Lusin area integral

Remarks

The power $t^{1 - n}$ makes $S(f)$ invariant under dilations: $S(f(\delta \cdot))(x) = S(f)(\delta x)$. For $1 < p < \infty$ one has $c \norm{f}_p \le \norm{S(f)}_p \le C \norm{f}_p$ with constants depending only on $n$ and $p$. Replacing the cone by the vertical half-line over $x$, with the weight $t$, gives the Littlewood–Paley $g$-function.

Example: the $L^2$ identity. Integrating over $x$ first, $\int_{\R^n} S(f)(x)^2 \, dx = \omega_n \iint_{\R^{n+1}_+} t \abs{\nabla u(y, t)}^2 \, dy \, dt$, where $\omega_n$ is the volume of the unit ball of $\R^n$. With $\hat f(\xi) = \int f(x) e^{-2\pi i x \cdot \xi} dx$, one has $\hat u(\xi, t) = e^{-2\pi t \abs{\xi}} \hat f(\xi)$, so $\int \abs{\nabla u(y, t)}^2 dy = 8\pi^2 \int \abs{\xi}^2 e^{-4\pi t \abs{\xi}} \abs{\hat f(\xi)}^2 \, d\xi$. Since $\int_0^\infty t e^{-4\pi t \abs{\xi}} \, dt = (4\pi\abs{\xi})^{-2}$, this gives $\norm{S(f)}_2^2 = \frac{\omega_n}{2} \norm{f}_2^2$.

Non-example: $f = 1$ is bounded but in no $L^p$ with $p < \infty$. Its Poisson integral is $u = 1$, so $\nabla u = 0$ and $S(f) = 0$ although $f \neq 0$. So the lower bound $c \norm{f}_p \le \norm{S(f)}_p$ cannot hold for $p = \infty$.