Cfrgtkky

Let $f$ a continuous function, moderate decrease and satisfying

begin{equation}
int_{-infty}^{infty}f(y)e^{-y^{2}}e^{2xy}dy=0
end{equation}
for all $xinmathbb{R}$ I need to prove that $f=0$.

The hint is to consider $f*e^{-x^2}$. I know that both $f$ and $e^{-x^{2}}$ are moderate decrease, so the convolution is too, i.e., $exists A>0$ such that $|(f*e^{-x^2})(x)|leqfrac{A}{1+x^{2}}$ for all $xinmathbb{R}.$ So,

$$|(f*e^{-x^2})(x)|=left|int_{-infty}^{infty}f(x-y)e^{-y^{2}}dyright|leqfrac{A}{1+x^{2}} $$

How can I use the $int_{-infty}^{infty}f(y)e^{-y^{2}}e^{2xy}dy=0$ condition?

Noting $e^{-y^2 + 2xy} = e^{-(y - x)^2 + x^2} = e^{-(x - y)^2}e^{x^2}$, the integral condition implies $f * e^{-x^2} = 0$ for all $x$. Taking the Fourier transform of $f * e^{-x^2}$, using the fact that the Fourier transform of a convolution is the product of the Fourier transforms, deduce that the Fourier transform $hat{f} = 0$. Since $lvert f(x)rvert$ is bounded by a constant times $1/(1 + x^2)$, then $int_{-infty}^infty lvert f(x)rvert , dx < infty$. Since both $f$ and $hat{f}$ are in $L^1(Bbb R)$, by the Fourier inversion theorem $f = 0$.