Making sense of the limit $\lim\limits_{x \to y} T(x,y)$ for a tempered distribution $T$ on $\mathbb{R}^{2n}$

I already posted a similar question on MO and looked into the references therein.

However, I cannot find a satisfactory answer for my question..So I ask here again in a more refined form.

Let $$T \in \mathcal{S}'(\mathbb{R}^{2n})$$ be a tempered distribution. For $$x, y \in \mathbb{R}^n$$, we may write as $$T(x,y)$$ and understand that $$$$T(f) = \int_{\mathbb{R}^n \times \mathbb{R}^n } T(x,y)f(x,y)d^nxd^ny$$$$ for $$f \in \mathcal{S}(\mathbb{R}^{2n})$$.

Now, in the context of defining the Wick products, I frequently run into limits of the form $$$$\lim\limits_{x \to y} T(x,y)$$$$ I do not clearly see how to make sense of such expressions...

Perhaps, does it mean $$$$\lim\limits_{g \to h} T\bigl( g \otimes h)= \lim\limits_{g \to h}\int_{\mathbb{R}^n \times \mathbb{R}^n } T(x,y)g(x)h(y)d^nxd^ny$$$$ for each $$h \in \mathcal{S}(\mathbb{R}^{n})$$, where the limit $$g \to h$$ is in the Frechet topology of $$\mathcal{S}(\mathbb{R}^{n})$$?

If so, I see that $$$$\lim\limits_{x \to y} \delta^n(x-y)$$$$ must be characterized by the expression $$$$\lim\limits_{g \to h}\int_{\mathbb{R}^n \times \mathbb{R}^n } \delta^n(x-y)g(x)h(y)d^nxd^ny = \lim\limits_{g \to h}\int_{\mathbb{R}^n } g(x)h(x)d^nx =\int_{\mathbb{R}^n } h^2(x)d^nx$$$$ for each $$h \in \mathcal{S}(\mathbb{R}^{n})$$.

However, $$h \to \int_{\mathbb{R}^n } h^2(x)d^nx$$ does NOT define a linear functional..so I am confused..

• I am missing your point; in the equation starting as $\lim_{g\rightarrow h}$ you are not taking the limit $x\rightarrow y$ of $\delta(x-y)$, which would be singular; this equation requires no regularization. Commented Dec 8, 2023 at 12:44
• @CarloBeenakker I am asking for the formal mathematical definition of such limits for general tempered distributions. So, my guess using $g \to h$ is correct? Commented Dec 8, 2023 at 13:05
• @CarloBeenakker And for the case of $\delta(x-y)$, the limit exists in the language of $g \to h$. You are saying that it does not define a tempered distributions nevertheless? Commented Dec 8, 2023 at 13:07
• I think you are completely justified in feeling uneasy about this. In general, distributions do not restrict well to submanifolds. The key issue is about wave-front sets... also for attempting to multiply distributions. What is your version of "Wick products", if you don't mind? Commented Dec 8, 2023 at 20:16
• @Isaac, ah, yes, this stuff is rough to justify, and many sources give glib-and-fake explanations. So far as I know, at some point in QFT, things cannot quite be made "squeaky clean" :) Much of what I know about this is from G. Folland's book about it... where he does make fairly clear the bad things happening at the edges. Beyond that, I would have to say that I've not learned enough "modern analysis" to have any hope of rigorizing QFT... though, yes, that is a very interesting issue! :) Commented Dec 9, 2023 at 0:11

$$\newcommand\R{\mathbb R}\newcommand\S{\mathcal S}\newcommand\ep{\varepsilon}\newcommand\th{\theta}\newcommand\de{\delta}\newcommand\De{\Delta}$$Take any $$\tau\in\S(\R^n\times\R^n)$$. Then for any $$\phi\in\S(\R^n)$$ and any mollifier $$\psi$$ on $$\R^n$$ and any real $$\ep>0$$ we have $$\int_{\R^n\times\R^n}dx\,dy\,\tau(x,y)\phi\Big(\frac{x+y}2\Big) \frac1{\ep^n}\,\psi\Big(\frac{x-y}\ep\Big) \\ =\int_{\R^n\times\R^n}du\,dv\,\tau\Big(u+\frac v2,u-\frac v2\Big) \phi(u)\frac1{\ep^n}\,\psi\Big(\frac v\ep\Big) \\ \underset{\ep\downarrow0}\longrightarrow\int_{\R^n}du\,\tau(u,u)\phi(u) =\int_{\R^n}dy\,\lim_{x\to y}\tau(x,y)\phi(y).$$ So, for any $$T\in\S'(\R^n\times\R^n)$$ it makes sense to denote the tempered distribution in $$\S'(\R^n)$$ in question by $$T_2$$ and define it by the condition $$T_2(\phi):=\lim_{\ep\downarrow0}T(\th_{\phi;\psi;\ep}) \tag{1}\label{1}$$ for all $$\phi\in\S(\R^n)$$ and some (or all) mollifiers $$\psi$$ on $$\R^n$$, where $$\th_{\phi;\psi;\ep}(x,y):=\phi\Big(\frac{x+y}2\Big) \frac1{\ep^n}\,\psi\Big(\frac{x-y}\ep\Big)$$ for $$(x,y)\in\R^n\times\R^n$$, provided that the limit in \eqref{1} exists for such $$\phi$$ and $$\psi$$. In particular, this limit will exist if $$T$$ is determined in the standard manner by a locally integrable function bounded by a polynomial.

The limit in \eqref{1} does not have to exist in general. E.g., consider the tempered distribution $$\De\in\S'(\R^n\times\R^n)$$ defined by the formula $$\De(\th):=\int_{\R^n}dx\,\th(x,x)$$ for $$\th\in\S(\R^n\times\R^n)$$; I guess this $$\De$$ is what you might mean by "$$\de(x-y)$$": informally, $$\int_{\R^n\times\R^n}dx\,dy\,\de(x-y)\th(x,y) =\int_{\R^n}dx\,\int_{\R^n}dy\,\de(x-y)\th(x,y) =\int_{\R^n}dx\,\th(x,x).$$ Then $$\De(\th_{\phi;\psi;\ep})=\int_{\R^n}dx\,\phi(x) \frac1{\ep^n}\,\psi(0)$$ does not converge to a finite limit as $$\ep\downarrow0$$ if $$\int_{\R^n}dx\,\phi(x) \psi(0)\ne0$$; that is, then the limit in \eqref{1} with $$\De$$ in place of $$T$$ does not exist.

• @Isaac : I have added a paragraph on possible nonexistence of the limit in (1). Commented Dec 8, 2023 at 15:45
• @Isaac : This is a good question, to which I don't have an answer at the moment. You may want to post it separately. Commented Dec 9, 2023 at 22:41
• @Isaac : This is another good question, to which I do not have an answer at the moment. Commented Dec 23, 2023 at 22:46
• @Isaac : In this case, you can similarly use $\phi(\frac{x+y+z}2)$, $\psi(\frac{x-y}\varepsilon)$, $\theta(\frac{y-z}\varepsilon)$ instead of $\phi(\frac{x+y}2)$, $\psi(\frac{x-y}\varepsilon)$. Commented Dec 26, 2023 at 3:54
• @Isaac : The coefficient there does not really matter. Commented Dec 26, 2023 at 22:14

This a comment (really warning) but too long. Firstly, the good news: if we consider the two dimensional distribution $$\delta(x-y)$$, this is perfectly well-defined on the plane as is $$\delta (x_0-y)$$ (a distribution on the line). It is the delta distribution with singularity at the fixed value $$x_0$$ and this is the same as the restriction of the original distribution to the line $$x=x_0$$ (I am assuming $$n=1$$ for simplicity). Then the limit exists (distributional convergence) and is the one dimensional delta distribution with singularity at the origin.

For a rigorous approach to the general situation (I am hoping you are looking for this and not the formal manipulations without recourse to definitions, which are the usual bill of fare on this site), you need two concepts, that of the restriction of a distribution to a subspace and that of convergence of a parametrised family of distributions. Both of these are considered in some detail in the literature (I can supply references on request). As one would expect, sometimes you can carry out these operations, sometimes not--it depends on the distribution you start with. As we saw above, the Dirac distribution works fine but one would have to know your candidate explicitly in order to give a precise argument.

Your toy example is computed as follows: let $$f(x,y)$$ be the characteristic function of $$y>x$$. Then it converges (e.g. on the local $$L^1$$ and hence distributionally) to the Heaviside functions $$H(y)$$. Now differentiate both side w.r.t. $$x$$ and exchange differentiation with taking the limit as every freshman would like to do. Not allowed in an Analysis course but quite legitimate in the distributional sense--consult the theory developed in the reference I gave.

This often works in concrete examples--take primitives until you get a limit which works in some classical sense, then use the above trick.

• Yes, I am looking for a rigorous definition. Could you provide any relevant references? And if possible, could you briefly tell me how the general framework applies for $\lim\limits_{x \to y} \delta(x-y)$? Commented Dec 8, 2023 at 14:56
• You will find everything you require in the text "Theory of Distributions" by José Sebastião e Silva, available online at his memorial site jss100.campus.ciencias.ulisboa.pt, all at the level of a freshman analysis course, without functional analysis. Commented Dec 8, 2023 at 15:10
• The two steps I outlined above for the Dirac function are perfectly obvious intuitively and can easily be made rigorous using the elementary definitions which you can find in the text I mentioned above. Commented Dec 8, 2023 at 15:15
• I looked it up but it is written in French or Spanish mostly.. Could you tell me which page I should go to? Commented Dec 9, 2023 at 16:34
• You specifically want the limit concept which is presented in Def. 8.1.6 on p. 157 (chapter 8). Commented Dec 9, 2023 at 19:33