\[ \DeclareMathOperator{\diam}{diam} \]

Consider a measurable convex bounded set \(K\subseteq \mathbb{R}^d\) with positive measure and \(0\in K\), and a function \(f\in L^p(K)\). Assume that \(f\) has an analytic extension on the polydisk

\[ \tilde{D}=D_1(0,6 \diam K)\times \cdots \times D_d(0,6\diam K). \]

Then there is a constant \(C>0\) depending only on \(d\) such that for every measurable subset \(S\subseteq K\) with nonzero measure and \(p\ge 1\) the following estimate hold

\[ \lVert f\rVert_{L^p(K)}\le (C(\diam K)^d)^{2 \frac{\log M}{\log 2}} \lvert K\rvert^{1/p}\biggl(\frac{2}{\lvert S \rvert}\biggr)^{2\frac{\log M}{\log 2} + \frac{1}{p}} \lVert f\rVert_{L^p(S)} \]

with

\[ M=\sup_{z\in \tilde{D}} \frac{\lvert K\rvert^{1/p}}{\lVert f\rVert_{L^p(K)}} \lvert f(z)\rvert. \]
Proof

We choose a point \(x\in K\) such that

\[ |f(x)|\ge \frac{\lVert f\rVert_{L^p(K)}}{\lvert K\rvert^{1/p}}. \]

Such a point exists according to (0x669e0e75) .

Suppose \(S\subseteq K\) is a subset with nonzero measure. Using (0x6698db36) we find a line segment \(I\subseteq K\) starting in \(x\) and satisfying

\[ \lvert S\rvert\le C(d) (\diam K)^{d-1} \lvert S\cap I\rvert. \]

{#eq:eq1} We denote the direction of \(I\) with \(\eta\in \mathbb{S}^{d-1}\).

If \(z\in D(0,5)\) then \(x+z\lvert I\rvert\eta\in \tilde{D}\) and therefore the function

\[ F(z)= \frac{\lvert K\rvert^{1/p}}{\lVert f\rVert_{L^p(K)}} f(x+z\lvert I\rvert\eta) \]

is well defined. The function \(F\) is analytic on \(D(0,5)\) and it satisfies \(F(0)\ge 1\) according to the selection of the point \(x\).

Applying (0x6698f5c7) on \(F\) yields

\[ 1 \le \sup_{t\in [0,1]}\lvert F(t)\rvert\le \biggl(\frac{12}{\lvert A\rvert}\biggr)^{2 \frac{\log M'}{\log 2}} \sup_{t\in A} \lvert F(t)\rvert, \]

{#eq:eq2} with \(M'=\max_{\lvert z\rvert \le 4} \lvert F(z)\rvert\) and \(A=\{t\in [0,1]\mid x+ t \lvert I\rvert \eta \in S\}\).

Using [@eq:eq1] and the estimate \(\lvert I\rvert \le \diam K\) yields

\[ \frac{1}{\lvert A\rvert} = \frac{\lvert I\rvert}{\lvert S\cap I\rvert} \le C\frac{(\diam K)^d}{\lvert S\rvert}. \]

{#eq:eq3}

We also note that \(M'\le M\). Combining this with [@eq:eq3], [@eq:eq2] and the estimate

\[ \sup_{t\in A} \lvert F(t)\rvert \le \frac{\lvert K\rvert^{1/p}}{\lVert f\rVert_{L^p(K)}}\sup_{y\in S} \lvert f(y)\rvert \]

gives us

\[ \lVert f\rVert_{L^p(K)}\le \lvert K\rvert^{1/p} \biggl(C \frac{(\diam K)^d}{\lvert S\rvert}\biggr)^{2\frac{\log M}{\log 2}} \sup_{x\in S} \lvert f(x)\rvert, \]

{#eq:eq4} where we transfered the norm of \(f\) to the left-hand side of the estimate.

Since \(S\) was arbitrary, we can apply (0x6698f127) , which yields

\[ \lVert f\rVert_{L^p(K)}\le (C(\diam K)^d)^{2 \frac{\log M}{\log 2}} \lvert K\rvert^{1/p}\biggl(\frac{2}{\lvert S \rvert}\biggr)^{2\frac{\log M}{\log 2} + \frac{1}{p}} \lVert f\rVert_{L^p(S)} \]

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