Does the following integral converge? Let $a\in(-1,1)$, let:
$$ P(x,y)=C_{n,a}\frac{y^{1-a}}{(|x|^2+y^2)^{(n+1-a)/2}},\quad\forall (x,y)\in \mathbb{R}^n\times(0,\infty),$$
let $f\in \mathcal{S}(\mathbb{R}^n)$, i.e. $f$ is a Schwartz function, let:
$$ u(x,y)=\int_{\mathbb{R}^n}P(x-\xi,y)f(\xi)\,d\xi,\quad\forall (x,y)\in \mathbb{R}^n\times(0,\infty),$$
is true that:
$$\int_0^\infty\int_{\mathbb{R}^n}y^a|u(x,y)|^2\,dx\,dy<\infty?$$
I heve no idea on how to proceed, any help is appreciated.
 A: In dimension one, this is false even when $a = 0$. In higher dimensions, however, the integral is indeed finite.
The proof involves just Plancherel's theorem: when $a = 0$ , the double integral is equal to
$$\int_0^\infty \int_{\mathbb R^n} |\hat u(\xi, y)|^2 d\xi dy = \int_0^\infty \int_{\mathbb R^n} |e^{-|\xi| y} \hat f(\xi)|^2 d\xi dy = \int_{\mathbb R^n} (2 |\xi|)^{-1} |\hat f(\xi)|^2 d\xi.$$
(Here $\hat u$ is the Fourier transform with respect to $x$ alone.)
Now $|\hat f(\xi)|^2$ is bounded and integrable. In dimension 2 and higher, $|\xi|^{-1}$ is a sum of a bounded function and an integrable function; hence the integral converges. In dimension 1, however, $|\xi|^{-1}$ is not locally integrable, and hence the integral diverges, unles $\hat f(0) = 0$, that is, $f$ has integral zero.
When $a \ne 0$, the argument is essentially the same, but the Fourier transform of $P(x, y)$ is no longer $e^{-|\xi| y}$, but rather $c_a (y |\xi|)^{(1-a)/2} K_{(1-a)/2}(y |\xi|)$  (or something very similar), where $K$ is the Bessel function.
