proof of the product: x times m derivative of delta function.

Question

The question is in one dimension and is : Prove that $$x\delta^{(m)}=-m\delta^{(m-1)},\ m \in \mathbb{N},$$ where $\delta^{(m)}$ is the $m$-derivative of $\delta.$

As I know, I got through this way: $$\langle x\delta^{(m)},\psi \rangle=\langle\delta^{(m)},x\psi\rangle$$ $$ -\langle\delta^{(m-1)},\frac{d(x\psi)}{dx}\rangle=-\langle\delta^{(m-1)},x\psi'+\psi\rangle$$ $$= -\langle \delta^{(m-1)},\psi\rangle -\langle \delta^{(m-1)},x\psi'\rangle=-\delta^{(m-1)} -\langle \delta^{(m-1)},x\psi'\rangle$$

I can get away with developing the term with $(\psi)',$ but then the $(m-1)$- derivative would disappear, any hints?

I thought about the possibility of error in writing the book, I don´t know.

Answer 1

You have, for any test function $\psi\in \mathcal D(\mathbb R)$ : \begin{align} \langle x\delta^{(m)},\psi\rangle &= \langle \delta^{(m)},x\psi\rangle\\ &= (-1)^m \langle \delta, (x\psi)^{(m)}\rangle \end{align}

Now, using the Leibniz formula, we have : $$(x\psi)^{(m)} = \sum_{k=0}^m {m\choose k}x^{(k)}\psi^{(m-k)} = x\psi^{(m)} + m\psi^{(m-1)}$$ Therefore : \begin{align} \langle x\delta^{(m)},\psi\rangle &= (-1)^m\langle \delta, x\psi^{(m)} + m\psi^{(m-1)}\rangle \\ &= (-1)^m\langle \delta, m\psi^{(m-1)}\rangle \\ &= \langle - m\delta^{(m-1)},\psi\rangle \end{align} i.e. $x \delta^{(m)} = -m\delta^{(m-1)}$.

Remark : You can also prove this by induction without using test functions. This is true for $m = 0$ and if $m>0$ and the formula is true for $m-1$ : $$x\delta^{(m)} = \left(x\delta^{(m-1)}\right)' - \delta^{(m-1)} = \left(-(m-1)\delta^{(m-2)}\right)' - \delta^{(m-1)} = -m\delta^{(m-1)}$$