It seems to me that in a paper of Charles Fefferman (open access), it is claimed in the introduction that (3rd page of the PDF file, 'page 11', $\lambda\in(0,n)$)

$$\sup_{\xi\in\mathbb R^n}\left| \int_{2^k\le |x| <2^{k+1}} \frac{\sin |x|}{|x|^\lambda} e^{2\pi i x \cdot\xi} \, dx \right| \le 2^{(\frac{n+1}2-\lambda)k}.$$ (He claims that Plancharel's formula should be used, I expect this to mean that we should compute the $L^\infty$ norm of the symbol.)

By using asymptotics of Bessel functions, I can get the upper bound for $\xi\gg 1$, but I am worried about $\xi\approx 0$. At $\xi=0$, the Fourier transform is just $\int\frac{\sin\dots}{\dots} dx$ leading to the estimate ($C_n$ is the volume of the unit $n$-sphere) $$ \left| \int_{2^k\le |x| <2^{k+1}} \frac{\sin |x|}{|x|^\lambda} \, dx \right| \le C_n 2^{(n-\lambda)k} $$ which is bigger for $n>1$? Am I missing something simple? Calculation details can be provided on request.

  • $\begingroup$ In deriving your bound did you use the obvious bound on sin? $\endgroup$
    – lcv
    May 14, 2020 at 11:38
  • $\begingroup$ @lcv yes, all I did was $|\sin r|\le 1$ $\endgroup$ May 14, 2020 at 11:39

1 Answer 1


You're ignoring the oscillations of $\sin\vert x\vert$; by the way, in the paper you can replace $[2^k,2^{k+1}]$ by a smooth version $\zeta_k(x) = \zeta(x/2^k)$, which simplify a lot the computations, so I'll do that.

By rotational symmetry we can assume that $\xi = se_n$, for $s\ge 0$. If $s\gg 1$ then it is sufficient to use Bessel functions because the oscillation of $e^{2\pi isx_n}$ are dominant. If $s\ll 1$ then the oscillation of $\sin\vert x\vert$ are dominant.

By dilation $$ \int\zeta_k\frac{\sin\vert x\vert}{\vert x\vert^\lambda}e^{2\pi isx_n}\,dx = 2^{k(n-\lambda)}\int_1^2\zeta\frac{\sin 2^k\vert x\vert}{\vert x\vert^\lambda}e^{2\pi i2^ksx_n}\,dx. $$ Since $\sin r = \frac{1}{2i}(e^{ir}-e^{-ir})$, it suffices to get the bound $$ \left\vert\int a(x)e^{2\pi i\omega(sx_n-\vert x\vert/(2\pi))}\,dx\right\vert\le C\omega^{-\frac{n-1}{2}} $$ uniformly in $s\ge 0$; here $a = \zeta\frac{1}{\vert x\vert^\lambda}$ is a smooth function supported in $[1,2]$. The term with the factor $e^{i2^k\vert x\vert}$ is bounded similarly.

After a change of variables we can write the integral also as $$ \int ae^{2\pi i\omega(sx_n-\vert x\vert/(2\pi))}\,dx = \iint_{S^{n-1}}a r^{n-1}e^{2\pi ir\omega(s\theta_n-1/(2\pi))}\,dS(\theta)dr. $$ We evaluate the integral in different regions.

$\sin\vert x\vert$ is dominant: If $s\le \frac{1}{2\pi}-c$, for some $0<c\ll 1$, then $\vert s\theta_n-\frac1{2\pi}\vert\ge c$ for every $\theta$, and then integrating by parts $N$ times we get $$ \left\vert\int\frac{ar^{n-1}}{(2\pi i \omega(s\theta_n-1/(2\pi)))^N} \partial_r^N(e^{2\pi i\omega r(s\theta_n-1/(2\pi))})\,dr\right\vert \le C_N\frac{1}{\omega^N}, $$ so let's take $N>\frac{n-1}{2}$ and integrate over the sphere.

$e^{2\pi isx_n}$ is dominant: If $s\ge \frac1{2\pi}+c$, then $\vert s-\frac{x_n}{2\pi\vert x\vert}\vert \ge c$ and then again using integration by parts we get $$ \left\vert\int a e^{2\pi i\omega(sx_n-\vert x\vert/(2\pi))}\,dx\right\vert\le \frac{c}{\omega^N}; $$ here, we used $\partial_ne^{2\pi i\omega(sx_n-\vert x\vert/(2\pi))} = 2\pi i\omega(s-x_n/(2\pi\vert x\vert))e^{2\pi i\omega(sx_n-\vert x\vert/(2\pi))}$.

The hard part: The case $\vert s-\frac1{2\pi}\vert \le c\ll 1$ is hard and here you should expect the upper bound $C/\omega^\frac{n-1}{2}$. The phase of your oscillatory integral is $f(x) = sx_n-\vert x\vert/(2\pi)$, and Stein's book Harmonic Analysis: real-variable methods, orthogonality, and oscillatory integrals, Ch. VIII-IX is the standard reference.

Since $\nabla f = (-\frac{x_1}{2\pi\vert x\vert},\cdots,-\frac{x_{n-1}}{2\pi\vert x\vert}, s-\frac{x_n}{2\pi\vert x\vert})$, then the principle of non-stationary phase suggests that we should divide $\mathbb{R}^d$ into a tube $T = \{(x',x_n)\mid \vert x'\vert\le c, x_n>0\}$ and its complement. Let $\psi$ be a cut-off of $T$, so by non-stationary phase, i.e. repeated integration by parts as above, we get $$ \left\vert\int a(1-\psi)e^{2\pi i \omega f}\,dx\right\vert \le C_N\frac{1}{\omega^N}. $$ To bound the last contribution from the region $T$, fix $x_n>0$ and write $$ \int_{\mathbb{R}^{n-1}} a\psi e^{2\pi i \omega f}\,dx' = c_ne^{2\pi i\omega (s-1/(2\pi))x_n} \int_0^\infty a\psi r^{n-2} e^{-i\omega (\sqrt{r^2+x_n^2}-x_n)}\,dr. $$ The coefficient $e^{2\pi i\omega (s-1/(2\pi))x_n}$ has absolute value 1, so we can ignore it. Now make the change of variables $t = \omega(\sqrt{r^2+x_n^2}-x_n)$ to see that $$ \int_0^\infty a\psi r^{n-2} e^{-i\omega (\sqrt{r^2+x_n^2}-x_n)}\,dr = \frac{1}{\omega^\frac{n-1}{2}}\int_0^\infty \tilde{a}(t/\omega)t^\frac{n-3}{2}e^{-it}\,dt; $$ here $\tilde{a}$ is a smooth function supported in $\vert x'\vert\le c$. Hence, it suffices to get the upper bound $$ \left\vert \int_0^\infty \tilde{a}(t/\omega)t^\frac{n-3}{2}e^{-it}\,dt\right\vert\le C, $$ uniformly in $\omega$, which after integration in $x_n$, suffices to get the desired result. Using repeated integrations by parts we get $$ \vert \int_0^\infty \tilde{a}(t/\omega)t^\frac{n-3}{2}e^{-it}\,dt\vert = \vert\int_0^\infty \partial^N_t(\tilde{a}(t/\omega)t^\frac{n-3}{2})e^{-it}\,dt\vert, $$ whenever $N<\frac{n-3}{2}$. When you reach $N = \lfloor\frac{n-3}{2}\rfloor$, you have to be careful at the next integration by parts because you get a boundary term at 0. The typical term in the derivative after $N_1+N_2 = N$ integration by parts is $\omega^{-N_1}\partial^{N_1}\tilde{a}(t/\omega)\cdot t^{\frac{n-3}{2}-N_2}$, whose absolute value is less than $\omega^{\frac{n-3}{2}-N}$ in the support of $\tilde{a}(t/\omega)$. Hence, $$ \left\vert \int_0^\infty \tilde{a}(t/\omega)t^\frac{n-3}{2}e^{-it}\,dt\right\vert \le C, $$ with $C$ uniform in $\omega$. Taking $N\gg 1$ we conclude the estimations.

  • $\begingroup$ Thank you very much, especially for finding the time to finish the proof :) I'm still reading and will take some time to process $\endgroup$ May 15, 2020 at 1:03
  • 1
    $\begingroup$ Please, take your time, there are many computations. $\endgroup$
    – user90189
    May 15, 2020 at 9:32

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