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

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

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

How can I continue? Can anybody help me please?

Thanks...

• It might be useful to mention where the problem comes from. – Martin Sleziak Oct 14 at 4:32

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

• Thank you very much friend, I already proved that $\varphi$ is a bijective function and an isometry with respect to hyperbolic metrics. I still have two questions, we know that $d\mu(z)=d\mu(\varphi(z))$ for being $\varphi$ an isometry. (First) Why is equality fulfilled $\int_{\mathbb{D}\cup\infty}f(\varphi(z))d\mu(\varphi(z))=\int_{\mathbb{D}\cup\infty}f(z)d\mu(z)$? – J.rafa Oct 14 at 22:16
• Thank you very much. It is very clear to me. (Second) Can you explain how you got that $\lim_{z\to\infty} |f(\varphi(z)d\mu(\varphi(z))|<\infty$? – J.rafa Oct 15 at 2:44
• @J.rafa $\varphi(\infty)=\frac{\alpha}{\bar{\beta}}$ and then plug it in you will find $|f(\varphi(\infty))d\mu(\varphi(\infty))|<\infty$ – Yourong Zang Oct 15 at 2:49
• Perfect thank you friend. – J.rafa Oct 15 at 2:51
• What book advises me to study these things? – J.rafa Oct 15 at 2:55