Consider the following integral

$\int_{-\infty}^{+\infty} dx \frac{\partial}{\partial x} \left[ f(x) \delta(x-y) \right]$

where f is a continuous function and x and y are real variables. Intuitively, I would say that the integral is zero, because $\left[ f(x) \delta(x-y) \right] = 0$ when $x=\pm \infty$ (unless also y goes to infinity).

Am I correct? Is there a more formal way to justify it?

  • $\begingroup$ i would shift $x-y=z$. then use $\delta'(x)f(x)"="-f'(0)$ $\endgroup$ – tired Aug 8 '17 at 23:04
  • $\begingroup$ let us define the distribution $\phi(x,y)=\delta(x-y)\partial_x+\partial_x\delta(x-y)$ and interpet it as a linear functional acting on the space of suitable test functions (Schwartz class $S$ or whatever). Let us denote by $f(x)$ one of them, we get ( the derivative operator is invariant under translations) $$ \langle\phi(x,y),f(x)\rangle=\langle\phi(x,0),f(x+y)\rangle=\langle\delta(x),\partial_xf(x+y)\rangle+\langle\partial_x\delta(x),f(x+y)\rangle $$ now we use the well known property $\delta'(x)f(x)=-f'(0)$ to conclude that $$ \langle\phi(x,y),f(x)\rangle=f'(y)-f'(y)=0 $$ $\endgroup$ – tired Aug 9 '17 at 0:11
  • $\begingroup$ Clear explanation, thanks! :) $\endgroup$ – Michele Cotrufo Aug 9 '17 at 17:30

Notice that $f(x)\delta(x-y)$ defines a compactly supported distribution. (This roughly means that $f(x)\delta(x-y)$ "vanishes" for large $|x|$.)

Here, a compactly supported distribution is a continuous linear functional on the sapce $\mathcal{E}(\mathbb{R})$, the set $C^{\infty}(\mathbb{R})$ endowed with a certain topology. Now for any compactly supported distribution $\eta$, we can regard the integral of $\eta$ as the pairing

$$ \int_{-\infty}^{\infty} \eta(x) \, dx \quad \text{$``$}=\text{''} \quad \langle \eta, 1 \rangle, $$

where $1 \in \mathcal{E}(\mathbb{R})$ is the constant function with value 1. Of course, this coincides with the usual integral when $\eta$ is a compactly supported integrable function, justifying the notation.

Then it follows that

$$ \int_{-\infty}^{\infty} \eta'(x) \, dx = \left\langle \frac{d}{dx}\eta', 1 \right\rangle = -\left\langle \eta, \frac{d}{dx}1 \right\rangle = -\left\langle \eta, 0 \right\rangle = 0. $$


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