Example of a pair which is not weakly annihilating - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T02:31:23Z http://mathoverflow.net/feeds/question/32468 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32468/example-of-a-pair-which-is-not-weakly-annihilating Example of a pair which is not weakly annihilating Vagabond 2010-07-19T10:58:35Z 2010-07-20T01:21:43Z <p>Let $\mathcal F$ denotes the Fourier transform $\mathcal{F} :L^2(\mathbb R)\rightarrow L^2(\mathbb R)$ and $E, \Sigma$ be two measurable sets in $\mathbb R$.</p> <p>The pair $(E,\Sigma)$ is called a weakly annihilating pair, if for any $f \in L^2(\mathbb{R})$, $support (f) \subseteq E$, $support(\mathcal F f)\subseteq \Sigma$, implies $f \equiv 0$.</p> <p>The pair $(E,\Sigma)$ is called a strongly annihilating pair, if there exists a constant $C$ such that for any $f \in L^2(\mathbb{R})$,</p> <p><code>$$\|f\|_2^2 \leq C \left(\int_{\mathbb R \setminus E} |f|^2 dx + \int_{\mathbb R \setminus \Sigma} |\mathcal F f|^2 d\xi \right).$$</code> The notion of annihilating pair arises in the study of uncertainty property in Fourier Analysis. For example Benedicks's Theorem says if $E$, $\Sigma$ are both sets of finite measures then they form a weakly annihilating pair, whereas Theorem of Amrein and Berthier says they form a strongly annihilating pair.</p> <p>I am looking for examples of</p> <p><strike> 1) A weakly annihilating pair which is $\underline{not}$ a strongly annihilating pair. </strike> ( Willie Wong has already answered this and I realise this was rather easy and I should have been able to figure it out myself, so my apologies.)</p> <p>1') Sets $E$, $\Sigma$ both have infinite measure such that $(E,\Sigma)$ is a strongly annihilating pair.</p> <p>2) Sets $E$, $\Sigma$ such that $E^c$ and $\Sigma^c$ have nonzero measure and $(E,\Sigma)$ is $\underline{not}$ a weakly annihilating pair.</p> <p>I realise the standard reference for this topic is the book by Havin and Jöricke, which unfortunately our library does not have a copy of!! Is there any alternative reference someone can suggest ? </p> <p>Thankyou.</p> http://mathoverflow.net/questions/32468/example-of-a-pair-which-is-not-weakly-annihilating/32480#32480 Answer by Willie Wong for Example of a pair which is not weakly annihilating Willie Wong 2010-07-19T12:41:52Z 2010-07-19T12:41:52Z <p>I think the following may work for (1). </p> <p>Let $E$ be the interval $[-1,1]$, and $\Sigma$ be the set $\mathbb{R}\setminus [-1,1]$. If $f$ has compact support, its Fourier transform can be extended analytically to $\mathbb{C}$, and so if $\mathcal{F}f$ vanishes on any interval ($\mathbb{R}\setminus \Sigma$), $f$ must be identically 0. So $E,\Sigma$ form a weakly annihilating pair. </p> <p>Now consider an arbitrary odd Schwarz function $g$ (so that $g(x) = - g(-x)$) with $L^2$ norm 1. Write $\hat{g}$ for its fourier transform. Consider the family $g_\lambda(x) = \lambda^{1/2} g(\lambda x)$. It is clear that $g_\lambda$ is in $L^2$, and that as $\lambda\to\infty$ we have $\int_{E^c} g_\lambda^2 dx \to 0$. </p> <p>One easily checks that $\hat{g_\lambda}(\xi) = \frac{1}{\lambda^{1/2}}\hat{g}(\frac\xi\lambda)$. We estimate <code>$\int_{-1}^1 \hat{g}_\lambda^2 d\xi$</code> by <code>$$2 \sup_{\Sigma^c} \hat{g}_\lambda^2 = \frac{2}{\lambda^{1/2}} \sup_{(-\lambda^{-1},\lambda^{-1})} \hat{g}$$</code> which using that $\hat{g}(0) = 0$ and its derivatives are uniformly bounded, gives that as $\lambda\to\infty$ the integral $\int_{\Sigma^c}\hat{g_\lambda}^2d\xi\to 0$ also. </p> <p>Combining we have that there cannot be a constant $C$ for the "strongly annihilating condition" to hold. </p>