Let $f \in L^2(\mathbb{R}^n)$, and consider its harmonic extension in the upper half-plane given by $u(x,y) = (P(\cdot,y) * f)(x)$, where $$P(x,y) = C \frac{y}{(\lvert x \rvert^2 + y^2)^{\frac{n+1}{2}}}$$ is the Poisson kernel. My problem is to prove (or disprove) that $u \in L^2(\mathbb{R^n} \times (0, \infty))$.
The motivation comes from this paper, where in Theorem 9 the above problem is part of a more general argument. However, there is no proof in the paper.
My strategy was to split the integral $\lVert u \rVert_{L^2(\mathbb{R}^n \times (0,\infty))}^2$ into two parts for $0<y<1$ and $1<y<\infty$ and find appropriate estimates, but I've had no success in estimating the $L^2(\mathbb{R}^n)$-norm of $u$ when $y > 1$. Intuitively it seems clear that the $L^2(\mathbb{R}^n)$-norm diminishes as $y \to \infty$ but by what rate?
Any help is greatly appreciated!
Say $u_y(t)=u(t+iy)$. If you look at the Fourier transform of $P(.,y)$ you see that $$\widehat u_y(\xi)=e^{-y|\xi|}\widehat f(\xi).$$
Hence $$||u_y||_2^2=\int e^{-2y|\xi|}|\widehat f(\xi)|^2\,d\xi,$$so $$\int_0^\infty||u_y||_2^2\,dy=\int\frac1{2|\xi|}|\widehat f(\xi)|^2\,d\xi;$$ so $u\in L^2$ if and only if that last integral is finite.
(This even makes sense. The reason $||u_y||_2$ should tend to zero is cancellation in the integral; saying $\widehat f(0)=0$ says $\int f=0$, cancellation.)