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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?

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  • $\begingroup$ Indeed, I would say it's undefined as well. $\endgroup$ – Crostul Dec 30 '18 at 14:48
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    $\begingroup$ Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $\delta(x-z)f(x)$ $\endgroup$ – Winther Dec 30 '18 at 14:50
  • $\begingroup$ 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. $\endgroup$ – GEdgar Dec 30 '18 at 14:51
  • $\begingroup$ @Winther It's occuring implicitly in the definition of $\kappa_n^\circ$ on page 4 here. $\endgroup$ – 0xbadf00d Dec 30 '18 at 14:52
  • $\begingroup$ 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. $\endgroup$ – Winther Dec 30 '18 at 15:15
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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, $$ \int \delta(x-y) \delta(y-z) f(y) \, dy = f(x) \delta(z-x). $$

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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.

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$$\color{green}{\delta(x\!-\!z)}\tag{1a}$$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.

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