$|\int_{\mathbb R} e^{-t^{2}} e^{-(t/\lambda -x)^{2}} e^{-2\pi i w\cdot t}| dt \leq G(x,w), G\in L^{1} ? $

Question

Put $\lambda >0,$ and we define,

$$F_{\lambda}(x, w)= \int_{\mathbb R} e^{-t^{2}} e^{-(t/\lambda -x)^{2}} e^{-2\pi i w\cdot t} dt;(x,w) \in \mathbb R^{2}$$ we note that, $F_{\lambda} \in L^{1}(\mathbb R^{2}).$

My Question is: Can we expect to find, $G\in L^{1}(\mathbb R^{2})$ such that, $$|F_{\lambda}(x, w)| \leq G(x,w); (x,w) \in \mathbb R^{2} ?$$ (Bit roughly speaking, how to dominate sequence functions uniformly by a $L^{1}-$ integrable function )

Thanks,

Answer 1

I do not know if this solution is what you are looking for! First notice that

$$ (t/\lambda-x)^2 \geq 0 \implies -(t/\lambda-x)^2 \leq 0 \implies e^{-(t/\lambda-x)^2}\leq 1. $$

and

$$ |e^{-2\pi i w t}| =1. $$

So we have

$$ |F_{\lambda}(x, w)| \leq \int_{\mathbb{R}}e^{-t^2}dt < \infty. $$