Let $V$ be the normalized volume measure on $\mathbb D^2$ and $k : \mathbb D \times \mathbb D \longrightarrow \mathbb C$ be the Bergman kernel on $\mathbb D^2$ given by $$k(z,w) = \frac {1} {\left (1 - z_1 \overline {w_1} \right )^2 \left (1 - z_2 \overline {w_2} \right )^2}$$ for $z = (z_1, z_2)$ and $w = (w_1, w_2)$ in $\mathbb D^2.$ For $0 \lt \varepsilon \lt \frac {1} {2}$ define $$I(w_1, w_2) = \int_{\mathbb D^2} \left (\left (1 - |z_1|^2 \right ) \left (1 - |z_2|^2 \right ) \left \lvert 1 - z_1 \overline {z_2} \right \rvert^2 \right )^{-2 \varepsilon} |z_1 - z_2|\ |k(z,w) - k(z, \sigma \cdot w)|\ dV(z)$$ where $\sigma . w = (w_2, w_1).$ I would like to know whether the following estimate holds $:$
$$I(w_1, w_2) \leq \left (\left (1 - |w_1|^2 \right ) \left (1 - |w_2|^2 \right ) \left \lvert 1 - w_1 \overline {w_2} \right \rvert^2 \right )^{-2 \varepsilon} |w_1 - w_2|.$$ I am able to show that $$I(w_1, w_2) \leq \left (\left (1 - |w_1|^2 \right ) \left (1 - |w_2|^2 \right ) \left \lvert 1 - w_1 \overline {w_2} \right \rvert^2 \right )^{-2 \varepsilon}$$ whenever $0 \lt \varepsilon \lt \frac {1} {4}.$ The reason I am requiring that interval of $\varepsilon$ because then we can bound the term $\frac {\left \lvert z_1 - z_2 \right \rvert} {\left \lvert 1 - z_1 \overline {z_2} \right \rvert^{4 \varepsilon}}$ by $2^{1 - 4 \varepsilon}$ as $\left \lvert \frac {z_1 - z_2} {1 - z_1 \overline {z_2}} \right \rvert < 1.$ But I don't have any idea as to how to achieve the required bound.
Any suggestion in this regard would be greatly appreciated. Thanks for your time.
