Is there an example of a causally supported Schwartz function on $\mathbb{R}^4$ invariant under the Lorentz transform?

Question

I am working on $\mathbb{R}^4$ with the sign convention $(1,-1,-1,-1)$.

I wonder if there is Schwartz function $f(x)$ on $\mathbb{R}^4$ such that the support satisfies the condition $0<x^2 < 4m^2$ for some given positive constant $m$.

Here $x^2$ is of course with respect to the above metric, and $x=(x_0,x_1,x_2,x_3)$.

The first idea that comes to me is a function of the form \begin{equation} f(x)=g(x^2) \end{equation} where $g : \mathbb{R} \to \mathbb{R}$ is the smooth function defined by $g(t):=e^{-\frac{1}{4m^2-t}-\frac{1}{t}}$ for $0<t<4m^2$ and $0$ otherwise.

However, such $f$ does not show decay in the case $\lvert x_1 \rvert \to \infty$ while $x^2=2m^2$ and as a result $\sup_{x \in \mathbb{R}^4} \lvert x_1 f(x) \rvert = \infty$.

I wonder if there exists a Schwartz function with causal support. Moreover, it would be better if $f(\Lambda x) =f(x)$ for all elements $\Lambda$ of the restricted Lorentz group.

Could anyone please provide an example?

Answer 1

If Lorentz invariance is not required:

Let $\phi$ be any smooth bump function $\phi:\mathbb{R}\to\mathbb{R}$ that is non-zero precisely on $(0,4m^2)$ (including the one you used in the question statement).

Let $f:\mathbb{R}^4\to \mathbb{R}$ be given by

$$ f(x_0, x_1, x_2, x_3) = \phi(x_0^2 - x_1^2 - x_2^2 - x_3^2) \exp(-x_0^2 - x_1^2 - x_2^3 - x_3^3) $$

Since the Gaussian term is nowhere vanishing, $f$ has the desired support property.

Since $\phi$ is a smooth bump function, there is a sequence of numbers $M_k$ such that $|\phi^{(k)}| \leq M_k$. This shows that the $k$th order partial derivatives of $f$ are uniformly bounded by a $k$th degree polynomial in $(x_0, x_1, x_2, x_3)$ times the Gaussian, which shows that $f$ is Schwartz.