\!v.}(\frac{1}{x})$ is continuous, we shall use condition (3) on this page (http://planetmath.org/Distribution4). For this, suppose $K$ is a compact subset of $\mathbb{R}$ and $u\in\mathcal{D}_{K}$. Again, we can assume that $K\subset[-k,k]$ for some $k>0$. For $x>0$, we have

 $\displaystyle|\frac{u(x)-u(-x)}{x}|$ $\displaystyle=$ $\displaystyle|\frac{1}{x}\int_{(-x,x)}u^{\prime}(t)dt|$ $\displaystyle\leq$ $\displaystyle 2||u^{\prime}||_{\infty},$

where $||\cdot||_{\infty}$ is the supremum norm. In the first equality we have used the fundamental theorem of calculus for the Lebesgue integral (valid since $u$ is absolutely continuous on $[-k,k]$). Thus

 $|\operatorname{p.\!v.}(\frac{1}{x})(u)|\leq 2k||u^{\prime}||_{\infty}$

and $\operatorname{p.\!v.}(\frac{1}{x})$ is a distribution of first order (http://planetmath.org/Distribution4) as claimed. $\Box$

## References

Title $\operatorname{p.\tmspace-{.1667em}v.}(\frac{1}{x})$ is a distribution of first order operatornamepvfrac1xIsADistributionOfFirstOrder 2013-03-22 13:46:07 2013-03-22 13:46:07 Koro (127) Koro (127) 7 Koro (127) Proof msc 46F05 msc 46-00