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56 votes
Accepted

Is $1/F$ Schwartz if $F$ is "reverse Schwartz"?

If one sets $F(x) := \sum_{n=1}^\infty a_n e^{a_n (x^2-n^2)}$, where $1 < a_1 < a_2 < \dots$ goes to infinity sufficiently quickly, then one can verify that $F$ is reverse Schwartz (all ...
Terry Tao's user avatar
  • 114k
54 votes

A roadmap to Hairer's theory for taming infinities

Let me try to expand a little bit on Ofer's answer, in particular on points 1-3. These functions (or rather distributions in general) are essentially the multilinear functionals of the driving noise ...
Martin Hairer's user avatar
30 votes

A roadmap to Hairer's theory for taming infinities

Let me comment on points 4) and 5). The problem with infinities in QFT or traditional equilibrium statistical field theory is related to the one addressed by Martin's theory but there are some ...
Abdelmalek Abdesselam's user avatar
28 votes
Accepted

Nice applications for Schwartz distributions

Unsurprisingly, the topics that occur to me have various connections to number theory (and related harmonic analysis) (and unclear to me what might have already been done in your course...): EDIT: ...
paul garrett's user avatar
28 votes

Anti-delta function?

Let $\beta\mathbb{R}$ be the Stone-Čech compactification of $\mathbb{R}$: any bounded continuous function $f$ on $\mathbb{R}$ extends uniquely to a (necessarily bounded) continuous function $f^\beta$ ...
Gro-Tsen's user avatar
  • 32.5k
21 votes

A roadmap to Hairer's theory for taming infinities

There are several treatments of Hairer's theory, including lecture notes of his that try to give the "big picture". Brief answers to some of your questions: 1) Those are the solutions to the ...
ofer zeitouni's user avatar
21 votes

How to generalize the various vector calculus theorems to distributions?

This may be borderline for this site, but here goes: What you are looking for may be the notion of weak derivative. A function $f$ defined on some $d$-dimensional set $\Omega$ has a weak partial ...
Dirk's user avatar
  • 12.7k
20 votes

Can distribution theory be developed Riemann-free?

I will only consider the temperate situation (involving the spaces $\mathscr{S}$, $\mathscr{S}'$ and $\mathscr{O}_{\rm M}$) and I will only discuss the first example as well as a "dual" ...
Abdelmalek Abdesselam's user avatar
18 votes
Accepted

Research topics in distribution theory

While I do not know much about current development of the general theory of distributions, I can say something about the current research topics in a special class of distributions, the theory of ...
Piotr Hajlasz's user avatar
15 votes
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Is every continuous endomorphism of the Schwartz space a pseudo-differential operator?

No, the most obvious example is the reflection operator: $Rf(x) = f(-x)$ this is not pseudolocal (in fact the $\xi$-compotent of the wavefront set gets a sign flip). Also the Fourier transform. More ...
mcd's user avatar
  • 591
15 votes

The "Spaces of Schwartz distributions are finite dimensional" challenge

Every short exact sequence $$0\to \mathscr S'\to Y\to \mathscr S'\to 0$$ (where $Y$ is any locally convex space) splits, i.e., the quotient map $Y\to \mathscr S'$ has a continuous linear right inverse....
Jochen Wengenroth's user avatar
15 votes

The "Spaces of Schwartz distributions are finite dimensional" challenge

Like others have pointed out the key, concept is that of nuclear spaces and a good presentation can be found in volume 4 of Gelfand's Generalized Functions. Kolmogorov has introduced a concept ...
Liviu Nicolaescu's user avatar
15 votes

Early successes of Schwartz distribution theory

Following the citation for the 1950 Fields medal I would argue that putting the Dirac delta function on a firm ground was the early success of the theory of distributions.$^\ast$ An extensive list of (...
Carlo Beenakker's user avatar
13 votes

How to generalize the various vector calculus theorems to distributions?

This is an addition to Dirk's answer. It is of course impossible to give an all-embracing answer to your question—there are so many identities. But the general answer is a resounding YES. The ...
burlington's user avatar
13 votes
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Why are distributions "tempered"?

Can someone explain, why in English the name "tempered" wins? Presumably because that’s how the inventor himself translated it (French past participle to English past participle), on e.g. p....
Francois Ziegler's user avatar
12 votes

Nice applications for Schwartz distributions

Two (edit: now four) not-so-usual examples come to my mind: There is the proof of the central limit theorem using Fourier analysis, as done in Chapter 7 of Hörmander's book. It's a cornerstone of ...
Pedro Lauridsen Ribeiro's user avatar
12 votes
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Representation of the Dirac delta function

As usual in such examples, there is no need to integrate against a test function. One can simply use the fact that if a sequence (or net) of distributions converges in the distributional sense, then ...
klempner's user avatar
  • 216
11 votes
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Chain rule in Sobolev space

The main difficulty in the proof of the rule is to prove that $\nabla u=0$ a.e. on the set $u^{-1}(\Sigma)$, where $\Sigma$ is the set where $F$ is not differentiable; and where $\nabla u=0$ one ...
Piero D'Ancona's user avatar
10 votes

Research topics in distribution theory

I think this is a good question. I am no longer working in this field. When I was working in this field, there is no general program and I do not know what are the important open problems. Instead of ...
10 votes

The "Spaces of Schwartz distributions are finite dimensional" challenge

The Schwartz kernel theorem: For a manifold $M$ we have \begin{multline*} \mathcal D'(M\times M) = (\mathcal D(M)\hat{\hat\otimes} \mathcal D(M))' = (\mathcal D(M)\hat\otimes \mathcal D(M))' = L(\...
Peter Michor's user avatar
  • 25.3k
10 votes

About the Fourier transform of the logarithm function

I thought that it might be instructive to present an approach to deriving the Fourier transform of $\log(|x|)$. The result includes a distributional interpretation of $\frac1{|x|}$. Finally, we show ...
Mark Viola's user avatar
10 votes
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Prove that a given distribution is tempered

In Laurent Schwartz's Théorie des Distributions (page 245, chap. VII, §5) you can find something similar: A distribution $T\in \mathscr D'(\mathbb R^d)$ is tempered if and only if all regularizations $...
Jochen Wengenroth's user avatar
10 votes

Early successes of Schwartz distribution theory

This is a very broad topic. If you want a nice little book on the use of distributions in mathematical physics, I suggest this one by Demidov. In the Preface and in Chapter 1, Section 1, the author ...
Alessandro Della Corte's user avatar
10 votes

Anti-delta function?

To elaborate on the comment, I would suggest to take $F(x)=x^{-2}\delta(1/x)$. Let me check for the representation $\delta_{\epsilon}(x)=(2\pi\epsilon)^{-1/2}e^{-x^2/2\epsilon}$, and $F_\epsilon(x)=x^{...
Carlo Beenakker's user avatar
10 votes
Accepted

Propagators and PDEs

You'll find some more info about the fundamental solutions of the wave equation in chapter 5.D of Folland and chapter I.7 of Trèves. The trick you use is the idea that (tempered) distributions, even ...
Daniel Shapero's user avatar
10 votes
Accepted

Non-Schwartz test functions for the explicit formula for L-functions

The class of "weights" $\phi(t)$ for which the explicit formula holds is pretty flexible. The particular choice of weight depends on the problem one studies. On one hand, sums over primes (...
2734364041's user avatar
  • 5,089
9 votes
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No kernel of the form $\lvert x - y\rvert^{-1}$ on tempered distributions?

Yes it is possible. Let $K\in S'(\mathbb{R}^2)$ be the distribution acting on test functions $f(x,y)$ by $$ K(f)=\int_{|x-y|\ge 1} \frac{f(x,y)}{|x-y|}\ dx\ dy \ + \int_{|x-y|< 1} \frac{f(x,y)-f(x,...
Abdelmalek Abdesselam's user avatar
9 votes

Function of moderate growth: history, motivation, and uses

Since you've linked to my question, let me give a first answer. The $M$ in the notation $\mathcal{O}_M$ comes from "multiplication". $\mathcal{O}_M$ is the space of functions that can be multiplied ...
Johannes Hahn's user avatar

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