Rate of Fourier decay of indicator functions 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?
 A: 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 $n$.
