Alternate Proof of Kovrijkine One-Dimensional Uncertainty Principle

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

We repeat the proof step by step without the first step:

1. Apply complex lemma on intervals to get an estimate where some supremum is involved Link to heading

Let \(I\) be an interval of length \(a\). We may apply (0x6698f0dc) on \(\phi=\frac{a^{1/p}g}{\lVert f\rVert_{L^p(I)}}\) since \(f\) is analytic on the whole real line and due to (0x669a1456) a point \(x_0\in I\) exists and such that \(f(x_0)\ge\frac{\lVert f\rVert_{L^p(I)}}{a^{1/p}}\). Thus we obtain

\begin{equation*} \lVert f\rVert_{L^p(I)} \le \biggl(\frac{24a}{|S\cap I|}\biggr)^{2 \frac{\log M}{\log 2} + \frac{1}{p}} \lVert f\rVert_{L^p(S\cap I)}, \end{equation*}

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

2. 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 &= \sup_{z\in D_{4a}(x_0)}\frac{a^{1/p}|f(z)|}{\lVert f\rVert_{L^p(I)}} \\ &= \sup_{z\in D_{5a}(x)}\frac{a^{1/p}|f(z)|}{\lVert f\rVert_{L^p(I)}} \\ &\le 2^{1/p}\exp(5Cab), \end{align}

where \(x\) is the expansion point mentioned in the lemma. Note, that the \(a\) appears in exponential, because \(R=5a\).

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

Combining both, we obtain

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

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