\documentclass[CRMATH,Unicode,biblatex,XML,published]{cedram} \TopicFR{Analyse harmonique} \TopicEN{Harmonic analysis} \addbibresource{CRMATH_Lerner_20240189.bib} %\makeatletter %\def\@setafterauthor{% % \vglue3mm% %% \hspace*{0pt}% %\begingroup\hsize=12cm\advance\hsize\abstractmarginL\raggedright %\noindent %%\hspace*{\abstractmarginL}\begin{minipage}[t]{10cm} % \leftskip\abstractmarginL % \normalfont\Small % \@afterauthor\par %\endgroup %\vskip2pt plus 3pt minus 1pt %} %\makeatother \usepackage{mathrsfs,amssymb} \usepackage{mathtools} \usepackage{pgfplots} \pgfplotsset{every axis/.append style={tick label style={/pgf/number format/fixed},font=\scriptsize,ylabel near ticks,xlabel near ticks,grid=major}} \DeclareMathOperator*{\esssup}{ess\,sup} \DeclareMathOperator*{\essinf}{ess\,inf} \newcommand{\ddn}{\mathrm{d}} \let \a=\alpha \newenvironment{claim}{\begin{enonce}{Claim}}{\end{enonce}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \makeatletter \let\save@mathaccent\mathaccent \newcommand*\if@single[3]{% \setbox0\hbox{${\mathaccent"0362{#1}}^H$}% \setbox2\hbox{${\mathaccent"0362{\kern0pt#1}}^H$}% \ifdim\ht0=\ht2 #3\else #2\fi } %The bar will be moved to the right by a half of \macc@kerna, which is computed by amsmath: \newcommand*\rel@kern[1]{\kern#1\dimexpr\macc@kerna} %If there's a superscript following the bar, then no negative kern may follow the bar; %an additional {} makes sure that the superscript is high enough in this case: \newcommand*\widebar[1]{\@ifnextchar^{{\wide@bar{#1}{0}}}{\wide@bar{#1}{1}}} %Use a separate algorithm for single symbols: \newcommand*\wide@bar[2]{\if@single{#1}{\wide@bar@{#1}{#2}{1}}{\wide@bar@{#1}{#2}{2}}} \newcommand*\wide@bar@[3]{% \begingroup \def\mathaccent##1##2{% %Enable nesting of accents: \let\mathaccent\save@mathaccent %If there's more than a single symbol, use the first character instead (see below): \if#32 \let\macc@nucleus\first@char \fi %Determine the italic correction: \setbox\z@\hbox{$\macc@style{\macc@nucleus}_{}$}% \setbox\tw@\hbox{$\macc@style{\macc@nucleus}{}_{}$}% \dimen@\wd\tw@ \advance\dimen@-\wd\z@ %Now \dimen@ is the italic correction of the symbol. \divide\dimen@ 3 \@tempdima\wd\tw@ \advance\@tempdima-\scriptspace %Now \@tempdima is the width of the symbol. \divide\@tempdima 10 \advance\dimen@-\@tempdima %Now \dimen@ = (italic correction / 3) - (Breite / 10) \ifdim\dimen@>\z@ \dimen@0pt\fi %The bar will be shortened in the case \dimen@<0 ! \rel@kern{0.6}\kern-\dimen@ \if#31 \overline{\rel@kern{-0.6}\kern\dimen@\macc@nucleus\rel@kern{0.4}\kern\dimen@}% \advance\dimen@0.4\dimexpr\macc@kerna %Place the combined final kern (-\dimen@) if it is >0 or if a superscript follows: \let\final@kern#2% \ifdim\dimen@<\z@ \let\final@kern1\fi \if\final@kern1 \kern-\dimen@\fi \else \overline{\rel@kern{-0.6}\kern\dimen@#1}% \fi }% \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax}% \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a %The following initialises \macc@kerna and calls \mathaccent: \if#31 \macc@nested@a\relax111{#1}% \else %If the argument consists of more than one symbol, and if the first token is %a letter, use that letter for the computations: \def\gobble@till@marker##1\endmarker{}% \futurelet\first@char\gobble@till@marker#1\endmarker \ifcat\noexpand\first@char A\else \def\first@char{}% \fi \macc@nested@a\relax111{\first@char}% \fi \endgroup } \makeatother \let\oldbar\bar \renewcommand*{\bar}[1]{\mathchoice{\widebar{#1}}{\widebar{#1}}{\widebar{#1}}{\oldbar{#1}}} \let\oldtilde\tilde \renewcommand*{\tilde}[1]{\mathchoice{\widetilde{#1}}{\widetilde{#1}}{\oldtilde{#1}}{\oldtilde{#1}}} %\let\tilde\widetilde %\let\hat\widehat \let\oldhat\hat \renewcommand*{\hat}[1]{\mathchoice{\widehat{#1}}{\widehat{#1}}{\oldhat{#1}}{\oldhat{#1}}} %\let\tilde\widetilde \renewcommand*{\to}{\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}} \let\oldmapsto\mapsto \renewcommand*{\mapsto}{\mathchoice{\longmapsto}{\oldmapsto}{\oldmapsto}{\oldmapsto}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} \hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}} \newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}} \newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}} \newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title[Sharpness of some quantitative Muckenhoupt--Wheeden inequalities] {On the sharpness of some quantitative Muckenhoupt--Wheeden inequalities} \alttitle{Optimalité des inégalités quantitatives de Muckenhoupt--Wheeden} %%%1 \author{\firstname{Andrei K.} \lastname{Lerner}\IsCorresp} \address{Department of Mathematics, Bar-Ilan University, 5290002 Ramat Gan, Israel} \email{lernera@math.biu.ac.il} %%%2 \author{\firstname{Kangwei} \lastname{Li}} \address{Center for Applied Mathematics, Tianjin University, Weijin Road 92, 300072 Tianjin, China} \email{kli@tju.edu.cn} %%%3 \author{\firstname{Sheldy} \lastname{Ombrosi}} \address{Departamento de Análisis Matemático y Matemática Aplicada\\ Universidad Complutense, Spain} \address{Departamento de Matemática e Instituto de Matemática. Universidad Nacional del Sur - CONICET Argentina} \email{sombrosi@ucm.es} %%4 \author{\firstname{Israel P.} \lastname{Rivera-R\'ios}} \address{Departamento de An\'alisis Matem\'atico, Estad\'istica e Investigaci\'on Operativa y Matem\'atica Aplicada. Facultad de Ciencias. Universidad de M\'alaga (M\'alaga, Spain).} \email{israelpriverarios@uma.es} \thanks{The first author was supported by ISF grant no. 1035/21. The second author was supported by the National Natural Science Foundation of China through project numbers 12222114 and 12001400. The fourth author was supported by Spanish Ministerio de Ciencia e Innovaci\'on grant PID2022-136619NB-I00 and by Junta de Andaluc\'ia grant FQM-354.} \CDRGrant[ISF]{1035/21} \CDRGrant[Natural Science Foundation of China]{12222114} \CDRGrant[Natural Science Foundation of China]{12001400} \CDRGrant[Spanish Ministerio de Ciencia e Innovaci\'on] {PID2022-136619NB-I00} \CDRGrant[Junta de Andaluc\'ia]{FQM-354} \keywords{\kwd{Matrix weights} \kwd{quantitative bounds} \kwd{endpoint estimates}} \altkeywords{\kwd{Poids matriciel} \kwd{borne quantitative} \kwd{estimations de type faible $(1,1)$}} \subjclass{42B20, 42B25} \begin{abstract} In the recent work [Cruz-Uribe et al. (2021)] it was obtained that \[ |\{x\in{\mathbb{R}^d}:w(x)|G(fw^{-1})(x)|>\alpha\}|\lesssim\frac{[w]_{A_1}^2}{\alpha}\int_{{\mathbb{R}^d}}|f|\,\mathrm{d} x \] both in the matrix and scalar settings, where $G$ is either the Hardy--Littlewood maximal function or any Calder\'on--Zygmund operator. In this note we show that the quadratic dependence on $[w]_{A_1}$ is sharp. This is done by constructing a sequence of scalar-valued weights with blowing up characteristics so that the corresponding bounds for the Hilbert transform and maximal function are exactly quadratic. \end{abstract} \begin{altabstract} Dans le récent travail [Cruz-Uribe et al. (2021)], il a été démontré \[ |\{x\in{\mathbb{R}^d}:w(x)|G(fw^{-1})(x)|>\alpha\}|\lesssim\frac{[w]_{A_1}^2}{\alpha}\int_{{\mathbb{R}^d}}|f|\,\mathrm{d} x \] à la fois dans les contextes matriciel et scalaire, où $G$ est soit la fonction maximale de Hardy-Littlewood ou tout opérateur de Calderón-Zygmund. Dans cette note, nous démontrons que la dépendance quadratique par rapport à $[w]_{A_1}$ est optimale. Cela est réalisé en construisant une séquence de poids à valeurs scalaires avec des caractéristiques d'éclatements, de sorte que les bornes correspondantes à la transformation de Hilbert et la fonction maximale soient exactement quadratiques. \end{altabstract} \dateposted{2024-11-05} \begin{document} \maketitle \section{Introduction} Recall that a non-negative locally integrable function $w$ satisfies the $A_1$ condition if there exists a constant $C>0$ such that for every cube $Q\subset {\mathbb R}^d$, \[ \frac{1}{|Q|}\int_Qw\le C\essinf_{Q}w. \] The smallest constant $C$ for which this property holds is denoted by~$[w]_{A_1}$. In the 70s, Muckenhoupt and Wheeden \cite{MW77} established weighted weak type $(1,1)$ bounds of the form \begin{equation} |\{x\in{\mathbb{R}}:w(x)|T(fw^{-1})(x)|>\a\}|\lesssim\frac{C_{w}}{\a}\int_{{\mathbb{R}}}|f|\,\ddn x,\label{weakscalar} \end{equation} where $T$ is either the Hilbert transform $H$ or the Hardy--Littlewood maximal operator~$M$. They showed as well that $w\in A_{1}$ is a sufficient condition for those inequalities to hold, even though it is not necessary. Muckenhoupt and Wheeden observed that (\ref{weakscalar}) could be regarded as a first step to settle inequalities of the form \begin{equation} u^{1-r}\left(\left\{u^r|Tf|>\a\right\} \right)\lesssim\frac{1}{\a}\int_{\mathbb{R}}|f|w\,\ddn x\label{eq:2wMW}, \end{equation} where $u,w$ are non-negative functions and $r\in[0,1]$. Note that for $u=w$, for $r=0$ this inequality is the standard weak type inequality and in the case $r=1$ it reduces to (\ref{weakscalar}). Their idea was to combine (\ref{eq:2wMW}) with interpolation with change of measures in order to obtain two weighted estimates. Pushing forward that idea, Sawyer \cite{S85} showed that \begin{equation} uv\left(\left\{ \frac{M(fv)}{v}>\a\right\} \right)\lesssim C_{u,v}\frac{1}{\a}\int_{\mathbb{R}}|f|uv\,\ddn x,\label{eq:S} \end{equation} where $u,v\in A_{1}$. This estimate combined with interpolation with change of measures allowed him to reprove Muckenhoupt's maximal theorem. Since the aforementioned papers a number of works have been devoted to estimates related to the ones above, that are known in the literature as mixed weak type estimates. Some worth mentioning are \cite{CUMP05} where (\ref{eq:S}) is extended to higher dimensions and further operators such as Calderón--Zygmund operators via extrapolation, or \cite{LOP19} where it is shown that $u\in A_{1}$ and $v\in A_\infty$ is sufficient for (\ref{eq:S}) to hold. In terms of quantitative estimates for $C_{u,v}$ in (\ref{eq:S}), and up to very recently for $C_w$ in \eqref{weakscalar}, as we will mention soon, not very much is known. Some results are provided in the aforementioned work \cite{LOP19} or in some other papers such as \cite{OPR} or \cite{CRR} for $C_{u,v}$ in (\ref{eq:S}). The purpose of this note is to provide some insight on $C_w$ in \eqref{weakscalar}. However, before presenting our main result we would like to connect this problem with the matrix weighted setting. We devote the following lines to that purpose. In the last years quantitative matrix weighted estimates have attracted the attention of a number of authors. Up until now only few sharp quantitative results in the matrix weight setting are known. Among them the sharp $L^p(W)$ bounds for the maximal operator~\cite{IM19}, the sharp $L^2(W)$ bound for the square function \cite{HPV19}, and also the sharp $L^p(W)$ bounds in terms of the $[W]_{A_q}$ constants with $1\leq q\a\bigr\}\bigr|\lesssim \frac{[W]_{A_1}^2}{\a}\int_{{\mathbb R}^d}|\vec{f}|\, \ddn x, \end{equation} and the same holds for $M_W$. \end{theo} \endgroup \addtocounter{cdrthm}{-1} At this point we are in the position to state the main result of this note. \begin{theorem}\label{mainresult} In the scalar-valued setting the quadratic dependence on $[w]_{A_1}$ in (\ref{weakmatrix}) is sharp for $T_w$ and for $M_w$. \end{theorem} As a direct consequence of Lemma \ref{lem:connect} below, this result shows the sharpness of $[W]_{A_1}^2$ in Theorem A in the matrix setting as well. An interesting phenomenon here is the contrast between the strong $L^2(W)$ and the weak $L^1(W)$ bounds for Calder\'on--Zygmund operators. As we mentioned above, the recent work {\cite{DPTV}} establishes the sharpness of $[W]_{A_2}^{3/2}$ in the matrix setting. Comparing this with the linear $A_2$ bound in the scalar case \cite{H12}, we see that the sharp weighted $L^2$ bounds for Calder\'on--Zygmund operators are different in the matrix and scalar settings. However, Theorem \ref{mainresult} shows that the sharp weighted weak $L^1$ bounds are the same in both settings. \section{Proof of Theorem \ref{mainresult}} \subsection{Connection between the scalar and the matrix weighted estimates} \begin{lemma}\label{lem:connect} Assume that $G_{W}$ stands either for $T_{W}$ or for $M_{W}$. Then if \[ \bigl|\bigl\{ x\in\mathbb{R}^{d}\ :\ |G_{W}\vec{f}(x)|>\alpha\bigr\} \bigr|\leq c\varphi\left([W]_{A_{1}}\right)\frac{1}{\alpha}\int_{{\mathbb R}^d}|\vec{f}|\, \ddn x, \] we have that for every $w\in A_{1}$ \[ \bigl|\bigl\{ x\in\mathbb{R}^{d}\ :\ |J_{w}f(x)|>\alpha\bigr\} \bigr|\leq c\varphi\left([w]_{A_{1}}\right)\frac{1}{\alpha}\int_{{\mathbb R}^d}|f|\, \ddn x, \] where $J_{w}$ stands, respectively, for $wT(fw^{-1})$ or for $wM(fw^{-1})$. \end{lemma} \begin{proof} Let $w\in A_{1}$. It is clear that $W=wI_{n}$ is a matrix $A_{1}$ weight. Furthermore, \[ [W]_{A_{1}}=[w]_{A_{1}}. \] Now given a scalar function $f$, we build $\vec{f}=(f,0,\dots,0)^{t}.$ Note that for these choices of $\vec{f}$ and $W$, clearly \[ \big|G_{W}\vec{f}(x)\big|=|J_{w}f(x)|. \] This ends the proof. \end{proof} \subsection{Construction of a family of weights providing the lower bound} We prove Theorem \ref{mainresult} via the following result. \begin{theorem}\label{thm:m} For any integer $N>20$, there exists a scalar weight $w\in A_1(\mathbb R)$ satisfying the following properties: \begin{enumerate} \item $\int_0^1w=1$; \item $[w]_{A_1}\simeq N$; \item $|\{x\in (1, \infty): w(x)> x\}|\gtrsim N^2$. \end{enumerate} \end{theorem} Observe that Theorem \ref{thm:m} implies Theorem \ref{mainresult} immediately because if we take $f=\chi_{[0,1]}$, then for $x>1$, \[ Mf(x)= \frac 1x,\qquad Hf(x)= \int \frac{f(y)}{x-y} \,\ddn y >\frac 1x. \] Hence, if $T$ is either $M$ or $H$, then \begin{align*} \| w Tf\|_{L^{1,\infty}}&\ge \bigl|\bigl\{x\in (1, \infty): w(x)|Tf(x)|>1 \bigr\} \bigr|\\ & \ge \left|\left\{x\in (1, \infty): w(x)\cdot\frac 1x>1 \right\} \right|\\ &\gtrsim N^2 \simeq [w]_{A_1}^2 \|f\|_{L^1(w)}. \end{align*} The rest of this section will be devoted to proving Theorem \ref{thm:m}. \begin{proof}[Proof of Theorem \ref{thm:m}] For $k=2,3,\ldots, N$ we denote $J_k=[2^k, 2^{k+1})$. We will split $J_k$ into small intervals. Set $I_k=[2^k, 2^k+k)$ and $L_k= J_k\setminus I_k=[2^k+k, 2^{k+1})$. Let $L_k^{-}$ and $L_k^{+}$ be the left and right halves of $L_k$, respectively. Next we define $(L_k^{-})^1$ to be the right half of $L_k^{-}$ and $(L_k^{+})^1$ the left half of $L_k^{+}$. Then \begin{itemize} \item when $\big(L_k^{-}\big)^j=\big[a_{k}^j, b_k^j\big)$ is defined, let $\big(L_k^{-}\big)^{j+1}=\big[a_{k}^{j+1}, b_{k}^{j+1}\big)$ satisfy that \[ b_{k}^{j+1}=a_k^j,\qquad \bigl|(L_k^{-})^{j+1}\bigr|= \frac 12 \bigl|(L_k^{-})^j\bigr|; \] \item when $\big(L_k^{+}\big)^j=\big[c_{k}^j, d_k^j\big)$ is defined, let $\big(L_k^{+}\big)^{j+1}=\big[c_{k}^{j+1}, d_{k}^{j+1}\big)$ satisfy that \[ c_{k}^{j+1}=d_k^j,\qquad \bigl|(L_k^{+})^{j+1}\bigr|= \frac 12 \bigl|(L_k^{+})^j\bigr|. \] \end{itemize} The process is stopped when we have $(L_k^{-})^{k-1}$ and $(L_k^{+})^{k-1}$ defined, and we simply define \[ (L_k^{-})^{k}= \left[2^k+k, 2^k+k+ \frac{|L_k^-|}{2^{k-1}}\right),\,\, (L_k^{+})^{k}= \left[2^{k+1}-\frac{|L_k^+|}{2^{k-1}}, 2^{k+1}\right). \] Now we have split $J_k$ into disjoint intervals, namely, \[ J_k= I_k \cup \bigcup_{j=1}^k (L_k^{-})^j \cup \bigcup_{j=1}^k (L_k^{+})^j. \] \begin{figure}[h!] \centering \makebox[\textwidth][c]{ \begin{tikzpicture}[scale=0.84] \draw (16,0.25)-- (32,0.25); %Axis \draw[line width=2.5pt, blue] (16,0.25)-- (20,0.25); \foreach \x in {16,20,32} { \draw (\x,0.75) -- (\x,-0.25); %\draw (\x,0.5) -- (\x,-0.5) node[below] {\x}; } \draw (18,0.4) node[above] {$I_4$}; \draw (26,0.4) node[above] {$L_4$}; \draw[line width=2.5pt, olive] (20,0.25)-- (26,0.25); \draw[line width=2.5pt, orange] (26,0.25)-- (32,0.25); \foreach \x in {20,26,32} { \draw (\x,0.5) -- (\x,0); %\draw (\x,0.5) -- (\x,-0.5) node[below] {\x}; } % \draw (23,0.25) node[below] {$L_4^-$}; % \draw (29,0.25) node[below] {$L_4^+$}; \draw[line width=2.5pt, olive] (23,0.25)-- (26,0.25); \draw[line width=2.5pt, orange] (26,0.25)-- (29,0.25); \foreach \x in {23,26,29} { \draw (\x,0.5) -- (\x,0); %\draw (\x,0.5) -- (\x,-0.5) node[below] {\x}; } \draw (24.5,0.25) node[below] {$(L_4^-)^1$}; \draw (27.5,0.25) node[below] {$(L_4^+)^1$}; \draw[line width=2.5pt, olive] (21.5,0.25)-- (23,0.25); \draw[line width=2.5pt, orange] (29,0.25)-- (30.5,0.25); \foreach \x in {21.5,23,29,30.5} { \draw (\x,0.5) -- (\x,0); %\draw (\x,0.5) -- (\x,-0.5) node[below] {\x}; } \draw (22.25,0.25) node[below] {$(L_4^-)^2$}; \draw(29.75,0.25) node[below] {$(L_4^+)^2$}; % \draw[fill=black] (27,-5) circle (0.1); % \draw[fill=black] (25,-5) circle (0.1); % \draw[fill=black] (26,-5) circle (0.1); % \draw[line width=2.5pt, olive] (20.75,0.25)-- (21.5,0.25); \draw[line width=2.5pt, orange] (30.5,0.25)-- (31.25,0.25); \foreach \x in {21.5,20.75,31.25,30.5} { \draw (\x,0.5) -- (\x,0); %\draw (\x,0.5) -- (\x,-0.5) node[below] {\x}; } %\draw [->, thick] (21.125,-0.1) -- (21.125,0.25); %\draw [->, thick] (30.875,-0.1) -- (30.875,0.25); \draw (21.125,-0.1) node[below] {$(L_4^-)^3$}; \draw (30.875,-0.1) node[below] {$(L_4^+)^3$}; \draw[line width=2.5pt, olive] (20,0.25)-- (20.75,0.25); \draw[line width=2.5pt, orange] (31.25,0.25)-- (32,0.25); \foreach \x in {20,20.75,31.25,32} { \draw (\x,0.5) -- (\x,0); %\draw (\x,0.5) -- (\x,-0.5) node[below] {\x}; } % \draw [->, thick] (20.375,-0.6) -- (20.375,0.25); % \draw [->, thick] (31.625,-0.6) -- (31.625,0.25); \draw (20.375,-0.6) node[below] {$(L_4^-)^4$}; \draw (31.625,-0.6) node[below] {$(L_4^+)^4$}; %Axis % \foreach \x in {16,20,26,32} { % \draw (\x,0.5) -- (\x,-0.5) node[below] {\x}; % } % \draw (-10,1) -- (5,1); % \draw[fill=white] (-10,1) circle (0.25); % \fill (5,1) circle (0.25); % \draw (1,2) -- (7,2); % \fill (1,2) circle (0.25); % \draw[fill=white] (7,2) circle (0.25); \end{tikzpicture} } \caption{Component intervals of $J_4$.} \end{figure} Define \[ w_k = 2^{k+1} \chi_{I_k}+ \sum_{j=1}^k 2^j \chi_{ (L_k^{-})^j \cup (L_k^{+})^j } \] and our weight on $[0, 2^{N+2}]$ is \begin{equation*} w(x)=\begin{cases} \chi_{[0, 4)}(x)+ \sum_{k=2}^N w_k(x), \quad & x\in [0, 2^{N+1}),\\ 2^{N}, \quad & x=2^{N+1},\\ w(2^{N+2}-x), \quad & x\in [2^{N+1}, 2^{N+2}]. \end{cases} \end{equation*} \begin{figure}[h!] \centering \makebox[\textwidth][c]{ \begin{tikzpicture} \begin{axis}[ width=15cm,height=6cm, ymin=0,ymax=34,xmin=15,xmax=33, xtick={0,1,...,32}, ytick={0,2,...,32}, %legend style={at={(0.9,0.9)},anchor=north}, y axis line style={draw opacity=0}, % grid=none, grid style={dotted,gray!30!black}, yticklabels={,,}, xticklabels={,,}, axis x line=center, axis y line=center, axis line style={-}, ticks=none ] %\addlegendimage{line width=2.5pt,blue} %\addlegendimage{line width=2.5pt,olive} % \addlegendimage{line width=2.5pt,orange} % \addlegendentry{$I_4$} % \addlegendentry{$(L_4^-)^i$} % \addlegendentry{$(L_4^+)^i$} \addplot[line width=2.5pt,olive,samples at={20,20.75}] {16}; \addplot[line width=2.5pt,orange,samples at={31.25,32}] {16}; \addplot[line width=2.5pt,blue,samples at={16,20}] {32}; \foreach \j [evaluate=\j as \u using 4+16+(32-4-16)/2-(32-4-16)/2*(1-1/2^(\j+1)), evaluate=\j as \v using 4+16+(32-4-16)/2-(32-4-16)/2*(1-1/2^(\j)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2} \addplot[line width=2.5pt,olive,samples at={\u,\v}] {\w}; \foreach \j [evaluate=\j as \u using 4+16+(32-4-16)*(1-1/2^(\j+2)), evaluate=\j as \v using 4+16+(32-4-16)*(1-1/2^(\j+1)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2} \addplot[line width=2.5pt,orange,samples at={\u,\v}] {\w}; %\addplot[line width=2.5pt,blue,mark=*,samples at={0,1.1}] {0.5}; \end{axis} \end{tikzpicture} } \caption{Graph of $w_4$} \end{figure} \begin{figure}[h!] \centering \makebox[\textwidth][c]{ \begin{tikzpicture} \begin{axis}[width=15cm,height=6cm, ymin=0,ymax=130,xmin=15, xmax=130, xtick={0,8,...,128}, ytick={0,8,...,128}, legend style={at={(0.9,0.9)},anchor=north}, % grid=none, grid style={dotted,gray!30!black}, yticklabels={,,}, xticklabels={,,}, axis x line=center, axis y line=center, y axis line style={draw opacity=0}, axis line style={-}, ticks=none] % \addlegendimage{line width=2.5pt,blue} % \addlegendimage{line width=2.5pt,olive} % \addlegendimage{line width=2.5pt,orange} % \addlegendentry{$I_k$} % \addlegendentry{$(L_k^-)^i$} % \addlegendentry{$(L_k^+)^i$} \addplot[line width=2.5pt,olive,samples at={20,20.75}] {16}; \addplot[line width=2.5pt,orange,samples at={31.25,32}] {16}; \addplot[line width=2.5pt,blue,samples at={16,20}] {32}; \foreach \j [evaluate=\j as \u using 4+16+(32-4-16)/2-(32-4-16)/2*(1-1/2^(\j+1)), evaluate=\j as \v using 4+16+(32-4-16)/2-(32-4-16)/2*(1-1/2^(\j)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2} \addplot[line width=2.5pt,olive,samples at={\u,\v}] {\w}; \foreach \j [evaluate=\j as \u using 4+16+(32-4-16)*(1-1/2^(\j+2)), evaluate=\j as \v using 4+16+(32-4-16)*(1-1/2^(\j+1)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2} \addplot[line width=2.5pt,orange,samples at={\u,\v}] {\w}; \addplot[line width=2.5pt,olive,samples at={37,37.84375}] {32}; \addplot[line width=2.5pt,orange,samples at={63.15625,64}] {32}; \addplot[line width=2.5pt,blue,samples at={32,37}] {64}; \foreach \j [evaluate=\j as \u using 5+32+(64-5-32)/2-(64-5-32)/2*(1-1/2^(\j+1)), evaluate=\j as \v using 5+32+(64-5-32)/2-(64-5-32)/2*(1-1/2^(\j)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2,3} \addplot[line width=2.5pt,olive,samples at={\u,\v}] {\w}; \foreach \j [evaluate=\j as \u using 5+32+(64-5-32)*(1-1/2^(\j+2)), evaluate=\j as \v using 5+32+(64-5-32)*(1-1/2^(\j+1)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2,3} \addplot[line width=2.5pt,orange,samples at={\u,\v}] {\w}; \addplot[line width=2.5pt,olive,samples at={70,70.90625}] {64}; \addplot[line width=2.5pt,orange,samples at={127.09375,128}] {64}; \addplot[line width=2.5pt,blue,samples at={64,70}] {128}; \foreach \j [evaluate=\j as \u using 6+64+(128-6-64)/2-(128-6-64)/2*(1-1/2^(\j+1)), evaluate=\j as \v using 6+64+(128-6-64)/2-(128-6-64)/2*(1-1/2^(\j)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2,3,4} \addplot[line width=2.5pt,olive,samples at={\u,\v}] {\w}; \foreach \j [evaluate=\j as \u using 6+64+(128-6-64)*(1-1/2^(\j+2)), evaluate=\j as \v using 6+64+(128-6-64)*(1-1/2^(\j+1)), evaluate=\j as \w using 2^(\j+1)] in {0,1,2,3,4} \addplot[line width=2.5pt,orange,samples at={\u,\v}] {\w}; \end{axis} \end{tikzpicture} } \caption{Joint graph of $w_4$, $w_5$ and $w_6$.} \end{figure} Finally we extend $w(x)$ from $[0, 2^{N+2}]$ to $\mathbb R$ periodically with period $2^{N+2}$. Such a weight trivially satisfies $\int_0^1w=1$. Moreover, since $w(x)>x$ on $I_k$, we have \begin{align*} \big|\{x\in (1, \infty): w(x)> x\}\big|\ge \sum_{k=2}^N |I_k|=\sum_{k=2}^Nk\simeq N^2. \end{align*} Hence it remains to check that $[w]_{A_1} \simeq N$. Since $w$ is periodic on $\mathbb R$ and symmetrical on $[0, 2^{N+2}]$, it suffices to prove that \[ \postdisplaypenalty1000000 \sup_{I \subset [0, 2^{N+1}]}\frac {w(I)}{|I| \essinf\limits_{x\in I}w(x)} \simeq N \] (we use the standard notation $w(E)=\int_Ew$). Observe that $|L_k^{-}|=|L_k^{+}|=\frac{1}{2}(2^k-k)$. Further, \[ \big|(L_k^{-})^j\big|=\big|(L_k^{+})^j\big|=\frac{1}{2^{j+1}}(2^k-k),\quad j=1,\dots, k-1, \] and \[ \big|(L_k^{-})^k\big|=\big|(L_k^{+})^k\big|=\frac{1}{2^k}(2^k-k). \] Hence, \begin{align*} w_k(J_k)&=2^{k+1}|I_k|+\sum_{j=1}^k2^j\left(\big|(L_k^{-})^j\big|+\big|(L_k^{+})^j\big|\right)\\ &=2^{k+1}k+(k-1)(2^k-k)+2(2^k-k)\simeq k2^k. \end{align*} From this, when $I = [0, 2^{N+1}]$ we have \begin{align*} \frac {w(I)}{|I|}= 2^{-(N+1)}\left( 4+\sum_{k=2}^N w_k(J_k) \right) &\simeq 2^{-(N+1)} \left( 4+\sum_{k=2}^N k 2^{k}\right) \\&\simeq N = N \essinf\limits_{x\in I}w(x). \end{align*} Therefore, we are left to prove that for any $I\subset [0, 2^{N+1}]$, \begin{equation}\label{eq:a1up} \frac{w(I)}{|I|}\lesssim N \essinf\limits_{x\in I}w(x). \end{equation} At this point we will make the following elementary observation. Our weight $w$ is a step function, and for each two adjacent intervals from its definition, the ``jump" of $w$ is at most 2. Since the ``jumps" are multiplicative we have the following. \begingroup \renewcommand{\thecdrthm}{\Alph{cdrthm}} \setcounter{cdrthm}{0} \begin{claim} If $I$ intersects at most $m$ intervals from the definition of $w$, then \[ \frac{w(I)}{|I|}\le \max_{x\in I} w(x)\le 2^m \min_{x\in I} w(x)= 2^m\essinf\limits_{x\in I}w(x). \] \end{claim} \endgroup In what follows we will prove \eqref{eq:a1up} according to the size of $I$. \begin{proof}[Case 1] $|I|\le 4$. In this case, note that in each $J_k$ $(k\ge 2)$, $(L_k^-)^{k-1}, (L_k^-)^k$ and $(L_k^+)^{k-1},(L_k^+)^k$ are the smallest intervals, and \[ \big|(L_k^-)^{k-1}\big| =\big|(L_k^-)^k\big| =\big|(L_k^+)^{k-1}\big| = \big|(L_k^+)^k\big| = 1-\frac k{2^k}\ge \frac 12. \] Hence $I$ intersects at most $9$ intervals from the definition of $w$, and we are in position to apply Claim A with $m=9$. \let\qed\relax \end{proof} \begin{proof}[Case 2] $|I|> 4$. In this case, we may assume $|I|\in (2^{k_0}, 2^{k_0+1}]$ with some $k_0\ge 2$. We may further assume $k_0 2^{k_0+1}, \] so that $I$ will intersect at most $(L_\ell^+)^{j}$ and $(L_\ell^+)^{j+1}$ and we again apply Claim A with $m=2$. If $j \ge \ell- k_0-3$, note that then \[ I \subset \bigcup_{j=\ell-k_0-3}^\ell (L_\ell^+)^j \cup \bigcup_{i=\ell-k_0-2}^{\ell+1} (L_{\ell+1}^-)^i \cup I_{\ell+1}. \] Here $i\ge \ell-k_0-2$ since \[ \big|(L_{\ell+1}^-)^{\ell-k_0-2} \big|= \frac{2^{\ell+1}-(\ell+1)}{2^{\ell-k_0-1 }}> \frac{2^{\ell}}{2^{\ell-k_0-1 }}=2^{k_0+1}\ge \ell(I). \] Hence we have \begin{align*} \frac {w(I)}{|I| \essinf\limits_{x\in I} w(x)}&\le \frac{\sum_{j=\ell-k_0-3}^\ell w\big((L_\ell^+)^j \big)+ \sum_{i=\ell-k_0-2}^{\ell+1} w\big((L_{\ell+1}^-)^i \big)+ w(I_{\ell+1}) }{2^{k_0} 2^{\ell-k_0-3}}\\ &\lesssim \frac{\sum_{j=\ell-k_0-3}^\ell 2^j\cdot 2^{\ell-j}+ \sum_{i=\ell-k_0-2}^{\ell+1} 2^i \cdot 2^{\ell+1-i} + (\ell+1)2^{\ell+2}}{2^\ell } \lesssim \ell. \end{align*} It remains to consider the case $a\in I_{\ell+1}\cup L_{\ell+1}^-$. However, in this case we just need to discuss whether $b\in (L_{\ell+1}^-)^{j}$ with some $j \le \ell- k_0-3$ or not, which is completely similar. This completes the proof. \end{proof} \let\qed\relax \end{proof} \section*{Acknowledgement} We would like to thank the anonymous referee for his/her careful reading that helped improving the presentation of the paper. \section*{Declaration of interests} The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations. \printbibliography \end{document}