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Let $\Omega = \mathbb B_n,$ the unit ball in $\mathbb C^n$ and $L^2_a(\Omega)$ be the Bergman space endowed with the normalized volume measure on $\Omega.$ Let $k_{\lambda}$ be the associated Bergman reproducing kernel. Let $\varphi_{\lambda}$ be an analytic automorphism of $\Omega$ having the properties $:$

$(1)$ $\varphi_{\lambda} (\lambda) = 0$

$(2)$ $\varphi_{\lambda} \circ \varphi_{\lambda} = \text{Id}_{\Omega}.$

It is proved in section 2.2 in Rudin's Function Theory in the Unit Ball of $\mathbb C^n$ that the real Jacobian of $\varphi_{\lambda}$ has the following form $:$ $$J_{\mathbb R} (\varphi_{\lambda}) (z) = \frac {\left \lvert k_{\lambda} (z) \right \rvert^2} {k_{\lambda} (\lambda)}.$$ Hence for any Lebesgue integrable or non-negative Lebesgue measurable function $h$ on $\Omega$ we have the following change of variable formula $:$ $$\int_{\Omega} h(\varphi_{\lambda} (w))\ dV(w) = \frac {1} {k_{\lambda} (\lambda)} \int_{\Omega} h(z)\ \left \lvert k_{\lambda} (z) \right \rvert^2\ dV(z).$$

Using the above formula I am trying to show that $\langle(f\circ\varphi_{\lambda})k_{\lambda}, (g\circ\varphi_{\lambda})k_{\lambda}\rangle=k_{\lambda}(\lambda)\langle f,g\rangle.$ But I am having hard time showing that. By simplifying we get $$\begin{align*} \langle(f\circ\varphi_{\lambda})k_{\lambda}, (g\circ\varphi_{\lambda})k_{\lambda}\rangle & = \int_{\Omega} f(\varphi_{\lambda} (w)) \overline {g (\varphi_{\lambda} (w))} \left \lvert k_{\lambda} (w) \right \rvert^2\ dV(w) \\ & = \frac {1} {k_{\lambda} (\lambda)} \int_{\Omega} f(z) \overline {g(z)} \left \lvert k_{\lambda} (\varphi_{\lambda} (z)) \right \rvert^2 \left \lvert k_{\lambda} (z) \right \rvert^2\ dV(z). \end{align*}$$

This does not seem like the desired expression for the inner product that I intended to show. Where do I make mistake? Any suggestion would be warmly appreciated.

Thanks for your time.

Anacardium
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1 Answers1

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$\varphi_{\lambda} \circ \varphi_{\lambda} = \text{Id}_{\Omega}$ implies that, for all $z \in \Omega$ : $$1 = J_\mathbb{R}(\text{Id}_{\Omega})(z) = J_\mathbb{R}(\varphi_{\lambda} \circ \varphi_{\lambda})(z) = J_\mathbb{R}(\varphi_\lambda)(\varphi_\lambda(z)) \cdot J_\mathbb{R}(\varphi_\lambda)(z) = \frac {\left \lvert k_{\lambda} (\varphi_\lambda(z)) \right \rvert^2} {k_{\lambda} (\lambda)} \frac {\left \lvert k_{\lambda} (z) \right \rvert^2} {k_{\lambda} (\lambda)}$$ Thus, going back to your calculation: $$\begin{split} \langle(f\circ\varphi_{\lambda})k_{\lambda}, (g\circ\varphi_{\lambda})k_{\lambda}\rangle &= \int_{\Omega} f(\varphi_{\lambda} (w)) \overline {g (\varphi_{\lambda} (w))} \left \lvert k_{\lambda} (w) \right \rvert^2 \mathrm{d}V(w)\\ & = \frac {1} {k_{\lambda} (\lambda)} \int_{\Omega} f(z) \overline {g(z)} \left \lvert k_{\lambda} (\varphi_{\lambda} (z)) \right \rvert^2 \left \lvert k_{\lambda} (z) \right \rvert^2 \mathrm{d}V(z)\\ &= k_\lambda(\lambda) \int_{\Omega} f(z) \overline {g(z)} \frac{\left \lvert k_{\lambda} (\varphi_{\lambda} (z)) \right \rvert^2 \left \lvert k_{\lambda} (z) \right \rvert^2}{k_\lambda(\lambda)^2} \mathrm{d}V(z)\\ &= k_\lambda(\lambda) \int_\Omega f(z)\overline{g(z)} \mathrm{d}V(z)\\ &= k_\lambda(\lambda) \langle f, g\rangle\end{split}$$

Bruno B
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