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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 859680, 7 pages
Almost Everywhere Convergence of Riesz Means Related to Schrödinger Operator with Constant Magnetic Fields
1College of Finance and Statistics, Hunan University, Changsha 410082, China
2College of Science and Information Engineering, Jiaxing University, Jiaxing, Zhejiang 314001, China
Received 25 September 2012; Accepted 28 January 2013
Academic Editor: Chuanzhi Bai
Copyright © 2013 Liurui Deng and Bolin Ma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We study almost everywhere convergence for Riesz means related to Schrödinger operator with constant magnetic fields. Through researching the weighted norm estimates for the maximal operator with power-weight functions, we obtain the desired result, which is similar to the work given by Anthony Carbery, Jose L. Rubio de Francia, and Luis Vega.
The magnetic Schrödinger operator (MSO) with constant magnetic fields in is of the form where is a real antisymmetric matrix. If is degenerated (this requires to be odd, and is the rank of the matrix ), then its eigenvalues have the form , . In properly chosen coordinates , the operator becomes where is the Laplacian in . The spectrum of is in the semiaxis starting from the point , and its spectral expansion is continuous (see ).
Let ’s be positive. Denote by the spectral function of . It is an integral operator with a kernel , which is skew-translation invariant; that is, .
For , set the Riesz means of order as and the kernel of as with the same skew-translation invariance. We will denote by the corresponding maximal operator; that is,
As an indispensable part in harmonic analysis, many people investigate the convergence of Bochner-Riesz means for Fourier transform in norm and almost everywhere, which is defined as Since the convergence of in -norm is equivalent to the boundedness of in , persons look for the boundedness of it. For and , Carleson, Cordoba, and Fefferman turn out the boundedness in (see [2–4]). When , it is only proven for (see [4, 5]). Returning the problem about almost everywhere convergence of , Carbery has finished it for and in two dimensions in 1983 (see ). For higher dimensions, it is completed by Christ only for (see ). In the special case, , Fefferman studies the -boundeness of for (see ). Not until 1988 was it solved by Carbery et al. for , , and (see ).
In , Rozenblum and Tashchiyan investigated the -norm convergence for Riesz means for Schrödinger operator with constant magnetic fields. They showed that under the restriction theorem similar to one for Fourier transform in , it is of the same results as Bochner-Riesz means in . However, very few results are considered for almost everywhere convergence of Riesz means for Schrödinger operator with constant magnetic fields. In the paper, we are interested in it. Through researching the boundedness of the maximal operator in , we get the desired result.
2. Main Results
Theorem 1. Set and . Write . If with , then almost everywhere.
Usually, we replace the almost everywhere convergence of Riesz means with estimates for the maximal operator. However, we only need to think about weighted estimates for the maximal operator , as follows. In fact, based on the idea in , for , there exists a number with such that . We have gotten boundedness of the maximal operator in another paper.
Theorem 2. Suppose that and . Then,
In order to prove the theorem, we introduce the following lemmas, which are the essential tools. In the following lemmas, we reduce the maximal operator to the one generated by a multiplier with compact support. Since it is easy to see that the boundedness of the maximal operator generated by the multipliers is independent of the index , the dimension will not play an important role in the following estimates.
Lemma 3 (see ). If , then where
For a small , let be a function supported in and satisfy Define Let supported in and for . Set
Denoting by the character function of the set , write Let be a function with Fourier transform as Set Accordingly, define the operator as where is the Fourier transform of . Apparently, since we decompose
Lemma 4. For , one has where the constant is independent of and .
Proof. With the method similar to the proof of Lemma 4 in , we write and expand into a Taylor series around . Then,
where the remainder satisfies
But is a Schwartz function and can be integrated against . Hence,
Since is a resolution of the identity, we see
Denote by the kernel of . For almost all , we let . Decompose
It is easy to show
Lemma 5. If and , then where is independent of and .
Proof. Suppose that is supported in . Write . If , then . Since
we only need to prove that
For the case of , with Lemma 4, it follows that
It is easy to see that
Thus, we have
Choosing , we get
On the other hand, is self-adjoint. So, Hence, With we get and it implies that is also self-adjoint; that is, Therefore, by duality, Taking , we can establish the inequality
Nextly, we consider . By the definition of and , we see that the kernel of is supported in and is supported in So, the support of is contained in With Lemma 4, we have where .
At last, we turn to the case of . Similar to the aforementioned, we have By duality again, we see Through we choose , it is not difficult to get Combining (28), (40),(44) with (47), when is supported in , we come to the conclusion
Now, we hope to establish (48) for all . Decompose , where are disjoint cubes of common side with centered at . Since have essentially disjoint supports, it suffices to prove (48) for every . When , we have proved it. If , then Therefore, we only need to confirm In fact, it follows from Lemma 4 that
At present, we complete the proof of Lemma 5.
Lemma 6. For , , and , one gets
Proof. Applying Minkowski and Cauchy-Schwartz’s inequalities into the left hand side of (52), we get Now, it suffices to prove that where is uniform in . For , it is equivalent to It is just as the result of Lemma 5.
Now, we come back to the proof of Theorem 2.
Proof. As in , we can decompose
Consequently, we consider
and be defined in the same way but using instead of the function
which satisfies the same estimates as . Then, by the fundamental theorem in calculus and Hölder’s inequality, we have
Take a Schwartz function such that and if and . Then, when , we have Using Lemma 6, we get
From Theorem 3.1 in page 411 of  and the density of in , we induce that is bounded in . As a result, At last, with (61) and Hölder’s inequality, we come to the result that
The authors deeply thank the referee for reviewing their paper. The first author was supported by the NSF of China (71201051, 71031004, and 71073047), NSF of Hunan Province (2012RS4028) and Postdoctoral Science Foundation of China (2012M521513). The second author was supported by the NSF of China (10771054) and NSF of Hunan Province (09JJ5002).
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