# $L^1$ norm of the Fourier transform of a truncated Gaussian

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 $\left \{x\in\mathbb{R}: \frac{1}{2}<|x|\leq2 \right\}$ such that $0\leq\phi\leq 1,$ and consider $$G_k(x):=\phi(2^{-k}x)G(x).$$ It is straightforward to check that $\|G_k\|_{L^1}\lesssim 2^ke^{-4^k}$. My question is: what can be said about $\|\widehat{G_k}\|_{L^1}$? In particular, what decay (if any) do you get in terms of $k$?

Thank you.

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Not to detract from the question being asked which is of interest in its own right, but most people take $(1/\sqrt{2\pi})\exp(-x^2/2)$ to be the standard Gaussian (density) on the real line. –  Dilip Sarwate Oct 12 '11 at 19:26
@DJC: I would think that your comment contains enough information to post it is an answer. –  Jonas Teuwen Oct 12 '11 at 22:17
@Dilip: Edited, thanks! –  user17240 Oct 13 '11 at 3:37

Hint: If we write $\phi_k(x) = \phi(2^{-k} x)$, then note that $$\widehat{G_k}(x) = \widehat{\phi_k}(x) \ast \widehat G(x).$$ Applying Young's inequality for convolutions gives $$\| \widehat{G_k}\|_{1} \leq \| \widehat{\phi_k} \|_1 \| \widehat{G} \|_1 \lesssim \| \widehat{\phi_k} \|_1.$$ Note that $\| \widehat{G} \|_1 \lesssim 1$ since $G$ is a Schwarz function and hence $\widehat{G}$ is too. It remains to find a good bound for $\| \widehat{\phi_k} \|_1$, which I will leave up to you.
Since $\phi(0)=0$ in a neighborhood of $0$, there is cancellation so that $\int\hat{\phi}(\xi)\xi^k\mathrm{d}\xi=0$ for all $k\ge0$ in $\mathbb{Z}$. –  robjohn Oct 13 '11 at 0:01
However, $\|\hat{\phi}_k\|_{L^1}=\|\hat{\phi}\|_{L^1}$. –  robjohn Oct 13 '11 at 0:02
Right, so this argument gives no decay in $k$, but that still doesn't answer my second question. It just shows that Young's inequality is too crude for our purposes (unless one could somehow show that the first inequality is actually an equality, which is in general far from being true). Any ideas? –  user17240 Oct 13 '11 at 0:29