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

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.

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Original MSE question: math.stackexchange.com/questions/72097 – Yemon Choi Oct 13 2011 at 21:37
Note that much of the phrasing above comes from DJC's answer to that MSE question – Yemon Choi Oct 13 2011 at 21:40
@Yemon: Yes, I should have mentioned that. Thanks! – user17240 Oct 13 2011 at 22:27

If I didn't miss anything, the dominated convergence argument turns out to work well. The following shows that $\widehat{G_k}$ can be dominated by a rapidly decaying function.

Denote $\psi=\widehat{\phi}, \psi_k=\widehat{\phi_k}$, then $\psi_k(y)=2^k\psi(2^ky), \int \psi_k=0$ and $\int |\psi_k|=\int |\psi|$, hence

$$|\widehat{G_k}(-x)| =|\widehat{\phi_k}\ast G(-x)| =|\int \psi_k(y)G(y+x)dy| =|\int \psi_k(y)[G(y+x)-G(x)]dy|$$ $$=|\int_{|y|\le |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|+|\int_{|y|\ge |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|$$ $$\le \frac{|x|}{2}\sup_{B(x,|x|/2)}|G'|\cdot \int |\psi|+C\int_{|y|\ge |x|/2}|\psi|dy$$

The last line is a function in $x$ decaying at any rate.

In view of the above argument, the function $\phi$ and $G$ can be replaced by any Schwartz functions where in addition $\phi$ vanishes at $0$, and we will still have $\|\widehat{G_k}\|_1 \rightarrow 0$. An estimate of the rate of decay follows in the specified case.

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Since $\phi$ is $C^\infty_c$ we know that $\widehat{\phi} \in \mathcal{S}(\mathbb{R})$, in particular $\widehat{\phi}(\xi) \lesssim (1+|\xi|)^{-100}$. On the other hand $\widehat{\phi_k}(\xi) = 2^k \widehat{\phi}(\xi 2^k)$ which by our previous estimate satisfies $|\widehat{\phi_k}(\xi)| \lesssim 2^k (1+|2^k \xi|)^{-100}$. Now, $\widehat{G_k} = G \ast \widehat{\phi_k}$ so by Cauchy-Schwarz we can estimate the $L^\infty$ norm of $\widehat{G_k}$ by $\lesssim 2^{-90k} \|G\|_{L^2}$. So by the dominated convergence theorem $\|\widehat{G_k}\|_{L^1} \to 0$ as $k \to \infty$.

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 I agree that the $L^\infty$ norm of $\widehat{G_k}$ is small (it is bounded by $\|G_k\|_{L^1}$). But how will this imply bounds on the $L^1$ norm of $\widehat{G_k}$? I don't think dominated convergence will do it... – user17240 Oct 26 2011 at 1:04 it doesn't imply bounds, just decay to 0 – anonymous student Oct 26 2011 at 22:34 it is a qualitative statement – anonymous student Oct 26 2011 at 23:05 Well, "decay to 0" is equivalent to some kind of bound. My second sentence can be rephrased as "how will this imply that the $L^1$ norm of $\widehat{G_k}$ tends to $0$ as $k\rightarrow\infty$?". $L^1$ and $L^\infty$ norms are in general not comparable... – user17240 Oct 27 2011 at 0:06 i dont think you can justify the use of dom conv in the last line – Otis Chodosh Nov 8 2011 at 20:18