Span of functions dense in $L^2$

Let $t\in\Bbb R$. We have for $t\in\Bbb R$ that $$\sum_{n=0}^{+\infty}\int_{\Bbb R}f(x)e^{-x^2/2}\frac{(itx)^n}{n!}=0.$$ As for all integer $n$, $$\left|f(x)e^{-x^2/2}\frac{(itx)^n}{n!}\right|= |f(x)|e^{-x^2/2}\frac{|x|^n|t|^n}{n!},$$ and the RHS is integrable for the product of Lebesgue and counting measure (because for each fixed $t$, $$\sum_{n=0}^{+\infty}|f(x)|e^{-x^2/2}\frac{|x|^n|t|^n}{n!}=|f(x)|\exp\left(-\frac{x^2}2+|x||t|\right)\leqslant 2f(x)^2+2\exp\left(- x^2 +2|x||t|\right)$$), we can switch the integral and the sum, which gives that for all $t\in\Bbb R$, $$\int_{\Bbb R}f(x)e^{-x^2/2}e^{itx}dx=0.$$ As $x\mapsto f(x)e^{-x^2/2}$ is integrable, by uniqueness of Fourier transform $f(x)e^{-x^2/2}=0$ almost everywhere, hence $f\equiv 0$.