How can we prove that the hyperbolic half space $\displaystyle\left(M=\mathbb{H}^{n+1} \ , \ g=\frac{|dy|^2+dx^2}{x^2}\right)$ where $x>0$ and $y\in\mathbb{R}^n$ is a conformally compact Einstein manifold ?
I know that in order to define a conformally compact manifold , one needs to define a defining function $x$ of $\partial M$ (which is $\mathbb{R}^n$ in our case) in $M$ such that \begin{cases}x>0 \ \ \text{in} \ M \\x=0 \ \ \text{on} \ \partial M \\dx\neq0 \ \ \text{on} \ \partial M \end{cases} But in our case $\partial M=\mathbb{R}^n$ is not compact .
Then how it is possible to define $(M,g)$ as a conformally compact manifold ? Any help is appreciated .
The fact that the hyperbolic space is Einstein is a standard fact, so I'll focus on the conformal compactification.
Let me emphasize the fact that conformal compactifications are built to bring "infinity" back at finite distance (thanks to the conformal factor). This procedure should then take care of all the different directions that are pointing toward infinity. The problem with considering the topological boundary of the half space model of the hyperbolic space is that does not capture the conformal compactification: you are missing the point at infinity with $x \to \infty$. If you add it, you obtain as a boundary the one point compactification of $\Bbb R^n$, which is just a sphere. The ball model is more suitable to work with.
Let's then look at the ball model $(B(0;1),g)$, with $g = \frac{4}{(1-|x|^2)^2}g_{eucl}$, where $g_{eucl}$ is the Euclidean metric. Consider $\rho(x) = \frac{1-|x|^2}{2}$ which is a defining function of the sphere $S(0;1) = \partial B(0;1)$. Then $\rho^2 g = g_{eucl}$, and it follows that the hyperbolic space is conformally compact, with conformal compactification the closed ball $\bar{B}(0;1)$ and with conformal boundary $(S(0;1), [g_{eucl}|_S])$, the conformal class of the round sphere.