Case \(p\neq \infty\) Link to heading

The proof can be divided in the following steps:

1. Simplification of the problem Link to heading

Without loss of generality we assume \(\supp \hat{f} \subset [-b/2, b/2]\), otherwise consider \(f(x)\exp(\mathrm{i}cx)\) for suitable \(c\in \mathbb{R}\) which does not change the overall norm. In the following we consider \(g(x)=f(ax)\) and \(\widetilde{E}=E/a=\{x/a \mid x\in E\}\), which is a \((\gamma,1)\)-thick set. Note, that \(\supp \hat{g} \subset [-ab/2, ab/2]\).

2. Apply complex lemma on intervals with unit length to get an estimate where some supremum is involved Link to heading

We decompose the real line into intervals of length 1. On every interval \(I\) we may apply (0x6698f0dc) on \(\phi=\frac{g}{\lVert g\rVert_{L^p(I)}}\) since \(g\) is analytic on the whole real line and due to (0x669e0e75) a point \(x_0\in I\) exists and such that \(g(x_0)\ge \lVert g\rVert_{L^p(I)}\). Thus we obtain

\[ \lVert g\rVert_{L^p(I)} \le \biggl(\frac{24}{|\widetilde{E}\cap I|}\biggr)^{2 \frac{\log M}{\log 2} + \frac{1}{p}} \lVert g\rVert_{L^p(\widetilde{E}\cap I)}, \]

where \(M=\max_{|z-z_0|\le 4} \frac{|g(z)|}{\lVert g\rVert_{L^p(I)}}\). Note that the scaling factor \(\frac{1}{\lVert g\rVert_{L^p(I)}}\) disappears in the inequality.

3. Estimate of the supremum on good intervals Link to heading

Using Kovrijkine’s Lemma we estimate \(M\) on each good interval \(I\) with

\begin{align} M &= \frac{1}{\lVert g\rVert_{L^p(I)}}\sup_{z\in D_4(x_0)}|g(z)| \\ &= \frac{1}{\lVert g\rVert_{L^p(I)}}\sup_{z\in D_5(x_1)}|g(z)| \\ &\le 2^{1/p}\exp(5Cab), \end{align}

where \(x_1\) is the expansion point mentioned in the lemma. Note, that the \(a\) appears in the constant, because \(\supp g \subset [-ab/2, ab/2]\).

4. Combine results of step 2 and 3 to obtain the result. Link to heading

Combining both, we obtain

\begin{align} \lVert g\rVert_{L^p(\widetilde{E})} &\ge \sum_{\text{\(I\) is good}} \lVert g\rVert_{L^p(\widetilde{E}\cap I)} \\ &\ge \sum_{\text{\(I\) is good}} \biggl(\frac{|\widetilde{E}\cap I|}{24}\biggr)^{C(ab + 1/p)} \lVert g\rVert_{L^p(I)}, \\ &\ge \sum_{\text{\(I\) is good}} \biggl(\frac{\gamma}{24}\biggr)^{C(ab + 1/p)} \lVert g\rVert_{L^p(I)}, \\ &\ge 2^{-1/p}\biggl(\frac{\gamma}{24}\biggr)^{C(ab + 1/p)} \lVert g\rVert_{L^p(\mathbb{R})}, \end{align}

where we also used the thickness property of \(\widetilde{E}\) and the fact, that the family of good intervals is sufficiently large. Note, that \(C>0\) is some numerical constant.

Finally using \(\lVert g\rVert_{L^p(\widetilde{E})}=1/a^p\lVert f\rVert_{L^p(E)}\) and \(\lVert g\rVert_{L^p(\mathbb{R})}=1/a^p\lVert f\rVert_{L^p(\mathbb{R})}\) we obtain the desired result.