Hilbert transform along polynomial curves, I: maximal function

Here we prove the ${L^p}$ boundedness of the Hilbert transform along a polynomial curve in ${\mathbb{R}^1}$. The result is taken from

Maximal functions and Hilbert transforms associated to polynomials,

Carbery, Ricci, Wright,

Rev.Mat.Ibero.,1998].

The proof is split into 2 parts due to its length.

Given a real polynomial ${p(t)}$ one defines the maximal function

$\displaystyle \mathcal{M}_p f(x):=\sup_{h>0}\frac{1}{2h}\int_{-h}^{h}|f(x-p(t))|\,dt,$

and the Hilbert transform analogue

$\displaystyle \mathcal{H}_p f(x):=p.v.\int_{-\infty}^{\infty}{f(x-p(t))}\,\frac{dt}{t}.$

Here we prove the first assertion in the following

Theorem 1 ${\mathcal{M}_p}$ and ${\mathcal{H}_p}$ are both bounded on ${L^q(\mathbb{R})}$ for ${1 (and $q=\infty$ too for $\mathcal{M}_p$) with constants depending only on the degree ${d}$ of ${p(t)}$.

Actually they are weak${(1,1)}$ too, which we’ll prove and from which the result follows.

We’ll make a fundamental use of the result proven in this previous post, which we recall briefly. In the following we adopt the notation that ${[a,b]=\emptyset}$ if ${a>b}$.

Let ${d}$ be the degree of ${p}$ and suppose wlog ${p(0)=0}$ and its leading coefficient is ${a_d=1}$, moreover suppose the roots are all distinct and ordered in such a way that ${0=|\zeta_1|<|\zeta_2|<\ldots<|\zeta_d|}$. There exists a constant ${A\gg 1}$ depending only on ${d}$ such that we can partition ${[0,+\infty[}$ (and thus ${\mathbb{R}}$ by taking the symmetric intervals with respect to ${0}$) into

$\displaystyle [0,A^{-1}|\zeta_2|]\cup[A^{-1}|\zeta_2|,A|\zeta_2|]\cup [A|\zeta_2|,A^{-1}|\zeta_3|]\cup\ldots \cup [A|\zeta_d|,+\infty[$

where the intervals of the form ${[A|\zeta_i|,A^{-1}|\zeta_{i+1}|]}$ (when non empty) are said to be of Type 2 (we’ll also call them “gaps” to stress the absence of roots of ${p}$ and ${p'}$ in them) if

$\displaystyle \frac{|\zeta_{i+1}|}{|\zeta_i|}>4 A^2,$

and all the other intervals are said to be of Type 1. Notice these last intervals are either of the form ${[A^{-1}|\zeta_i|,A|\zeta_i|]}$ or ${[A|\zeta_i|,A^{-1}|\zeta_{i+1}|]}$, but in this latter case the measure is less than ${3 A |\zeta_i|}$, while an interval of Type 2 has measure bounded from below by the same amount (for the relevant root ${\zeta_i}$). Finally, on gaps the polynomial is monotonic and ${\left|t \frac{p'(t)}{p(t)}\right|}$ is bounded from below by ${c_d}$, a constant depending only on the degree.

The heuristic behind the proof is that if we partition the integral defining ${\mathcal{H}_p f}$ according to this Type 1/Type 2 interval subdivision, then Type 1 intervals contribute with an amount which is dominated by a constant multiple of ${\mathcal{M}_p f}$, and the contribution from Type 2 intervals is dominated by a constant multiple of ${H^\ast f}$, i.e. the maximal Hilbert transform, defined by

$\displaystyle H^\ast f(x):= \sup_{0

i.e. we prove that pointwise

$\displaystyle |\mathcal{H}_p f(x)| \lesssim \mathcal{M}_p f(x) + H^{\ast}f(x).$

Separately, we prove that ${\mathcal{M}_p}$ is weak${(1,1)}$, and then the result follows. We’ll start with this latter case in here and conclude with the former in a following post.

Lemma 2 Call ${G}$ the union of all the gaps, or Type 2 intervals. Then

$\displaystyle \sup_{h>0}{\left|\frac{1}{h}\int_{[h,2h]\cap G}{f(x-p(t))}\,dt\right|}\lesssim_d Mf(x),$

where ${M}$ is the ordinary Hardy-Littlewood maximal operator.

Proof: We prove it in the case ${h=1}$. It is enough to do so: with the change of variable ${hs\rightarrow t}$ one has

$\displaystyle \int_{[h,2h]\cap G}{f(x-p(t))}\,\frac{dt}{h}=\int_{[1,2]\cap{G/h}}{f(x-p(hs))}\,ds.$

What’s ${G/h}$? Well, if we look at the polynomial with roots ${\zeta_i/h}$ we see that

$\displaystyle \prod_{i=1}^{d}{\left(t-\frac{\zeta_i}{h}\right)}=h^{-d}p(ht),$

that is ${G':=G/h}$ are the gaps of the polynomial $p(hs)$. This isn’t monic but it’s easy to see that if we prove the theorem for monic polynomials, it will follow for polynomial with any leading coefficient. Thus assume $p$ monic and $h=1$ in the following. ${p}$ is strictly monotonic on the gaps, and therefore we can use a further change of variables ${p^{-1}(u)\rightarrow t}$,

$\displaystyle \int_{[1,2]\cap G'}{f(x-p(s))}\,ds=\int_{p\left([1,2]\cap G'\right)}{f(x-u)}\,\frac{du}{p'\left(p^{-1}(u)\right)}.$

Now, ${|p'|}$ is bounded from below on ${[1,2]\cap G'}$ and thus will have a bounded minimum in, say, ${t_0}$, and ${|p|}$ will have a maximum in ${t_1}$. Therefore

$\displaystyle \left|\int_{p\left([1,2]\cap G'\right)}{f(x-u)}\,\frac{du}{p'\left(p^{-1}(u)\right)}\right|\leq \frac{1}{|p'(t_0)|}\int_{-|p(t_1)|}^{|p(t_1)|}{|f(x-u)|}\,du.$

By the properties of ${p}$ on the gaps we have actually that all the values are comparable, in particular ${|p(t_0)|\sim_d |p(t_1)|}$, because (see again this previous post)

$\displaystyle |p(t_1)|\lesssim_d \left(\prod_{i=m+1}^{d}|\zeta_i|\right)|t_1|^m\leq 2^d \left(\prod_{i=m+1}^{d}|\zeta_i|\right),$

where ${m}$ depends on what gap ${t_1}$ belongs to, and suppose ${t_0}$ is in a lower gap than ${t_1}$ (cases of the same or a upper gap are essentially the same), say the gap of root ${m-l}$, so

$\displaystyle |p(t_0)|\gtrsim_d \left(\prod_{i=m+1-l}^{d}|\zeta_i|\right)|t_0|^{m-l},$

but since ${|t_0| one has

$\displaystyle |p(t_0)|\gtrsim_d \left(\prod_{i=m+1}^{d}|\zeta_i|\right)|t_0|^{m-l} A^{l}|t_0|^l\gtrsim_d \left(\prod_{i=m+1}^{d}|\zeta_i|\right),$

since ${|t_0|\geq 1}$, and thus we have ${|p(t_0)|\gtrsim_d |p(t_1)|}$.

On the other hand

$\displaystyle |p'(t_0)|\geq c_d \frac{|p(t_0)|}{|t_0|}\geq \frac{c_d}{2}|p(t_0)|\gtrsim_d |p(t_1)|,$

and therefore the integral we were estimating is dominated by

$\displaystyle \lesssim_d \frac{1}{|p(t_1)|}\int_{-|p(t_1)|}^{|p(t_1)|}{|f(x-u)|}\,du \lesssim Mf(x).$

$\Box$

With this lemma we control only the contribution of the gaps to the maximal function ${\mathcal{M}_p f}$. However it turns out that the intervals of Type 1 are essentially irrelevant:

Proposition 3 ${\mathcal{M}_p }$ is weak${(1,1)}$, with constant depending only on ${d}$ and not on the coefficients of ${p}$.

Proof: Instead of ${\mathcal{M}_p}$ we prove the result for a related dyadic version of it, namely

$\displaystyle \tilde{\mathcal{M}}_p f(x):= \sup_{k\in\mathbb{Z}}{2^{-k}\left|\int_{2^k<|t|<2^{k+1}}{f(x-p(t))}\,dt\right|}.$

We prove that ${\tilde{\mathcal{M}}_p f}$ is pointwise dominated by ${Mf}$, which implies ${\tilde{\mathcal{M}}_p}$ is itself weak${(1,1)}$, and that implies the same for ${\mathcal{M}_p }$ since if ${2^i\leq h < 2^{i+1}}$

$\displaystyle \begin{array}{rcl} && \frac{1}{2h}\left|\int_{-h}^{h}{f(x-p(t))}\,dt\right| \leq \frac{1}{2\cdot 2^{i}}\int_{-2^{i+1}}^{2^{i+1}}{|f(x-p(t))|}\,dt \\ && \leq \frac{1}{2\cdot 2^{i}}\sum_{k=-\infty}^{i}{\frac{2^k}{2^k}\int_{2^k<|t|<2^{k+1}}{|f(x-p(t))|}\,dt}\\ && \leq \frac{1}{2^{i+1}}\left(\sum_{k=-\infty}^{i}2^k\right)\tilde{\mathcal{M}}_p f(x)=\tilde{\mathcal{M}}_p f(x). \end{array}$

(Our notation is a little incoherent in that we put the absolute value either on the integral or on the integrand, but for a maximal function there isn’t any real difference.)

Now pick ${k\in\mathbb{Z}}$. We split the integral over ${[2^k,2^{k+1}]}$ into two integrals, ${\int_{[2^k,2^{k+1}]}=\int_{[2^k,2^{k+1}]\cap G}+\int_{[2^k,2^{k+1}]\cap G^c}}$. By the previous lemma we have

$\displaystyle \frac{1}{2^k}\left|\int_{[2^k,2^{k+1}]\cap G}{f(x-p(t))}\,dt\right|\lesssim_d Mf(x),$

so we are left with the integral over ${[2^k,2^{k+1}]\cap G^c}$. Notice though that this can be non empty only for a finite number of ${k}$‘s, a number bounded in terms of ${d}$. Infact, Type 1 intervals are either of the form ${[A^{-1}|\zeta|,A|\zeta|]}$ or ${[A|\zeta_i|,A^{-1}|\zeta_{i+1}]}$, and there’s at most ${2d}$ of them. Suppose we’ve taken ${A = 2^N}$ (which we can do). In the first case one sees that if ${|\zeta|\sim 2^{i_0}}$ then the intersection is non empty for ${i_0 \leq k \leq i_0+2N}$, and

$\displaystyle 2^{-k}\int_{|t|\sim 2^k}{|f(x-p(t))|}dt \lesssim Mf(x)$

since it’s just convolution with a measure of total mass 1. In the second case remember ${A^{-1}|\zeta_{i+1}-A|\zeta_i| \leq 3 A |\zeta_i|}$, so that if ${|\zeta_i|\sim 2^{j_0}}$ then intersection with ${[2^k,2^{k+1}]}$ is non-empty for at most those ${k}$ s.t. ${N+j_0 \leq k \leq N+j_0+2}$. Again, for these ${k}$ the integral is dominated by the Hardy-Littlewood maximal function, and thus we’re done: for every ${k}$ the integral is dominated by a constant multiple of ${Mf}$, and these constants are uniformly bounded (in terms of the degree ${d}$ only). $\Box$

Thus ${\mathcal{M}_p f}$ is dominated by a weak${(1,1)}$ operator and is therefore weak${(1,1)}$ as well. Interpolating with Marcinkiewicz theorem between this and the trivial bound ${\|\mathcal{M}_p f\|_{L^\infty}\leq \|f\|_{L^\infty} }$ one gets the full result we stated at the beginning.

2nd part of the post will come soon.