Dirac distribution verifies $x\delta(x)=0$

Question

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!

Answer 1

1. For a distribution $u$ and polynomials $P,Q$, we define the distribution $P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u$ as $$\tag{1} \left(P(M_{{\rm id}_{\mathbb{R}}}) Q(\partial)u\right) [\varphi]~:=~ u\left[ Q(-\partial)P(M_{{\rm id}_{\mathbb{R}}})\varphi\right], $$ where $\varphi$ is a test function; where $M_f(g):=fg$ denotes the multiplication operator with the function $f$; where $\partial(f)=f^{\prime}$ is differentiation; and where ${\rm id}_{\mathbb{R}}$ is the identity function: $\mathbb{R}\to \mathbb{R}$. Note that the multiplication operator $M_{{\rm id}_{\mathbb{R}}}$ is often denoted by the indeterminate $x$ by a slight abuse of notation.

2. The Dirac delta distribution $u=\delta$ is defined as $$\tag{2}\delta [\varphi]~:=~\varphi(0).$$

3. The distribution identities $$\tag{3} M_{{\rm id}_{\mathbb{R}}}\delta~=~0\qquad\text{and}\qquad M_{{\rm id}_{\mathbb{R}}}\partial\delta~=~-\delta,$$ (which is often written as $$\tag{4} x\delta(x)~=~0\qquad\text{and}\qquad x\delta^{\prime}(x)~=~-\delta(x),$$ with a slight abuse of notation), follow from $$\tag{5} \left(M_{{\rm id}_{\mathbb{R}}}\delta\right)[\varphi]~\stackrel{(1)}{=}~\delta\left[M_{{\rm id}_{\mathbb{R}}} \varphi\right]~\stackrel{(2)}{=}~0\varphi(0)~=~0~=~0[\varphi],$$ and $$ \left(M_{{\rm id}_{\mathbb{R}}}\partial\delta\right)[\varphi]~\stackrel{(1)}{=}~\delta\left[-\partial M_{{\rm id}_{\mathbb{R}}} \varphi\right]~\stackrel{\text{Leibniz' rule}}{=}~\delta\left[-\varphi- M_{{\rm id}_{\mathbb{R}}} \partial\varphi\right]$$ $$\tag{6}~\stackrel{(2)}{=}~-\varphi(0)-0\varphi^{\prime}(0)~\stackrel{(2)}{=}~-\delta[\varphi],$$ respectively.

Answer 2

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.