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,


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<q< \infty} (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<a<b<\infty}\left|\int_{a<|t|<b}{f(x-t))}\,\frac{dt}{t}\right|;

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|<A^{-1}|\zeta_{m-l+1}|<\ldots} 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).


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.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s