For a tempered distribution $F$ on $\mathbb{R}^2$, what exactly does it mean by $\lvert F(x,y) \rvert \leq \lvert x-y \rvert^{-n}$?

Question

Let $F$ be a tempered distribution on $\mathbb{R}^2$ and $n \in \mathbb{N}$ be a fixed natural number. I wonder what exactly it means by

$\lvert F(x,y) \rvert \leq \lvert x-y \rvert^{-n}$ where $x,y \in \mathbb{R}$.

I run into such notations when dealing with a sequence of tempered distributions and their integral representation.

By definition, $F$ is a continuous linear functional on the Schwartz space $\mathcal{S}(\mathbb{R}^2)$. So, I guess the above bound should be expressed in terms of Schwartz functions.

Could anyone please clarify for me?

Add : A reference context is this paper, with p.211 Eqn (6) in particular.

Add 2 : For those who down-voted it, could you please tell me what you don't like about this problem, for the sake of being constructive?

Answer 1

I'll try to explain what Igor meant in his comments in a different way, maybe it helps.

1. Of course, any tempered distribution is a distribution in the broader sense - more precisely, any compactly supported test function is tempered, and the seminorms defining the topology of $\mathscr{S}(\mathbb{R}^n)$ are majorized by seminorms defining the topology of $\mathscr{C}^\infty(K)$ for all $K\subset\mathbb{R}^n$ compact with nonvoid interior.

2. Any distribution $u$ in $\mathbb{R}^n$ (tempered or not, by 1.) may be restricted to a nonvoid open subset $U$ thereof by restricting to test functions supported in $U$. The result is, of course, a distribution $u|_U$ in $U$ (which is no longer tempered because temperedness only has meaning if $U=\mathbb{R}^n$), for the needed seminorm bounds guaranteeing the continuity of $u|_U$ are clearly inherited from those of $u$.

3. A basic theorem in distribution theory tells us that if two continuous functions $f,g$ on $U$ as in 2. are such that $$ \int_U f\phi\,dx=\int_U g\phi\,dx$$ for all test functions $\phi$ compactly supported in $U$, then $f=g$. This means that if $f$ is a continuous function on $U$ such that $$u|_U(\phi)=u(\phi)=\int_U f\phi\,dx$$ for all such test functions, then $f$ is the only continuous function on $U$ with such a property, and therefore we may identify $u|_U$ with (the smearing of compactly supported test functions on $U$ with) $f$. More generally, this can be done with $f$ any locally integrable function on $U$, but then the identification is only up to a zero-measure subset of $U$ because of the integral in the rhs. This freedom disappears if $f$ is required to be continuous. In the case of $n$-point Schwinger functions on $\mathbb{R}^d$, $$U=\mathbb{R}^{nd}\smallsetminus\{(x_1,\cdots,x_n)\in\mathbb{R}^{nd}\ |\ x_i\in\mathbb{R}^d\ ,i=1,\ldots,n\ ,\,x_j=x_k\text{ for some }j\neq k\}$$ and $f$ is a real analytic (hence continuous) function, so $f$ is the unique such function.

1.-3. together explain what your inequality means - to wit, it implicitly states that the restriction $F|_U$ of $F$ to the open subset $U=\mathbb{R}^2\smallsetminus\{(x,x)\ |\ x\in\mathbb{R}\}$ is identified with (the smearing of compactly supported test functions on $U$ with) a(n at least locally integrable) function $f$ on $U$ which then satisfies the given inequality, since the latter's rhs is clearly only defined in $U$. In the context of the paper, $f$ is real analytic and therefore continuous as in the end of 3., so $f$ is unique and thus the given inequality has a clear, unambiguous meaning.

Hope it helped somehow to understand Igor's comments. Sorry if my answer sounds pedantic at places, but I wanted to be sure not to leave any stone unturned.

Answer 2

Distributions and ordinary functions are two different things, however, there are bridges between them, going both ways, modulo some technical hypotheses. If $f$ is an ordinary function one can associate to it a distribution that sends a smooth test function g to $\int fg$, i.e., by "integrating against $f$" as mentioned in the article. For this to make sense, i.e., for the integral to converge and the resulting linear form to be continuous, the needed technical hypothesis is that $f$ is locally $L^1$. Conversely, given a distribution $T$ on $\Omega$, it has a singular support $K$, which allows one to produce a smooth function $f$ by "restriction" to $\Omega\backslash K$. Namely, that means that if $g$ is a test function with support in $\Omega\backslash K$, then $T(g)=\int_{\Omega\backslash K} fg$.

Traditionally, Schwinger functions are defined as ordinary functions with domain given by the complement of the big diagonal, i.e., for non-coinciding points. Then, one can ask if one can associate a distribution to that. This is a bit tricky because this does not use the standard spaces of distributions say PDE folks work with.

Answer 3

This is just a comment (but I am not entitled) to put on record that there is a simple and natural way to answer the question in the title (the meaning of the claim $|F(x,y)|\leq |x-y|^{-n}$ for any (tempered) distribution $F$ of two variables). First note that any bounded, measurable function defines a unique distribution (true even for a locally integrable function). Secondly, one can always multiply a distribution by a smooth function. Hence it is natural to say that an (arbitrary) distribution $F$ satisfies the inequality whenever the distribution $F(x,y)(x-y)^n$ corresponds to a function in the ball of $L^\infty$ in the above sense.

I would emphasise that this is a reply to the question of how to ascribe meaning to the statement in the title, independent of any explicit context.