Two-weight Bergman inequality without the pointwise weight comparison condition

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Consider the upper half-plane R+2={z∈C:Im⁡z>0}\mathbb{R}_{+}^2=\{z\in\mathbb{C}:\operatorname{Im}z>0\}, two weights μ1\mu_1 and μ2\mu_2 on R+2\mathbb{R}_{+}^2, and the two-weight inequality

∫R+2∣∫R+2−f(w)(z−w‾)2 dA(w)∣pμ1(z) dA(z)≤C∫R+2∣f(z)∣pμ2(z) dA(z).\int_{\mathbb{R}_+^2}\left|\int_{\mathbb{R}_+^2}-\frac{f(w)}{(z-\overline{w})^2}\,dA(w)\right|^p\mu_1(z)\,dA(z)\le C\int_{\mathbb{R}_+^2}|f(z)|^p\mu_2(z)\,dA(z).

Here (μ1,μ2)∈Ap+(R+2)(\mu_1,\mu_2)\in A_p^+(\mathbb{R}_+^2) denotes the relevant two-weight Muckenhoupt condition. The two-weight Bergman inequality conjecture. For p>1p>1, if μ1\mu_1 and μ2\mu_2 satisfy (μ1,μ2)∈Ap+(R+2)(\mu_1,\mu_2)\in A_p^+(\mathbb{R}_{+}^2), then the displayed inequality holds for some C>0C>0.

This asks whether the pointwise hypothesis cμ1≥μ2c\mu_1\geq\mu_2 used in the proposition can be removed from the sufficient direction. It is presented as an open problem related to the two-weight inequality.

References

Primary source

Liwei Chen, “Weighted Bergman Projection on the Hartogs Triangle”, arXiv:1410.6205 (2015).

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