I asked this question on Math StackExchange recently but the only useful comment I got was that this could be a good question for Math Overflow. Here it goes:

Consider the Gaussian $G(x):=e^{-x^2}$ on the real line, and localize it to the region $|x|\sim 2^k$ by multiplying it by an appropriate smooth cut-off. More precisely, take $\phi\in C_0^\infty(\mathbb{R})$ supported in the region $$\{x\in\mathbb{R}: \frac{1}{2}<|x|\leq2\}$$ such that $0\leq\phi\leq 1,$ and let $\phi_k(x):=\phi(2^{-k}x)$. Consider: $$G_k(x):=\phi_k(x)G(x).$$ It is straightforward to check that $\|G_k\|_{L^1}\lesssim 2^ke^{-4^k}$, which tends (very quickly) to $0$ as $k\rightarrow\infty$. Also, using Young's convolution inequality one can easily show that $\|\widehat{G_k}\|_{L^1}\leq \|\phi\|_1\|G\|_1$, but this gives no decay in terms of $k$.

My question is: does $\|\widehat{G_k}\|_{L^1}$ decay as $k\rightarrow\infty$? If so, how fast? Can you prove sharp bounds (in $k$)?

Thank you.