# Almost Everywhere Convergence of Riesz Means Related to Schrödinger Operator with Constant Magnetic Fields

Authors: Liurui Deng; Bolin Ma
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

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.


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## Body

## 1. Introduction

The magnetic Schrödinger operator (MSO) with constant magnetic fieldsHb in ℝn is of the form
(1)Hb=-(∇+iBz2)2,z∈Rn,
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=1l∂2/∂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 [1]).Letbj’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′)expi〈B(z-z′),ω〉.Forβ>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(ℝn), persons look for the Lp boundedness of it. For 2n/(n+1+2β)<p<2n/(n-1-2β) and 0≤β≤(n-1/2), Carleson, Cordoba, and Fefferman turn out the boundedness in ℝ2 (see [2–4]). 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 [6]). For higher dimensions, it is completed by Christ only for β≥(n-1)/2(n+1) (see [7]). In the special case, β=0, Fefferman studies the L2-boundeness of Sλ0 for n≥2 (see [8]). Not until 1988 was it solved by Carbery et al. for β>0, n≥2, and 2≤p<2n/(n-1-2β) (see [9]).In [1], 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 [5], it is of the same results as Bochner-Riesz means in ℝn. 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(ℝn,|x|αdx), we get the desired result.


## 2. Main Results

Theorem 1.
Setβ>0 and n≥2. Write pβ=2n/(n-1-2β). If f∈Lp(ℝn) with 2≤p<pβ, then limλ→∞Sλβf(x)=f(x) almost everywhere.Usually, we replace the almost everywhere convergence of Riesz means withLp 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 [9], for 2≤p<pβ, there exists a number α with 0≤α<1+2β such that Lp⊆L2+L2(|x|α). We have gotten L2 boundedness of the maximal operator in another paper.Theorem 2.
Suppose thatβ>0 and 0≤α<1+2β≤n. Then,
(7)∫‍|S*βf(x)|2|x|-αdx≤Cα,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 indexβ, the dimension n=2d+l will not play an important role in the following estimates.Lemma 3 (see [9]).
If0<ε<1/2, then
(8)∫||x|-1|≤ε|f(x)|2dx≤CαBα(ε)∫|f(x)|2|x|αdx,
where
(9)Bα(ε)={εα,ε|logε|,0≤α<1,ε|logε|,asiα=1,ε,ε|logε|,1<α<n.For a smallδ>0, let mδ(t) be a C∞ function supported in [1-δ,1] and satisfy
(10)0≤mδ(t)≤1,|Dkmδ(t)|≤Cδ-k∀k∈ℕ.
Define
(11)Kλδf(x)=∫‍mδ(λt)dEtf(x),K*δf(x)=supλ>0|Kλδf(x)|.
Let ϕ(t)∈C∞(ℝ) supported in [-1,2] and ϕ(t)=1 for 0≤t≤1. Set
(12)hl(t)={ϕ(t),l=0,ϕ(2-lt)-ϕ(21-lt),l≥1.Denoting byχE 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|),j≥1.
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)=∫ℝnet(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=0∞‍∑l=0∞‍Kλδ,l,jf(x)=∑j=0∞‍∑l=0∞‍∫‍k^λδ,l,j(t)dEtjf(x).Lemma 4.
Forλ>0, one has
(19)∥Kλδ,l,jf(x)∥22≤C2-2M(j+l)δ2M∥f∥22,
where the constant C is independent of λ and δ.Proof.
With the method similar to the proof of Lemma  4 in [9], 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|M≤2-M(l+j)δM|r|M.
Buth^ is a Schwartz function and can be integrated against |r|M. Hence,
(22)|k^λδ,l,j(t)|≤CM2-M(j+l)δM.
SinceEℝ is a resolution of the identity, we see
(23)Eℝf(x)=∫0∞dEtf(x)=f(x).
Denote byeℝ(x,y) the kernel of Eℝ. For almost all x0∈ℝn, we let S={y∈ℝn:eℝ(x0,y)f(y)>0}. Decompose
(24)f(x)=f(x)χS(x)+f(x)χℝn/S(x)=f1(x)+f2(x).
We have(25)|Eℝjf(x0)|=|∫ℝn‍eℝ(x0,y)Hj(x0,y)f(y)dy|=|∫ℝn‍eℝ(x0,y)Hj(x0,y)f1(y)dy+∫ℝn‍eℝ(x0,y)Hj(x0,y)f2(y)dy|≤|∫ℝn‍eℝ(x0,y)Hj(x0,y)f1(y)dy|+|∫ℝn‍eℝ(x0,y)Hj(x0,y)f2(y)dy|=∫ℝn‍|eℝ(x0,y)Hj(x0,y)f1(y)|dy+∫ℝn‍|eℝ(x0,y)Hj(x0,y)f2(y)|dy≤∫ℝn‍|eℝ(x0,y)f1(y)|dy+∫ℝn‍|eℝ(x0,y)f2(y)|dy=|∫ℝn‍eℝ(x0,y)f1(y)dy|+|∫ℝn‍eℝ(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=〈Eℝj(k^λδ,l,j(Hb)f),Eℝj(k^λδ,l,j(Hb)f)〉≤∥k^λδ,l,j(Hb)f∥22≤∫0∞|k^λδ,l,j(t)|2d(Etf,f)≤CM2-2M(j+l)δ2M∫0∞d(Etf,f)≤CM2-2M(j+l)δ2M∥f∥22.Lemma 5.
If0<α<n and λ>0, then
(27)∫|∫‍k^λδ(t)dEtf(x)|2dx|x|α≤Cαδ∫‍|f(x)|2dx|x|α,
where Cα is independent of δ and λ.Proof.
Suppose thatf 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 C≤1/4, 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|α≤Cαδ∫‍|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)|2dx≤C2-l2-jMδM∥f2∥22≤C2-l2-jMδM∫C≤|x|≤C2j‍|f(x)|2dx≤C2-l2-jMδM×∑k=0j-1‍∫C2k≤|x|≤C2k+1‍|f(x)|2dx.
It is easy to see that
(31)∑k=0j-1‍∫C2k<|x|≤C2k+1‍|f(x)|2dx≤C∑k=0j-1‍2-kα∫C2k<|x|≤C2k+1‍|f(x)|2|x|αdx≤C∫‍|f(x)|2|x|αdx.
Thus, we have
(32)∫‍|Kλδ,j,lf2(x)|2dx≤C2-l2-jMδM∫|f(x)|2|x|αdx.
Choosing M>1, we get
(33)∫‍|Kλδ,l,jf2(x)|2dx≤Cα2-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|α≤Cα2-l2-jMδ∫|f(x)|2dx≤Cα2-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|α≤Cα2-l2-jδ∫‍|f(x)|2dx|x|α.
Nextly, we consideri=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){x∈ℝn:2j+14≥|x|≥2j-1-14≥20-1-14=14}.
With Lemma 4, we have
(44)∫‍|Kλδ,l,jf1(x)|2dx|x|α≤∫‍|Kλδ,l,jf1(x)|24αdx≤C∫‍|Kλδ,l,jf1(x)|2dx≤C2-l2-jMδM∥f1∥22≤C2-l2-jMδM∥f∥22≤C2-l2-jMδM(2j)α∫‍|f(x)|2dx|x|α≤C2-l2-jδ∫‍|f(x)|2dx|x|α,
where M>α+1.
At last, we turn to the case ofi=3. Similar to the aforementioned, we have
(45)∫‍|Kλδ,l,jf3(x)|2dx≤C2-l2-jMδM∥f3∥22≤C2-l2-jMδM∫1/4≤|x|≤C‍|f(x)|2dx≤C2-l2-jMδM(14)-α∫1/4≤|x|≤C‍|f(x)|2(14)αdx≤C2-l2-jMδM4α∫1/4≤|x|≤C‍|f(x)|2|x|αdx≤C2-l2-jMδM∫‍|f(x)|2|x|αdx.
By duality again, we see
(46)∫‍|Kλδ,l,jf3(x)|2dx|x|α≤C2-l2-jMδM∫‍|f(x)|2dx≤C2-l2-jMδM(2j)α∫‍|f(x)|2dx|x|α≤C2-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|α≤C2-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|α≤C2-l2-jδ∫‍|f(x)|2dx|x|α.
Now, we hope to establish (48) for all f∈L2(ℝn). Decompose f=∑i∈ℤn‍χQif=∑i∈ℤn‍fi, 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(l≥0).
Therefore, we only need to confirm
(50)∫|Kλδ,l,jf(x)|2dx≤Cα2-l2-jδ∫|f(x)|2dx.
In fact, it follows from Lemma 4 that
(51)∫‍|Kλδ,l,jf(x)|2dx≤C2-2M(j+l)δ2M∫‍|f(x)|2dx≤C2-l2-jδ∫|f(x)|2dx.
At present, we complete the proof of Lemma5.Lemma 6.
Forδ>0, k∈ℤ, and 0≤α<n, one gets
(52)∫ℝn‍∫2k-12k‍|Kλδf(x)|2dλλdx|x|α≤Cαδ∫|f(x)|2dx|x|α.Proof.
Applying Minkowski and Cauchy-Schwartz’s inequalities into the left hand side of (52), we get
(53)∫ℝn‍|∫2k-12k‍Kλδf(x)dλλ|2|x|-αdx≤C(∫2k-12k‍(∫ℝn‍|Kλδf(x)|2|x|-αdx)1/2dλλ)2≤C((∫2k-12k‍∫ℝn‍|Kλδf(x)|2|x|-αdxdλλ)1/2×(∫2k-12k‍dλλ)1/2)2≤C∫2k-12k‍∫ℝn‍|Kλδf(x)|2|x|-αdxdλλ.
Now, it suffices to prove that
(54)∫ℝn‍|Kλδf(x)|2|x|-αdx≤Cαδ∫ℝn‍|f(x)|2|x|-αdx,
where Cαδ is uniform in 2k-1≤λ≤2k. For 0≤α<n, it is equivalent to
(55)∫ℝn|Kλδg(x)|2dx|x|α≤Cαδ∫ℝn|g(x)|2dx|x|α.
It is just as the result of Lemma 5.Now, we come back to the proof of Theorem2.Proof.
As in [3], we can decompose
(56)(1-tλ)+β=∑k=0∞2-kβm2-k(tλ).
Thus,
(57)S*βf(x)≤∑k=0∞2-kβK*2-kf(x).
Consequently, we consider
(58)∫ℝn|K*δf(x)|2dx|x|α≤Cαδ∫ℝn|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)|2≤∫0∞2|Kλδf(x)dKλδf(x)dλ|dλ≤2δ-1∫0∞‍|Kλδf(x)|λ1/2λδ|dKλδf(x)/dλ|λ1/2dλ≤2δ-1(∫0∞|Kλδf(x)|2dλλ)1/2×(∫0∞|λδdKλδf(x)dλ|2dλλ)1/2=2δ-1Gδf(x)G-f(x).
Take a Schwartz functionψ such that ψ(0)=0 and ψ(t)=1 if 1/2≤t≤2 and ψk(t)=ψ(2kt). Then, when 2k-1≤λ≤2k, we have
(62)Kλδf(x)=∫mδ(λt)ψk(t)dEtf(x).
Using Lemma 6, we get
(63)∫ℝn∫2k-12k|Kλδf(x)|2dλλdx|x|α=∫ℝn‍∫2k-12k‍|∫‍mδ(λt)ψk(t)dEtf(x)|2dλλdx|x|α=∫ℝn‍∫2k-12k‍|∫‍mδ(λt)dEt(ψk(Hb)f)(x)|2dλλdx|x|α≤Cαδ∫|ψk(Hb)f(x)|2dx|x|α=Cαδ∫|∫‍ψk(t)dEtf(x)|2dx|x|α.
From Theorem 3.1 in page 411 of [10] 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|α≤∑k=-∞∞‍Cαδ∫|∫ψk(t)dEtf(x)|2dx|x|α≤Cαδ∫∑k=-∞∞‍|∫‍ψk(t)dEtf(x)|2dx|x|α≤Cαδ∫|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|α≤Cα∫|f(x)|2dx|x|α.



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