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Suppose that a tempered distribution $f(x)$ on 4-dimensional Minkowski space with signature $+---$ vanishes for $x²<0$ and its Fourier transform $\tilde f(p)$ vanishes for $p²<0$.

Are then $f$ and $\tilde f$ necessarily Lorentz invariant? I am interested in a proof or a counterexample, and in the latter case ideally in a description of all possible $f$ with the above property.

Edit: I noticed that directional derivatives of distributions with the above property have this property, too, and are no longer Lorentz invariant. So my original question has a negative answer. In the light of this information, the question is now whether $f$ and $\tilde f$ can necessarily be written as polynomials with Lorentz invariant coefficients.

Suppose that a tempered distribution $f(x)$ on 4-dimensional Minkowski space with signature $+---$ vanishes for $x^2<0$ and its Fourier transform $\hat f(p)$ vanishes for $p^2<0$.

Are then $f$ and $\hat f$ necessarily Lorentz invariant? I am interested in a proof or a counterexample, and in the latter case ideally in a description of all possible $f$ with the above property.

Edit: I noticed that directional derivatives of distributions with the above property have this property, too, and are no longer Lorentz invariant. So my original question has a negative answer. In the light of this information, the question is now whether $f$ and $\hat f$ can necessarily be written as polynomials with Lorentz invariant coefficients.

Suppose that a tempered distribution $f(x)$ on 4-dimensional Minkowski space with signature $+---$ vanishes for $x²<0$ and its Fourier transform $\tilde f(p)$ vanishes for $p²<0$.

Are then $f$ and $\tilde f$ necessarily Lorentz invariant? I am interested in a proof or a counterexample, and in the latter case ideally in a description of all possible $f$ with the above property.

Suppose that a tempered distribution $f(x)$ on 4-dimensional Minkowski space with signature $+---$ vanishes for $x²<0$ and its Fourier transform $\tilde f(p)$ vanishes for $p²<0$.

Are then $f$ and $\tilde f$ necessarily Lorentz invariant? I am interested in a proof or a counterexample, and in the latter case ideally in a description of all possible $f$ with the above property.

Edit: I noticed that directional derivatives of distributions with the above property have this property, too, and are no longer Lorentz invariant. So my original question has a negative answer. In the light of this information, the question is now whether $f$ and $\tilde f$ can necessarily be written as polynomials with Lorentz invariant coefficients.