Littlewood–Paley square function

Remarks

The operator $f \mapsto Sf$ is not linear, but it is sublinear: $S(f + g) \le Sf + Sg$, by the triangle inequality in $\ell^2$. It depends on the choice of $\chi$, but the $L^p$ norms of square functions built from two choices are comparable for $1 < p < \infty$.

Example: assume moreover that $\chi$ is radial and nonincreasing in $\abs{\xi}$, so that $\psi \ge 0$. For each $\xi \ne 0$ at most two of the numbers $\psi(2^{-j}\xi)$ are nonzero, and they sum to $1$. So $\frac{1}{2} \le \sum_j \psi(2^{-j}\xi)^2 \le 1$. By Plancherel's theorem for the Fourier transform, $\norm{Sf}_2^2 = \int \sum_j \psi(2^{-j}\xi)^2 \abs{\hat f(\xi)}^2 \, d\xi$, hence $\frac{1}{2}\norm{f}_2^2 \le \norm{Sf}_2^2 \le \norm{f}_2^2$ for $f \in L^2(\R^n)$.

Non-example: the constant function $1$ is not in any $L^p(\R^n)$ with $p < \infty$. As a tempered distribution its Fourier transform is the point mass at $0$, and $\psi(0) = 0$, so every $\Delta_j 1$ vanishes and the formula gives $S1 = 0$ although $1 \ne 0$. This is why the comparison of $\norm{Sf}_p$ and $\norm{f}_p$ needs $p < \infty$.