Two Weight Inequalities for Riesz Transforms: Uniformly Full Dimension Weights

Fix an integer $n$ and number $d$, $0< d\neq n-1 \leq n$, and two weights $w$ and $σ$ on $\mathbb R ^{n}$. We two extra conditions (1) no common point masses and (2) the two weights separately are not concentrated on a set of codimension one, uniformly over locations and scales. (This condition holds for doubling weights.) Then, we characterize the two weight inequality for the $d$-dimensional Riesz transform on $\mathbb R ^{n}$, \begin{equation*} \sup_{0< a < b < \infty}\left\lVert \int_{a < \lvert x-y\rvert < b} f (y) \frac {x-y} {\lvert x-y\rvert ^{d+1}} \; σ(dy) \right\rVert_{L ^{2} (\mathbb{R}^n;w)} \le \mathscr N \lVert f\rVert_{L ^2 (\mathbb{R}^n;σ)} \end{equation*} in terms of these two conditions, and their duals: For finite constants $\mathscr A_2$ and $\mathscr T$, uniformly over all cubes $Q\subset \mathbb R ^{n}$ \begin{gather*} \frac {w (Q)} {\lvert Q\rvert ^{d/n}} \int_{\mathbb R ^{n}} \frac {\lvert Q\rvert ^{d/n}} {\lvert Q\rvert ^{2d/n} +{dist}(x, Q) ^{2d/n}} \; σ(dx) \leq \mathscr A_2 \\ \int_{Q} \lvert \mathsf R_σ \mathbf 1_{Q} (x)\rvert ^2 \; w(dx) \le \mathscr T ^2 σ(Q), \end{gather*} where $\mathsf R_σ$ denotes any of the truncations of the Riesz transform as above, the dual conditions are obtained by interchanging the roles of the two weights. Examples show that a key step of the proof fails in absence of the extra geometric condition imposed on the weights.

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