# Is the F.T of $\operatorname{Span}(\mathscr S(\mathbb R^2)\otimes\mathscr D(D))$, $D\subset\mathbb R^2$ dense in $L^2(\mathbb R^4)$?

Let $K$ be the real vector space generated by elements $f$ in $\mathscr S(\mathbb R^2,\mathbb R)\otimes\mathscr D(D, \mathbb R)$, where $D$ is any bounded subset of $\mathbb R^2$. Let $\hat K$ be the vector space generated by the Fourier transform of each $f\in K$, i.e. $\hat K = \{\hat f\ |\ f\in K\}$. Is $\hat K+i\hat K$ dense in $L^2(\mathbb R^4)$?

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Could you please clarify the notation? One of the spaces is Schwartz space, I presume; what's the other one? – Yemon Choi Jun 27 '11 at 9:18
I think the second factor is test functions with support in D. I am not sure if D is supposed to be fixed or arbitrary. Either way, this question has the flavor of homework. – Michael Renardy Jun 27 '11 at 10:01
I guess $\mathscr D$ are smooth functions with compact support. If $L^2(\RR^4)$ is the complex space i guess this is not true, because the Furiertransform of some real function has some symmetry, and otherwise the Fourier transform does not make sense to me. – Marcel Bischoff Jun 27 '11 at 20:11
Probably this question is not approriate for here. But if I know understand your notation correctly, if you take a function that has support outside of D then the Fourier transform is orthogonal to your space, or not? So it would not be dense. The situation is different if you restrict the fourier transform to something lying in some "cone". Then you end up to some situation similar to the Reeh-Schlieder Theorem in Quantum Field Theory. I recommened you to look into the books of Reed and Simon. – Marcel Bischoff Jun 28 '11 at 10:56
"Probably this question is not appropriate for here"... actually I don't see why. At most, one may ask the OP to provide some motivation and details, as explained in the section "how to ask". – Pietro Majer Jul 27 '11 at 11:30

## 1 Answer

I'm sorry if this is not the appropriate place to ask such questions. I'll be more careful next time. By the way my question is strongly related to an actual problem in QFT (actually QFT on QST). I'm not sure about M. Bischoff conclusion on orthogonality pointed out in one of the comments to the question.

There is a result from Araki that state that the real Hilbert subspace of the one-particle Hilbert space $K(O)=\{\hat f\big\vert_{\Omega_m^+}\ |\ f\in\mathscr D(O,\mathbb R)\}$, where $\Omega_m^+$ is the hyperboloid of mass $m$ in the future light-cone and $O\subset\mathbb R^4$ is a non-empty simply connected bounded open subset of $\mathbb R^4$, is standard, i.e. $\overline{K(O)+iK(O)}$ is dense and $K\cap iK=\{0\}$. The vector space $\hat K$ I defined in the question ought to contain such a vector space $K(O)$ associated with a region $O\subset\mathbb R^2\times D$, where $D$ satisfies some suitable regularity condition (e.g. regular boudary, simply connectedness,...) and therefore $K(O)\subset \hat K$, that would imply $\hat K$ standard, according to the result from Araki.

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But the important fact is the restriction to the mass shell, because then the "two-point function" (which is essentially the scalar product written is tempered distribution) is an analytic function in some tube region, because the Fourier transform has support in some cone. Then using the Edge of the Wedge theorem one can show that a vector orthogonal to your set is already the zero vector. This can be found eg. in Streater-Wightmann etc under Reeh-Schlieder theorem (the original article is german). If you do not restrict I think my argument says it will be not dense anymore. – Marcel Bischoff Jul 4 '11 at 19:41