Suppose that $f \in L^2(\mathbb{R}^n)$ and let $P_y(x)$ ($x \in \mathbb{R}^n$, $y > 0$) be the dilation of the Poisson kernel:
$$P_y(x) = \frac{C_n y}{(y^2 + |x|^2)^\frac{n+1}{2}},$$
where $C_n$ is a constant (depending on $n$) such that $\int_{\mathbb{R}^n} P_y(x) dx = \int_{\mathbb{R}^n} P_1(x) dx = 1$ for any $y > 0$. Define $u_f(x,y) = P_y \ast f (x)$, and let $T(f)$ be the function:
$$T(f)(x) = \left(\int_0 ^\infty \int_0 ^\infty \frac{|u_f(x,y_1) - u_f(x,y_2)|^2}{|y_1 - y_2|^2}dy_1 dy_2 \right)^\frac{1}{2}.$$
Prove that
$$||T(f)||_2 = C||f||_2,$$
where $C$ is some constant independent of $f$.
Here is my attempt so far. We use Fubini's theorem to reorder the integration, and Plancherel's theorem to equate the integrals of our original function and its Fourier transform:
$$||T(f)||^2_2 = \int_{\mathbb{R^n}} \int_0 ^\infty \int_0 ^\infty \frac{|u_f(x,y_1) - u_f(x,y_2)|^2}{|y_1 - y_2|^2}dy_1 dy_2 dx = \\ \int_0 ^\infty \int_0 ^\infty \frac{1}{|y_1 - y_2|^2} \int_{\mathbb{R}^n} |u_f(x,y_1) - u_f(x,y_2)|^2 dx dy_1 dy_2 =\\ \int_0 ^\infty \int_0 ^\infty \frac{1}{|y_1 - y_2|^2} \int_{\mathbb{R}^n} |\widehat{u_f( \cdot ,y_1)}(\xi) - \widehat{u_f(\cdot ,y_2)}(\xi)|^2 d\xi dy_1 dy_2. $$
Then we take advantage of how the Fourier transform behaves on convolutions, and use the fact that $\hat{P_y}(\xi) = e^{-2\pi |y| |\xi|}$
$$ \widehat{u_f( \cdot ,y)}(\xi) = \hat{P_y}(\xi) \hat{f}(\xi) = e^{-2\pi |y| |\xi|} \hat{f}(\xi).$$
So now we have:
$$ ||T(f)||_2 = \int_0 ^\infty \int_0 ^\infty \frac{1}{|y_1 - y_2|^2} \int_{\mathbb{R}^n} |\hat{f}(\xi)|^2|e^{-2\pi |y_1| |\xi|} - e^{-2\pi |y_2| |\xi|} |^2 d\xi dy_1 dy_2,$$
and the term $|e^{-2\pi |y_1| |\xi|} - e^{-2\pi |y_2| |\xi|} |^2$ is messy and preventing me from moving any further. Hints or solutions on how to deal with this term are greatly appreciated!
In the last formula, integrate over $y_1,y_2$ first, moving $|\hat{f}(\xi)|^2$ out of the integral as a constant. $$ \|T(f)\|_2 =\int_{\mathbb{R}^n} |\hat{f}(\xi)|^2 d\xi \int_0 ^\infty \int_0 ^\infty \left|\frac{e^{-2\pi y_1 |\xi|} - e^{-2\pi y_2 |\xi|}}{y_1 - y_2} \right|^2 \, dy_1 \,dy_2 $$ A change of variables $y_1|\xi| = z_1$, $y_2|\xi| = z_2$ makes $\xi$ disappear from the inner double integral. $$ \|T(f)\|_2 =\int_{\mathbb{R}^n} |\hat{f}(\xi)|^2 d\xi \int_0 ^\infty \int_0 ^\infty \left|\frac{e^{-2\pi z_1 } - e^{-2\pi z_2 }}{z_1 - z_2} \right|^2 \, dz_1 \,dz_2 $$ The inner double integral will be that multiplicative constant, but you should check that it converges.