Regarding a non-standard definition of tempered distributions in Schuller's lecture

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.

• An authoritative reference towards the definition/theorem in the way Schuller presents his definition would be highly appreciated as well. Aug 22, 2023 at 19:12
• If you have a single function you can make it into a family by setting all the other functions to zero! Aug 23, 2023 at 7:33
• @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. Aug 23, 2023 at 11:13

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...)
• 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. Aug 23, 2023 at 10:52
• 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? Aug 23, 2023 at 11:27
• 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. Aug 23, 2023 at 11:55
• 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. Aug 23, 2023 at 12:07