Density size and tree count for carleson tiles

Assume The Axiom of Choice as inherited from the packet and real-line analytic conventions. Use the tiles, plus trees, fixed packets and finite model of Carleson tiles wave packets and tile order, and a measurable selector $N:\mathbb{R}\to \mathbb{R}$ as in Carleson operator and measurable linearisation. Fix $f\in L^2(\mathbb{R})$, a measurable testing set E of finite measure, and the integer $\kappa =20$. Put ${\chi }_I(x)=\mathrm{\mid }I{\mathrm{\mid }}^{-1}(1+\mathrm{\mid }x-c(I)\mathrm{\mid }\mathrm{/}\mathrm{\mid }I\mathrm{\mid }{)}^{-\kappa }\mathrm{.}$ For a tile s and a finite tile set S, define ${\mathrm{dens}}_{E,N}(s)={\sup }_{t>s}{\int }_{E\cap \{x:N(x)\in {\omega }_t\}}{\chi }_{I_t}(x)dx,{\mathrm{dens}}_{E,N}(S)={\sup }_{s\in S}{\mathrm{dens}}_{E,N}(s),$ where $t>s$ means $s\le t$ and $s\mathrm{\ne }t$ in the tile order. The supremum over t ranges over all strict ancestors of s, not just members of S. This family is nonempty: a dyadic spatial parent of $I_s$, paired with either dyadic half of ${\omega }_s$ of reciprocal length, is a strict ancestor. The integral tests the whole frequency interval ${\omega }_t$. Every integrand is nonnegative measurable. Substitution $u=(x-c(I))\mathrm{/}\mathrm{\mid }I\mathrm{\mid }$ and the antiderivative of $(1+u{)}^{-20}$ on $[0,\mathrm{\infty })$ give $\int {\chi }_I=2\mathrm{/}19$, so densities are finite and between zero and $2\mathrm{/}19$. For empty S define density zero; for null E every integral is zero. Null modifications of N or E leave each integral, and hence its supremum, unchanged.

Define ${\mathrm{size}}_f(S)={\sup }_{(T,t)}{(\frac{1}{\mathrm{\mid }{I}_t\mathrm{\mid }}{\sum }_{s\in T}\mathrm{\mid }⟨f,{\varphi }_s⟩{\mathrm{\mid }}^2)}^{1\mathrm{/}2},$ where the supremum ranges over nonempty plus subtrees $T\subseteq S$ with designated top t. In particular singleton trees with their own tile as top are allowed. Empty S has size zero. This supremum is finite: if $\mathrm{\ell }={\min }_{s\in S}\mathrm{\mid }{I}_s\mathrm{\mid }>0$, each candidate top has length at least ell and the numerator is at most the finite sum over S of the finite squared coefficients. Size zero is equivalent to every coefficient being zero: the forward implication follows from singleton trees and the reverse from the displayed sum. The pairing depends only on the $L^2$ class of f.

A displayed forest is a finite family of disjoint tile subcollections, each equipped with a designated top and forming a tree. Its count is ${\sum }_T\mathrm{\mid }{I}_T\mathrm{\mid }$, where $I_T$ means its designated top interval. This counts top lengths with multiplicity even when intervals overlap. The empty forest has count zero. A different assignment of tops may change the count; no intrinsic count is attached to a tile set without a forest or an explicit existence assertion about such a decomposition. Restricting S can only decrease density and size, because it restricts the families in their suprema. No selection estimate is part of these definitions.