Paraproduct

Remarks

The function $S_m g$ is the convolution of $g$ with $2^{mn} \check\chi(2^m \cdot)$, a smooth average of $g$ at scale $2^{-m}$, so $\abs{S_m g} \le C \norm{g}_\infty$. The Fourier transform of the $j$-th term is supported in $\set{2^{j-2} \le \abs{\xi} \le 2^{j+2}}$, because $\Delta_j f$ has frequencies between $2^{j-1}$ and $2^{j+1}$ and $S_{j-3} g$ has frequencies below $2^{j-2}$. So the terms are almost orthogonal, which gives the convergence and the bound $\norm{\Pi_g f}_2 \le C \norm{g}_\infty \norm{f}_2$. The shift $3$ is a convention; other authors use a different shift or a different cut-off.

Grouping the terms of $fg = \sum_{j, k} \Delta_j f \, \Delta_k g$ according to $k \le j - 3$, $j \le k - 3$ or $\abs{j - k} \le 2$ gives Bony's decomposition $$fg = \Pi_g f + \Pi_f g + \sum_{\abs{j - k} \le 2} \Delta_j f \, \Delta_k g$$ for $f$ and $g$ both in $L^2(\R^n) \cap L^\infty(\R^n)$, since $\sum_{k \le m} \Delta_k g = S_m g$.

Example: if $g = c$ is a constant, then $\hat g$ is a multiple of the point mass at $0$ and $\chi(0) = 1$, so $S_m g = c$ for every $m$. Hence $\Pi_g f = c \sum_j \Delta_j f = c f$.

Non-example: the term $\Delta_j f \, \Delta_j g$, a product of two pieces at the same frequency, is not part of a paraproduct. Its Fourier transform can be supported near $0$, for instance when $\widehat{\Delta_j g}$ is the reflection of $\widehat{\Delta_j f}$, so high frequencies combine into low ones; such terms go into the remainder.