Let $\phi:\mathbb R \to \mathbb R$ be a test function. We denote $D(\mathbb R)$ the set of test functions. The dirac distribution
$$\delta :D(\mathbb R)\to \mathbb R$$ is defined by:
$$<\delta , \phi>=\phi(0)$$ Now take any smooth function $g:\mathbb R\to \mathbb R$, and define the product distribution $g.\delta$ by:
$$<g.\delta,\phi>=<\delta,g\phi>$$
From this definition we see that
$$<g.\delta,\phi>=g(0)\phi(0)$$
Now i'm reading the following problem prove that $x\delta(x)=0$ and $x\delta'(x)=-\delta(x)$ wihtout any firther information, and I want to understand the meaning of these formulas regarding the definitions above, I mean what is $x$ and what is $\delta(x)$ and what is the product $x\delta(x)$? It seems like he takes $g$ for being the identity $id :\mathbb R\to \mathbb R$ but this would give
$$<id.\delta, \phi>=id(0)\phi(0)=0*\phi(0)=0$$ but i don't see where $x\delta(x)=0$ come from. thank you for your help!
Best Answer
By definition we can write for any test function $\phi$ $$\langle x\delta_0,\phi(x)\rangle = \langle \delta_0, x\phi(x)\rangle = 0\cdot \phi(0)=0=\langle 0,\phi(x)\rangle,$$ which means that $x\delta_0=0$ in the sense of distributions.
In the same spirit, we write
$$\langle x\delta'_0,\phi(x)\rangle = \langle \delta_0', x\phi(x)\rangle =- \langle \delta_0,(x\phi(x))'\rangle = -\langle \delta_0,\phi(x)+x\phi'(x) \rangle = -\phi(0)-0\cdot\phi'(0) = -\phi(0) = -\langle \delta_0, \phi(x) \rangle,$$ which means that $x\delta'_0= -\delta_0$ in the sense of distributions.