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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 $\mathbb R$.

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 $f \equiv 0$.

The pair $(E,\Sigma)$ is called a strongly annihilating pair, if there exists a constant $C$ such that for any $f \in L^2(\mathbb{R})$,

$$\|f\|_2^2 \leq C \left(\int_{\mathbb R \setminus E} |f|^2 dx + \int_{\mathbb R \setminus \Sigma} |\mathcal F f|^2 d\xi \right).$$ 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.

I am looking for examples of

1) A weakly annihilating pair which is $\underline{not}$ a strongly annihilating pair. ( 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.)

1') Sets $E$, $\Sigma$ both have infinite measure such that $(E,\Sigma)$ is a strongly annihilating pair.

2) Sets $E$, $\Sigma$ such that $E^c$ and $\Sigma^c$ have nonzero measure and $(E,\Sigma)$ is $\underline{not}$ a weakly annihilating pair.

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 ?


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Theorem 2.1 in ams.org/mathscinet-getitem?mr=1650106 uses the notion of $\epsilon$-thinness. I think such sets can have infinite measure. –  Willie Wong Jul 19 '10 at 16:03

1 Answer 1

I think the following may work for (1).

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.

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 $\int_{E^c} g_\lambda^2 dx \to 0$.

One easily checks that $\hat{g_\lambda}(\xi) = \frac{1}{\lambda^{1/2}}\hat{g}(\frac\xi\lambda)$. We estimate $\int_{-1}^1 \hat{g}_\lambda^2 d\xi$ by $$2 \sup_{\Sigma^c} \hat{g}_\lambda^2 = \frac{2}{\lambda^{1/2}} \sup_{(-\lambda^{-1},\lambda^{-1})} \hat{g}$$ 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.

Combining we have that there cannot be a constant $C$ for the "strongly annihilating condition" to hold.

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Of course it does. (Should have thought of it). Now it seems to me that the condition for being strong is too strong, so as to make most' pairs ineligible, whereas the condition of being weak is so weak that most' pairs would be eligible. So I think a better question would be to ask a) when is a pair strongly annihilating pair ? b) when is a pair not weakly annihilating pair ? –  Vagabond Jul 19 '10 at 13:38

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