# What is $\int\delta(x-y)\delta(y-z)f(y)\:{\rm d}y$?

Let $$(\Omega,\mathcal A,\mu)$$ be a measurable space and $$\delta$$ denote the Dirac delta function. If $$f\in\mathcal L^1(\mu)$$ and $$x,z\in\Omega$$, what is $$\int\delta(x-y)\delta(y-z)f(y)\:\mu({\rm d}y)?$$ I've found that in a paper, but isn't that undefined?

• Indeed, I would say it's undefined as well. – Crostul Dec 30 '18 at 14:48
• Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $\delta(x-z)f(x)$ – Winther Dec 30 '18 at 14:50
• Unfortunately "a paper" is not enough for us to find it. Some papers are pure nonsense. Other papers have definitions in them revealing what their notation means. – GEdgar Dec 30 '18 at 14:51
• @Winther It's occuring implicitly in the definition of $\kappa_n^\circ$ on page 4 here. – 0xbadf00d Dec 30 '18 at 14:52
• Don't know how to rigorously define and prove the results above (products of distributions are troublesome, but my guess is that this one can make sense). One way of justifying a formula like the one above is to consider a smooth approximation to the Dirac $\delta$ like for example $\delta_\epsilon(x) \equiv \frac{1}{\sqrt{2\pi \epsilon}}e^{-\frac{x^2}{2\epsilon}}$. In this case it's easy to check that $\int \delta_\epsilon(x-y)\delta_\epsilon(y-z){\rm d}y = \delta_{\epsilon/2}(x-z)$ so atleast for cases where the delta is just an approximation for a sharp pulse the formula above should hold. – Winther Dec 30 '18 at 15:15

Talking non-rigorously, $$\delta(x-y) \delta(y-z) f(y)$$ will be non-zero only when $$x-y=0$$ and $$y-z=0$$, i.e. when $$x=y=z.$$ Therefore the integral over $$y$$ would be non-zero only when $$x=z.$$ We can thus expect the integral to be a multiple of $$\delta(x-z).$$

So, let $$\phi$$ be a nice function and study the formal integral $$\int \left( \int \delta(x-y) \delta(y-z) f(y) \, dy \right) \phi(z) \, dz.$$

Swapping the order of integration gives $$\int \delta(x-y) \left( \int \delta(y-z) \phi(z) \, dz \right) f(y) \, dy = \int \delta(x-y) \phi(y) f(y) \, dy = \phi(x) f(x).$$

Thus, $$\int \delta(x-y) \delta(y-z) f(y) \, dy = f(x) \delta(z-x).$$

Recall the usual $$\int\delta(x-y)g(y)dy=g(x)$$ (assuming it's a definite integral over $$A$$), in this case with $$g(y):=f(y)\delta(y-z)$$, giving $$\color{blue}{f(x)\delta(x-z)}$$ for your in-title integral (again, if it's definite over $$A$$). Here we've eliminated the $$\delta$$ containing $$x$$, effectively imposing $$y=x$$. Of course we could have used the other $$\delta$$ to impose $$y=z$$ instead, obtaining $$\color{blue}{f(z)\delta(x-z)}$$. But by inspection, these two answers are equivalent.

$$\color{green}{\delta(x\!-\!z)}\tag{1a}$$and$$\color{red}{\int_{\mathbb{R}} \!\mathrm{d}y ~f(y)\delta(x\!-\!y)\delta(y\!-\!z)}\tag{1b}$$ are informal notations for the distributions $$u,v\in D^{\prime}(\mathbb{R}^2)$$ given by$$^1$$ $$u[\varphi]~:=~\int_{\mathbb{R}} \!\mathrm{d}y~\varphi(y,y),\tag{2a}$$ $$v[\varphi]~:=~\int_{\mathbb{R}} \!\mathrm{d}y ~f(y)~\varphi(y,y),\tag{2b}$$ respectively, where $$\varphi\in D(\mathbb{R}^2)$$ is a test function. The distributions are written in the informal notation as $$u[\varphi]~=~\int_{\mathbb{R}^2} \!\mathrm{d}x~\mathrm{d}z~\color{green}{\delta(x\!-\!z)}~ \varphi(x,z), \tag{3a}$$ $$v[\varphi]~:=~\int_{\mathbb{R}^2} \!\mathrm{d}x~\mathrm{d}z~\color{red}{\int_{\mathbb{R}} \!\mathrm{d}y ~f(y)\delta(x\!-\!y)\delta(y\!-\!z)} ~\varphi(x,z),\tag{3b}$$ respectively.

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$$^1$$Disclaimer: Here we assume for simplicity that OP's space is $$\Omega=\mathbb{R}$$ and the measure $$\mu$$ is the Lebesque measure. Note that e.g. boundaries $$\partial \Omega$$ often requires extra attention in distribution theory.