Two Theorems on Convergence of Schrödinger Means

Localization and convergence almost everywhere of Schrödinger means are studied.


Introduction
For f ∈ L 2 (R n ), n 1 and a > 1 we set and S t f (x) = R n e iξ ·x e it|ξ | a f (ξ )dξ, x ∈ R n , t 0.
For a = 2 and f belonging to the Schwartz class S (R n ) we set u(x, t) = S t f (x)/(2π) n . It then follows that u(x, 0) = f (x) and u satisfies the Schrödinger equation i∂u/∂t = u.
We introduce Sobolev spaces H s = H s (R n ) by setting In the case n = 1 it is well-known (see Sjölin [7] and Vega [9] and in the case a = 2 Carleson [3] and Dahlberg and Kenig [4]) that almost everywhere if f ∈ H 1/4 . Also it is known that H 1/4 can not be replaced by H s if s < 1/4. Assuming n 2 and a = 2 Bourgain [1] has proved that (1) holds almost everywhere if f ∈ H s and s > 1/2 − 1/4n. On the other hand Bourgain [2] has proved that s n/2(n + 1) is necessary for convergence for a = 2 and all n 2. In the case n = 2 and a = 2, Du, Guth, and Li [5] proved that the condition s > 1/3 is sufficient. Recently Du and Zhang [6] proved that the condition s > n/2(n + 1) is sufficient for a = 2 and all n 3.
In the case a > 1, n = 2, it is known that (1) holds almost everywhere if f ∈ H 1/2 and in the case a > 1, n 3, convergence has been proved for f ∈ H s with s > 1/2 (see [7] and [9]).
almost everywhere. We shall here study the problem of deciding for which sequences (t k ) ∞ 1 one has almost everywhere if f ∈ H s . We have the following result.
Theorem 1 Assume n 1 and a > 1 and s > 0. We assume that (2) holds and that ∞ k=1 t 2s/a k < ∞ and f ∈ H s (R n ). Then for almost every x in R n . Now assume n = 1, a > 1, and 0 s < 1/4. In Sjölin [8] we studied the problem if there is localization or localization almost everywhere for the above operators S t and the functions f ∈ H s with compact support, that is, do we have lim t→0 S t f (x) = 0 everywhere or almost everywhere in R n \(supp f )? It is proved in [8] that there is no localization or localization almost everywhere of this type for 0 s < 1/4. In fact the following theorem was proved in Sjölin [8].
Theorem A There exist two disjoint compact intervals I and J in R and a function f which belongs to H s for all s < 1/4, with the properties that (supp f ) ⊂ I and for every x ∈ J one does not have Let ω be a continuous and decreasing function on We have the following result.

Theorem 2 The function f in theorem A can be chosen so that f ∈ H ω .
Theorem 2 shows that the sufficient condition f ∈ H 1/4 for convergence almost everywhere and localization almost everywhere of Schrödinger means is very sharp. In the case a = 2 Theorem 2 was obtained in 2009 (unpublished). After proving Theorem 2 we shall use Theorem 1 to make a remark on the Schrödinger means S t f (x) for the function f which was constructed in [8] to prove Theorem A.

Proofs
In the proof of Theorem 1 we shall need the following lemma.
Lemma 1 Assume n 1, a > 1, 0 < s < 1, and 0 < δ < 1. Set Then one has m ∞ Cδ s/a where the constant C does not depend on δ, and m ∞ denotes the norm of m in L ∞ (R n ).
In the case |ξ | δ −1/a one has Then assume 0 |ξ | 1. We obtain In the remaining case 1 < |ξ | < δ −1/a one obtains and the proof of Lemma 1 is complete.
We shall then give the proof of Theorem 1.

Proof of Theorem 1.
We may assume 0 < s < 1. We set We have f ∈ H s and we define g by taking It then follows that g ∈ L 2 (R n ).
We have According to Lemma 1 we have m ∞ t s/a k and applying the Plancherel theorem we obtain and applying the theorem on monotone convergence one also obtains We conclude that ∞ 1 |h k | 2 is convergent almost everywhere and hence lim k→∞ h k (x) = 0 and almost everywhere. Now assume n = 1 and a > 1. We set m(ξ ) = e i|ξ | a , ξ ∈ R, and let K denote the Fourier transform of m so that K ∈ S (R). According to Sjölin [8] p.142, K ∈ C ∞ (R) and there exists a number α 0 such that For t > 0 it is then clear that e it|ξ | a = m(t 1/a ξ) has the Fourier transform It follows that S t f (x) = K t f (x) for f ∈ L 2 (R m ) with compact support. We let q g denote the inverse Fourier transform of g and choose g ∈ S (R) such that suppq g ⊂ (−1, 1) and q g(0) = 0. We set According to [7], p.143, one has f v (ξ ) = vg(vξ + 1/v) and We shall state three lemmas from [8].

Lemma 2
There exist positive numbers c 0 , δ and v 0 such that In the remaining part of this paper δ and v 0 are given by Lemma 2. We may also assume that δ < 1.
We have where It follows that We have f = ∞ 1 f v k and it follows that We conclude that f ∈ H ω and the proof of Theorem 2 is complete. [8] the function f in Theorem A is given by the formula

Remark 1 In Sjölin
where v k is defined by taking 0 < v 1 < min(v 0 , δ/4) and v k = k v μ k−1 for k = 2, 3, 4, ... Here k = 2 −k and μ > 0 is given in the proof of Theorem 2. Also let the intervals I and J be defined as in the proof of Theorem 2. We then set t k (x) = xv 2a−2 k /a for x ∈ J and k = 1, 2, 3, ... It is proved in [8] that for every x 0 ∈ J one does not have lim We now fix x 0 ∈ J and shall use Theorem 1 to prove that although (3) holds one also has lim k→∞ S t k (x 0 ) f (x) = 0 for almost every x ∈ J . (4) We have v k < k and it follows that and ∞ 1 (t k (x 0 )) 2s/a ∞ 1 2 −k(2a−2)2s/a < ∞ for 0 < s < 1/4. Also f ∈ H s for 0 < s < 1/4 and (4) follows from an application of Theorem 1.

Remark 2
In the case a = 2 one has μ = 2 and v k = k v 2 k−1 , and we also have 0 < v 1 < 1/4. It can be proved that it follows that v k = 4 · 2 k−d2 k where d is a constant and d > 2.
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