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My question is about the $L^1_x$ norm of an oscillatory integral like $$ \int_{\mathbb{R}^n} e^{i(y\cdot x+\lambda \phi(y))}f(y)dy,$$ where $\lambda \in \mathbb{R}$, $f\in C^{\infty}_c(\mathbb{R}^n)$ is compactly supported and $\phi$ is real-valued and smooth (I have in mind $\phi(y)=\sqrt{1+|y|^2}$ but feel free to assume what you need). Of course, I am interested in the decay rate for $\lambda$ in this norm. Is there any known result or some hint?

The classical $L^{\infty}$ van der Corput bound is too "strong" to be integrated and I already used Bernstein-type lemmas, but I would like to have a refined estimate if any. I tried to look for results in this area following the traces of the classical papers by Hormander and Phong-Stein on $L^p \to L^{p'}$ boundedness, but nothing useful came out.‌

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  • $\begingroup$ Just a stupid question. If $\phi(y)$ is constant on $supp(f)$, the norm you are looking at does not depend on $\lambda$ at all (the Fourier transform just gets shifted), so you need to impose some non-degeneracy condition on the differential of $\phi$ to get any decay. What would that be? (or what is your $\phi$ really?) $\endgroup$
    – fedja
    Jun 21, 2019 at 20:03
  • $\begingroup$ In fact I have in mind $\phi(y) = \sqrt{1+y^2}$, but the general problem of giving suitable conditions on $\phi$ for bounding the $L^1$ norm seems worthy of interest too. $\endgroup$
    – Paul
    Jun 21, 2019 at 21:35
  • $\begingroup$ Erm.. Is $\lambda$ a vector then? $\endgroup$
    – fedja
    Jun 22, 2019 at 1:56
  • $\begingroup$ Sorry, I fixed the typo! $\endgroup$
    – Paul
    Jun 22, 2019 at 6:41
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    $\begingroup$ You know well that for typical real nonstationary phases the pointwise decay is arbitrarily fast and for degenerate phases (with finite order) it is algebraically fast. However, already the $L^2$ norm of the integral does not decay, by Plancherel. Thus I expect the $L^1$ norm to actually grow in $\lambda$. $\endgroup$ Jun 22, 2019 at 8:56

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