Let \(\mathbb{T}^d\) be the \(d\)-torus . If \(f\in L^p(\mathbb{T}^d)\) with \(p\in [1,\infty ]\) such that \(\supp \hat{f}\subseteq J\), where \(J\) is a \(d\)-dimensional rectangle with side length \(b\), and \(S\subseteq \mathbb{R}^d\) is a \((\gamma,a)\)-thick set with \(a=(a_1, \ldots , a_d)\) such that \(0 < a_i \le 2 \pi\) for every \(i=1,\ldots ,d\), then there is a constant \(C>0\) depending only on \(d\) such that (?).
Let \(x_0\in \mathbb{T}^d\) and \(Q = x_0+[0,a_1] \times \cdots \times [0,a_d]\). We want to use (0x67ac7785) .
To do this, we estimate the supremum of the analytic extension of \(f\) on
\[ \tilde{D}:=D_1(x_0, 6 \lVert a\rVert_2)\times \cdots \times D_d(x_0, 6 \lVert a\rVert_2). \]Rewriting \(f\) using a local Fourier basis gives us
\[ \begin{align*} f(x) &= \sum_{k \in J} \frac{1}{(2\pi)^d} \int_{\mathbb{T}^d} f(y) e^{-i k\cdot (y - x_0) } \, dy \cdot e^{i k\cdot (x - x_0)} \\ &= \sum_{k \in J} \frac{1}{2\pi^d} \int_{\mathbb{T}^d - x_0} f(y + x_0) e^{-i k\cdot y } \, dy \cdot e^{i \langle k, x - x_0 \rangle} \\ &= \sum_{k \in J} \widehat{\tau_{x_0}f}(k) \cdot e^{i k\cdot (x - x_0) } \end{align*} \]With Hölders inequality we obtain
\begin{equation} \sup_{z\in \tilde{D}}\lvert f(x)\rvert \leq \sup_{z\in \tilde{D}}\Bigl( \sum_{k\in J} \lvert \widehat{\tau_{x_0}f}(k)\rvert^2 \Bigr)^{1/2} \Bigl( \sum_{k \in J} \lvert e^{i k\cdot (x - x_0) }\rvert \Bigr)^{1/2}. \label{1} \end{equation}Since the \(L^2\)-norm is translation invariant, Plancherel’s theorem implies
\[ \Bigl( \sum_{k\in J} \lvert \widehat{\tau_{x_0}f}(k)\rvert^2 \Bigr)^{1/2} = \lVert f\rVert_{L^2(\mathbb{T}^d)}. \]We proceed by estimating the second sum in \eqref{1}
\begin{align*} \sup_{z\in \tilde{D}} \sum_{k \in J} \lvert e^{i k\cdot (x - x_0) }\rvert &\leq \sum_{k \in J} \sup_{z\in \tilde{D}} \left| e^{i \langle k, z - x_0 \rangle} \right|\\ &= \sum_{k\in J} \sup_{z\in \tilde{D}} \prod_{j = 1}^{d} \left| e^{i k_j (z_j - x_{0, j})} \right| \\ &\leq \sum_{k\in J} \prod_{j = 1}^{d} e^{6\lVert a\rVert_2|k_j|} \\ &= \sum_{k\in J} e^{6\lVert a\rVert_2 \lVert k\rVert_1} \\ &\leq b^d \cdot e^{6d \lVert a\rVert_2 b}, \end{align*}where we used \(|e^{i k_j (z_j - x_{0,j})}| = e^{-k_j \Im (z_j - x_{0,j})}\). Thus, we have
\[ \sup_{z\in \tilde{D}}\lvert f(x)\rvert \leq b^d e^{6d\lVert a\rVert_2 b}\lVert f\rVert_{L_2(\mathbb{T}^d)}. \]Applying (0x67ac7785) yields
\begin{equation} \lVert f\rVert_{L^2(Q)}\le \biggl(\frac{C \lvert Q\rvert}{\lvert Q\cap S \rvert}\biggr)^{C\log M + \frac{1}{2}} \lVert f\rVert_{L^2(Q\cap S)}, \label{eq:2} \end{equation}where \(C>0\) is a constant depending on \(d\) and
\[ M=\sup_{z\in \tilde{D}} \frac{\lVert a\rVert_2}{\lVert f\rVert_{L^2(Q)}} \lvert f(z)\rvert. \]The preceding considerations gives us
\[ M\le b^d \lVert a\rVert_2 e^{C' \lVert a\rVert_2 b} \frac{\lVert f\rVert_{L^2(\mathbb{T}^d)}}{\lVert f\rVert_{L^2(Q)}}, \]where \(C'\) is a constant depending on \(d\). Combining this with \eqref{eq:2} and using the thickness property of \(S\) gives us
\[ \lVert f\rVert_{L^2(Q)}\le \biggl(\frac{C}{\gamma}\biggr)^{C \Bigl(\lVert a\rVert_2 b + \log b \lVert a\rVert_2 + \log \frac{\lVert f\rVert_{L^2(\mathbb{T}^d)}}{\lVert f\rVert_{L^2(Q)}} + 1\Bigr)} \lVert f\rVert_{L^2(Q\cap S)}. \]Splitting the torus in suitable cubes, applying the above estimate and summing up gives us the desired result.