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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?

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  • $\begingroup$ Are you sure your parentheses are as they should? The LHS has $t^z(1-\psi\chi)$ but the RHS suggests that you might want $t^z(1-\psi)\chi$. $\endgroup$ – Joonas Ilmavirta Aug 21 '14 at 14:45
  • $\begingroup$ @JoonasIlmavirta Actually not, thanks for noticing :) $\endgroup$ – AlexR Aug 21 '14 at 16:13
  • $\begingroup$ For $N=0$ and $z\in\mathbb R$ your statement reads $\int_0^1t^z|1-\psi(t\tau)||\chi(t)|dt\leq C_0\int_0^1t^z|\chi(t)|dt$. You can make $\psi|_{[4/3,5/3]}$ arbitrarily large and choose $\tau$ large to violate this for any $C_0$. Are there perhaps more assumptions (or errors in what I wrote)? $\endgroup$ – Joonas Ilmavirta Aug 21 '14 at 16:29
  • $\begingroup$ @JoonasIlmavirta Actually, I forgot to mention that $\chi(\mathbb R) = \psi(\mathbb R) = [0,1]$. Editing that in. Again, thanks for looking at the problem :) $\endgroup$ – AlexR Aug 22 '14 at 7:21
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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.

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  • $\begingroup$ Actually the authors need the power of $z$ in the estimate to be $N$, at least for the $N=2$ case. They later use that $$\int_0^\infty e^{-ix_n t} \chi(t) t^{z+k-1} \mathrm dt \le C_2 \frac{(1+|z+k-1|)^2}{\Re z + k} (1+|x_n|)^{-\Re z -k} \le C (1+|z|)^2 (1+|x_n|)^{-\Re z-k}$$ $\endgroup$ – AlexR Aug 23 '14 at 12:53
  • $\begingroup$ I have finally come around the estimate using your suggestion with a little more care put into the $z$ dependency. Since your answer gave me the necessary hint (to look at specific terms and try to estimate them uniformly), I accept it :) $\endgroup$ – AlexR Sep 11 '14 at 11:39
  • $\begingroup$ @AlexR: Why don't you post your complete answer below now that you have it? $\endgroup$ – Hans Jan 13 '18 at 19:28
  • $\begingroup$ @Hans It‘s been a while. I will look in my notes when I find the time. $\endgroup$ – AlexR Jan 15 '18 at 5:51

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