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I came across an integral in the wild:

$$\int^{+\infty}_{-\infty}dx\int^{+\infty}_{-\infty}dy \, f(x,y) \, \delta(x-y)\,\frac{\partial}{\partial y}\delta(\, \epsilon- |x-y|\,).$$

In the expression, $\epsilon>0$ and $|...|$ is the absolute value. Further, $\delta(..)$ denotes the Dirac delta generalized function defined as $\int_{-\infty}^{\infty} F(z)\delta(z-\epsilon)=F(\epsilon)$. I suspect the expression evaluates to zero since the two distributions, the Dirac delta and its derivative have disjoint supports. However my knowledge of distribution theory is rather lacking and I am not entirely sure, could someone please explain? A rigorous proof along with an intuitive explanation and references would be much appreciated. Also, the integral appeared in an application of quantum mechanics so f(x,y) is smooth, falls off at infinity and is reasonably well behaved. Thanks!

PS: In my current understanding, the product of two distributions can make sense if their 'singular support' is disjoint. Naively, I think since both the support and the singular support of the Dirac delta and its derivative, as given in the integral above are disjoint, the integral should vanish. Is this true? I am still not sure as I am not very familiar with distribution theory. If I am mistaken, kindly let me know. A related integral is $I_2 = \int_{-\infty}^{\infty} \,dz \,\delta(z)\,\delta'(\epsilon-|z|)\, f(z) $. This should vanish too then?

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  • $\begingroup$ What’s the definition of a distribution that you know and use? $\endgroup$ Commented Apr 16 at 14:46
  • $\begingroup$ Thanks Mark and Deane. I use only the integral forms and the expression of the Dirac delta generalized function as defined above; this is loosely referred to as a Dirac delta distribution by physicists. I would be more than happy to refer to other more rigorous definitions if it helps me resolve the above integral. @MarkViola, I have read that the product of two distributions can make sense if their "singular support" is disjoint, for example here: arxiv.org/pdf/1404.1778 in theorem 1. I was hoping to use a result along those lines. Thanks! $\endgroup$ Commented Apr 16 at 15:01
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    $\begingroup$ Yes, that would be $0$ since indeed the support are disjoints ... $\endgroup$ Commented Apr 16 at 22:09
  • $\begingroup$ @LL3.14, thanks, can you provide details? Why does the integral exist? Why is it zero? $\endgroup$ Commented Apr 17 at 11:30
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    $\begingroup$ You are correct that for the distribution $T=\delta(x-y)\delta_y(|x-y|-\varepsilon)$ where $\varepsilon>0$, we have for any $\phi(x,y)\in C_C^\infty(\mathbb{R}^2)$ $$\langle T,\phi \rangle =0$$ The Dirac Delta $\delta(x-y)$ has support on $y=x$ and its derivative $\delta_y(|x-y|-\varepsilon)$ has support on $y=x\pm \varepsilon$. Inasmuch as these lines are disjoint for all $\varepsilon\ne0$, the distribution $T$ of their product is $0$. Hence, $\langle T,\phi\rangle=0$. $\endgroup$ Commented Apr 17 at 13:08

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You are correct that for the distribution $T=\delta(x-y)\delta_y(|x-y|-\varepsilon)$ where $\varepsilon>0$, we have for any $\phi(x,y)\in C_C^\infty(\mathbb{R}^2)$

$$\langle T,\phi \rangle =0$$

Let's take a closer look. First,note that for any $\varepsilon>0$, we have the distributional equality

$$\frac{\partial}{\partial y}\delta(|x-y|-\varepsilon)=\delta'(y-(x+\varepsilon))+\delta'(y-(x-\varepsilon))$$

Second, note that both the Dirac Delta and its derivative have supprt $\{0\}$. Hence, $\operatorname{supp}\delta(y-x)=\{y|y=x\}$ and $\operatorname{supp} \delta'(y-(x\pm \varepsilon))=\{y|y=x\pm \varepsilon\}$. Inasmuch as these sets are disjoint for all $\varepsilon>0$, the distribution $T$ of their product is $0$. Hence, $\langle T,\phi\rangle=0$.

Explantion Using Sheafs:

Here, we provide a rigorous way to define the product of distributions that have disjoint supports. To that end, we proceed.

Distributions are defined as sheafs. As such, a distribution is zero if it is zero in a neighborhood of every point.

Suppose we have two distributions, $T_1$ with $\operatorname{supp}\{T_1\}=S_1$ and $T_2$ with $\operatorname{supp}\{T_2\}=S_2$, where $S_1 \cap S_2=\emptyset$. We seek to define their product, $T_1T_2$.

Let $x$ be any point in the domain (e.g., $\mathbb{R}^2$). Obviously, it is impossible for $x$ to be in both $S_1$ and $S_2$.

So, suppose $x\notin S_1$. Then, there exists an open neighborhood $N_1$ of $x$ where $T_1$ is the zero distribution. To have a consistent definition of a product, we must have $T_1T_2=0\cdot T_2=0$ on $N_1$.

Similarly, suppose $x\notin S_2$, Then, there exists an open neighborhood $N_2$ of $x$ where $T_2$ is the zero distribution. Hence $T_1T_2=0\cdot T_1=0$ on $N_2$.

Finally, the entire space is covered by the open sets $\{N_1$} and $\{N_2\}$ and the distribution $T_1T_2$ is, therefore, zero everywhere.

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  • $\begingroup$ Great answer, thank you $\endgroup$ Commented Apr 18 at 8:53
  • $\begingroup$ @Cryo Thank you. Much appreciated $\endgroup$ Commented Apr 18 at 15:40
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You could try to tackle it by regularizing two delta-s. Let them be some gaussians of vanishing width $\alpha$, $\beta$. Then

$$ \begin{align} \int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta(x-y)\delta(\epsilon-|x-y|)\to I(\alpha,\beta)\\ I(\alpha,\beta,\epsilon)=\int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta_\alpha(x-y)\partial_y\delta_\beta(\epsilon-|x-y|) \end{align} $$

Replace coordinates to $q=x+y$ and $p=x-y$ and $\partial_y\delta_\beta(\epsilon-|x-y|)=\partial_\epsilon\delta_\beta(\epsilon-|x-y|)\,\text{sign}_\beta(x-y)$:

$$ \begin{align} 2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \\ &=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \end{align} $$

In the limit $\alpha \ll \beta,\epsilon$:

$$ \begin{align} \lim_{\alpha \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dq\, f(\frac{q}{2},-\frac{q}{2})\delta_\beta(\epsilon)\text{sign}_\beta(0) \\ &=0 \end{align} $$

In the limit $\beta \ll \alpha,\epsilon$:

$$ \begin{align} \lim_{\beta \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\left[\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)-\int dq f(\frac{-\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)\right]=\\ &=\partial_\epsilon\left(\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)\right) \end{align} $$

With $F(\epsilon)=\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})$. Now $\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)$ is a product of anti-symmetric function and a symmetric function. So the result is zero.

You could also explore what happens in the limit $\alpha=a\cdot\beta$ for finite $a$. If you stick to Gaussians, as definitions of delta functions, you should be able to evaluate the integrals. I believe you will still get zero.

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    $\begingroup$ Unfortunately, this is not adequate. Equality of the iterated limits does not imply that the double limit exists (this would include the case for which $\alpha =\beta$). Moreover, you need to show that this works for an arbitrary regularization, not just the Gaussian regularization. $\endgroup$ Commented Apr 17 at 12:38
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    $\begingroup$ @MarkViola . What I did above does rely on Gaussian reg. I give it as an example, but the only property I actually use is symmetry. In the end I do mention Gaussian, but only after I finished proving two limits. I would agree that it may not be enough to answer a mathematical question, of course. One would need more information to define this better. If it was a physics problem, I would know what regularisation is appropriate based on physics $\endgroup$ Commented Apr 17 at 13:52
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    $\begingroup$ I would argue that what my answer does is provide one way to approach the problem, provided one can choose a way (or ways) to regularise this $\endgroup$ Commented Apr 17 at 13:56
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    $\begingroup$ @MarkViola, thank you for your answer. If by 'missing', you mean I did not talk about distributions, then please note this was I conscious choose, explained in the first sentence of my post. I am aware of rigourous theory of distributions, but whenever I made the use of delta functions in the wild, they appeared as approximations to physical things, like point-particles. In such settings, the specification of regularisation is really not that difficult. In fact it is a valuable sanity-checking exercise $\endgroup$ Commented Apr 18 at 9:00
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    $\begingroup$ @cyro Well, as you know, equality of the iterated limits does not guarantee existence of the double limit. Perhaps you could address this. But use of regularization is a fine way of approaching this. $\endgroup$ Commented Apr 18 at 15:42
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The inner integral, $$ \int^{+\infty}_{-\infty}dy \, f(x,y) \, \delta(x-y)\,\frac{\partial}{\partial y}\delta(\, \epsilon- |x-y|\,), $$ can be seen as a convolution of $\delta$ and the other factors of the integrand, w.r.t. $y$. Its result is then $$ \left. f(x,y)\,\frac{\partial}{\partial y}\delta(\, \epsilon- |x-y|\,) \right|_{y=x} = f(x,x) \, \delta'(\epsilon) (-\operatorname{sign}(0)) = 0, $$ where the last equality comes from $\delta'(\epsilon)=0$ for $\epsilon\neq 0.$ Then, $$ \int^{+\infty}_{-\infty}dx \, \int^{+\infty}_{-\infty}dy \, f(x,y) \, \delta(x-y)\,\frac{\partial}{\partial y}\delta(\, \epsilon- |x-y|\,) = \int^{+\infty}_{-\infty}dx \, 0 = 0. $$

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    $\begingroup$ This solution lacks mathematical rigor as the Dirac Delta is not a function and the objects represented by integral signs are not integrals. The reason that the result is zero is that the product of singular distributions with disjoint support is zero. See my posted solution herein in which we explain this using sheafs to define the Dirac Delta and its derivative. $\endgroup$ Commented Apr 20 at 21:11

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