Proof.
If we have
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Thus we consider the case . Let . We decompose into . By maximal function estimate, Lemma 5.7, (5.37) and (5.38) in [10] one has
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If , we firstly manipulate the case . Similar to the former argument one has
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(4.1) |
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Thus we only need to estimate for . Let
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Then
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Note that on the support set of one has since , .
By the maximal function estimate and the Plancherel identity one has
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Define
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Let , .
Then
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Thus by the Strichartz estimate [8] one has
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Note that there exists constant such that
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By the Hölder inequality we have
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We conclude the proof.
∎
Proof.
By Lemma 4.1 we have
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Let . By the triangle inequality we have
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For the term we use Proposition 5.12 in [10].
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Then by the local smoothing estimate one has
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By symmetry we control similarly.
For we decompose the summation into four parts.
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For we show
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where are supported on . Let be the maximum, medium and minimum among . By Lemma 5.7 in [10] we have
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Then by the duality we have
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For we show
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In fact by Lemma 5.7 (5.38) in [10] we have
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Then by the duality we have
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By the symmetry we can control similarly.
Now, we estimate the term . Let , . Firstly we estimate
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Typically we control
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By high modulation and transversal estimates, we have
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On the other hand by the Strichartz estimate we have
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Then, by the interpolation [7], one has
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Thus,
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Secondly we estimate . By high modulation and transversal estimates, we have
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Then by the duality one has
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We finish the proof of this proposition.
∎
Proof.
By the duality, we only need to show
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and
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where are supported on . Let . Firstly, we control
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Since for , , , , one has . Typically, we need to control
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By transversal and high modulation estimates we can control this term by
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Recall that is supported on . Thus for any , one has .
By the Hölder inequality and Strichartz estimates,
we can also control this term by
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Then, by the interpolation [7] and Lemma 3.2, one has
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Similarly, one has
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For the other part, by the Hölder inequality,
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To conclude the proof of this lemma, we show
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To show the first one, we only need to show
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Let be intervals included in with length . Also, we can assume that for any , one has . It is equivalent to
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There exists a interval included in with length such that one has for any , . Then by Lemma 3.3 and
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(4.2) |
we can control this part. Thus we only need to show
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By the orthogonality, we can assume that are intervals with length .
There exists a interval included in with length at most such that one has for any , . By Lemma 3.3 and the former argument we only need to show
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where are intervals with length .
There exists a interval included in with length such that one has for any , . By Lemma 3.4 and following the former argument we only need to show
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where are intervals with length . Note that
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where . Note that . For , by (4.2) and the Plancherel identity we have
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Let be the center of , . Note that
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Thus by the Taylor expansion and the transversal estimate one has
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Thus we conclude the proof of the first inequality.
For the second one, similar to the former argument, we reduce it to
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where , are intervals included in , are intervals included in and
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(4.3) |
Recall the former argument. Let , we only need to show
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Let be the center of respectively. Note that
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By (4.3), , we obtain
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Following the same argument for the first inequality, by the Taylor expansion and the transversal estimate, we conclude the proof.
∎
Combining Lemmas 2.1–2.3, Corollary 3.6, Propositions 4.2, 4.4, and 4.5, one can construct the solution of (2.2). Let . By rescaling we obtain the local solution of initial system (S-KdV) with . For general , see the argument in [2]. One can use the method in this paper to manipulate the case with some modification. However there exists easier argument. For example if , we do not need rescaling. One can use the norms
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Then we can show the contraction for the iteration of the initial integral equation. If , , we can show . The argument is also much easier than the case . See also [11]. We omit the details.