High-power asymptotics of some weighted harmonic Bergman kernels

0.17 MB·

by  Miroslav Engliš

HIGH-POWER ASYMPTOTICS OF SOME
WEIGHTED HARMONIC BERGMAN KERNELS
6
1 Miroslav Engliˇs
0
2
Abstract. Forweightsρwhichareeitherradialontheunitballordepend onlyon
n
a theverticalcoordinateontheupperhalf-space,wedescribetheasymptoticbehaviour
J ofthecorrespondingweightedharmonicBergmankernels withrespect toρα asα→
4 +∞. This can be compared to the analogous situation for the holomorphic case,
1 which is of importance in the Berezin quantization as well as in complex geometry.

]
V
C 1. Introduction
.
h Let Ω be a domain in Cn, ρ a positive smooth (= C∞) weight on Ω, L2 (Ω,ρα)
hol
at thesubspaceofallholomorphicfunctionsintheweightedLebesguespaceL2(Ω,ρα),
m and K (x,y) the reproducing kernel for L2 (Ω,ρα), i.e. the weighted Bergman
α hol
[ kernel on Ω with respect to the weight ρα. Under suitable hypothesis on Ω and ρ

(namely, forΩboundedandpseudoconvex,log 1 strictly plurisubharmonic,andρa
1 ρ
v defining function for Ω, i.e. vanishing to precisely the first order at the boundary),
6
it is then known that
1
7
αn 1
3 (1) K (x,x) det ∂∂log as α + .
0 α ∼ πnρ(x)α ρ(x) ր ∞
h i
.
1
0 In fact, there is even a similar result for K (x,y) with y close to x, and one also
α
6
has a complete asymptotic expansion as α +
1 ր ∞
:
v αn ∞ b (x,y)
i (2) K (x,y) j , b (x,x) = det[∂∂log 1 ],
X α ≈ πnρ(x,y)α αj 0 ρ(x)
Xj=0
r
a
with some “sesqui-analytic extension” ρ(x,y) of ρ(x) and sesqui-analytic coefficient
functionsb (x,y). Furthermoreonecandifferentiate (1)and(2)termwiseanynum-
j
ber of times. There are, finally, variants also for the weighted Bergman spaces with
respect to ραψm, where ψ is another weight function satisfying the same hypothe-
ses as ρ and m 0 is a fixed real number. All these “high power asymptotics” can
≥
also be extended from functions on domains Ω to sections of holomorphic Hermit-
ian line bundles over a manifold Ω, and are then of central importance in certain
approaches to quantization (the Berezin-Toeplitz quantization procedure), as well
as in complex geometry (where (1) is sometimes known as the Tian-Yau-Zelditch
1991Mathematics Subject Classification. Primary 46E22;Secondary 31B05,32A36.
Key words and phrases. Bergman kernel, harmonic Bergman kernel, asymptotic expansion.
Research supported by GA CˇR grantno. 201/12/0426and RVO funding forICˇ 67985840.
Typeset by AMS-TEX
1
2 M. ENGLISˇ
expansion, and plays prominent role e.g. in connection with semistability and con-
stant scalar curvature metrics on Ω); see for instance Berezin [2], Engliˇs [7], [9],
Zelditch [16], Catlin [4], Donaldson [6], and the references therein.
While there exist several well-understood variants of methods how to prove (1)
(or (2)) nowadays, none of them makes it quite clear what does the holomorphy
of functions in L2 have to do with (1), (2) or with the coefficients b above;
hol j
in fact, a priori there is little reason to expect that holomorphic functions should
have anything to do either with quantization or with constant scalar curvature
metrics, and one is just left to wonder at Berezin’s original insight in noticing (1)
and its applications. In particular, it remains quite elusive what happens for other
reproducing kernel subspaces in L2(Ω,ρα).
Thegoalofthispaperistoexploretheanalogueof(1)forthespacesofharmonic,
rather than holomorphic, functions, i.e. for the reproducing kernels R (x,y) —
α
the harmonic Bergman kernels — of the subspaces L2 (Ω,ρα) of all harmonic
harm
functions in L2(Ω,ρα).
Intheholomorphicsetting,thesimplestexamplesfor(1)and(2)arethestandard
weighted Bergman spaces on the unit disc D in C with ρ(z) = 1 z 2, when
−| |
(3) K (x,y) = α+1(1 xy)−α−2;
α π −
or, equivalently (via the Cayley transform), on the upper half-plane z : Imz > 0
in C with ρ(z) = Imz and { }
α+1 x y −α−2
(4) K (x,y) = − .
α
4π 2i
(cid:16) (cid:17)
More generally, for the unit ball B2n of Cn =R2n with ρ(z) = 1 z 2 one gets
∼
−| |
Γ(α+n+1)
(5) K (x,y) = (1 xy)−α−n−1.
α Γ(α+1)πn −
Explicitformulasfortheharmonicanalogues of (3)–(5), namely forR (x,y)forthe
α
upper half-space Hn = x Rn : x > 0 with the weight ρ(x) = x , and for the
n n
unitballBn = x Rn{: x∈< 1 withρ(x}) = 1 x 2,havebeencomputedinmany
{ ∈ | | } −| |
places,seee.g.CoifmanandRochberg[5],JevticandPavlovic[14], Miao[15],orthe
bookbyAxler,BourdonandRamey[1]. ForΩ = Bn andα = 0(i.e.theunweighted
situation), the kernel is given by
Γ(n) (n 4)x 4 y 4 +(8 x,y 2n 4)x 2 y 2+n
R (x,y) = 2 − | | | | h i− − | | | | .
0 2πn/2 (1 2 x,y + x 2 y 2)n/2+1
− h i | | | |
For the weighted case with α > 1, one already gets the much more complicated
−
formula
Γ(α+ n +1) α+ n +1
R (x,y) = 2 F 2 ; n 1, n 1;z,z ,
α Γ(α+1)πn/2 1 n 1 2 − 2 −
(cid:16) 2 − (cid:17)
involvingAppel’shypergeometricfunctionF [3];herez = x y+i x 2 y 2 (x y)2.
1
· | | | | − ·
For x = y, this reduces to the ordinary hypergeometric functionp
Γ(α+ n +1) α+ n +1,n 2
Rα(x,x) = Γ(α+12)πn/2 2F1(cid:16) 2n2 −1 − (cid:12)(cid:12)|x|2(cid:17)
(cid:12)
HARMONIC BERGMAN KERNELS 3
from which one gets the asymptotics
2Γ(n) αn−1 x n−2
2 | | for x = 0
 πn/2Γ(n 1)(1 x 2)n+α 6
(6) Rα(x,x)  − −| | as α + ,
∼  αn/2 ր ∞
for x = 0
πn/2



as the simplest harmonic analogue of (1). Similarly, for the upper half-space and
α = 0, the unweighted kernel is given by
2Γ(n)(n 1)(x +y )2+(x y )2 x y 2
R (x,y) = 2 − n n n − n −| − | ,
0 πn/2 [(x +y )2 (x y )2+ x y 2]n/2+1
n n n n
− − | − |
while for general α one can compute e.g. from [13] that
Γ(n+α)23−2n αn−123−2n
(7) R (x,x) = ,
α πn−21Γ(α+1)Γ(n−21)xnn+α ∼ πn−21Γ(n−21)xnn+α
giving the harmonic analogue of (1) for the upper half-space with ρ(x) = x .
n
Finally, onecan also consider theentireRn withtheGaussian weightρ(x) = e−|x|2
(theharmonicFock, orSegal-Bargmann, space), inwhichcaseitwasderivedin[11]
that
αn/2 n 1, n 1
Rα(x,y) = πn/2 Φ2(cid:16) 2 −n2 −21− (cid:12)(cid:12)αz,αz(cid:17),
with Horn’s hypergeometric function Φ and again z =(cid:12) x y+i x 2 y 2 (x y)2;
2
· | | | | − ·
for x = y this reduces to the confluent hypergeometric functionp
αn/2 n 2
Rα(x,x) = πn/2 1F1(cid:16) n2 −−1(cid:12)(cid:12)α|x|2(cid:17),
yielding (cid:12)
2Γ(n)
2 eα|x|2 x n−2αn−1 for x = 0,
 Γ(n 1)πn/2 | | 6
(8) Rα(x,x)  −
∼  αn/2
for x = 0,
πn/2



as α + . Note thatin (6) and(8), weget the“Stokes phenomenon”of different
ր ∞
asymptotics at x = 0 and x = 0, which is unparalleled in the holomorphic case as
6
well as in the case of the upper half-space in (7).
Our result here is a rather coarse description for the asymptotics of R (x,x) for
α
fairly general ρand Ω, and more precise descriptions on the level of (1) for domains
and weights of a particular form.
Theorem 1. Let Ω Rn, n 2, be a bounded domain and ρ a bounded positive
⊂ ≥
continuous function on Ω such that log 1 is convex. Then
ρ
1
lim R (x,x)1/α = .
α
α→∞ ρ(x)
Keeping the usualdefinition from complex analysis, we call ρ a definingfunction
for Ω if ρ > 0 on Ω and ρ vanishes precisely to the first order at the boundary ∂Ω,
i.e. ρ = 0< ρ on ∂Ω.
k∇ k
4 M. ENGLISˇ
Theorem 2. Let Ω = Bn, n 2, and let ρ be radial, i.e. ρ(x) = φ(x 2) for
≥ | |
some positive φ C∞[0,1]. Assume that ρ is a defining function (i.e. φ(1) = 0 an
∈
φ′(1) < 0) and that (tφ′)′ < 0. Then for any x = 0,
φ 6
2Γ(n)tn2−1 φ′ n−2 tφ′ ′
(9) lim α1−nρ(x)αR (x,x) = 2 .
α→∞ α πn/2Γ(n 1)(cid:16)− φ(cid:17) (cid:16)− φ (cid:17) (cid:12)t=|x|2
− (cid:12)
(cid:12)
Let us call a positive function g on (0,+ ) admissible if ∞etxg(x)dx = +
∞ 0 ∞
for all t > 0; this means that g should not decay too rapidly aRt infinity.
Theorem 3. Let Ω = Bn, n 2, and assume that ρ(x) = ρ(x ) depends only
n
≥
on the vertical coordinate, is admissible, vanishes at x = 0 precisely to the first
n
order, and ρ′ > 0, (ρ′/ρ)′ < 0 on (0,+ ). Then
∞
23−2n ρ′ n−2 ρ′ ′
(10) lim α1−nρ(x)αR (x,x) = .
α→∞ α πn−21Γ(n−1)(cid:16)ρ(cid:17) (cid:16)− ρ(cid:17)
2
Notethatthechoicesφ(t) = 1 tandρ(x) =x recover(6)and(8),respectively.
n
−
In fact, φ(t) = e−t recovers also (7).
The proof of Theorem 1 appears in Section 2, and those of Theorems 2 and 3
in Sections 3 and 4, respectively. Our main idea is to reduce the harmonic case to
the holomorphic one (on a different domain) and then use (1). Some concluding
remarks are given in the final Section 5.
Throughout the rest of the paper, we write just K (x), R (x) for K (x,x)
α α α
and R (x,x), respectively; and, as usual, “A(x) B(x) as α + ” means that
α
∼ → ∞
A(x)/B(x) 1 as α + . The norm in L2(Ω,ρα) is denoted by , and
α
→ → ∞ k ·k
“plurisubharmonic” will be abbreviated to “psh”.
2. Coarse asymptotics
Theproof ofour firsttheorem is actually almost thesameas for theholomorphic
case in [8].
Proof of Theorem 1. Let D(x,r) be the polydisc with center x and radius r, where
r > 0 is so small that D(x,r) Ω. By the mean value property, for any h
⊂ ∈
L2 (Ω,ρα),
harm
h(x) = (πr2)−n h(y)dy,
Z
D(x,r)
where dy stands for the Lebesgue measure. The Cauchy-Schwarz inequality gives
1/2 1/2
h(x) (πr2)−n h2ραdy ρ−αdy
| | ≤ (cid:16)ZD(x,r)| | (cid:17) (cid:16)ZD(x,r) (cid:17)
(πr2)−n/2 h ( sup 1)α/2.
≤ k kα ρ
D(x,r)
Now by the extremal property of reproducing kernels, R (x)1/2 is the norm of the
α
evaluation functional f f(x) on L2 (Ω,ρα), that is,
7→ harm
(11) R (x) = sup h(x)2 :h L2 (Ω,ρα), h 1 .
α {| | ∈ harm k k ≤ }
HARMONIC BERGMAN KERNELS 5
Thus
R (x) (πr2)−n( sup 1)α.
α ≤ ρ
D(x,r)
Taking α-th roots on both sides and letting α + gives
ր ∞
limsupR (x)1/α sup 1.
α ≤ ρ
α→∞ D(x,r)
Letting r 0, the continuity of ρ implies
ց
1
(12) limsupR (x)1/α .
α
≤ ρ(x)
α→∞
On the other hand, by (11), for any h L2 (Ω,ρα) not identically zero we
∈ harm
have
h(x)2
R (x) | | .
α ≥ h 2
k kα
Take in particular h(x) = eα(x·z+c), with arbitrary c R and z Cn satisfying
∈ ∈
z2 + z2 + + z2 =: z z = 0; clearly this is a bounded function and hence in
1 2 ··· n ·
L2(Ω,ρα) (since Ω and ρ are bounded by hypothesis), while the condition z z = 0
·
ensures that h is harmonic. Thus
e2α(x·Rez+c)
R (x)
α ≥ ex·z+c 2ρ α
k| | kLα(Ω)
and
e2(x·Rez+c)
R (x)1/α .
α ≥ ex·z+c 2ρ
Lα(Ω)
k| | k
Nowsince ex·z+c 2ρ(x)isboundedandΩhasfiniteLebesguemeasure,itisstandard
| |
that ex·z+c 2ρ ex·z+c 2ρ as α + . We thus obtain
Lα(Ω) L∞(Ω)
k| | k → k| | k ր ∞
(13) liminfR (x)1/α sup e2(x·Rez+c) : z z = 0,c R, e•·z+c 2ρ 1 .
α
α→+∞ ≥ { · ∈ | | ≤ }
Writing z = a+bi for the real and imaginary parts of z, the condition z z = 0
·
becomes
(14) a a = b b, a b = 0.
· · ·
Since n 2, we can for any a Rn find b Rn such that (14) holds (just take
≥ ∈ ∈
any b orthogonal to a and of the same length). Thus (13) translates into
logliminfR (x)1/α sup ψ : ψ affine on Rn,ψ log 1 .
α→+∞ α ≥ { ≤ ρ}
If log 1 is convex, then the right-hand side equals 1/ρ, whence
ρ
1
(15) liminfR (x)1/α .
α
α→+∞ ≥ ρ(x)
Combining (15) and (12), the assertion follows. (cid:3)
6 M. ENGLISˇ
3. The case of the ball
Let us recall some prerequisites, available e.g. in [1]. For k = 0,1,2,..., let
k(Rn) denote the space of all harmonic polynomials on Rn homogeneous of de-
H
gree k. Each such polynomial is uniquely determined by its restriction to the unit
sphere Sn−1 := ∂Bn Rn, and we denote by k the space of all such restrictions
(called “spherical har⊂monics of degree k”), vieHwed as a subspace of L2(Sn−1,dσ),
where dσ stands for the normalized surface measure on Sn−1. Then k l for
k = l, and the span of all k, k 0, is dense in L2(Sn−1,dσ). FurtherHmo⊥reH, each
6 H ≥
k is a reproducing kernel space, with reproducing kernel Z (x,y) given by the
k
H
so-called zonal harmonic; this is a certain Gegenbauer polynomial in x y. Finally,
any harmonic function f on Bn can be uniquely expressed in the form·
∞
f = f , f k(Rn),
k k
∈ H
kX=0
with the sum converging uniformly on compact subsets of Bn.
For any radial weight ρ(x) = φ(x 2) on Bn and 0 < r < 1, we therefore get
| |
(recalling that the volume of Sn−1 equals 2πn/2/Γ(n))
2
f 2ρdx = f f ρdx
ZrBn| | Xj,k ZrBn j k
2πn/2 r
= rj+kφ(r2)rn−1 f (ζ)f (ζ)dσ(ζ)dr
Xj,k Γ(n2) Z0 ZSn−1 j k
2πn/2 r
= r2k+n−1φ(r2)dr f 2 ,
Γ(n2) Xk (cid:16)Z0 (cid:17)k kkL2(Sn−1,dσ)
by the orthogonality of k and l for k = l. Denoting
H H 6
1
(16) rkφ(r2)dr =:ρ ,
k
Z
0
it follows upon letting r 1 that
ր
2πn/2
f 2 = ρ f 2 ,
k kL2harm(Ω,ρ) Γ(n2) Xk 2k+n−1k kkL2harm(Ω,ρ)
and, consequently, the reproducing kernel of L2 (Ω,ρ) is given by
harm
Γ(n) Z (x,y)
R(x,y)= 2 k .
2πn/2 ρ
Xk 2k+n−1
For x = y, it is known that Z (x,x) = N x 2k, where N = dim k is given by
k k,n k,n
| | H
(n+k 3)!(n+2k 2)
N = − − .
k,n
k!(n 2)!
−
HARMONIC BERGMAN KERNELS 7
We thus obtain
Γ(n) N
(17) R(x) = 2 k,n x 2k.
2πn/2 ρ | |
Xk 2k+n−1
On the other hand, a completely similar formula is available also for the holo-
morphic Bergman kernel K(x,y) on the unit ball B2m of R2m = Cm, m 1, with
∼
≥
respect to a radial weight function w(z) = ψ(z 2). Namely, it is standard that the
| |
monomials zν, ν a multiindex, are then orthogonal, with norm squares
1
zν 2w(z)dz = r2|ν|+2m−1ψ(r2)dr ζν 2dζ
ZB2n⊂Cn| | (cid:16)Z0 (cid:17)(cid:16)ZS2n−1| | (cid:17)
=:w2|ν|+2m−1
| 2πmν! {z }
= w ,
2|ν|+2m−1
(ν +m 1)!
| | −
and by the familiar formula expressing the reproducing kernel in terms of an arbi-
trary orthonormal basis,
xνyν (ν +m 1)! 1 ∞ x,y k (k+m 1)!
K(x,y) = | | − = h i − .
2πmν! w 2πm k! w
Xν 2|ν|+2m−1 Xk=0 2k+2m−1
We thus obtain
∞
1 (k+m 1)!
(18) K(z) = − z 2k.
2πm k!w | |
Xk=0 2k+2m−1
Finally, note that for any function
∞
F(t) = c tk
k
Xk=0
holomorphic on the unit disc, we have
∞
(k+m 1)!
(19) − c tk = (tm−1F)(m−1),
k
k!
kX=0
and
∞ (tn−2F)(n−2) +t(tn−3F)(n−2)
(20) N c tk = ,
k,n k
(n 2)!
kX=0 −
as can be checked by elementary manipulations.
The last thing we will need is the fact that the condition (tφ′)′ < 0 is actually
φ
equivalent to the function log 1 being strictly-psh on B2m, see e.g. [7], Sec-
φ(|z|2)
tion 3. Since φ C∞[0,1] by hypothesis, this condition also implies that tφ′ > 0
∈ − φ
for t > 0, and hence φ′ < 0 for t > 0, that is, φ is decreasing on (0,1).
8 M. ENGLISˇ
Proof of Theorem 2. Assume first that n is even. Take m = n in (18), with
2
ψ = φ. Replacing φ by φα and denoting the corresponding ρ from (16) by ρ (α),
k k
we thus get
∞ ∞
Γ(m) N 1 (k+m 1)!
R (x) = k,2m tk, K (z) = − tk,
α 2πm ρ (α) α 2πm k!ρ (α)
Xk=0 2k+2m−1 Xk=0 2k+2m−1
where we have set t = x 2 and t = z 2, respectively. Thus 2πm R (x) =: r (t)
| | | | Γ(m) α α
and (2πm)K (z) =: k (t) are related as in (19) and (20), with
α α
∞ tk
F(t) F (t) = .
α
≡ ρ (α)
kX=0 2k+2m−1
By the Leibniz rule, rα(t) comes as a sum of terms of the form cjtmjkα(j)(t), j =
(m 1),...,m 1,withsomerealnumbersc andintegersm (independentofα),
j j
− − −
(j)
wherek denotes the j-th derivatives of k for j 0andthe j -th primitiveof k
α α α
≥ | |
(normalized to vanish to order j at t = 0) for j < 0. Now by (1),
| |
2αm 1
(21) k (z 2) det ∂∂log ,
α | | ∼ ρ(z)α ρ(z)
h i
and this also remains in force upon applying any derivative to both sides. Since a
short computation shows that
1 φ′ m−1 tφ′ ′
det ∂∂log = ,
h ρ(z)i (cid:16)− φ(cid:17) (cid:16)− φ (cid:17)(cid:12)t=|z|2
(cid:12)
(cid:12)
we have
2αm φ′ m−1 tφ′ ′
k .
α ∼ φα − φ − φ
(cid:16) (cid:17) (cid:16) (cid:17)
Differentiation gives
2αm αφ′ φ′ m−1 tφ′ ′ 2αm φ′ m−1 tφ′ ′ ′
k′ − + .
α ∼ φα φ − φ − φ φα − φ − φ
(cid:16) (cid:17) (cid:16) (cid:17) h(cid:16) (cid:17) (cid:16) (cid:17) i
As α + , the first term dominates the second. Thus by induction
ր ∞
αφ′ j
(22) k(j) − k
α ∼ φ α
(cid:16) (cid:17)
for any j 0. On the other hand,
≥
t t 1 φ′ m−1 tφ′ ′
k(−1)(t) = k 2αm .
α Z0 α ∼ Z0 φα(cid:16)− φ(cid:17) (cid:16)− φ (cid:17)
The right-hand side is a standard Laplace-type integral, that is, an integral of the
form
b
(23) I(α) = F(x)eαS(x)dx
Z
a
HARMONIC BERGMAN KERNELS 9
with real-valued function S. It is known that I(α) gets the largest contribution
from points where S attains its maximum, and, in particular, if the maximum is
attained at the endpoint x = b and S′(b) > 0, then (see e.g. [12], II.1.4)
§
F(b) eαS(b)
I(α) .
∼ S′(b) α
Since, as we have observed, φ is decreasing, 1 indeed attains its maximum at the
φ
endpoint t, and thus
2αm φ′ m−1 tφ′ ′ φ
k(−1) .
α ∼ φα − φ − φ · αφ′
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
−
Proceeding inductively, it follows that (22) in fact remains in force for j 0 as
≤
well. Thus the leading term in the asymptotics of r as α + will be the one
α
ր ∞
coming from (22) with j = m 1. In other words, k tm−1F(m−1) and
α
− ∼
2t2m−2F(2m−2) 2tm−1k(m−1) 2tm−1 αφ′ m−1
α α
r − k
α α
∼ (2m 2)! ∼ (2m 2)! ∼ (2m 2)! φ
(cid:16) (cid:17)
− − −
4α2m−1tm−1 φ′ 2m−2 tφ′ ′
,
∼ (2m 2)!φα − φ − φ
(cid:16) (cid:17) (cid:16) (cid:17)
−
or
Γ(m) 2αn−1tn2−1Γ(n) φ′ n−2 tφ′ ′
R (x) = r (t) 2 ,
α 2πm α ∼ πn/2(n 2)!φα (cid:16)− φ(cid:17) (cid:16)− φ (cid:17) (cid:12)t=|x|2
− (cid:12)
proving (9) for even n. (cid:12)
For n odd, take m = n−1 and replace φ,ψ by φα and φα√t, respectively. Then
2
φα√t = φα−1(φ√t) and φ√t again vanishes to exactly the first order at ∂B2m
and log 1 is still strictly-psh, except that φ(z 2)z now fails to be smooth at
φ(|z|2)|z| | | | |
the origin; however, it is known that (1) — or, more precisely, its variant for two
weight functions mentioned at the beginning of this paper — then still remains in
force at points where the weights are smooth, so in our case for z = 0. (See [10],
6
Theorem 1.) We thus have
αm 1
K (z) det ∂∂log
φ(|z|2)α|z| ∼ πmφ(z 2)α z φ(z 2)z
h i
| | | | | | | |
for z = 0. Arguing as in the case of even n, now with k (t) = 2πmK (z),
α φ(|z|2)α|z|
6
r (t) = 2πn/2R (x), and F (t) = tk , we arrive at
α Γ(n2) α α k ρk+2m(α)
P
2t2m−1F(2m−1) 2tmk(m) φ′ m 2αmtm
α
r − k
α α
∼ (2m 1)! ∼ (2m 1)! ∼ φ (2m 1)!
(cid:16) (cid:17)
− − −
4α2mtm φ′ 2m−1 tφ′ ′
∼ φαt1/2(2m 1)! − φ − φ
(cid:16) (cid:17) (cid:16) (cid:17)
−
and
2Γ(n)αn−1tn2−1 φ′ n−2 tφ′ ′
R (x) 2 ,
α ∼ πn/2φα(n 2)! − φ − φ
(cid:16) (cid:17) (cid:16) (cid:17)
−
which settles (9) also for odd n and thus completes the proof of the theorem. (cid:3)
10 M. ENGLISˇ
4. The upper half-space
We again begin by reviewing some standard prerequisites on harmonic functions
on Hn. Write points in Hn temporarily as (x,y), with x Rn−1 andy > 0, and let
∈
fˆ(ξ) = f(x,y)e−ix·ξdξ
y
ZRn−1
denote the Fourier transform of a function f(x,y) f (x) on Hn. The condition
y
≡
that f be harmonic then translates into ∂2 fˆ + ξ 2fˆ = 0, or
∂y2 y | | y
fˆ(ξ) =A(ξ)e−|ξ|y +B(ξ)e|ξ|y
y
for some functions A,B. Now for any weight ρ(x,y) = ρ(y) depending only on the
vertical coordinate y, we have by Parseval
∞ ∞
f 2ρdxdy = ρ(y) f 2dxdy = (2π)1−n ρ(y) fˆ 2dξdy.
y y
ZZHn| | Z0 ZRn−1| | Z0 ZRn−1 | |
Consequently, if in addition ρ is admissible, this can only be finite if B 0. Thus
for f L2 (Hn,ρ), ≡
∈ harm
fˆ(ξ) = fˆ(ξ)e−y|ξ|,
y 0
wherefˆ A hasthe obvious interpretation of the Fourier transform of thebound-
0
≡
ary value f of f at y = 0, and we can continue the last computation with
0
∞
f 2ρdxdy = (2π)1−n ρ(y) e−2y|ξ| fˆ(ξ)2dξdy
0
ZZHn| | Z0 ZRn−1 | |
(2π)1−n ρ˜(ξ )fˆ(ξ)2dξ,
0
≡ ZRn−1 | | | |
where
∞
(24) ρ˜(t) := ρ(y)e−2tydy.
Z
0
Comparing this with the Fourier inversion formula
f(a,b)= (2π)1−n fˆ(ξ)eia·ξdξ = (2π)1−n fˆ(ξ)eia·ξ−b|ξ|dξ,
b 0
ZRn−1 ZRn−1
weseethatthereproducingkernelR(x,y;a,b) R (x,y)ofL2 (Hn,ρ)satisfies
≡ a,b harm
ρ˜(ξ )(R )∧(ξ) = e−b|ξ|−ia·ξ, or
| | a,b 0
ei(x−a)·ξ−(b+y)|ξ|
R(x,y;a,b) = (2π)1−n dξ.
ZRn−1 ρ˜(ξ )
| |
In particular, for (a,b) = (x,y),
e−2y|ξ| 22−n ∞ e−2yr
(25) R(x,y) = (2π)1−n dξ = rn−2dr.
ZRn−1 ρ˜(|ξ|) πn−21Γ(n−21) Z0 ρ˜(r)