On the embedding of \(A_1\) into \(A_\infty \) (Q2817021)

The author considers the extreme case of the inclusion \(A_{p_1} \subseteq A_{p_2}, p_1 \leq p_2\), namely \(A_1 \subseteq A_{\infty}\). A weight \(w \geq 0\) belongs to \(A_1\) if there is a constant \(C\) such that \(Mf(x) \leq Cw(x), x \in \mathbb R^n\), where \(M\) is the uncentered Hardy-Littlewood maximal function, and it belongs to \(A_{\infty}\) if there is an \(\epsilon>0\), and a constant \(C_{\epsilon}\) such that NEWLINE\[NEWLINE \frac{ w(E)}{w(P)} \leq C_{\epsilon} \left( \frac{ E|}{ |P|} \right)^{\epsilon}, NEWLINE\]NEWLINE for all cubes \(P\), and measurable subsets \(E \subseteq P\). The best constants in the \(A_1\) inequality is denoted \([w]_{A_1}\), and there is a corresponding \([w]_{A_{\infty}}\) defined as NEWLINE\[NEWLINE [w]_{A_{\infty}} = \sup_{x \in Q} \frac{1}{w(Q)} \int_Q M(w \mathbf{1}_Q) \, dx. NEWLINE\]NEWLINE Finally when one considers only dyadic cubes and the dyadic maximal function, there are classes \(A_1^d\) and \(A_{\infty}^d\)NEWLINENEWLINEThe question under investigation is the optimal choice of \(\epsilon\). Previous research by \textit{A. D. Melas} [Bull. Lond. Math. Soc. 37, No. 6, 919--926 (2005; Zbl 1087.42015)] had shown that if \(w \in A_1^d\), the \(A_{\infty}^d\) inequality holds for NEWLINE\[NEWLINE 0 < \epsilon < \frac{\log(1 - \frac{2^n - 1}{2^n [w]_{A_1^d}})}{n \log 2} := \epsilon([w]_{A_1^d}, n) , NEWLINE\]NEWLINE and raising the question of whether \(\epsilon\) could attain the maximal value. \textit{A. Osękowski} [Arch. Math. 101, No. 2, 181--190 (2013; Zbl 1435.42014)] showed that this was possible by proving a weak type inequality; however, it only showed that this optimal choice held if \([w]_{A_1^d} \leq M\), and the constant \(C_{\epsilon_{(M, n)}}\) blows up if if \(M \to \infty\).NEWLINENEWLINEThe author calculates the Bellman function for the problem and uses it to strengthen the weak type inequality of Osękowski and show that you can have the optimal choice \(\epsilon([w]_{A_1^d}, n)\) in the \(A_{\infty}\) inequality even as \(M \to \infty\).NEWLINENEWLINEReviewer's remark: The author uses \(d\) as a superscript of the dyadic version of weights and for the dimension, and uses \(Q\) for cubes and for the upper bound of the \(A_1^d\) constant. For this review, I have substituted \(n\) for the dimension and replaced the number \(Q\) by \(M\).