$L^1$ norm of the Fourier transform of a truncated Gaussian - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:21:50Z http://mathoverflow.net/feeds/question/78067 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian $L^1$ norm of the Fourier transform of a truncated Gaussian user17240 2011-10-13T21:28:23Z 2012-02-22T00:06:10Z <p>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:</p> <p>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 $${x\in\mathbb{R}: \frac{1}{2}&lt;|x|\leq2}$$ 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 $k$. </p> <p>My question is: does $\|\widehat{G_k}\|_{L^1}$ decay as $k\rightarrow\infty$? If so, how fast? Can you prove sharp bounds (in $k$)?</p> <p>Thank you.</p> http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian/79061#79061 Answer by anonymous student for $L^1$ norm of the Fourier transform of a truncated Gaussian anonymous student 2011-10-25T10:14:52Z 2011-10-25T10:59:59Z <p>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 $k \to \infty$.</p> http://mathoverflow.net/questions/78067/l1-norm-of-the-fourier-transform-of-a-truncated-gaussian/88911#88911 Answer by Syang Chen for $L^1$ norm of the Fourier transform of a truncated Gaussian Syang Chen 2012-02-19T09:03:53Z 2012-02-22T00:06:10Z <p>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.</p> <p>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</p> <p>$$|\widehat{G_k}(-x)| =|\widehat{\phi_k}\ast G(-x)| =|\int \psi_k(y)G(y+x)dy| =|\int \psi_k(y)[G(y+x)-G(x)]dy|$$ $$=|\int_{|y|\le |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|+|\int_{|y|\ge |x|/2} \psi_k(y)[G(y+x)-G(x)]dy|$$ $$\le \frac{|x|}{2}\sup_{B(x,|x|/2)}|G'|\cdot \int |\psi|+C\int_{|y|\ge |x|/2}|\psi|dy$$</p> <p>The last line is a function in $x$ decaying at any rate. </p> <p>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.</p>