This is a cross-post from [Math.SE](https://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?



Controling mixed derivatives
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:


1. A term where all derivatives fall on the weight $t^z$. This gives the first term in your expansion.
2. Terms, excluding the one above, where no derivatives hit $\psi$.
3. 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. 


