This is a cross-post from [Math.SE](http://math.stackexchange.com/questions/893974/odd-form-of-controlling-derivatves) 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?