Show $\int_ {\mathbb{D}}f(\frac{\alpha(x_1+ix_2)+\beta}{\bar{\beta}(x_1+ix_2)+\bar{\alpha}})d\mu(x_1,x_2)=\int_ {\mathbb{D}}f(x_1+ix_2)d\mu(x_1,x_2)$

Question

Definitions needed in the problem:

$\text{SU}(1,1)=\left\{\left( \begin{array}{ccc}
\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right)\mid \alpha,\beta\in \mathbb C,|\alpha|^2-|\beta|^2=1\right\}$

$\mathbb{D}={\{(x_1+ix_2)\in\mathbb{C}:(x_1^2+x_2^2)<1}\}$

Problem:

Let $d\mu(x_1, x_2) =\frac{4}{(1- (x_1 ^ 2 + x_2 ^ 2)) ^ 2}d\lambda(x_1, x_2)$ where $\lambda$ denotes the measure of Lebesgue on the unit disk $\mathbb{D} $. Show that for all $\left( \begin{array}{ccc}
\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right)\in\text{SU}(1,1)$, $f\in\text{L}_1(\mathbb{D},\mu)$:

$\int_ {\mathbb{D}}f(\frac{\alpha(x_1+ix_2)+\beta}{\bar{\beta}(x_1+ix_2)+\bar{\alpha}})d\mu(x_1,x_2)=\int_ {\mathbb{D}}f(x_1+ix_2)d\mu(x_1,x_2)$

Idea (proof): Let $\left( \begin{array}{ccc}
\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right)\in\text{SU}(1,1)$, $f\in\text{L}_1(\mathbb{D},\mu)$ then $|\alpha|^2-|\beta|^2=1$ and

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$\int_ {\mathbb{D}}f(\frac{\alpha(x_1+ix_2)+\beta}{\bar{\beta}(x_1+ix_2)+\bar{\alpha}})d\mu(x_1,x_2)=\int_ {\mathbb{D}}f(\frac{\beta\bar{\beta}-\alpha\bar{\alpha}}{\bar{\beta}^2}((x_1+ix_2)+\frac{\bar{\alpha}}{\bar{\beta}})^{-1}+\frac{\alpha}{\bar{\beta}})d\mu(x_1,x_2)=\int_ {\mathbb{D}}f(\frac{-1}{\bar{\beta}^2}((x_1+ix_2)+\frac{\bar{\alpha}}{\bar{\beta}})^{-1}+\frac{\alpha}{\bar{\beta}})d\mu(x_1,x_2)$

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How can I continue?
Can anybody help me please?

Thanks…

Answer 1

Here are some hints for you.

You can show that $\varphi:z\to\frac{z\alpha+\beta}{z\bar{\beta}+\bar{\alpha}}$ is an isometry based on the hyperbolic metric, and you can also prove that it is a bijective function of $\mathbb{D}$ onto $\mathbb{D}$. Therefore, $d\mu(z)=d\mu(\varphi(z))$ and thus $$\int_{\mathbb{D}}f(\varphi(z))d\mu(\varphi(z))=\int_{\mathbb{D}}f(z)d\mu(z)$$ since $\varphi(\mathbb{D})=\mathbb{D}$. Because $|\varphi(\infty)|>1$ and $|\varphi^{-1}(\infty)|>1$, this Mobius transformation is holomorphic on $\mathbb{D}$.

If you need more details of the proof, please check the link below or make a comment.

Prove that $\varphi$ is an isometry