2
$\begingroup$

Is the below definition of tempered distributions correct? $\newcommand{\rr}{\mathbb{R}} \newcommand{\cc}{\mathbb{C}} \newcommand{\nn}{\mathbb{N}_0} \newcommand{\schwartz}{\mathcal{S}} \newcommand{\schwartzr}{\schwartz(\rr)} \newcommand{\dist}{\schwartz^\times} \newcommand{\distr}{\dist(\rr)} \newcommand{\distar}{\schwartz^*(\rr)} \renewcommand{\i}{\mathrm{i}} \newcommand{\e}{\mathrm{e}} \newcommand{\cinfinity}{\mathrm{C}^\infty} \newcommand{\family}{\mathcal{F}} \newcommand{\hilbert}{\mathcal{H}} \newcommand{\der}{\operatorname{d\!}{}}$

The space of tempered distributions $\distr$ is defined as follows. Let $\schwartzr$ denote the Schwartz space and $\distar$ denote the algebraic dual of $\schwartzr$. Let $\family$ denote a set of piecewise(?) continuous, polynomially bounded functions from $\rr$ to $\cc$. It is not required that elements of $\family$ be linear. We say that a $\Phi \in \distar$ is a tempered distribution if there exists such a family $\family = \{\Phi_m \mid m \in \nn\}$, such that for each $f\in \schwartzr$, $$\Phi[f] = \int_{\rr} \der x \sum_{m \in \nn} \Phi_m (x) \cdot f^{(m)} (x).$$

I am following a lecture [1] and the lecturer provides the above definition. Then, he attempts to show that $L^2(\mathbb{R})$ is embedded in $\distr$ by identifying an element $\phi \in L^2(\mathbb{R})$ with $\Phi_\phi \in \distr$ as follows. $$\Phi_\phi [f] \:= \int_{\rr} \phi(x) \cdot f(x) \der x,$$ so that $(\Phi_\phi)_0(x) = \phi(x)$ and $(\Phi_\phi)_m(x) = 0$ for all other $m \in \nn$. But $\phi \in L^2(\mathbb{R})$ may not be continuous. So, he conjectures that we might need the $\Phi_m$'s to be only piecewise continuous, instead of continuous. He says that he shall tell the fix in the next lecture, but I was not able to find anything related to this in the next lec.


Such a definition seems non-standard. Most of the sources I have seen define a tempered distribution as a continuous linear functional on the Schwartz space. I need to confirm or correct the above definition.

I was pointed towards a regularity theorem for distributions in Reed and Simon [2, Theorem V.10, p. 139]. If $\Phi \in \schwartzr$, then $\Phi = D^m g$ (weak derivative) for some polynomially bounded continuous function $g$ and some $\beta \in \nn$, i.e. $$\Phi[f] = \int (-1)^m g(x) (D^m f)(x) dx$$ for all $f \in \schwartzr$.

But this theorem tells that atleast one such polynomially bounded, continuous function exists, not a family of functions. How does one create a family out of this one function? I feel like what is required here is somehow decomposing the $g$ into "basis" of functions with first derivative, second derivative, and so on until the $m$-th derivative.

[1] https://youtu.be/FNJOyxOp3Ik. Defines tempred distributions at around 33:15. Discusses embeddings at around 1:00:01. (German) Eight lecture (Aufgetakelte Hilberträume or Rigged Hilbert Spaces) of Frederic Schuller in the Theoretische Quantenmechanik series. I used YouTube's subtitle auto-translate tool.

[2] Michael Reed and Barry Simon. Methods of Modern Mathematical Analysis I: Functional Analysis. Revised and enlarged editon. Academic Press, Inc., 1980. isbn: 978-0-080-57048-8.


I think problem 24 on p. 176, chp. 5 of Reed and Simon is essentially what is needed. The problem says that let $\Phi \in \schwartzr$ with $|\Phi(f)| \leq C\sum_{\alpha, \beta = 0}^n ||x^{\alpha} (d/dx)^\beta f||_\infty$. Then $$\Phi[f] = \sum_{\beta = 0}^{n} \int D^{\beta} f \,d\mu_\beta$$ where $\mu_0, \ldots, \mu_n$ are measures of polynomial growth.

Now, since $f \in \schwartzr$, $\sup_{x \in \mathbb{R}} |x^\alpha \cdot f^{(\beta)}(x)| < \infty$ implies that $|\Phi(f)| \leq C\sum_{\alpha, \beta = 0}^n ||x^{\alpha} (d/dx)^\beta f||_\infty$ for all $f$. I think this is said as every tempered distribution is of a finite order. Please tell me if my understanding of this problem is correct.

$\endgroup$
3
  • $\begingroup$ An authoritative reference towards the definition/theorem in the way Schuller presents his definition would be highly appreciated as well. $\endgroup$ Aug 22, 2023 at 19:12
  • $\begingroup$ If you have a single function you can make it into a family by setting all the other functions to zero! $\endgroup$
    – PhoemueX
    Aug 23, 2023 at 7:33
  • $\begingroup$ @PhoemueX Yes. That is indeed the case for two examples Schuller gives --- the Dirac distribution and plane waves. But there must a reason for him talking about a family, right? Otherwise he would have simply said a function. $\endgroup$ Aug 23, 2023 at 11:13

1 Answer 1

1
$\begingroup$

I reckon this is equivalent to the usual definition in terms of the topology on $\mathcal S(\Bbb R)$ via the structure theorem for distributions, here stated in the case of a generic distribution (you can find its proof in Rudin, Functional Analysis (1991), Theorem 6.28):

Theorem. If $u \in \mathscr D'(\Bbb R)$, then for all $m\in \Bbb N$ there exists $g_m \in C^0(\Bbb R)$ such that every compact subset $K\Subset \Bbb R$ intersects the support of $g_m$ for only finitely many indices $m$, and $$u \equiv \sum_m \frac{d^m g_m}{dx^m}. $$

Here, the derivative is intended in the distributional sense, so it is meant to be discharged onto the test function as in your definition. Of course, this is only a local theorem, given the nature of objects in $\mathscr D'(\Bbb R)$; for tempered distributions, the situation improves due to their subexponential growth at infinity (cf. this quick summary, top of last slide). (I do not understand the remarks concerning piecewise continuity – after all, every piecewise continuous function can be written as the derivative of a continuous function...)

$\endgroup$
5
  • $\begingroup$ Isn't the structure theorem on the top of the last slide the same as the regularity theorem in Reed--Simon? As you say, the theorem in Rudin is local; it involves the compact subset $K$. I fully understand how the situation improves for tempered distributions, in the sense that we retrieve Schuller's definition. $\endgroup$ Aug 23, 2023 at 10:52
  • $\begingroup$ I found out that problem 24, p. 176, chp. 5, of Reed--Simon, vol. 1 might be related to what I am asking, but I am not sure. imgur.com/QEKLVKo $\endgroup$ Aug 23, 2023 at 11:00
  • $\begingroup$ Am I right in concluding that if every tempered distribution satisfies $|T(f)| \leq C \sum_{\alpha, \beta = 0}^n ||x^{\alpha} (d/dx)^{\beta} f||_\infty$, then Schuller's definition is correct. That is essentially problem 24, right? $\endgroup$ Aug 23, 2023 at 11:27
  • $\begingroup$ It seems to me that the Schwartz representation theorem as given on p. 5 of math.mit.edu/~rbm/iml/Chapter1.pdf is what is needed, but there the $u_{\alpha, \beta}$ need not be polynomially bounded. $\endgroup$ Aug 23, 2023 at 11:55
  • $\begingroup$ And the condition on $|T(f)|$ is satisfied since $\sup_{x \in \mathbb{R}} |x^\alpha \cdot f^{(\beta)}(x)|$, as $f$ belongs to the Schwartz space. $\endgroup$ Aug 23, 2023 at 12:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .