Solve $xu = 0$ in the sense of distribution

Question

I got stuck at showing $\delta_0$ solves $xu=0$ in the sense of distribution (up to some constant).

The hint states to decompose $\phi = \phi(0)g(x)+x\varphi(x)$ for some function $g(x),\varphi(x) \in \mathbf{D}(\mathbb{R})$. However I can only think of $g(x)= 1,\varphi = {\phi}^\prime$ which does not seem to show the result (After plugging this into $\left<xu,\phi\right>$ and integrating by part).

Any suggestion is greatly appreciated!

Answer 1

To show $x\delta_0=0$, just calculate $$\left<x\delta_0,\phi\right>=\left<\delta_0,x\phi\right>=0$$ as $x\phi$ vanishes at zero.

But I presume you really want to prove that the only distribution solutions of $xu=0$ are the scalar multiples of $\delta_0$. So suppose $xu=0$.

Let $g$ be a test function with $g(0)=1$. We can write an arbitrary test function as $\phi=\phi(0)g+\psi$ where $\psi$ is a test function with $\psi(0)=0$. Then $\psi=x\theta$ where $\theta$ is also a test function ($\theta(0)=\psi'(0)$).

Then $$\left<u,\phi\right>=\left<u,\phi(0)g+x\theta\right> =\phi(0)\left<u,g\right>+\left<xu,\theta\right>=k\left<\delta_0,\phi\right>$$ where $k=\left<u,g\right>$.

Answer 2

I think that the most difficult part of this question is actually showing that if $\psi$ is a test function such that $\psi(0)=0$, then so is $\theta$, where $\psi = x\theta$, which was not proved in Angina's answer. So I decided to fill in that gap. We can write $$\psi(x) = \psi(0) + \int_0^1 \frac{d}{dt} \psi(tx) dt = x\int_0^1 \psi'(tx) dt $$ Hence we can put $\theta(x) := \int_0^1 \psi'(tx) dt$, which is clearly smooth. Its support is contained in support of $\psi$, because for $x \ne 0$ we have $\theta(x) = \frac{\psi(x)}{x}$. Therefore $\theta(x)$ is indeed a test function.