Is Schwartz space canonical in any sense?

Question

Schwartz space $\mathcal S(\mathbb R^n, \mathbb C)$ is a dense subspace of $L^2(\mathbb R^n,\mathbb C)$ that appears frequently in physics, because the position operators $X_i$ and momentum operators $P_i$, defined by $(X_if)(x) = x_if(x)$ and $P_if=-i (\partial/\partial x_i)f$ are defined and well behaved on it.

However, there are many other spaces of test functions that could be used (and sometimes are used) in this context, and elsewhere in functional analysis. Still the Schwartz functions seem to appear the most often, so I wonder if they are "natural" or "canonical" in any sense. Are there any natural characterizations of $\pmb{\mathcal S(\mathbb R^n, \mathbb C)}$? For example, something like "Schwartz space is the unique subspace of $L^2$ such that ..." or "Schwarz space is the largest subspace of $L^2$ such that ...".

I think I once saw a such a characterization in terms of the position and momentum operators above, along the lines of "the largest subspace such that $X_i$ and $P_i$ are simultaneously defined," or something like that. But I cannot find it now, and I may be misremembering completely.

Answer 1

Yes, there are two such properties (in fact, they might be equivalent, but I'm not 100% sure of that)

• The Schwartz space is the "canonical" space for Fourier transform

Fourier Transforms are defined over the space of all integrable functions. However, it is not (in the function-sens) bijective; meaning some functions do not have an inverse Fourier transform. However, like for any operation you can restrict the Fourier Transform to a subspace over which it is a bijection, and the biggest such space is $\mathcal{S}$.

• The dual of $\mathcal{S}$ is isomorphic to $\mathcal{S}$. (Dual as in the space of linear forms.)

In deed, you can show that when studying integrable functions the dual of the dual is isomorphic to the original space. This is what allows to construct distribution theory (dual of test function space). Some subspaces of integrable functions (like $L_1$ or test functions, or $L_2$) have duals that are either "bigger" or "smaller" than themselves. Schwartz is the biggest space whose dual is basically itself.

Finally (to get back to physics) a great property of Schwarz space is also that you can use Stokes formula to the infinite limit: $$\int_\Omega \partial f = \int_{\partial \Omega} f = 0$$ And like you mentioned, it is dense in $L_2$. This allows (you can make it rigorous by a lot of painful proofs, or just assume it "physics' style") to extend many properties of Schwartz (especially the previous one) to $L_2$ (the actual physical space) by considering a sequence of functions of $\mathcal{S}$ converging towards something in $L_2$.

So sometimes, in physics, we assume being in Schwarz wile in fact not being in it, simply because, in the limit, it doesn't really matter, so we might as well for shortening the proofs.

Answer 2

I don't know about a canonical definition of Schwartz space $\mathcal{S}(\mathbb{R}^d)$, and I in fact asked a similar question on MO

https://mathoverflow.net/questions/217663/uniform-definition-of-s-mathbbr-and-s-mathbbq-p

There is however a canonical definition of the dual space $\mathcal{S}'(\mathbb{R}^d)$ which is Schwartz's original definition of the notion of "spherical distributions". Use the stereographic projection to see $\mathbb{R}^d$ as the $d$-dimensional sphere $\mathbb{S}^d$ minus the point at infinity. A distribution on the compact smooth manifold $\mathbb{S}^d$ restricts on the open subset $\mathbb{R}^d\subset\mathbb{S}^d$ to a distribution in $\mathcal{D}'(\mathbb{R}^d)$. Now one may ask which distributions are obtained in this way and the answer is exactly the temperate distributions, i.e., the elements of $\mathcal{S}'(\mathbb{R}^d)$. I don't if from this characterization as a set there is a canonical way to obtain the LCTVS topology which makes this into a complete space, i.e., a way to recover the canonical strong dual topology of $\mathcal{S}'(\mathbb{R}^d)$. If this is done, then one can take the strong dual of $\mathcal{S}'(\mathbb{R}^d)$ and recover Schwartz space at least as an abstract LCTVS.

Answer 3

Mathematically, I think the main reason $\mathcal S(\mathbb R^n)$ is natural to use is that it is a space which is densely contained in many interesting function spaces, while at the same time is left invariant by the Fourier transform. If the Fourier transform mapped $\mathcal D(\mathbb R^n)$ to itself, then $\mathcal D(\mathbb R^n)$ would be used instead of $\mathcal S(\mathbb R^n)$, but it is of course not the case.

So it is often much more natural and convenient to use $\mathcal S(\mathbb R^n)$ if you work with the Fourier transform. But of course it does not mean that you use it all the time. As you said, different spaces are used depending on the setting and the applications.

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Edit 1: I was wondering if $\mathcal S(\mathbb R^n)$ is the smallest (if not the only…?) space which is a translation and dilation invariant, dense, proper subset of $L^2$ and is left invariant by the Fourier transform… Of course this is not true, one could simply take $\mathcal D(\mathbb R^n)+\mathcal F(\mathcal D(\mathbb R^n))$, but maybe adding some more hypotheses it becomes true.

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Edit 2 (the actual answer): What about

• “$\mathcal S(\mathbb R^n)$ is the largest subspace of $L^2$ which is invariant by the Fourier transform and by the operators $\partial_i$” This is true almost by definition of $\mathcal S(\mathbb R^n)$, and it is a nice characterization. Some equivalent characterizations are
• “$\mathcal S(\mathbb R^n)$ is the largest subspace of $L^2$ which is invariant by the operators $X_i$, $P_i$”
• “$\mathcal S(\mathbb R^n)$ is the largest subspace of $L^2$ which is invariant by the Fourier transform and by the operators $X_i$, $P_i$”.

Rough explanation. The Schwartz space consists of all functions $f$ such that $X^\alpha P^\beta f\in L^\infty$ for all multiindices $\alpha,\beta$ (rephrasing the mathematical definition with physics notation). It is known (although I don’t remember the details of the proof) that one can equivalently substitute $L^\infty$ with $L^2$, and the two definitions coincide. This way (with some straightforward adjustments) we prove the second characterization. The first and third one are equivalent to the second one essentially because $P_i$ is the coniugation of $X_i$ via the Fourier transform, and viceversa (e.g., $X_i=\mathcal F P_i\mathcal F$ up to reflection and multiplication by constant).

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Edit 3: Another interesting one would be “$\mathcal S(\mathbb R^n)$ is the largest subspace of $L^2$ of smooth functions which is left invariant by the Fourier transform”, but maybe this one is false.

Answer 4

I haven't seen this remarked in the other answers yet, even though to me it seems the most natural: In QM algebras of operators are whats crucial. If the operators are unbounded one has a problem that the composition of operators is not necessarily well defined. Take the following definition of a representation of a $*$-algebra by possibly unbounded operators:

Def. Let $A$ be a $*$-algebra, $H$ a Hilbert space. A representation of $A$ on $H$ is a pair $(D_\rho, \rho)$ where$D_\rho\subseteq H$ is a dense subspace and $\rho: A\to \mathrm{Lin}(D_\rho, H)$ is linear with $\rho(a)(D_\rho)\subseteq D_\rho$ for all $a\in A$ and $\rho(ab)=\rho(a)\rho(b)$ as well as $\rho(a^*)=\rho(a)^*\lvert_{D_\rho}$.

Lets say a representation $(D_\rho, \rho)$ is maximal if there is no representation $(D_{\rho'},\rho')$ with $D_{\rho}\subsetneq D_{\rho'}$ and $\rho'(a)\lvert_{D_\rho}=\rho(a)$.

$\mathcal S(\Bbb R)$ is then the maximal domain of the standard representation of the Heisenberg algebra, being the intersection of the domains $D((1+x^2)^n(i\partial_x)^m)$.