AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 859680 10.1155/2013/859680 859680 Research Article Almost Everywhere Convergence of Riesz Means Related to Schrödinger Operator with Constant Magnetic Fields Deng Liurui 1 Ma Bolin 2 Bai Chuanzhi 1 College of Finance and Statistics Hunan University Changsha 410082 China hnu.edu.cn 2 College of Science and Information Engineering Jiaxing University Jiaxing Zhejiang 314001 China zjxu.edu.cn 2013 25 3 2013 2013 25 09 2012 28 01 2013 2013 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.

1. Introduction

The magnetic Schrödinger operator (MSO) with constant magnetic fields Hb in n is of the form (1)Hb=-(+iBz2)2,zRn, where B is a real antisymmetric matrix. If B is degenerated (this requires n=2d+l to be odd, and 2d is the rank of the matrix B), then its eigenvalues have the form ±ibj, j=1,2,,d. In properly chosen coordinates z=(z1,z2,,zn)=(x1,y1,x2,y2,,xd,yd,z2d+1,,z2d+l), the operator becomes (2)Hb=-j=1d((xj+ibj2yj)2+(yj+ibj2xj))-Δl, where Δl=j=1l2/z2d+j2 is the Laplacian in Rl. The spectrum of Hb is in the semiaxis starting from the point c=bj, and its spectral expansion is continuous (see ).

Let bj’s be positive. Denote by Et the spectral function of Hb. It is an integral operator with a kernel et(z,z), which is skew-translation invariant; that is, et(z+ω,z+ω)=et(z,z)expiB(z-z),ω.

For β>0, set the Riesz means of order β as (3)Sλβf(x)=0(1-tλ)+βdEtf(x)=c(1-tλ)+βdEtf(x), and the kernel of Sλβ as (4)sλβ(z,z)=sβ(λ,z,z)=0(1-tλ)+βdet(z,z), with the same skew-translation invariance. We will denote by S*β the corresponding maximal operator; that is, (5)S*βf(x)=sup0<λ<|Sλβf(x)|.

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 (6)(TRβf)(ξ)=(1-|ξ|2R)+βf^(ξ). Since the convergence of TRβ in Lp-norm is equivalent to the boundedness of T1β in LP(n), persons look for the Lp boundedness of it. For 2n/(n+1+2β)<p<2n/(n-1-2β) and 0β(n-1/2), Carleson, Cordoba, and Fefferman turn out the boundedness in 2 (see ). When n>2, it is only proven for β(n-1)/2(n+1) (see [4, 5]). Returning the problem about almost everywhere convergence of TRβ, Carbery has finished it for β>0 and 2n/(n+1+2β)<p<2n/(n-1-2β) in two dimensions in 1983 (see ). For higher dimensions, it is completed by Christ only for β(n-1)/2(n+1) (see ). In the special case, β=0, Fefferman studies the L2-boundeness of Sλ0 for n2 (see ). Not until 1988 was it solved by Carbery et al. for β>0, n2, and 2p<2n/(n-1-2β) (see ).

In , Rozenblum and Tashchiyan investigated the Lp-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 n. 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 L(n,|x|αdx), we get the desired result.

2. Main Results Theorem 1.

Set β>0 and n2. Write pβ=2n/(n-1-2β). If fLp(n) with 2p<pβ, then limλSλβf(x)=f(x) almost everywhere.

Usually, we replace the almost everywhere convergence of Riesz means with Lp estimates for the maximal operator. However, we only need to think about weighted L2 estimates for the maximal operator S*β, as follows. In fact, based on the idea in , for 2p<pβ, there exists a number α with 0α<1+2β such that LpL2+L2(|x|α). We have gotten L2 boundedness of the maximal operator in another paper.

Theorem 2.

Suppose that β>0 and 0α<1+2βn. Then, (7)|S*βf(x)|2|x|-αdxCα,β|f(x)|2|x|-αdx.

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 β, the dimension n=2d+l will not play an important role in the following estimates.

Lemma 3 (see [<xref ref-type="bibr" rid="B2">9</xref>]).

If  0<ε<1/2, then (8)||x|-1|ε|f(x)|2dxCαBα(ε)|f(x)|2|x|αdx, where (9)Bα(ε)={εα,ε|logε|,0α<1,ε|logε|,asiα=1,ε,ε|logε|,1<α<n.

For a small δ>0, let mδ(t) be a C function supported in [1-δ,1] and satisfy (10)0mδ(t)1,|Dkmδ(t)|Cδ-kk. Define (11)Kλδf(x)=mδ(λt)dEtf(x),K*δf(x)=supλ>0|Kλδf(x)|. Let ϕ(t)C() supported in [-1,2] and ϕ(t)=1 for 0t1. Set (12)hl(t)={ϕ(t),l=0,ϕ(2-lt)-ϕ(21-lt),l1.

Denoting by χE the character function of the set E, write (13)Hj(x,y)={χ[0,1)(|x-y|),j=0,χ[2j-1,2j)(|x-y|),j1. Let kδ(t) be a function with Fourier transform as (14)k^λδ(t)=mδ(λt). Set (15)kλδ,l,j(t)=kλδ(t)hl(δ22-jtλ),Etjf(x)=net(x,y)Hj(x,y)f(y)dy. Accordingly, define the operator Kλδ,l,j as (16)Kλδ,l,jf(x)=k^λδ,l,j(t)dEtjf(x), where k^λδ,l,j is the Fourier transform of kλδ,l,j. Apparently, since (17)Kλδf(x)=k^λδ(t)dEtf(x), we decompose (18)Kλδf(x)=k^λδ(t)dEtf(x)=j=0l=0Kλδ,l,jf(x)=j=0l=0k^λδ,l,j(t)dEtjf(x).

Lemma 4.

For λ>0, one has (19)Kλδ,l,jf(x)22C2-2M(j+l)δ2Mf22, where the constant C is independent of λ and δ.

Proof.

With the method similar to the proof of Lemma  4 in , we write h(t)=ϕ(t)-ϕ(2t) and expand mδ into a Taylor series around λt. Then, (20)k^λδ,l,j(t)=mδ(λ(t-2-(j+l)δ2rλ))h^(r)dr=mδ(λt-2-(j+l)δ2r)h^(r)dr=RM(t,r)h^(r)dr, where the remainder RM satisfies (21)|RM(t,r)||DMmδ||2-(j+l)δ2r|M2-M(l+j)δM|r|M.

But h^ is a Schwartz function and can be integrated against |r|M. Hence, (22)|k^λδ,l,j(t)|CM2-M(j+l)δM.

Since E is a resolution of the identity, we see (23)Ef(x)=0dEtf(x)=f(x).

Denote by e(x,y) the kernel of E. For almost all x0n, we let S={yn:e(x0,y)f(y)>0}. Decompose (24)f(x)=f(x)χS(x)+f(x)χn/S(x)=f1(x)+f2(x).

We have (25)|Ejf(x0)|=|ne(x0,y)Hj(x0,y)f(y)dy|=|ne(x0,y)Hj(x0,y)f1(y)dy+ne(x0,y)Hj(x0,y)f2(y)dy||ne(x0,y)Hj(x0,y)f1(y)dy|+|ne(x0,y)Hj(x0,y)f2(y)dy|=n|e(x0,y)Hj(x0,y)f1(y)|dy+n|e(x0,y)Hj(x0,y)f2(y)|dyn|e(x0,y)f1(y)|dy+n|e(x0,y)f2(y)|dy=|ne(x0,y)f1(y)dy|+|ne(x0,y)f2(y)dy||f1(x0)|+|f2(x0)|C|f(x0)|.

It is easy to show (26)Kλδ,l,jf(x)22=k^λδ,l,j(t)dEtjf(x)22=dEtj(k^λδ,l,j(Hb)f(x))22=Ej(k^λδ,l,j(Hb)f),Ej(k^λδ,l,j(Hb)f)k^λδ,l,j(Hb)f220|k^λδ,l,j(t)|2d(Etf,f)CM2-2M(j+l)δ2M0d(Etf,f)CM2-2M(j+l)δ2Mf22.

Lemma 5.

If  0<α<n and λ>0, then (27)|k^λδ(t)dEtf(x)|2dx|x|αCαδ|f(x)|2dx|x|α, where Cα is independent of δ and λ.

Proof.

Suppose that f is supported in {|x|C2j}. Write f=χ{0|x|1/4}(x)f(x)+χ{C<|x|C2j}(x)f(x)+χ{1/4<|x|C}(x)f(x)=f1+f2+f3. If C1/4, then f3=0. Since (28)(|Kλδf(x)|2dx|x|α)1/2=(|Kλδ(f1+f2+f3)(x)|2dx|x|α)1/2(|Kλδf1|2dx|x|α)1/2+(|Kλδf2|2dx|x|α)1/2+(|Kλδf3|2dx|x|α)1/2, we only need to prove that (29)|k^λδ(t)dEtfi(x)|2dx|x|αCαδ|f(x)|2dx|x|α,(i=1,2,3). For the case of i=2, with Lemma 4, it follows that (30)|Kλδ,j,lf2(x)|2dxC2-l2-jMδMf222C2-l2-jMδMC|x|C2j|f(x)|2dxC2-l2-jMδM×k=0j-1C2k|x|C2k+1|f(x)|2dx. It is easy to see that (31)k=0j-1C2k<|x|C2k+1|f(x)|2dxCk=0j-12-kαC2k<|x|C2k+1|f(x)|2|x|αdxC|f(x)|2|x|αdx. Thus, we have (32)|Kλδ,j,lf2(x)|2dxC2-l2-jMδM|f(x)|2|x|αdx. Choosing M>1, we get (33)|Kλδ,l,jf2(x)|2dxCα2-l2-jMδ|f(x)|2|x|αdx.

On the other hand, Et is self-adjoint. So, (34)Etf,g=et(x,y)f(y)dyg(x)dx=f(y)et(x,y)g(x)dxdy=f,Etg=f(y)et(y,x)g(x)dxdy. Hence, (35)et(x,y)=et(y,x). With (36)Hj(x,y)=Hj(y,x), we get (37)etj(x,y)=et(x,y)Hj(x,y)=et(y,x)Hj(y,x)=etj(y,x), and it implies that Kλδ,l,j is also self-adjoint; that is, (38)Etjf,g=f,Etjg. Therefore, by duality, (39)|Kλδ,l,jf2(x)|2dx|x|αCα2-l2-jMδ|f(x)|2dxCα2-l2j(α-M)δ|f(x)|2dx|x|α. Taking M>α+1, we can establish the inequality (40)|Kλδ,l,jf2(x)|2dx|x|αCα2-l2-jδ|f(x)|2dx|x|α.

Nextly, we consider i=1. By the definition of Kλδ,l,j and f1, we see that the kernel of Kλδ,l,j is supported in (41){(x,y):2j-1|x-y|2j} and f(y) is supported in (42)0|y|14. So, the support of Kλδ,l,jf(x) is contained in (43){xn:2j+14|x|2j-1-1420-1-14=14}. With Lemma 4, we have (44)  |Kλδ,l,jf1(x)|2dx|x|α|Kλδ,l,jf1(x)|24αdxC|Kλδ,l,jf1(x)|2dxC2-l2-jMδMf122C2-l2-jMδMf22C2-l2-jMδM(2j)α|f(x)|2dx|x|αC2-l2-jδ|f(x)|2dx|x|α, where M>α+1.

At last, we turn to the case of i=3. Similar to the aforementioned, we have (45)  |Kλδ,l,jf3(x)|2dxC2-l2-jMδMf322C2-l2-jMδM1/4|x|C|f(x)|2dxC2-l2-jMδM(14)-α1/4|x|C|f(x)|2(14)αdxC2-l2-jMδM4α1/4|x|C|f(x)|2|x|αdxC2-l2-jMδM|f(x)|2|x|αdx. By duality again, we see (46)|Kλδ,l,jf3(x)|2dx|x|αC2-l2-jMδM|f(x)|2dxC2-l2-jMδM(2j)α|f(x)|2dx|x|αC2-l2-j(M-α)δM|f(x)|2dx|x|α. Through we choose M>α+1, it is not difficult to get (47)|Kλδ,l,jf3(x)|2dx|x|αC2-l2-jδ|f(x)|2dx|x|α. Combining (28), (40),(44) with (47), when f is supported in {|x|C2j}, we come to the conclusion (48)|Kλδ,l,jf(x)|2dx|x|αC2-l2-jδ|f(x)|2dx|x|α.

Now, we hope to establish (48) for all fL2(n). Decompose f=inχQif=infi, where {Qi} are disjoint cubes of common side 10·2j with Q0 centered at 0. Since {Kλδ,l,jfi(x)} have essentially disjoint supports, it suffices to prove (48) for every fi. When i=0, we have proved it. If i>1, then (49)0<(12+l)10·2j<|x|-α<(12+l+1)10·2j(l0). Therefore, we only need to confirm (50)|Kλδ,l,jf(x)|2dxCα2-l2-jδ|f(x)|2dx. In fact, it follows from Lemma 4 that (51)|Kλδ,l,jf(x)|2dxC2-2M(j+l)δ2M|f(x)|2dxC2-l2-jδ|f(x)|2dx.

At present, we complete the proof of Lemma 5.

Lemma 6.

For δ>0, k, and 0α<n, one gets (52)n2k-12k|Kλδf(x)|2dλλdx|x|αCαδ|f(x)|2dx|x|α.

Proof.

Applying Minkowski and Cauchy-Schwartz’s inequalities into the left hand side of (52), we get (53)n|2k-12kKλδf(x)dλλ|2|x|-αdxC(2k-12k(n|Kλδf(x)|2|x|-αdx)1/2dλλ)2C((2k-12kn|Kλδf(x)|2|x|-αdxdλλ)1/2×(2k-12kdλλ)1/2)2C2k-12kn|Kλδf(x)|2|x|-αdxdλλ. Now, it suffices to prove that (54)n|Kλδf(x)|2|x|-αdxCαδn|f(x)|2|x|-αdx, where Cαδ is uniform in 2k-1λ2k. For 0α<n, it is equivalent to (55)n|Kλδg(x)|2dx|x|αCαδn|g(x)|2dx|x|α. 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 (56)(1-tλ)+β=k=02-kβm2-k(tλ). Thus, (57)S*βf(x)k=02-kβK*2-kf(x). Consequently, we consider (58)n|K*δf(x)|2dx|x|αCαδn|f(x)|2dx|x|α. Let (59)Gδf(x)=(0|Kλδf(x)|2dλλ)1/2 and G-δ be defined in the same way but using instead of mδ the function (60)m-δ(λ)=δλdmδ(λ)λ which satisfies the same estimates as mδ. Then, by the fundamental theorem in calculus and Hölder’s inequality, we have (61)|K*δf(x)|202|Kλδf(x)dKλδf(x)dλ|dλ2δ-10|Kλδf(x)|λ1/2λδ|dKλδf(x)/dλ|  λ1/2dλ2δ-1(0|Kλδf(x)|2dλλ)1/2×(0|λδdKλδf(x)dλ|2dλλ)1/2=2δ-1Gδf(x)G-f(x).

Take a Schwartz function ψ such that ψ(0)=0 and ψ(t)=1 if 1/2t2 and ψk(t)=ψ(2kt). Then, when 2k-1λ2k, we have (62)Kλδf(x)=mδ(λt)ψk(t)dEtf(x). Using Lemma 6, we get (63)n2k-12k|Kλδf(x)|2dλλdx|x|α=n2k-12k|mδ(λt)ψk(t)dEtf(x)|2dλλdx|x|α=n2k-12k|mδ(λt)dEt(ψk(Hb)f)(x)|2dλλdx|x|αCαδ|ψk(Hb)f(x)|2dx|x|α=Cαδ|ψk(t)dEtf(x)|2dx|x|α.

From Theorem 3.1 in page 411 of  and the density of Lc in Lp(w), we induce that (64)(k=-|ψk(t)dEtf(x)|2)1/2 is bounded in L2(dx/|x|α). As a result, (65)(Gδf(x))2dx|x|α=0|Kλδf(x)|2dλλdx|x|α=k=-2k-12k|Kλδf(x)|2dλλdx|x|α=k=-2k-12k|Kλδf(x)|2dλλdx|x|αk=-Cαδ|ψk(t)dEtf(x)|2dx|x|αCαδk=-|ψk(t)dEtf(x)|2dx|x|αCαδ|f(x)|2dx|x|α. At last, with (61) and Hölder’s inequality, we come to the result that (66)(K*δf(x))2dx|x|αCα|f(x)|2dx|x|α.

Acknowledgments

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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