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Hello,

I am stuck with the following (hopefully not too trivial) problem. I want to know, if the map $${\cal D}(\mathbb{R}^2)\to L^2(H_m,d\Omega_m)\qquad f \mapsto \hat{f}|_{H_m}$$ has dense range.

Here $H_m$ is the "upper mass shell" $\{ p\in \mathbb{R}^2:p_0>0, p^2=m \}$ in the 2-dimensional Minkowski space, $d\Omega_m=dp_1/\sqrt{p_1^2+m^2}$ the Lorentz invariant measure, $\hat{f}$ is the Fourier Transform.

In Reed-Simon II, Chapter X, Exercise 44 one has to show that the Schwartz functions have dense range. I think I have proved this, but I don't know how to extend it, to $\mathcal{D}$, if this is possible at all.

For the Schwartz functions I would show, that there are functions s.t. $\hat{f}, p_1\hat{f},\dots,p_1^n\hat{f},\cdots$ is a complete system in $L^2(H_m,d\Omega_m)$. But for these functions I need $| \hat{f}(\sqrt{p_1^2 + m^2}, p_1) | \leq c \exp(-a |p_1|)$, with $c,a>0$ some constants.

Is there any function in $\mathcal{D}$ that fullfills this? Is there another way to prove this?

Any suggestions are very welcome.

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Well, I found this Blog blogs.ethz.ch/kowalski/2009/11/02/vade-retro-test-function which suggests I will not find a compactly supported function, whose fourier transform will decay exponentially, am I right? So, maybe there is another way to prove density or maybe the fourier transforms of compactly supported functions restricted to the mass shell is not dense in the L^2 functions on the mass-shell, which would surprise me. –  Jan S Aug 9 '10 at 13:22
    
Isn't this even true if one takes smooth functions with support in some fixed open region $O$? I mean this is basically the Reeh-Schlieder theorem applied for the "one-particle space". –  Marcel Bischoff Sep 8 '11 at 7:36
    
@Marcel: Thanks for your interest in this problem. Actually I encountered it, when proving the Reeh-Schlieder property, i.e. cyclicity of the vacuum for local algebras. Following Baumgärtel's book (Operatoralgebraic methods in quantum field theory) one can show that local algebras have this property, if the global algebra has it. That is in one-particle space language $E(D(O))$ is dense if $E(D(R^2))$ is. That is why I still need to show $E(D(R^2))$ is dense. Edit: Spelling and clarity. –  Jan S Sep 9 '11 at 8:56
    
@Jan: that $E(D(\mathbb R^2))$ $\Longrightarrow$ $E(D(O))$ is interesting. –  Marcel Bischoff Sep 12 '11 at 10:38
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3 Answers

up vote 1 down vote accepted

I don't have any book in front of me at the moment, so I am a bit improvising here, but you'll find this in eg. Reed Simon 2, Streater Wightman, or Josts book.

Let us call the map $$E:{\cal D}(\mathbb{R}^2)\to H=L^2(H_m,d\Omega_m)\qquad f \mapsto \hat{f}|_{H_m}.$$ I claim that already $E(\mathcal D(O))$ with $O$, let's say, compact with non-trivial interior is dense in the Hilbert space, where $\mathcal D(O)$ are the smooth functions with support in $O$. Let us take a vector $v\in H$ orthogonal to $\mathcal D(O)$, i.e. $(v,E(\mathcal D(O)))=0$. We have to show that $v=0$ for which it is actually enough to show that $(v,E(\mathcal S(\mathbb R))=0$, because the Schwartz functions are dense. Now let us define the map $$ T:\mathcal S(\mathbb R^2) \to H: f \mapsto T(f):=(v,E(f)) $$ which turns out to be continous, i.e. a tempered distribution. We have to show that $T=0$.

The Fourier transform of this distribution has support on the forward light cone (even on the mass shell), so it turn out that $T$ is the boundary value of an analytic function $\tilde T$ on $\mathcal T=\mathbb R^2+ i V_+$ (this depends a bit on your convention) where $V_+$ is the forward light-cone. It means that $$T(f) = \lim_{\mathcal T \ni z\to 0} \int_{\mathbb R^2} \tilde T(x+z)f(x)\mathrm dx.$$ But because $T \restriction_{\mathcal D(O)}=0$ it follows that $\tilde T \restriction_O=0$. By the edge of the wedge theorem $\tilde T \equiv 0$ on whole $\mathcal D$ and therefore $T=0$, what we wanted to show.

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Ah yes, this is a direct path to show that $E(\mathscr{D}(O))$ is dense. My answer above actually suggests that an argument similar to the one used for Schwartz functions (which is essentially equivalent to showing that Hermite polynomials are an orthonormal basis in the weighted $L^2$ space whose weight is given by a Gaussian) cannot possibly work for $\mathscr{D}$. –  Pedro Lauridsen Ribeiro Sep 8 '11 at 14:47
    
@Marcel: There is one thing troubling me. Obviously $\mathcal{D}(\mathcal{O})\subset \mathcal{S}(\mathbb{R}^2)$. Now you have shown that all vectors orthogonal to $E(\mathcal{S}(\mathbb{R}^2))$ are also orthogonal to $E(\mathcal{D}(\mathcal{O}))$, but not that there are no nontrivial vectors orthogonal to $E(\mathcal{D}(\mathcal{O}))$. Or otherwise stated, why does your proof not show that any linear subspace of $\mathcal{S}(\mathbb{R}^2)$ is mapped into a dense st in $H$, which is obvioulsy false for the space spanned by only one function? –  Jan S Sep 9 '11 at 8:45
    
Actually what Marcel shows goes the other way round, that is, any $v$ orthogonal to $E(\mathscr{D}(\mathscr{O}))$ is also orthogonal to $E(\mathscr{S}(\mathbb{R}^2))$. As the latter subspace is dense, we must have $v=0$. –  Pedro Lauridsen Ribeiro Sep 9 '11 at 14:33
    
Exactly like Pedro wrote... –  Marcel Bischoff Sep 11 '11 at 9:59
    
Yes, sorry I didn't read carefully enough. Thanks for your answer. –  Jan S Sep 11 '11 at 20:12
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Take your Schwartz functions to be Gaussian times polynomial. The Fourier transform takes this space to itself, and these functions decrease really fast.

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These functions are great for the case of Schwartz functions. In fact it was the Gaussian I used for Schwarz functions. But I want them to have compact support in "position space". Maybe I am just to "blind", but I dont see how the gaussian is helpful there. –  Jan S Aug 9 '10 at 12:34
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I doubt it. My reason is related to your comment: take your favorite function $f$ on $\mathscr{D}$, multiply it by a Gaussian with covariance $\sigma$ and centered around a point $x$ in the support of $f$, and take the Fourier transform. This will have an exponential decay in $\frac{1}{\sigma}$ if and only if there is an open neighbourhood of $x$ where $f$ is real analytic. Since points at the boundary of the support of $f$ are "essential singularities" from a real-analytic viewpoint, the above operation will lead to a function which does not decay exponentially if $x$ as above is in the boundary of the support of $f$. Since the resulting function is a Gaussian convolution in momentum space, there is no reason to expect that the Fourier transform of $f$ will itself decay exponentially.

The proof of the above claim uses the Fourier-Bros-Iagolnitzer (FBI) transform, which is a sort of "Gaussian-convoluted" Fourier transform. Check out Section 9.6 in Hörmander's "The Analysis of Linear Partial Differential Operators I" or Daniel Iagolnitzer's appendix to "Hyperfunctions and Theoretical Physics" (F. Pham, ed.), Springer Lecture Notes in Mathematics 449 (1975), pp. 121-131.

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A couple of other useful references for the above are: - J.-M. Delort, "F.B.I. Transform - Second Microlocalization and Semilinear Caustics". Springer Lecture Notes in Mathematics 1522 (1992), particularly pp. 11-13; - S. G. Krantz, H. R. Parks, "A Primer on Real-Analytic Functions". Birkhäuser (1992), particularly Section 4.3. In any case, there is the direct argument for answering your first question given below by Marcel Bischoff. –  Pedro Lauridsen Ribeiro Sep 8 '11 at 14:54
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