$xT' = 1$ in $D'(\mathbb{R})$

Question

I need help solving the following problem: I want to show that all solutions of $$xT' = 1\ , T \in D'(\mathbb{R})$$ take the following form: $c_{1} + c_{2}1_{[0, \infty)} + ln|.|$ What I tried so far is: $$xT' = 1 = x P[\frac{1}{x}] \Rightarrow x(T'-P\left[\frac{1}{x}\right]) = 0 \Rightarrow T'-P[\frac{1}{x}] = c \delta_{0}$$ From this I could derive $\forall \phi \in D(\mathbb{R}):$ $$T(\partial\phi) = c\int_{0}^{\infty}\partial\phi + \int_{\mathbb{R}} \ln|.|\partial\phi$$ This entails the result for all $\phi \in D(\mathbb{R})$ that can be written as a derivative of another element of $D(\mathbb{R})$ But I'm still unable to extend the result to all $\phi \in D(\mathbb{R})$. I would be very thankful for any input on this.

Answer 1

You look for a $T\in\mathscr D'(\mathbb R)$ satisfying $$-T(\varphi+x\varphi')=T'(x\varphi)=(xT')(\varphi)=\int\varphi\,dx,$$ for all $\varphi\in\mathscr D(\mathbb R)$.

Hint. Try $$T(\varphi)=\frac{1}{2}\int_{\mathbb R}(\log x^2+c)\,\varphi(x)\,dx.$$

Note that $\log x^2\in L^1_{\mathrm{loc}}(\mathbb R)$.

Answer 2

First, we want to solve the equation $xS=0$ in $ D'(\mathbb{R})$. The solution, obviously, is $c\delta_0$. Then we study $xS=1$. It's easy to see that distribution $vp[1/x]$ (principal value, I believe that's what meant by $P(1/x)$) satisfies this equation. Therefore, the family of solutions is $vp[1/x] +c\delta_0$.

Thus, $$T' = vp[1/x] +c\delta_0.$$

Integrating $\delta_0$ is easy, we obtain Heavyside function $H(x)$ (plus an arbitrary constant, of course).

The only tricky part is to integrate $vp[1/x]$. But here the intuition from calculus really helps: the antiderivative of $1/x$ is $\ln x$, so we check that indeed $\ln |x|$ seen as a distribution has a derivative (in the sense $D'(\mathbb{R})$) $vp[1/x]$. Note that $\ln |x|\in L^1_{loc}(\mathbb{R})$.

Finally, we obtain the family of solutions

$$\ln |x|+cH(x)+c_2.$$