2 Proof of Theorem 1.4
It suffices to establish the -bounded estimate of when , where
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(2.1) |
where the cutoff function is supported in .
We choose a smooth cutoff function which is supported in , and on , and decompose
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(2.2) |
where
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We first consider the remainder term . Denote
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Then
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It is clear that provided that or , and there holds
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for any multi-index . Then integrating by parts implies that
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for any positive integer .
Therefore, Sobolev embedding and Young’s inequality yield that
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(2.3) |
whenever . So we are reduced to investigating the main contribution term .
A dyadic decomposition of the -variable and a standard scaling argument imply that
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(2.4) |
where
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and
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the cutoff function is supported in .
When and , integration by parts with respect to the variable in the inner integral of implies that
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for each non-negative multi-index and non-negative integer , then the method of handling the remainder
term can be applied here to complete the proof. Next we consider the case when and . Notice that in this case, the inner integral in can be put into the amplitude.
If , for any and that belong to the interval , there holds
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By performing integration by parts on the variable , we obtain
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(2.5) |
for any positive integer and multi-index . By employing the method of handling the remainder term , we can complete the proof when . When , there exists a solution for the equation provided that .
Then by the method of stationary phase, we only need to consider the main contribution term given by
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where is a smooth cutoff function supported in ,
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which is a symbol of order zero in . We will deal with this main contribution term later.
Next, we will consider the case when . According to the definition of , we have . If , for any , and that belong to the interval , we have
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Therefore, we still maintain the validity of inequality (2.5). Furthermore, if or , there holds the uniform estimate
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for every , and that belong to the interval . By utilizing integration by parts in the inner integral of , we get
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(2.6) |
for any positive integer and multi-index . Since , we can also finish the proof when or as in the similar proof for the remainder term .
Now we turn to consider the subcase when . Firstly, we treat the inner integral of . Since is sufficiently large and , then the equation has a unique solution according to the implicit function theorem.
Applying the method of stationary phase to the -integration, the inner integral in can be replaced by its leading term,
whose phase function can be written as . Here,
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which is a small perturbation of . Secondly, we apply the method of stationary phase to the -integration, the phase function of which is now given by
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The equation
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has a unique solution since is sufficiently large, and by Taylor’s expansion, the phase function can be written as
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where
and
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Indeed, is a small perturbation of
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Therefore, when , we reduce our problem to estimating the main contribution term
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with
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(2.7) |
Here, the support of the cutoff functions and are uniformly contained in for each .
Based on the above stationary phase method, by Sobolev embedding, there holds
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(2.8) |
where the bump function is supported in and the Fourier integral operator
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For the same reason, we have
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(2.9) |
with the Fourier integral operator
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(2.10) |
Theorem 1.4 can be proved by the following estimates for , .
Theorem 2.1.
There holds the following estimates for , .
(E1) For each , ,
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(2.11) |
(E2) For each ,
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(2.12) |
and for each ,
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(2.13) |
(E3) For each ,
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(2.14) |
and for each ,
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(2.15) |
(E4) For each ,
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(2.16) |
and for each ,
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(2.17) |
(E5) For each ,
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(2.18) |
and for each ,
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(2.19) |
(E6) For each ,
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(2.20) |
and for each ,
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(2.21) |
Now we apply Theorem 2.1 to prove Theorem 1.4, and leave the proof of Theorem 2.1 to the next section. According to inequalities (2.4), (2.8) and (2.9), there holds
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(2.22) |
Firstly, we show the following diagonal results for . By the estimate (E1) in Theorem 2.1, there holds
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(2.23) |
By the estimates (E4) in Theorem 2.1, we have
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Interpolation between the estimates in (E1) and (E5) implies
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Then we give the following off-diagonal results for satisfying . When , we interpolate the estimates in (E4) with the estimates in (E3) to get
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When , the following estimates can be easily obtained by interpolation, for each ,
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(2.24) |
and for each ,
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(2.25) |
then combining the estimate in (E3) and inequality (2), there holds
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Furthermore, we show the following off-diagonal estimates for satisfying . When , by interpolating the estimates in (E2) with the estimates in (E4), we get
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In order to obtain the results for the case , we give the following easy estimates by interpolation, for each ,
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(2.26) |
and for each ,
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(2.27) |
then combining the estimates from (E2) and inequality (2), we obtain
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Finally, by interpolation, it is easy to see that for each , , there exists such that
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This completes the proof when . For the case , since the curve is non-degenerate near , after conducting stationary phase analysis, we are reduced to considering Fourier integral operators related to hypersurfaces with two non-vanishing principal curvatures. Hence, we can get the estimates using a similar argument as in [14]. Then we finish the proof of Theorem 1.4.
3 Proof of Theorem 2.1
It is clear that the estimates in (E1) follows directly from Plancherel’s theorem. So we first consider the estimates in (E2).
Notice that for each , the hypersurface
and the hypersurface
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for large enough have at leas one non-vanishing principal curvature. Then according to [[13], Proposition 3.4], there holds the inequalities (2.12), (2.13) for and , respectively.
Now we explain how to establish estimates for the operator in (E3)-(E5). Interpolating between and estimates for , then we arrive at
inequality (2.14). It is noted that
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where
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We rewrite as
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By employing the angular decomposition of the variable as utilized in [Lemma 2.7, [11]] and [Lemma 2.6, [12]], see also the proof in Lemma 3.2 below, we can obtain that for each and positive integer , there holds
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It follows that
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and
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for each . Then inequalities (2.16), (2.18), (2.20) can be proved by Young’s inequality, we omit their proofs here.
Next, we proceed to establish the estimates for the operator in (E4)-(E5).
We rewrite the operator as
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where
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Then ineqalities (2.17) and (2.19) follow from Young’s inequality and part (2) of Theorem 3.1 below. Additionally, the estimate (2.21) follows directly from part (1) of Theorem 3.1.
Finally, we are left with building the desired estimate for the operator in (E3). We may replace in by according to Plancherel’s theorem, then by the dual method as in the proof of [Proposition 3.4, [13]], inequality (2.15) will follow if we can obtain for arbitrary ,
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(3.1) |
where denotes convolution with respect to the variable , and
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For fixed , , by Plancherel’s theorem, there holds
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(3.2) |
Theorem 3.1 below implies that for each , the following estimate holds true
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(3.3) |
Interpolation between (3.2) and (3.3) yields
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(3.4) |
By combining (3.4) and the Hardy-Littlewood-Sobolev inequality, we obtain
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Then we arrive at inequality (3.1).
Therefore, in order to complete the proof of Theorem 2.1, we are left to investigate the estimate for the kernel related to the Fourier integral operator .
Theorem 3.1.
We have the following estimates for , which also hold true for .
(1) For fixed and , there holds the pointwise estimate
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(3.5) |
where the implied constant is independent of and .
(2) For each , we have the uniform estimate
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(3.6) |
Theorem 3.1 follows from Lemma 3.2 below.
Lemma 3.2.
There holds the following pointwise estimates for .
(1) If , then for any positive integer , there holds
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(3.7) |
(2) If , we have
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(3.8) |
where .
(3) If , we obtain
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(3.9) |
where ,
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We firstly give the proof of Lemma 3.2, and show the proof how Theorem 3.1 can be deduced by Lemma 3.2 at the end of Section 3.
Proof.
We first prove the part (1). By rescaling, we have
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(3.10) |
Since , integrating by parts implies that
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To prove the part (2), we recall the definition of in equality (2), and decompose given by equality (3.10) as , where
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and . For fixed , changing variables , its phase function becomes
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By Taylor’s expansion of at , the phase function can be written as
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where we have put all the remainder terms into . Then integration by parts yields
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This implies the inequality (3.10) since .
Similarly, we can get the part (3). We decompose , where
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and
.
Then we change variables , , . We get the phase function in as follows
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Then by Taylor’s expansion, we get the phase function
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where all the remainder terms are included in . Then the estimate (3) follows by integration by parts.
Proof of Theorem 3.1:
(1) When , according to the estimates in (1) and (2) of Lemma 3.2, it is easy to obtain inequality (3.5).
Next we will show that (3.5) holds for . Observing the part (3) in Lemma 3.2, for fixed , and , the term
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can be viewed as a characteristic function over the set whose projection on the -plane is a rectangle centered at with side length . Hence, in order to obtain (3.5), it suffices to prove that if , then we have either or . Indeed, if , there holds
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Conversely, if , we have
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Then we obtain inequality (3.5).
(2) For each , inequality (3.6) follows from part (3) of Lemma 3.2, we omit the proof here.
4 Proofs of Theorem 1.2 and Theorem 1.3
Firstly, we introduce some necessary notations in our proof. For which is given by (1.2),
we denote
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Notice that since , for , there holds
Next we define according to the following four cases:
(C1) if contains but does not contain , then we define
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(C2) if contains but does not contain , then we define
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(C3) if contains both and , then we define
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(C4) if contains neither nor , then we define
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The following lemma will show that the height of the homogeneous polynomial is the maximal element of . To be convenient, let us denote by the maximal element of .
Lemma 4.1.
For the homogeneous polynomial given by (1.2), we have .
We have the following observations before the proof of Lemma 4.1.
For fixed , if , it is clear that .
Otherwise, without loss of generality, we may assume that , where corresponds to the factor in (1.2) with , . A simple caculation implies that . Moreover, for a homogeneous polynomial , the supremum in the definition of can always be achieved at some . Conversely, for fixed straight line passing through the origin which is contained in , without loss of generality, we may again consider corresponding to the factor in (1.2), there is an such that . In particular, if , then for each , there exists an such that .
Proof.
We first consider the case when . If , then it is clear that
; if , there holds
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then it follows that .
In the case when , there exists an such that , without loss of generality, we may assume that , where corresponds to the factor in (1.2). Then we write as
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where is a homogeneous polynomial such that for . It is easy to claim that when , .
When and , then we must have . Therefore, when , we have and
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Hence there holds .
Now we are left with the case when or . When and , the conclusion follows directly by the previous claim. When and , we may write
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where is a linear polynomial that is homogeneous of degree one. A direct calculation implies that , then we have . When , we must have by the previous observation, then
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It is obvious that , so and . This finishes the proof of Lemma 4.1.
The proof of Theorem 1.2 is based on the following Lemma 4.2.
Lemma 4.2.
If vanishes along the straight line for some and such that , we define the local maximal operator
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(4.1) |
where , supp, supp, and is sufficiently small, such that only vanishes along in the angle . It follows that when ,
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(4.2) |
holds for and .
When , we get
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(4.3) |
provided that and .
Proof.
After some linear transformation (which does not change the norm of the maximal operator), we are reduced to estimating the maximal operator defined by
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Notice that
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(4.4) |
where
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and since and . Moreover, , .
By isometric transform, we obtain
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where denotes the averaging operator along the hypersurface
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(4.5) |
Now it is clear that the surface has non-vanishing Garssian curvature, then
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(4.6) |
provided that .
The estimate (4.6) arises from a general conclusion. Let be a compact set in , be a function with the non-degenerate Heissian matrix, be a smooth hypersurface, . Then the norm of the local maximal operator along is bounded by provided that . Here we outline the main idea behind the proof of this conclusion, which will be directly referenced in the rest of this paper.
Upon Littlewood-Paley decomposition, when is constrained to , , if , it can be proven that the Fourier transform along decays rapidly to complete the corresponding maximal function estimate. Next, we present the application of the method of the stationary phase in
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where , is a smooth cutoff function supported in . According to Theorem 1.2.1 in [15],
can be expressed as the sum of a finite number of terms of the following form
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where the function is homogeneous of degree one and its Heissian matrix has a rank of , is a symbol of order zero in . Through Sobolev embedding, we transform the estimate of the maximal operator into the estimate of the following Fourier integral operator
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It is worth noting that the surface has two non-vanishing principal curvatures, which implies that the method of estimate for the corresponding Fourier integral operator is consistent with the proof described in Section 4 of reference [14].
Now let’s proceed to the proof of Lemma 4.2. Inequality (4.6) yields
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(4.7) |
It follows that when , the right side of (4.7) is convergent provided that .
Then inequality (4.2) holds true for and .
When , the right side of (4.7) converges to if . Hence, there holds inequality (4.3)
provided that and .
Proof of Theorem 1.2:
Let be the operator defined by (1.1).
A dyadic decomposition and scaling argument imply
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where
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(4.8) |
with supp. Notice that since is a homogeneous function, either contains the origin only, or contain some straight lines passing through the origin. In the first case, our results follows directly since does not vanish on the support of . In the later case, by the assumption of Theorem 1.2, we are reduced to considering the maximal operators which are localized near the straight lines containing in , then Lemma 4.2 and Remark 4.3 can be applied to get the following conclusions.
When , for
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(4.9) |
there holds
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(4.10) |
When , for
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(4.11) |
we get
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(4.12) |
This finishes the proof of Theorem 1.2.
Now we are left with the proof of Theorem 1.3, which follows from Lemma 4.4 below.
Lemma 4.4.
If vanishes along for some non-zero real number such that for any . We consider the maximal operator
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where is sufficiently small, such that does not vanish in the angle . Then there holds
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(4.13) |
for .
Proof.
As in Lemma 4.2, we are reduced to estimating the maximal operator defined by
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Here
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(4.14) |
where
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and since and . It is clear that when and ,
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Hence under some transformations (which does not change the norm of up to a constant),
it is sufficient to consider the estimate of the maximal operator along the hypersurface
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(4.15) |
where , is a smooth function and , is a small constant. Here we applied the assumption that the level set determines a curve of finite type near the straight line .
Finally, the boundedness of follows from Theorem 1.4 directly.
Proof of Theorem 1.3: As in the beginning of the proof of Theorem 1.2, then we are reduced to estimating the maximal operator , where is given by equality (4.8). Since , there exists at least one straight line passing through the origin which is contained in but not contained in . Then Lemma 4.4 and Remark 4.5 can be applied.
When , for
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(4.16) |
we obtain
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(4.17) |
When , for
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(4.18) |
there holds
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(4.19) |
Then we finish the proof of Theorem 1.3.
5 Proofs of Theorem 1.6 and Theorem 1.7
Proof of Theorem 1.6: A simple calculation implies that may vanish along the curve and the straight line . Hence we decompose the averaging operator as follows
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Here, we define the operators , respectively as (without of generality, we may assume that ):
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and
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where is a smooth cutoff function, on [-1,1] and supp, and the positive integer can be sufficiently large. Next we will demonstrate that the maximal operators are -bounded for , .
We first consider the maximal operator . By using the dyadic decomposition and isometric transformation, we obtain
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(5.1) |
where the averaging operator is given by
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and is a smooth cutoff function supported in .
It is easy to see that the Heissian matrix of the function degenerates only along the curve and the straight line . Therefore, the hypersurface
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possesses a non-vanishing Gaussian curvature everywhere when is restricted to the support of the function
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It follows that the maximal operator is -bounded if . According to (5.1), the result also holds for the maximal operator .
So it is sufficient to estimate the maximal operators
and . Theorem 1.6 follows from the two lemmas below.
Lemma 5.1.
For each and , there exists a constant such that
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(5.2) |
Lemma 5.2.
For each , there exists and a constant such that
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(5.3) |
Combining the above lemmas with the similar arguments as in [12], we can get the weighted results for the global maximal operators . More details can be found in the appendix of this paper. In what follows, we first prove Lemma 5.1 and Lemma 5.2 for . By dyadic decomposition, isometric transform and a simple changing coordinates, for each , there holds
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(5.4) |
where
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and
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and is a smooth function supported in .
By Littlewood-Paley decomposition, we define as in equality (2.1), and we are reduced to considering the maximal operator , . Let
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When , we can obtain
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and by performing integration by parts, we achieve rapid decay of .
Therefore, we only need to consider the main contribution term where . By an abuse of notation, we define the main contribution term by
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By scaling, there holds
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(5.5) |
Here,
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and . By the stationary phase method and Sobolev embedding, it suffices to estimate the corresponding Fourier integral operators. More concretely, when ,
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(5.6) |
where the bump function is supported in and the Fourier integral operator is defined by
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and is a symbol of order zero. When ,
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(5.7) |
where
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is a small perturbation of , and is a symbol of order zero.
By replacing with , a similar proof implies that the results in Theorem 2.1 remain valid for Fourier integral operators and .
By the estimate (E1) in Theorem 2.1, for , we have
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(5.8) |
and for ,
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(5.9) |
The estimates (E5) and (E4) in Theorem 2.1 imply that for each , there hold
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(5.10) |
and
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(5.11) |
The interpolation argument implies that inequality (5.2) holds for and .
By the estimate (E2) in Theorem 2.1, for each , there holds
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(5.12) |
Interpolation with the -estimates (5.8) and (5.9) implies that when , inequality (5.3) for holds.
This completes the proof of Lemma 5.2 for .
Now let’s turn to consider the maximal operator , there holds
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(5.13) |
Here,
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and
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By an argument similar to that of , we can get -boundedness of provided that , and -boundedness of for .
Then we completed the proof of the positive results in Theorem 1.6.
Finally, we will show that cannot be -bounded for when .
By a change of coordinates, it is equivalent to consider the maximal operator , where is the averaging operator along the hypersurface ,
is a small neighborhood of the origin. Let , . Let
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It is obvious that , where the implied constants depend on .
We define
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and which is the characteristic function over . Then
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(5.14) |
Notice that for each , there exists a and (which depend on ) such that . Then there holds
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(5.15) |
where . It is clear that . Moreover, for each , there holds
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which implies that .
Hence for each , there holds
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Furthermore,
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(5.16) |
If is bounded, then the inequalities (5.14) and (5.16) imply that
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Then it follows that since can be sufficiently small.
When , we can get the necessity of by modifying the above arguments, i.e., just choose , , , and as the characteristic function over with .
We will prove Theorem 1.7 in the rest of this section.
Proof of Theorem 1.7: Our construction of was inspired by Example 6.1.1 in [16]. Let be a smooth cutoff function, , on and supp . For each , set the dyadic interval with center . Put
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It is clear that is a smooth function and for every integer . We notice that for each , on .
For arbitrary small neighborhood of the origin, we choose sufficiently large such that , where which is allowed to be sufficiently small. Let the vectors
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and the rectangle
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For each , set
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Notice that if and , then .
Let be the characteristic function over the set . Then
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(5.17) |
where the implied constant depends on (notice that is fixed). Moreover, for fixed and every , , there holds
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Hence,
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We choose a sequence such that , there holds
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(5.18) |
Finally, Theorem 1.7 follows from inequality (5.17) and (5.18) since can be sufficiently small.