$\mathrm{p}.-.1667\mathrm{emv}.(\frac{1}{x})$ is a distribution of first order

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

$\mathrm{p}.\mathrm{v}.({\displaystyle \frac{1}{x}})(u)$$=$$\underset{\epsilon \to 0+}{lim}{\displaystyle {\int }_{[\epsilon ,k]}}{\displaystyle \frac{u(x)-u(-x)}{x}}𝑑x.$

Now it is clear that the integrand is continuous for all $x\in ℝ\setminus \{0\}$.What is more, the integrand approaches $2u^{\prime }(0)$ for $x\to 0$, so theintegrand 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$f_n(x)={\chi }_{[1/n,k]}\left(u(x)-u(-x)\right)/x$(where $\chi$ is the characteristic function), andapplying the Lebesgue dominated convergence theorem, we have

$\mathrm{p}.\mathrm{v}.({\displaystyle \frac{1}{x}})(u)$$=$${\displaystyle {\int }_{[0,k]}}{\displaystyle \frac{u(x)-u(-x)}{x}}𝑑x.$

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

To prove that $\mathrm{p}.\mathrm{v}.(\frac{1}{x})$ is continuous, we shall use condition (3) onthis page (http://planetmath.org/Distribution4).For this, suppose $K$ is a compact subset of $ℝ$ and$u\in {𝒟}_K$. 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 \frac{1}{x}}{\displaystyle {\int }_{(-x,x)}}{u}^{\prime }(t)𝑑t|$

$\le$$2{||u^{\prime }||}_{\mathrm{\infty }},$

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

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

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