Controling mixed derivatives 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?
 A: As stated, the inequality indeed fails. But the final estimate (11.27) in the book is essentially correct. From the product rule the derivative
$$ (\frac{d}{dt})^n (t^z (1 - \psi(t\tau)) \chi(t) $$
contains three types of terms:


*

*A term where all derivatives fall on the weight $t^z$. This gives the first term in your expansion. 

*Terms, excluding the one above, where no derivatives hit $\psi$. 

*Terms where some number of derivatives hit $\psi$. 
The first term you estimate exactly as the authors do, which gives you
$$ \leq \int_{1/\tau}^\infty \prod (z-k) t^{\Re z - N}~\mathrm{d}t = \prod(z-k)\cdot \frac{1}{\Re z - N + 1} \cdot \tau^{N -1 -\Re z}  $$
assuming, as they did, that $\Re z - N < -1$. 
For the second types of terms, note that since $\chi$ is support on $[-1,1]$ this means that $|t^z| < 1$ on its support. So every single term of type two is estimated by 
$$ \leq \sup_{k \in \{1, \ldots, N\}} \sup_{t \in [0,1]} | \chi^{(k)}(t) | $$
in the context of the estimates we will take this to just be a constant depending on $N$. So all terms of type two together is estimated by 
$$ \leq C_N$$
We are left with terms of type three. Since a derivative hits $\psi$, the terms of type three are supported in the region 
$$ t \in [1/\tau, 2/\tau ] $$
which has width $1/\tau$. In this region $t \approx 1/\tau$. So a typical term can be controlled by 
$$ \leq C\cdot \frac{1}{\tau} \cdot \prod(z-k) \cdot \frac{1}{\tau}^{\Re z - N_1} \cdot \tau^{1 + N_2} \cdot |\psi^{(1+N_2)}|_\infty \cdot |\chi^{(N_3)}|_\infty $$
where $1 + N_1 + N_2 + N_3 = N$. We can simplify a little bit putting the $L^\infty$ norms of derivative of $\psi$ and $\chi$ into a constant 
$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z - k) \cdot \tau^{N_1 + N_2 - \Re z} $$
since we are interested in the $\tau \gg 1$ case 
$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z-k)  \cdot \tau^{N - 1 - \Re z} $$
So up to some factors of products of the form $z(z-1)\ldots (z-N_1)$ the estimate (11.27) is correct. Since I haven't read the rest of the chapters to see why the authors chose to keep track of the $z$ dependence, I don't know how much this changes the exposition. 
