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My question is about the van der Corput lemma for $$ \int_a^b e^{i t \phi(x)} \psi(x) dx $$ The version you find everywhere, e.g. on

http://www.tricki.org/article/The_van_der_Corput_lemma_for_oscillatory_integrals

requires $\psi$ to be absolutely continuous. However, what are weaker conditions on $\psi$ such that the lemma still holds?

I was able to prove it when $\psi$ is either of bounded variation or in the Wiener class (summable Fourier coefficients).

Now I found a paper

Boldrighini, Carlo and Pellegrinotti, Alessandro and Triolo, Livio, Convergence to stationary states for infinite harmonic systems, J. Stat. Phys., 30 (1983), 123-155

where the authors claim in Prop A.3 that continuity of $\psi$ is sufficient for the case k=2. The proof refers to a book by T. Kawata "Fourier Analysis in Probabity Theory" (Academic Press, New York, 1972) but in this book I could only find the case $\psi\equiv 1$.

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    $\begingroup$ What is the precise statement? Van der Corput's inequality as is found in your link reads $$\left\lvert \int_a^b e^{i t \phi(x)}\, \psi(x)\, dx \right\rvert \le \frac{C}{\sqrt{t}}\left( \lvert\psi(b)\rvert + \int_a^b\lvert \psi'(x)\rvert\, dx\right), $$ which cannot be interpreted literally if $\psi$ is not absolutely continuous. $\endgroup$
    – Giuseppe Negro
    Commented Mar 16, 2014 at 16:19
  • $\begingroup$ Yes, I don't mean it literally: An estimate of the type $\frac{C}{\sqrt{t}}$, where $C$ depends on $\psi$. $\endgroup$
    – gerald
    Commented Mar 16, 2014 at 19:02
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    $\begingroup$ I might be wrong, but I think bounded variation is required on $\psi$ in order to have decay $\endgroup$
    – Piero D'Ancona
    Commented Mar 16, 2014 at 20:49
  • $\begingroup$ Yes, I also feel that continuity alone is not sufficient. Clearly there cannot be a bound of the form $\frac{C_k \|f\|_\infty}{t^{1/k}}$ with $C_k$ independent on $f$ (just take $\psi=e^{-i t \phi}$). However note that the Wiener algebra contains functions which don't have bounded variation. $\endgroup$
    – gerald
    Commented Mar 17, 2014 at 14:35

1 Answer 1

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Andreas Seeger pointed out that the fact that an estimate of the form $\frac{C_{k,\psi}}{t^{1/k}}$ implies an estimate of the form $\frac{C_k\|\psi\|_\infty}{t^{1/k}}$ via Banach-Steinhaus. Since the latter is wrong, so is the first and thus the claim from the paper.

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