Schrödinger equation on the Heisenberg group

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STUDIA MATHEMATICA 161 (2) (2004) Schrödinger equation on the Heisenberg group by Jacek Zienkiewicz (Wrocław) Abstract. Let L be the full laplacian on the Heisenberg group H n of arbitrary dimension n. Then for f L 2 (H n ) such that (I L) s/2 f L 2 (H n ) for some s > 1/2 and for every φ C c (H n ) we have φ(x) sup 1 e tlf(x) 2 dx Cφ f 2 W s. 0<t 1 H n Introduction. Let V t be the Schrödinger unitary group generated by a self-adjoint, positive differential operator L on R d. The degree of smoothness needed for the almost everywhere convergence of V t f to f as t 0 has been extensively studied. In general, the result of Cowling [Cw] says that if (1 + L) s/2 f L 2 < for some s > 1, then ( ) lim V t f(x) = f(x) t 0 This does not depend on any other properties of L. For L being the Laplace operator on R d, s > 1/2 suffices for all d, and for d = 2, s > 1/2 δ is also sufficient. See [B], [Mo]. In our previous paper [Z] the Laplace operator L on the Heisenberg group H n has been studied from this point of view, and we have proved that s > 3/4 implies ( ). In this paper we simplify the proof of the result in [Z] and decrease the needed regularity of f to f W s, s > 1/2. The author would like to express his gratitude to M. Derencz and A. Hulanicki for their help in editing the paper. a.e. 0. Preliminaries. We identify R 2 with C and consequently R 2n with C n. Denote by S(z, w) = 2I(z w) the standard symplectic form on R 2n. 2000 Mathematics Subject Classification: 35S10, 42B25. [99]

100 J. Zienkiewicz For m = 0, 1, 2,... let L m (x) = m k=0 ( ) m ( x) k k k! be the Laguerre polynomial of degree m and for a 0 let l m,a (z) = e a z 2 L m (2 a z 2 ) be the corresponding Laguerre function. Let m = (m 1,..., m n ) and z = (z 1,..., z n ). We write q m,a (z) = l m1,a(z 1 )l m2,a(z 2 )... l mn,a(z n ). It is well known that the q m,1 form an orthonormal basis of the space of polyradial functions on C n. We denote by dz the Lebesgue measure on C n and for a 0 we define the twisted convolution f a g(z) = f(z w)g(w)e ias(z,w) dw, f, g C c (C n ). We have the following orthogonality relation for the Laguerre functions (cf. [M]): (0.1) a n q k,a a q m,a (z) = δ k,m q m,a (z). Fix a real a 0 and let (0.2) Q m,a f(z) = a n q m,a a f(z). It follows from (0.1) that for a fixed a 0 the operators Q m,a are mutually orthogonal projectors. Moreover m Q m,a = Id (cf. [M]). We introduce a separate notation for the operators Q m,a in the case m = m N, i.e. n = 1. We then write Q m,a f = P m,a f = a l m,a a f. The Heisenberg group H n is defined as C n R, with the group product (z, s)(w, t) = (z + w, s + t + 2I(z w)) where z = (z 1,..., z n ), z j = x j + iy j. Then the Lebesgue measure on C n R is the Haar measure on H n. Let and let X i = xi + 2y i t, Y i = yi 2x i t for 1 i n, T = t, L = n i=1 (X 2 i + Y 2 i ) + T 2 be the elliptic laplacian on H n. The closure of L on Cc (H n ) is a self-adjoint operator (see [NS]). Therefore il generates a group {V t } t R of unitary operators on L 2 (H n ). We will use the following formula for V t, valid for f S(H n )

Schrödinger equation on the Heisenberg group 101 (cf. [M]): (0.3) V t f(z, u) = m R e iua e itλ m (a) Q m,a f a (z) da, where λ m (a) = (2 m + n) a + a 2, m = m 1 +... + m k and f a denotes the Fourier transform with respect to the central variable. Let s 0. We define a scale of Sobolev spaces by putting f W s = (I L) s/2 f L 2. Since Q m,a are mutually orthogonal projectors and Lf(z, u) = e iua λ m (a)q m,a f a (z) da the Plancherel theorem applied to the variable u implies R m (0.4) f 2 W s = m R (1 + λ m (a)) s Q m,a f a 2 L 2 (C n ) da. For a more detailed exposition of the preliminary facts we refer the reader to [M] and [Z]. 1. Basic lemmas. Let 0 < α < 1. The fractional derivative of order α is defined by α f(s) = (f(s t) f(s)) t (1+α) dt. R Lemma 1 (Sobolev). Let γ > 0 be a Schwartz function and 1/2 < α < 1. Then ( ) sup f(t) 2 C α α f(t) 2 γ(t) dt + f(t) 2 γ(t) dt. 1 t 1 R R For a function φ, let M φ denote the operator of multiplication by φ. Set B(r) = {z : z r}. Fix φ Cc (C) with supp φ B(1) and φ(z) 1, and define T m,a f(z) = M φ P m,a f(z) = φ(z) a l m,a a f(z). 1. The fol- Since P m,a is an orthogonal projector we have T m,a L2 L 2 lowing two lemmas have been proved in [Z]. Lemma 2. For 4 a m + 1 we have ( ) 1/2 a T m,a 2 L 2 L C. 2 m + 1 Lemma 3. For a 4 we have T m,a 2 C ( ) 1/2 a. m + 1

102 J. Zienkiewicz For the reader s convenience we include the proofs of Lemmas 2 and 3. To do this we need a number of consequences of the classical estimates for Laguerre functions, collected below. Lemma 4. (1.1) L m (λx) = m k=0 ( ) m L k (x)λ k (1 λ) m k. k Lemma 5. Let 1 z (m + 1) 1/2. Then l m,1 (z) C(m + 1) 1/4 z 1/2. Proof. Let 0 < ε ϕ π/2 ε(m + 1) 1/2. Then by a theorem of Szegő [Sz], for x = (4m + 2) cos 2 ϕ, we have e x/2 L m (x) = ( 1) m (π sin ϕ) 1/2 (sin((m + 1/2)(sin 2ϕ 2ϕ) + 3π/4) Lemma 6. Let z 1. Then (x(m + 1)) 1/4 + (x(m + 1)) 1/2 O(1)). l m,1 (z) = J 0 (2 1/2 z (m + 1/2) 1/2 ) + O((m + 1) 3/4 ), where J 0 is the zero Bessel function. Proof. Follows from an asymptotic formula for the Laguerre polynomials (cf. [Sz]): Also e x/2 L m (x) = J 0 ((2x(m + 1/2)) 1/2 ) + O((m + 1) 3/4 ). Lemma 7. There is a constant C such that for A 1 we have l m,1 (z) 2 e z 2 /A 2 dz CA(m + 1) 1/2. Proof. By Lemma 5, we obtain l m,1 (z) 2 e z 2 /A 2 dz 1 z (m+1) 1/2 C l m,1 (z) 2 e z 2 /A 2 dz e m/a2 z (m+1) 1/2 1 2 /A 2 e z dz CA(m + 1) 1/2. z (m + 1) 1/2 l m,1 (z) 2 dz CA(m+1) 1/2. On the other hand, by Lemma 6, using the estimate J 0 (x) C(1+ x ) 1/2 for the Bessel function (see [Sz]) we obtain Hence z 1 l m,1 (z) C(1 + z 1/2 (m + 1) 1/4 ) 1. l m,1 (z) 2 e z 2 /A 2 dz C(m + 1) 1/2.

Schrödinger equation on the Heisenberg group 103 Proof of Lemma 2. Since P m,a is an orthogonal projector, T m,a T m,a = M φ P m,a M φ. Hence, the kernel K of T m,a T m,a is given by the formula (1.1) K(z 1, z 2 ) = φ(z 1 ) a l m,a (z 1 z 2 )e ias(z 1,z 2 ) φ(z 2 ). We write Thus (1.2) K(z 1, z 2 ) = α 1 = e z 1 z 2 2 e z 1 z 2 2 = e z 1 z 2 2 α c α z α 1 1 zα 3 1 zα 2 2 zα 4 2. c α z α 1 1 zα 3 1 φ(z 1)e z 1 z 2 2 a l m,a (z 1 z 2 )e ias(z 1,z 2 ) φ(z 2 )z α 2 2 zα 4 2. Consequently, the operator T m,a T m,a is the sum over α of operators where c α M φ M z α 1 1 zα 3 1 T K 1 M z α 2 2 zα 4 2 M φ, T K1 = f a K 1, K 1 (z) = e z 2 a l m,a (z). Since c α converges to zero faster than exponentially, it suffices to estimate the norm of T K1. Dilating we see that the norm of T K1 is the same as the norm of the 1-twisted convolution operator by K(z) = e z 2 a 1 l m,1 (z). The radial function K(z) has a decomposition K(z) = c k,m,a l k,1 (z), where k=0 ( ) c k,m,a = e z 2 a 1 l k,1 (z)l m,1 (z) dz. So K(z) 1 f(z) = c k,m,a P m,1 f(z). k=0 Since P m,1, m = 0, 1,..., are mutually orthogonal projectors the norm of the operator f K 1 f is equal to By the Schwarz inequality, we obtain sup c k,m,a. k c k,m,a e z 2 /2 a l k,1 (z) L 2 e z 2 /2 a l m,1 (z) L 2.

104 J. Zienkiewicz Now, by Lemma 7, if 10k m, then ( a c k,m,a C m + 1 ) 1/4 ( a k + 1 ) 1/4 ( ) 1/2 a C. m + 1 It remains to estimate the coefficients c k,m,a for 10k m. Observe that by the definition of l m,1 (z), for λ = (1 + (2 a ) 1 ) 1, ( ) turns into Then c k,m,a = C 0 e λ 1x L m (x)l k (x) dx. c k,m,a = Cλ e x L m (xλ)l k (xλ) dx, 0 whence, in virtue of (1.1), because the L k form an orthonormal basis with the weight e x, we obtain m k ( )( ) m k c k,m,a = Cλ λ (s 1+s 2 ) (1 λ) m+k (s 1+s 2 ) = Cλ s 1 =0 s 2 =0 0 s 1 s 2 e x L s1 (x)l s2 (x) dx k s=0 ( )( ) m k λ 2s (1 λ) m+k 2s. s s Now, if a 4 then 2/3 λ 1 so for 10k m we have k c k,m,a 2 m 2 k (1 λ) m+k 2k k2 m 2 k 3 m k s=0 for some positive constant ε. ( ) m+k 2 k 2 εm 2 εm a 3 Proof of Lemma 3. In order to estimate the norm of T m,a T m,a we use (1.1) and the asymptotic formula for the Laguerre functions given in Lemma 6. Let a 4. By the Taylor series expansion for e ias(z 1,z 2 ) we have K(z 1, z 2 ) = α z α 1 1 zα 3 1 φ(z 1) a l m,a (z 1 z 2 )φ(z 2 )z α 2 2 zα 4 2 a α a α /2 = α a α a α /2 K α (z 1, z 2 ). Since the a α s decay faster than exponentially, and the norms of the operators M φ M z α grow at most exponentially, it suffices to estimate the norm of

the operator K given by the kernel Schrödinger equation on the Heisenberg group 105 A(z 1, z 2 ) = ψ(z 1 ) a l m,a (z 1 z 2 )ψ(z 2 ), Now using Lemma 6 we obtain where ψ C c with ψ(z) = 1 on supp φ. ψ(z 1 ) a l m,a (z 1 z 2 )ψ(z 2 ) = Cψ(z 1 ) a J 0 (2 a 1/2 z 1 z 2 (2m +1) 1/2 )ψ(z 2 ) + ψ(z 1 )ψ(z 2 )O( a (m + 1) 3/4 ). Observe that the error term in the last formula gives an operator with norm of order a (m + 1) 3/4, so it is negligible. Hence, for a function φ S(C) with φ = 1 on supp ψ supp ψ we write ψ(z 1 ) a J 0 ( a 1/2 z 1 z 2 (2m + 1) 1/2 )ψ(z 2 ) = φ(z 1 z 2 )ψ(z 1 ) a J 0 ( a 1/2 z 1 z 2 (2m + 1) 1/2 )ψ(z 2 ). Thus we may drop ψ(z 1 ), ψ(z 2 ) and we estimate the norm of the convolution operator by the function R = φ(z) a J 0 ( a 1/2 z (2m + 1) 1/2 ). By definition, J 0 is the Fourier transform of the normalized Lebesgue measure supported on the unit circle. Hence R = φ a µ, where µ is the normalized Lebesgue measure supported by the circle of radius a(2m + 1) 1/2. We write (using a smooth resolution of identity 1 = j Z 2 k(z j) with supp k B(2)) φ = j α j φ j, where j α j <, φ j L 1 and the support of φ j is contained in the disc of radius two. A trivial geometric argument shows that for (2m + 1)a 1, φ j µ L C (2m + 1)a 1/2. These imply that the L norm of R is bounded by C a 1/2 (m + 1) 1/2. If (2m + 1)a 1 then φ j µ L C and consequently R L C a C (m + 1) 1/2 a 1/2. This proves the lemma. 2. Main theorem. For a fixed φ C c (H n ) we define the local maximal function of the group V t by We have Mf(z, u) = φ(z, u) sup V t f(z, u). 0 t 1

106 J. Zienkiewicz Theorem 1. Let s > 1/2 and f W s. Then Mf L 2 C f W s. Proof. Let f L 2 (H n ). To estimate Mf L2 (H n ) we introduce a family of projections P α. Then we write Mf L2 (H n ) α MP α f L2 (H n ) and we estimate each MP α f L2 (H n ) separately. We will use the abbreviation For k, l N let P k,l f(z, u) = s 2 k iff 2 k s < 2 k+1. {m : m 2 k } { a 2 l } e iua Q m,a f a (z) da, P 0 f(z, u) = m { a 1} e iua Q m,a f a (z) da. Then obviously P 0 + k,l P k,l = Id. The maximal function of the theorem splits into the maximal functions (2.2) S k,l f(z, u) = S 0 f(z, u) = sup ψ(z)φ(u)p k,l V t f(z, u), 0 t 1 sup ψ(z)p 0 V t f(z, u), 0 t 1 where ψ C c (C n ), φ C c (R), supp φ B(1). We are going to estimate the norms S k,l W 1/2+ε L 2 and S 0 W 1/2+ε L 2. Then we sum up the estimates. With no loss of generality we may consider only the m s in I 1 = {m : m 1 = max(m 1,..., m n )}. Let A = {(m, r) : m 2 = r 2,..., m n = r n, m, r I 1 }. We fix a and we note that m = r and (m, r) A imply m = r. By the orthogonality relations (0.1) for P m,a we have Q m,a f(z) Q r,a f(z) dz 2... dz n = P m1,ap m2,a... P mn,af(z) P r1,ap r2,a... P rn,af(z) dz 2... dz n = 0 if (m 2,..., m n ) (r 2,..., r n ). In the formula above P mi,a acts on the variable z i.

Schrödinger equation on the Heisenberg group 107 We begin by estimating the norm of S 0, making use of the Sobolev lemma. We have ( ) S 0 f(z, u) 2 C 1/2+ε t V t P 0 f(z, u) 2 γ(t) dt+ V t P 0 f(z, u) 2 γ(t) dt ψ(z). R R In what follows we assume that γ is supported in the interval [ 1, 1]. Integrating with respect to dzdu, by the Plancherel theorem applied to the Fourier transform in the central variable, we have S 0 f(z, u) 2 dz du 1/2+ε t P 0 V t f(z, u) 2 ψ(z)γ(t) dz du dt + C f 2 L 2 = C I {0 a 1} (a)q m,a f a (z) 1/2+ε t e iλ m (a)t 2 da γ(t) dt ψ(z) dz m + C f 2 L 2 C I {C/ m a 1} (a)(λ m (a)) 1/2+ε m Q m,a f a (z)e iλ m (a)t 2 da γ(t) dt ψ(z) dz + f 2 L 2. In the last inequality we have used the fact that for a C m 1, we have λ m (a) C. In the above sum the multiindices m belong to I 1. We enlarge the last expression by replacing the ψ(z) by ψ(z 1 ), ψ Cc (C). Thus I {C/ m a 1} (a)(λ m (a)) 1/2+ε Q m,a f a (z)e iλ m (a)t 2 da γ(t) dt ψ(z) dz m = m I 1 r I 1 I {C/ m a 1} (a)i {C/ r a 1} (a)(λ m (a)λ r (a)) 1/2+ε Q m,a f a (z) Q r,a f a (z) γ(λ m (a) λ r (a)) da ψ(z 1 ) dz. By orthogonality of P m,a the last expression is equal to I {C max{ m 1, r 1 } a 1}(a)(λ m (a)λ r (a)) 1/2+ε (m,r) A = Q m,a f a (z) Q r,a f a (z) ψ(z 1 ) dz γ(λ m (a) λ r (a)) da I {C max{ m 1, r 1 } a 1}(a)(λ m (a)λ r (a)) 1/2+ε (m,r) A Q m,a f a (z) Q r,a f a (z) ψ(z 1 ) dz γ(2m 1 a 2r 1 a ) da

108 J. Zienkiewicz (m,r) A ( a m + 1 I {C max{ m 1, r 1 } a 1}(a)(λ m (a)λ r (a)) 1/2+ε ) 1/4 a r + 1 Q m,a f a Q r,a f a γ(2(m 1 r 1 ) a ) da. To verify the last inequality we use Lemma 3. The last expression is bounded by S = C I {C max{ m 1, r 1 } a 1}(a)(λ m (a)λ r (a)) 1/2+ε (m,r) A (( ) 1/2 ( ) 1/2 ) a a Q m,af a 2 + Q r,af a 2 γ(2(m 1 r 1 ) a ) da. m +1 r +1 For fixed r we have (2.3) I {C max{ m 1, r 1 } a 1}(a)(λ m (a)λ r (a)) 1/2+ε {m : (m,r) A} ( ) a 1/2 γ((m 1 r 1 ) a ) C(λ r (a)) 1/2+2ε. r + 1 In order to verify (2.3) we observe that for m, r, and a as in (2.3) one can write cλ m (a) ( m + 1) a Cλ m (a), c m r C m. cλ r (a) ( r + 1) a Cλ r (a), To show the last inequality we observe that the conditions γ((m 1 r 1 ) a ) 0, (m, r) A and C max{ m 1, r 1 } a 1 imply that m r Cmin{ r, m }. Also {m : (m, r) A, r 1 m 1 a supp γ} C/ a. Now (2.3) follows by an easy calculation. By (2.3), S is dominated by 2 I {Cr 1 a 1}(a)(λ r (a)) 1/2+ε Q r,a f a 2 da f 2 W. 1/2+ε r We are going to estimate S k,l f(z, u) L 2 in a similar way. Without loss of generality, we can consider only where Sk,lf(z, 1 u) = sup ψ(z 1 )φ(u)pk,lv 1 t f(z, u), 0 t 1 P 1 k,lf(z, u) = {m I 1 : m 2 k } { a 2 l } e iua Q m,a f a (z) da.

Schrödinger equation on the Heisenberg group 109 Again by the Sobolev lemma, the norm Sk,l 1 f(z, u) 2 L is controlled by 2 e iua+itλ m (a) λ 1/2+ε m (a)q m,a f a (z) da 2 {m I 1 : m 2 k } { a 2 l } = {m,r : m, r 2 k, r I 1 } e iu(a 1 a 2 )+it(λ m (a 1 ) λ r (a 2 )) I { a 2l }(a 1 )I { a 2l }(a 2 )(λ m (a 1 )λ r (a 2 )) 1/2+ε ψ(z 1 )φ(u) dz du γ(t) dt = Q m,a1 f a 1 (z)q r,a2 f a 2 (z)ψ(z 1 ) dz φ(u) du γ(t) dt da 1 da 2 I { a 2l }(a 1 )I { a 2l }(a 2 )(λ m (a 1 )λ r (a 2 )) 1/2+ε {m,r : r 2 k, m,r I 1 } Q m,a1 f a 1 (z)q r,a2 f a 2 (z)ψ(z 1 ) dz = φ(a 1 a 2 ) γ(λ m (a 1 ) λ r (a 2 )) da 1 da 2 I { a 2l }(a 1 )I { a 2l }(a 2 ) {(m,r) A : m,r I 1, m, r 2 k } (λ m (a 1 )λ r (a 2 )) 1/2+ε Q m,a f a (z)q r,a f a (z)ψ(z 1 ) dz φ(a 1 a 2 ) γ(λ m (a 1 ) λ r (a 2 )) da 1 da 2 2 I { a 2 }(a l 1 )I { a 2 }(a l 2 ) {r I 1 : r 2 k } {m : (m,r) A, m 2 k } (λ m (a 1 )λ r (a 2 )) 1/2+ε γ(λ m (a 1 ) λ r (a 2 )) φ(a 1 a 2 ) da 1 Q r,a2 f a 2 (z) 2 ψ(z 1 ) dz da 2 = J. For fixed r and a 2 we have (2.4) {m : (m,r) A, m 2 k } I { a 2 l }(a 1 )I { a 2 l }(a 2 ) (λ m (a 1 )λ r (a 2 )) 1/2+ε φ(a1 a 2 ) γ(λ m (a 1 ) λ r (a 2 )) da 1 (2 k + 2 l ) (1+2ε) 2 2lε. d To see (2.4) we observe that da 1 λ m (a 1 ) = ((2 m +n)+2 a 1 )sgn(a 1 ). So the measure of {a 1 : λ m (a 1 ) λ r (a 2 ) supp γ} is dominated by C/(2 k + 2 l ). Hence (2.5) γ(λ m (a 1 ) λ r (a 2 )) da 1 C 2 k + 2 l.

110 J. Zienkiewicz Also (2.6) {m 2 k : (m, r) A and a1 2 lλ m (a 1 ) λ r (a 2 ) supp γ and a 1 a 2 supp φ} C max{1, r /2 l }. Combining (2.5) and (2.6) gives (2.4). Hence by Lemma 2 and (2.4) we obtain the desired estimate for J: ( ) 2 J 2 2lε l 1/2 2 k + 2 l (2 k + 2 l ) 1+2ε Q r,af a 2 L da 2 {r I 1 : r 2 k } (2 k 2 l + 2 2l ) 1/2+2ε Q r,a f a 2 L da C f 2 W 1/2+8ε2 (k+l)ε {r I 1 : r 2 k } Summing up the estimates for S 0 and S k,l we get the theorem. Remark. The above theorem combined with the estimates obtained in [Z] allows one to state a slightly sharper result. This requires a different definition of the scale of Sobolev spaces. We do not go into details here. References [B] J. Bourgain, A remark on Schrödinger operators, Israel J. Math. 77 (1992), 1 16. [Car] A. Carbery, Radial Fourier multipliers and associated maximal functions, in: Recent Progress in Fourier Analysis, North-Holland Math. Stud. 111, North- Holland, 1985, 49 56. [C] L. Carleson, Some analytical problems related to statistical mechanics, in: Euclidean Harmonic Analysis, Lecture Notes in Math. 779, Springer, 1979, 5 45. [Cw] M. Cowling, Pointwise behavior of solutions to Schrödinger equations, in: Harmonic Analysis, Lecture Notes in Math. 992, Springer, 1983, 83 90. [DK] B. E. J. Dahlberg and C. E. Kenig, A note on the almost everywhere behavior of solutions to the Schrödinger equation, in: Harmonic Analysis, Lecture Notes in Math. 908, Springer, 1982, 205 209. [E1] A. Erdélyi, Asymptotic forms for Laguerre polynomials, J. Indian Math. Soc. 24 (1960), 235 250. [E2] A. Erdélyi, W. Magnus, F. Oberhettinger and G. F. Tricomi, Higher Transcendental Functions, McGraw-Hill, New York, 1953. [HR] A. Hulanicki and F. Ricci, A Tauberian theorem and tangential convergence of bounded harmonic functions on balls in C n, Invent. Math. 62 (1980), 325 331. [KPV1] C. E. Kenig, G. Ponce and L. Vega, Oscillatory integrals and regularity of dispersive equations, Indiana Univ. Math. J. 40 (1991), 33 69. [KPV2],,, Well-posedness of the initial value problem for the Korteweg de Vries equation, J. Amer. Math. Soc. 4 (1991), 323 347. [KR] C. E. Kenig and A. Ruiz, A strong type (2, 2) estimate for a maximal operator associated to the Schrödinger equation, Trans. Amer. Math. Soc. 280 (1983), 239 246.

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