Fourier Transforms restricted to mass shell

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Take your Schwartz functions to be Gaussian times polynomial. The Fourier transform takes this space to itself, and these functions decrease really fast.