MAXIMUM REGULARITY OF HUA HARMONIC FUNCTIONS

Transcription

1 MAXIMUM REGULARITY OF UA ARMONIC FUNCTIONS P. JAMING 1. The Siegel upper half space 1.1. Setting. On C 2 consider r(z) = Im(z 2 ) z 1 2. The Siegel upper half space and its boundary are given by U ={z = (z 1, z 2 ) : r(z) > 0} U ={z = (z 1, z 2 ) : r(z) = 0} The eisenberg group is given by = C R with the action given by [ζ, t][η, s] = [ζ + η, t + s + 2Im(ζ η)] acts on U and U bu [ζ, t]z = (z 1 + ζ, z 2 + t + 2iz 2 ζ + i ζ 2 ). This action is simply transitive on U but not on U. For this, one needs to add dilations: if a > 0 then a z = (a 1/2 z 1, az 2 ). One can then identify S = R + with U by identifying [ζ, t, a] S with [ζ, t, a] i where the action of S on U is given by The group law is then [ζ, t, a] z = [ζ, t] a z. [ζ, t, a][η, s, b] = [ [ζ, t][a 1/2 η, as], ab ] =... One can then define left invariant vector field : on by X = x + 2y t, Y = y + 2x t, Z 1 = 1 2 (X iy ), T = t ; on S by a 1/2 X, a 1/2 Y, a 1/2 Z 1, at, a a, az 2 = a 2 (T i a). One may then identify functions on U with functions on S and extend notions such as holomorphy, pluriharmonicity... from functions on U to functions on S. For instance on U on S f holomorphic f z 1 = f z 2 = 0 Z 1 f = Z 2 f = 0 f pluri-harmonic 2 f z k z j = 0 Z 2 Z 1 f = Z 1 Z 2 f = Z 2 Z 2 f = 0 and (Z 1 Z 1 + 2iZ 2 )f = 0. Joint work with A. Bonami (Orléans), D. Buraczewski, E. Damek, A. ulanicki (Wroc law). 1

2 2 P. JAMING Finally we will consider the following family of second order differential operators (α R) L α = 1 2 (Z 1Z 1 + Z 1 Z 1 ) + iαt, on (we simply write L = L 0 ) and L = 1 2 a(l + a) + a 2 ( 2 a + T 2 ), on S. This operator has a Poisson kernel : recall that the aar measure on R 3 is the Lebesgue measure of R 3 and that convolution is given by f g[ζ, t] = f[η, s]g([η, s] 1 [ζ, t])dηds. For exists a > 0, there exists a unique function P a on with P a = 1, called the Poisson kernel, such that F is L harmonic and bounded if and only if F = f P a armonicity and pluri-harmonicity 1. Theorem 1.1. Assume that F = f P a with f bounded. Then (1) F is holomorphic if and only if L 1 f = 0, (2) F is anti-holomorphic if and only if L 1 f = 0, (3) F is pluri-harmonic if and only if (L 2 + T 2 )f = L 1 L 1 f = 0. Sketch of proof. Denote by C the Cauchy-Szegö projection. There is a well-known function Φ such that, if f is smooth and fastly decreasing, then f C(f) = L 1 (f Φ). Apply this to fϕ R with ϕ R a dilation of a cutoff function, compose with Z 1 (this removes C(fϕ R )) estimate the convolution, noting that it is an integration over a shell and let R 0, this gives Z 1 f = 0. The second point is obtained by conjugation. For the 3rd point, note that L 1 F is antiholomorphic, one obtains pluri-harmonicity by computing commutators Automatic regularity of harmonic functions. We will say that a function G on S has a boundary distribution if, for every ψ Cc (), G(ω, a)ψ(ω)dω lim a 0 exists. It is obvious that if G has a boundary distribution, so do LG and T G. Theorem 1.2. If F is harmonic with a boundary distribution, then a F has also a boundary distribution. Sketch of proof. Let ψ Cc () and set ϕ(a) = a F (ω, a)ψ(ω)dw. We want to show that ffi has a limit as a 0. We have a 2 af (a) 1 2 af = ( 1 2 L + at 2 )F

3 MAXIMUM REGULARITY OF UA ARMONIC FUNCTIONS 3 and the right hand side has a boundary distribution. Multiplying by ψ and integrating over gives a ϕ 1 2 ϕ = g(a) for some function g that has a limit as a 0. Solving this differential equation, we get ϕ(a) = λa 1/2 + a 1/2 a 1 g(t) dt t1+1/2 and, as 1/2 > 0 it is easy to show that this has a limit as a Maximal regularity of t-invariant harmonic functions. If F is L-harmonic and t-invariant (i.e T F = 0), then ΛF = 0 with where is the Euclidean Laplacian. Λ = 1 2 a( a) + a 2 2 a Proposition 1.3. Assume that F is bounded and Λ-harmonic such that, for every ψ S(C n ) and k = 0, 1 or 2, (BR) sup k af (ζ, a)ψ(ζ)dζ < +. a 1 C n Then F is constant. Proof. Let Q a be the Poisson kernel for Λ, so that F = Q a f with f bounded. Let ϕ S(C n ) such that 0 / supp ˆϕ, we will prove that ˆf, ˆϕ ˆQa = F a (ζ)ϕ(ζ)dζ = 0. Once this is done, ˆf is a distribution supported in {0} so f is bounded polynomial, thus a constant. To prove the claim, we first identify Q a : ˆQa (ξ) = z(a ξ 2 /2) where z is the unique bounded solution of with z(0) = 1 (z is the Legendre function). Then we study the regularity of z : (a 2 a 1 2 a 1)z(a) = 0 with γ, γ smooth functions, γ(0) 0. The condition (BR) implies the claim. 2 z(a ξ 2 /2) = γ(a ξ 2 ) + a 1/2 ξ 1 γ(a ξ 2 ) 1.5. Fourier analysis on. The represntations f on L 2 (R) are given by (λ R ) : R λ [u, v, t]φ(x) = e 2iπλ(ux+uv/2+t/4) Φ(x + v). The fourier transform of a function f on is the operator defined by ˆf(λ)φ, ψ = R λ (w) φ, ψ f(w)dw.

4 4 P. JAMING It is an ilbert-schmidt operator with kernel Kf λ(x, v)f 1,3f(λ(x + v)/2, x v, λ/4). inversion formula is given by f(u, v, t) = tr ( ˆf(λ)R λ (u, v, t) ) λ n dλ. R The trace in this formula may be written using h λ k, a properly scaled ermite basis, and e λ k R (ω) = λ (ω)h λ k, hλ k Identification of P a. Notons ω f(w) = f(ωw). Alors Lemma 1.4. For λ 0 and k N, set gk ωp λ (ω, a) = a (λ)h λ k, hλ k = e λ k ˇP a (ω 1 ). Then gk λ(ω, a) = eλ k (ω 1 )g( λ a) where g is the unique bounded solution of ( ( )) k + 1 a g(a) = 0 a with g(0) = 1. Such a g is a confluent hypergeometric function. The proof uses the fact that e λ k are eigenfunctions of L and T An orthogonality property. For ψ S(R) and k N, write e ψ k (ω) = e λ k (ω)ψ(λ)dλ. R Proposition 1.5. Let f be a bounded function on and F = f P a. Assume that F satisfies (BR) sup af k (ω, a)ψ(ω)dω < + a 1 for k = 0, 1, 2 and ψ S(). Then for k 0, w, ψ with ˆψ Cc (R \ {0}), wf(ω)e ψ k (ω)dω = 0. The 1.8. Final result. Theorem 1.6. Let f be a bounded function on and F = f P a. Assume that F satisfies (BR) sup af k (ω, a)ψ(ω)dω < + a 1 for k = 0, 1, 2 and ψ S(). Then F is pluri-harmonic.

5 MAXIMUM REGULARITY OF UA ARMONIC FUNCTIONS 5 Sketch of proof. By replacing f by ψ f with ψ smooth, we may assume that f is smooth and has 4 bounded derivatives. Write g = (L 2 + T 2 )f and G = g P a = (L 2 + T 2 )F. Then G still satisfies the hypothesis of the theorem. According to Theorem 1.1, it is enough to prove that g = 0 and, with arnack, that g is a constant. With Proposition 1.3, it is enough to prove that G does not depend on t. But Proposition 1.5 states that g(ω)e ψ k (ω)dω = 0 for k = 0 and ψ with 0 / supp ˆψ. As g = (L 2 + T 2 )f, the same is true for k = 0, then, by Fourier inversion on the eisenberg group, g is a polynomial in the t variable, and by boundedness, does not depend on t. So G does not depend on the t variable. 2. ua operators on tube type domains 2.1. Jordan algebras. Let V = Sym(2, R) the symetric 2 2 matrices with product A B = 1 2 (AB + BA) ( and ) scalar product ( A, ) B = tr(abt ). This is an Euclidean Jordan algebra Let c 1 = and c =. Then 0 1 c 2 i = c i, c i c j = 0, c 1 + c 2 = I and is maximal for this property. This is a Jordan frame of V. Let Ω = Int{A A, A V } = Sym + (2, R) the set of positive definite symmetric 2 2 matrices. This is an irreducible cone in V. Write L(A)B = AB an endomorphism of V and let V (c i, λ) be the eigenspaces of L(c i ). The only eigenvalues are 0, 1/2 and 1. ere ( V i,i ) := V (c i, 1) = Rc i i = 1, 2 and V i,j := V (c i, 1/2) V (c j, 1/2) = Rc 1,2 with c 1,2 = , so that V = i j r V 1,1 V 1,2 V 2,2. This is the Peirce decomposition of V Automorphisms of Ω. Let G be the component of I of the group of g GL(V ) s.t g Ω Ω. Let G be its Lie algebra. Let A be the abelian part of G, then A = V ect{l(c 1 ), L(c 2 )} and note that this are given by L(c 1 ) = /2 0 L(c 2 ) = / Let N be the nilpotent part of G, N = i<j N i,j where N i,j = V ect{c i,j c i } and the operation is defined by A B = L(A B) + [L(A), L(B)]. ere N = N 1,2 = Rc 1,2 c 1 = R / /2 0

6 6 P. JAMING Let S 0 = A N and S 0 = exp S 0 = a2 0 0 γa ab 0, a, b > 0γ R γ 2 /2 γb b 2. This acts simply transitively on Ω Tube domains. We now consider the tube domain T Ω = V + iω V C. Then V S 0 acts on T Ω by (x, I) (u + iv) = x + u + uv and (0, s) (u + iv) = s u + is v. The group operation is therefore (v, s)(v, s ) = (v + s v, s s ). The action is simply transitive, so we may identify V S 0 with T Ω by (u + iv) (v, s) ii. The Lie algebra of V S 0 is V S 0 = V A i,j N i,j. A basis of order 1 invariant differential operators is given by X 1, X 2, X 1,2 that correspond to c 1, c 2, c 1,2 V, Y 1,2 that corresponds to 2e 1,2 N and 1, 2 that correspond to L(c 1 ), L(c 2 ) in A ua operators and main theorem. We that set 1 = X , 2 = X , 1,2 = X 2 1,2 + Y 2 1,2 2. At the point ii, these correspond to 2 z 1 z 1, 2 z 2 z 2 and 2 z 1,2 z 1,2 function is pluriharmonic if and only if 1 f = 2 f = 1,2 f = 0. The (strongly diagonal) ua operators are then given by j = j + 1 k,j + 1 j,k 2 2 k<j k>j and one can show that a so that here they are 1 = ,2 and 2 = ,2. It is known that the bounded ua-harmonic functions are the Poisson integrals. Theorem 2.1. Every bounded ua-harmonic function f has a finite number of derivatives with a boundary distribution, unless it is pluri-harmonic. The proof is an induction on the rank of the cone (=the number of elements in the Jordan frame). One first isolates everything that contains the index 2 by averaging f over the other variables (against a test function ψ). This way, one obtains a function over a subgroup of S which turns out to be the extension of the eisenberg group presented earlier. Moreover, the function f ψ one obtains is L-harmonic, (the operator L has been built for that purpose). Further if f has a given number of boundary derivatives, so does the function f ψ one obtains by averaging. From the previous section, we know that if one asks for too many derivatives, this only occurs if f ψ is pluri-harmonic. This then implies that 2 f = 1,2 f = 0 so that 1 f = 0 reduces to 1 f = 0 and finally f is pluri-harmonic.