Why we have $f=0$

Question

Define the Fourier transform for a suitable function $f\in L^1(\Bbb R)$ by $\widehat{f}(\xi)=\int_{\Bbb R}f(x)e^{-ix\xi} dx$. Assume the condition $$\int_{\Bbb R}\int_{\Bbb R}|\widehat{f}(\xi)f(x)|^2e^{2|x\xi|} d\xi dx<\infty.$$ Put $f(x)=P(x) e^{-t x^2}$ where $t>0$ and $P$ is a polynomial and suppose $f$ verify the above condition. My question why $P=0$. Thank you in advance.

Answer 1

Fourier transform of $f(x)=e^{-tx^2}$ is $\hat{f}(\xi)=ce^{-\xi^2/(4t)}$. Multiplication of $f$ by a polynomial results in applying a differential operator with constant coefficients to $\hat{f}$, and for our $\hat{f}$ this is equivalent to multiplication on some polynomial. These polynomials play no role in convergence (unless they are both zero), since the expression under the exponent in the product $f(x)\hat{f}(\xi)e^{2|x\xi|}$ is $$-tx^2+2|x\xi|-\xi^2/(4t).$$ Since this quadratic expression is sometimes positive in the first quadrant, the integral diverges, unless $P=0$.