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4 deleted 900 characters in body

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 $k$)?

Update: Noting that $$\widehat{G_k}(\xi)=\widehat{\phi_k}\ast\widehat{G}(\xi)=\int\widehat{\phi_k}(\xi-\eta)\widehat{G}(\eta)d\eta=2^k\int\widehat{\phi}(2^k(\xi-\eta))\widehat{G}(\eta)d\eta$$ and using the fact that $\widehat{\phi}$ is a Schwartz function (and so $|\widehat{\phi}(\xi)|\lesssim_N\langle\xi\rangle^{-N}$ for every $N\in\mathbb{N}$), one can estimate:

$$\int|\widehat{G_k}(\xi)|d\xi\lesssim 2^k\int\widehat{G}(\eta)\int|\widehat{\phi}(2^k(\xi-\eta))|d\xi d\eta\lesssim_N 2^k\int\widehat{G}(\eta)\int\langle2^k(\xi-\eta)\rangle^{-N} d\xi d\eta.$$ This last quantity is $\lesssim2^{-(N-1)k}\|\widehat{G}\|_{L^1}$, and so $\|\widehat{G_k}\|_{L^1}$ does indeed decay as $k\rightarrow\infty$.

The last question remains unanswered though: is this decay sharp in $k$? Or can one hope for $\|\widehat{G_k}\|_{L^1}$ to decay like $\|G_k\|_{L^1}\lesssim 2^ke^{-4^k}$?

Thank you.

3 Answered the first question; added 95 characters in body

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 consider let $$G_k(x):=\phi(2^{-k}x)G(x).$$ \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$k$)? Update: Noting that $$\widehat{G_k}(\xi)=\widehat{\phi_k}\ast\widehat{G}(\xi)=\int\widehat{\phi_k}(\xi-\eta)\widehat{G}(\eta)d\eta=2^k\int\widehat{\phi}(2^k(\xi-\eta))\widehat{G}(\eta)d\eta$$ and using the fact that$\widehat{\phi}$is a Schwartz function (and so$|\widehat{\phi}(\xi)|\lesssim_N\langle\xi\rangle^{-N}$for every$N\in\mathbb{N}$), one can estimate: $$\int|\widehat{G_k}(\xi)|d\xi\lesssim 2^k\int\widehat{G}(\eta)\int|\widehat{\phi}(2^k(\xi-\eta))|d\xi d\eta\lesssim_N 2^k\int\widehat{G}(\eta)\int\langle2^k(\xi-\eta)\rangle^{-N} d\xi d\eta.$$ This last quantity is$\lesssim2^{-(N-1)k}\|\widehat{G}\|_{L^1}$, and so$\|\widehat{G_k}\|_{L^1}$does indeed decay as$k\rightarrow\infty$. The last question remains unanswered though: is this decay sharp in$k$? Or can one hope for$\|\widehat{G_k}\|_{L^1}$to decay like$\|G_k\|_{L^1}\lesssim 2^ke^{-4^k}$? Thank you. 2 added 37 characters in body 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 consider $$G_k(x):=\phi(2^{-k}x)G(x).$$ It is straightforward to check that$\|G_k\|_{L^1}\lesssim 2^ke^{-4^k}$, 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.

1