Here we prove the boundedness of the Hilbert transform along a polynomial curve in . The result is taken from
Carbery, Ricci, Wright,
The proof is split into 2 parts due to its length.
Given a real polynomial one defines the maximal function
and the Hilbert transform analogue
Here we prove the first assertion in the following
Theorem 1 and are both bounded on for (and too for ) with constants depending only on the degree of .
Actually they are weak 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 if .
Let be the degree of and suppose wlog and its leading coefficient is , moreover suppose the roots are all distinct and ordered in such a way that . There exists a constant depending only on such that we can partition (and thus by taking the symmetric intervals with respect to ) into
where the intervals of the form (when non empty) are said to be of Type 2 (we’ll also call them “gaps” to stress the absence of roots of and in them) if
and all the other intervals are said to be of Type 1. Notice these last intervals are either of the form or , but in this latter case the measure is less than , while an interval of Type 2 has measure bounded from below by the same amount (for the relevant root ). Finally, on gaps the polynomial is monotonic and is bounded from below by , a constant depending only on the degree.
The heuristic behind the proof is that if we partition the integral defining 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 , and the contribution from Type 2 intervals is dominated by a constant multiple of , i.e. the maximal Hilbert transform, defined by
i.e. we prove that pointwise
Separately, we prove that is weak, 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 the union of all the gaps, or Type 2 intervals. Then
where is the ordinary Hardy-Littlewood maximal operator.
Proof: We prove it in the case . It is enough to do so: with the change of variable one has
What’s ? Well, if we look at the polynomial with roots we see that
that is are the gaps of the polynomial . 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 monic and in the following. is strictly monotonic on the gaps, and therefore we can use a further change of variables ,
Now, is bounded from below on and thus will have a bounded minimum in, say, , and will have a maximum in . Therefore
By the properties of on the gaps we have actually that all the values are comparable, in particular , because (see again this previous post)
where depends on what gap belongs to, and suppose is in a lower gap than (cases of the same or a upper gap are essentially the same), say the gap of root , so
but since one has
since , and thus we have .
On the other hand
and therefore the integral we were estimating is dominated by
With this lemma we control only the contribution of the gaps to the maximal function . However it turns out that the intervals of Type 1 are essentially irrelevant:
Proposition 3 is weak, with constant depending only on and not on the coefficients of .
Proof: Instead of we prove the result for a related dyadic version of it, namely
We prove that is pointwise dominated by , which implies is itself weak, and that implies the same for since if
(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 . We split the integral over into two integrals, . By the previous lemma we have
so we are left with the integral over . Notice though that this can be non empty only for a finite number of ‘s, a number bounded in terms of . Infact, Type 1 intervals are either of the form or , and there’s at most of them. Suppose we’ve taken (which we can do). In the first case one sees that if then the intersection is non empty for , and
since it’s just convolution with a measure of total mass 1. In the second case remember , so that if then intersection with is non-empty for at most those s.t. . Again, for these the integral is dominated by the Hardy-Littlewood maximal function, and thus we’re done: for every the integral is dominated by a constant multiple of , and these constants are uniformly bounded (in terms of the degree only).
Thus is dominated by a weak operator and is therefore weak as well. Interpolating with Marcinkiewicz theorem between this and the trivial bound one gets the full result we stated at the beginning.
2nd part of the post will come soon.