Assume $\delta$ is the zero Dirac distribution (measure) on $\Bbb R$. namely,
$$(\delta,f)= \int_\Bbb R f\delta(dx) =f(0)$$
We know if $T\in \mathcal {D'}(\Bbb R)$ is a distribution then for every $ \phi \in C^\infty(\Bbb R)$ we have that $\phi T$ is also a distribution defines by
$$(\phi T, f) = ( T, \phi f)$$ Moreover, $$(\phi T)'= \phi'T+\phi T' $$ My problem. If we consiser the particular case $\phi= x$ or $\phi=x^2$ and $T= \delta$
we have, $$(x\delta, f)= (xf)(0)=0~~~\forall ~~f$$
Question: does this means that $\color{red}{x\delta \equiv 0}$ in $ \mathcal {D'}(\Bbb R)$?
- If yes how to explain the fact that $$\color{blue}{0 =(x\delta)' = \delta +x\delta'}$$
- If no what is $\color{red}{x\delta }$ as a distribution?
Patently if one consider the measure $$\mu(dx) = x^2 \delta(dx)$$ then it happens that $$\int f\mu(dx) = 0~~~\forall ~~f$$
Why is $\mu$ a trivial measure.?
As you defined it $\mu \ll \delta$ and $\frac{d \mu}{d \delta}(x) = x^{2}$. That's fine, but $x^{2} = 0$ $\delta$-almost everywhere. Thus, we can say $\frac{d \mu}{d \delta}(x) = 0$ $\delta$-a.e. Therefore, $\mu = 0$.
The equation $0 = \delta + x \delta'$ does indeed seem strange. Then again, $\langle \delta', f \rangle = - f'(0)$ (by the definition of distributional derivative) so \begin{align*} \forall \psi \in C_{c}^{\infty}(\mathbb{R}) \quad \langle x \delta', \psi \rangle &= \langle \delta', x \psi \rangle \\ &= -(x \psi)'(0) \\ &= - 0 \cdot \psi'(0) - (1) \cdot \psi(0) \\ &= - \psi(0) \\ &= -\langle \delta, \psi \rangle. \end{align*} This proves $\delta + x \delta' = 0$ directly.