My Problem is to find all solutions of the distribution equation $xT =1$.
I don't know how to solve it. The solutions should be $pv\frac1x+c$ , $c \in \mathbb{R}$.
My idea was to create a function like for: Let $T$ a Distribution. $xT=0$ then $T=c\delta$.
But I cant construct a function for my problem and I´m not sure if it´s the right way.
Distributions need something to act on, so let $\phi$ be a test function. Denote the action of $T$ on $\phi$ by $\langle T,\phi \rangle$.
Functions act on test functions by integration, so that $\langle 1,\phi \rangle = \displaystyle \int_{\mathbb R} 1 \cdot \phi(x) \, dx$.
The distribution $xT$ is defined via the rule $\langle xT,\phi \rangle = \langle T,x\phi \rangle$.
The question then becomes, what distribution $T$ satisfies $$\langle T,x\phi \rangle = \int_{\mathbb R} \phi(x) \, dx ?$$ You can get $x \phi$ in the integrand via $$\int_{\mathbb R} \phi(x) \, dx = \lim_{\epsilon \to 0^+} \int_{\mathbb R \setminus [-\epsilon,\epsilon]} \phi(x) \, dx = \lim_{\epsilon \to 0^+} \int_{\mathbb R \setminus [-\epsilon,\epsilon]} \frac{x\phi(x)}{x} \, dx.$$ The principal value distribution $\mathrm{pv} \frac 1x$ is defined via the formula $$\langle \mathrm{pv} \frac 1x,\phi \rangle = \lim_{\epsilon \to 0^+} \int_{\mathbb R \setminus [-\epsilon,\epsilon]} \frac{\phi(x)}{x} \, dx$$ so that $$ \int_{\mathbb R} \phi(x) \, dx = \langle \mathrm{pv} \frac 1x,x\phi \rangle.$$ Thus $\mathrm{pv} \frac 1x$ satisfies $xT = 1$.
If $T$ is an arbitrary distribution with $xT = 1$ it follows that $x(T - \mathrm{pv} \frac 1x) = 0$, implying that $T - \mathrm{pv} \frac 1x = c \delta$ for some $c$, where $\delta$ is the delta distribution.