This is a cross-post from Math.SE since the question got nothing (but upvotes) even after offering a decent bounty. If it is too trivial or in other ways not suited for this site, please let me know and I'll delete it.

In Muscalu, Schlag - Classical and Multilinear Harmonic Analysis (Cambridge Universitv Press 2013), page 299 there is a rather odd estimate for wich I cannot find any justification:

Functions used: $$\def\supp{\mathop{\rm supp}}\begin{align*} \psi & \in C_c^\infty(\mathbb R) \\ \supp \psi & \subset [-2,2] \\ \psi|_{[-1,1]} & \equiv 1 \\ \chi & \in C_c^\infty(\mathbb R) \\ \supp \chi & \subset [-1,1] \\ \chi(0) & = 1 \\ \psi(\mathbb R) = \chi(\mathbb R) & = [0,1]\\ z & \in\mathbb C\\ \tau & \in\mathbb R \end{align*}$$

The claim is that $$\begin{align*} \int_0^\infty \left| \frac{\mathrm d^N}{\mathrm dt^N} (t^z (1-\psi(t\tau)) \chi(t)) \right| \mathrm dt & \le C_N \int_0^\infty \left| \prod_{k=0}^{N-1} (z-k) t^{z - N} (1-\psi(t\tau)) \chi(t) \right| \\ & \qquad \qquad + \left| t^{z} (-\psi^{(N)}(t\tau) \tau^N) \chi(t) \right| \\ & \qquad \qquad + \left| t^{z} (1-\psi(t\tau)) \chi^{(N)}(t) \right| \mathrm dt \\ & = C_N \int_0^\infty \left| \prod_{k=0}^{N-1} (z-k) \right| t^{\Re z - N} (1-\psi(t\tau))\chi(t) \\ & \qquad \qquad + t^{\Re z} |\psi^{(N)}(t\tau)| \tau^N \chi(t) \\ & \qquad \qquad + t^{\Re z} (1-\psi(t\tau)) |\chi^{(N)}(t)| \mathrm dt \end{align*}$$

So basically that we can control $$\int |\partial^N (uvw)| \le C_N \int |\partial^N u vw| + |u \partial^N vw| + |uv\partial^N w|$$ Wich is certainly not true in general (chose $u=v=w=x$ and $N=3$, for example)

So how can we justify that estimate in this special case?