# Sup norm of Fourier transform of $\frac{\sin |x|}{|x|^\lambda} \mathbb 1_{\{2^k\le |x| <2^{k+1}\}}, \ 0<\lambda<n$

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.

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

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 $$0, then $$\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.

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