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The Fourier transform of the indicator function of an interval $$\widehat{\chi}_{[a,b]}(\xi)=\int^b_{a} e^{i \xi x}dx=\frac{e^{i\xi b}-e^{i\xi a}}{i\xi}$$ has decay $O(|\xi|^{-1})$ as $|\xi|\rightarrow \infty$. On the other hand, for general compact set $K$ of positive Lebesgue measure, by Riemann-Lebesgue lemma $\widehat{\chi}_{K}(\xi)=o(1), |\xi|\rightarrow \infty$.

My question is, does there exist compact set $K$ such that $\widehat{\chi}_{K}(\xi)$ does not decay as $O(|\xi|^{-1})$ as $|\xi|\rightarrow \infty$? Thank you.

EDIT: Is it possible to find such $K$ without interior point?

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A remark: $BV(\mathbf T) \subset L^\infty(\mathbf T)$ and $BV$ has decay $O(|\xi|^{-1})$ and the last space $o(1)$. But a compact set does not need to have measure $0$ as a boundary. You will need some strange thing like this. Of course, we also have Paley-Wiener for functions of compact support. – Jonas Teuwen May 25 '12 at 15:42
up vote 5 down vote accepted

Consider the union of intervals $[(2j+1) 2^{-n^2}, (2j+2) 2^{-n^2}]$, $j=0 \ldots 2^{2n-2}-1$, $n=1,2,\ldots$, and $\{0\}$.

EDIT: Let $I(j,n) = [(2j+1) 2^{-n^2}, (2j+2) 2^{-n^2}]$ and $U(N) = \bigcup_{n=N+1}^{\infty}\bigcup_{j=0}^{2^{2n-2}-1} I(j,n)$ Then $\widehat{\chi_{I(j,n)}}(2^{n^2} \pi) = 2^{1-n^2} i/\pi$. For $m < n$, $\widehat{\chi_{I(j,m)}}(2^{n^2} \pi) = 0$. On the other hand, $\left|\widehat{\chi_U(n)}(\xi)\right| \le m(U(n)) \le 2^{-n^2}$. So $\left|\widehat{\chi_K}(2^{n^2} \pi)\right|\cdot (2^{n^2} \pi) = \Omega(2^{n})$ as $n \to \infty$

EDIT: If you want a $K$ with empty interior, you can proceed as follows. Start with $K_0$ as above (which is the union of $\{0\}$ and intervals with rational endpoints) and a sequence $\xi_m$ such that $|\widehat{\chi_{K_0}}(\xi_m)| > m |\xi_m|$, and a sequence $\{r_n\}$ of irrationals dense in $K_0$. I'll choose rational numbers $a_n,b_n$ with $a_n < r_n < b_n$ and take $K_n = K_{n-1} \backslash (a_n, b_n)$, and then $K = \bigcup_n K_n$ will have empty interior.

By the fact that $r_n$ is irrational, either $r_n \notin K_{n-1}$ (in which case we take $a_n$ and $b_n$ so that $(a_n,b_n) \cup K_{n-1} = \emptyset$) or $r_n$ is in the interior of $K_{n-1}$, in which case we will want $[a_n,b_n] \subset K_{n-1}$. In the first case $\widehat{\chi_{K_n}} = \widehat{\chi_{K_{n-1}}}$, in the second $\widehat{\chi_{K_n}} = \widehat{\chi_{K_{n-1}}} - \widehat{\chi_{(a_n,b_n)}}$. Now $|\widehat{\chi_{(a_n,b_n)}}(\xi)| \le b_n - a_n$, so if $b_n - a_n$ is small enough and $|\widehat{\chi_{K_{n-1}}}(\xi_m) > (m/2) |\xi_m|^{-1}$ for $m = m_1, m_2, \ldots, m_{n-1}$, $|\widehat{\chi_{K_{n-1}}}(\xi_m)| > (m/2) |\xi_m|^{-1}$ for those same $m$, with $t_n < m_1$. Now since $|\widehat{\chi_{(a_n,b_n)}}(\xi)| = O(1/|\xi|)$, we can take $m_n$ to be any sufficiently large $m$ and we will have $|\widehat{\chi_{K_n}}(\xi_{m_n})| > (m_n/2) |\xi_{m_n}|^{-1}$. In the limit we will have $|\widehat{\chi_K}(\xi_{m_n})| \ge (m_n/2) |\xi_{m_n}|^{-1}$ for all $n$.

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Sorry, I'm slow on this. I computed $$\hat{\chi}_K(\xi)=\frac{1}{i\xi}\sum^\infty_{n=1} \frac{e^{i\xi 2^{-(n-1)^2}}-1}{e^{-i\xi 2^{-n^2}}+1}$$ But I can't see the exact rate of decay of this expression. – Syang Chen May 26 '12 at 1:50
What do you get at $\xi = 2^{n^2} \pi$? – Robert Israel May 29 '12 at 7:40
@RobertIsrael: Got it! Thanks. – Syang Chen May 29 '12 at 18:17
@WillieWong: The $n$-th term in the series is essentially $2^{2n}$ while the tail after that is bounded. So $|\hat{\chi}_K(2^{n^2}\pi)|\gtrsim\frac{2^{2n}}{2^{n^2}}$. – Syang Chen May 29 '12 at 18:24
@Syang + Robert: ah, thanks, I see now. – Willie Wong May 30 '12 at 7:50

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