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?



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


