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?

  • 1
    $\begingroup$ I suppose that it matters here that $v=0$ near one endpoint of integration, while $w=0$ near the other. One can imagine using integration by parts to better arrange what gets differentiated. However, I would expect the estimate to be $$\int |\partial^N (uvw)| \le C_N \int |\partial^N u vw| + |u \partial^N vw| + |uv\partial^N w|$$ that is, with absolute values inside of the sum. Maybe this is what they meant to write? (I wanted to take a look at page 299; naturally, Google Books showed 298 and 300). $\endgroup$ – user147263 Aug 22 '14 at 0:31
  • $\begingroup$ @900sit-upsaday Due to the nonnegativity of all the functions, you may arrange the absolute values as you wish as long as the $\prod$ is made nonnegative. I will put them accordingly. Note that $|t^z| = t^{\Re z}$ $\endgroup$ – AlexR Aug 22 '14 at 7:49
  • $\begingroup$ The functions are nonnegative, but their derivatives are another matter. $\endgroup$ – user147263 Aug 22 '14 at 14:07

I have found a solution with the help of the MO Answer provided on my cross-post.


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