$f$ continuous, moderate decrease and $\int_{-\infty}^{\infty}f(y)e^{-y^{2}}e^{2xy}dy=0$ implies $f=0$.

Question

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 $x\in\mathbb{R}$ 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 $|(f*e^{-x^2})(x)|\leq\frac{A}{1+x^{2}}$ for all $x\in\mathbb{R}.$ So,

$$|(f*e^{-x^2})(x)|=\left|\int_{-\infty}^{\infty}f(x-y)e^{-y^{2}}dy\right|\leq\frac{A}{1+x^{2}} $$

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

Answer 1

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 $f = 0$.

Answer 2

Define $g=f * e^{-x^{2}}$. \begin{align*} g(x)&=\int_{-\infty}^{\infty} f(y) e^{-(x-y)^{2}} d y\\ &=e^{-x^{2}} \int_{-\infty}^{\infty} f(y) e^{-y^{2}+2 x y} d y\\ &=0. \end{align*} Therefore \begin{align*} 0&=\hat{g}(\xi)\\ &=\hat f(\xi) \cdot \hat{e^{-x^{2}}}(\xi)\\ &=\hat{f(\xi)} \cdot \sqrt{\pi} e^{-\pi^{2} \xi^{2}} \end{align*} Therefore $$ \hat{f}(\xi)=0. $$ By Fourier inversion or Plancheral identity $$f\equiv 0.$$