Harmonic Bergman Spaces
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1 Holomorphic paces MRI Publications Volume 33, 998 Harmonic ergman paces KAREL TROETHOFF Abstract. We present a simple derivation of the explicit formula for the harmonic ergman reproducing kernel on the ball in euclidean space and give an elementary proof that the harmonic ergman projection is L p - bounded, for < p <. We furthermore discuss duality results.. Introduction Throughout the paper n is a positive integer greater than. We will be working with functions defined on all or part of R n. Let D j denote the partial derivative with respect to the j-th coordinate variable. Recall that u(x) = (D u(x),, D n u(x)). The Laplacian of u is u(x) = Du(x) Dnu(x). 2 A real- or complex-valued function u is harmonic on an open subset Ω of R n if u on Ω. The purpose of this article is to present an elementary treatment of some known results for the harmonic ergman spaces consisting of all harmonic functions on the unit ball in R n that are p-integrable with respect to volume measure. everal properties of these spaces are analogous to those of the ergman spaces of analytic functions on the unit ball in C n. As in the analytic case, there is a reproducing kernel and associated projection. Duality results follow once we know that the projection is L p -bounded. Coifman and Rochberg [98] used deep results from harmonic analysis to establish L p -boundedness of the harmonic ergman projection. An explicit formula for the harmonic ergman reproducing kernel has only been determined recently; see [Axler et al. 992]. We give a simple derivation for such a formula in ection 2. In ection 3 we give an elementary proof that the harmonic ergman projection is L p -bounded for < p <. Our proof is similar to Forelli and Rudin s proof of L p -boundedness of the analytic ergman projection [Forelli and Rudin 974/75; Rudin 98], but as in Axler s argument [988] 99 Mathematics ubject Classification. 4735, The author was partially supported by grants from the Montana University ystem and the University of Montana. 5
2 52 KAREL TROETHOFF in the context of the analytic ergman spaces on the unit disk, we will avoid the use of the binomial theorem, the gamma function and tirling s formula. In ection 4 we discuss the dual and the predual of the ergman space of integrable harmonic functions on the unit ball in R n. Analogous to the analytic case, the dual and predual are identified with the harmonic loch and little loch space, which are spaces of harmonic functions determined by growth conditions on the gradient. As in the analytic case, these duality results follow from the fact that the harmonic ergman projection maps L onto the harmonic loch space. The remainder of this section establishes some of the notation and contains the prerequisites from harmonic function theory needed in the paper. We will repeatedly make use of Green s identity, which states that if Ω is a bounded open subset of R n with smooth boundary Ω and u and v are continuously twice-differentiable functions in an open set containing Ω, the closure of Ω, then (u v v u) dv = (ud n v vd n u) ds, ( ) Ω where V denotes volume measure on R n, s denotes surface area measure on Ω, and the symbol D n denotes differentiation with respect to the outward unit normal vector n: D n w = w n. A special case of ( ) is obtained by taking v : if u is harmonic in an open neighborhood of Ω, then D n u ds =. ( 2) Ω For y R n and r > we write (y, r) = {x R n : x y < r} and (y, r) for its closure. We use to denote the unit ball (, ) and write for its boundary, the unit sphere = {x R n : x = }. The area of is easily determined: if we take Ω =, u and v(x) = x 2 in ( ), then v 2n and D n v 2, and we obtain 2nV () = 2s(); thus s() = nv (). It will be convenient to work with normalized surface area on, which we denote by σ; thus ds = nv () dσ on and σ() =. If u is harmonic on an open neighborhood of (y, r), the chain rule gives (d/dr)u(y + rζ) = u(y + rζ) ζ = D n u(y + rζ), where the normal derviative D n is taken with respect to (y, r), so that d u(y + rζ) dσ(ζ) = D n u(y + rζ) dσ(ζ) =, dr by ( 2). Hence u(y + rζ) dσ(ζ) does not depend on r, so that the function u satisfies the following so-called mean value property u(y + rζ) dσ(ζ) = u(y). ( 3) Ω
3 HARMONIC ERGMAN PACE 53 In particular, if u is harmonic on an open set containing, we have u() = u(ζ) dσ(ζ). In fact, for any y, u(y) is a weighted average of u over, namely u(y) = u(ζ) P (ζ, y) dσ(ζ), where P (ζ, y) is the so called Poisson kernel for the ball. ince this Poisson kernel will in the sequel play a fundamental role, we will derive an explicit formula for P. Fix y \ {}, choose any < r < y, and set Ω = {x R n : r < x y < }. We will only consider here the case n > 2. Put v(x) = x y 2 n. It is easy to verify that v is harmonic on R n \ {y} and v(x) = (2 n) x y n (x y); consequently, D n v = (2 n)r n on (y, r). For ζ it is easy to verify that ζ y 2 = y 2 ζ y/ y 2 2 ; thus, on the function v coincides with the function w(x) = y 2 n x y/ y 2 2 n. Using that w is harmonic on, an application of ( ) yields ud nw ds = wd nu ds = vd nu ds. It follows that (ud n v ud n w) ds = (ud n v vd n u) ds. y ( ) (ud n v vd n u) ds = (y,r) = (2 n)r n (ud n v vd n u) ds (y,r) u ds = (2 n)nv ()u(y). We conclude that u(y) = up y dσ, where P y = (2 n) (D n v D n w). A simple calculation shows that P y (ζ) = P (ζ, y) is given by the formula In particular we have P (ζ, y) = y 2 ζ y n. y 2 dσ(ζ) = for y. ( 4) y ζ n We extend the Poisson kernel P to a function on as follows: if x, y, then we set P (x, y) = P (x/ x, x y). This is called the extended Poisson kernel. Thus x 2 y 2 P (x, y) = for x, y. ( 2x y + x 2 y 2 ) n/2 For each fixed y the function x P (x, y) is harmonic, and by symmetry, for each fixed x the function y P (x, y) is harmonic.
4 54 KAREL TROETHOFF It will be convenient to use polar coordinates to integrate functions over balls. Heuristically, we have d f(x) dv (x) = f ds, dr (,r) r and because the sphere r has area nv ()r n this is equal to nv ()r n f(rζ) dσ(ζ). Integrating with respect to r we obtain the following formula ρ f(x) dv (x) = nv () r n f(rζ) dσ(ζ) dr. ( 5) (,ρ) If u is harmonic on, then we define its radial derivative Ru by Ru(x) = u(x) x. It is easy to verify that Ru is also harmonic on. In fact, this follows at once from the formula (Ru) = 2 u + R( u), which is easy to verify. The reader interested in learning more harmonic function theory should consult [Axler et al. 992]. 2. The Harmonic ergman paces For p < we denote by b p () the set of all harmonic functions u on for which ( /p u p = u(x) p dv (x)) <. The spaces b p () are called harmonic ergman spaces. The space b 2 () is a linear subspace of L 2 () with inner product given by f, g = f(x)g(x) dv (x) for f, g L 2 (). If u is harmonic on and y is fixed, it follows from ( 3) and Cauchy- chwarz s inequality that u(y) 2 u(y + rζ) 2 dσ(ζ) for < r < y. Applying ( 5) to the function f(x) = u(y + x) 2 with ρ = y we see that ρ nv () r n u(y + rζ) 2 dσ(ζ) dr = u 2 dv u 2 2, and conclude that u(y) (y,ρ) V () /2 ( y ) n/2 u 2 for u b 2 (). (2 ) It follows from (2 ) that b 2 () is a closed subspace of L 2 (), and thus it is a Hilbert space. Inequality (2 ) implies that for each fixed y the linear functional u u(y) is bounded. y the Riesz representation theorem there is a unique function R y b 2 (), called the reproducing kernel at y, such that u(y) = u, R y, for all u b 2 (). y considering real and imaginary parts, it is easily
5 HARMONIC ERGMAN PACE 55 seen that the function reproducing the value at y for the real ergman space, also reproduces the value at y for complex-valued functions of b 2 (), and thus (by uniqueness) we conclude that R y is real-valued. We write R(x, y) = R y (x), and call this the ergman reproducing kernel of b 2 (). ecause the ergman reproducing kernel is real valued we have R(y, x) = R x, R y = R y, R x = R y (x) = R(x, y), for all x, y. Thus R is a symmetric function on. The ergman kernel for the ball. In this subsection we will derive an explicit formula for the ergman kernel R of. We will not make use of so-called zonal and spherical harmonics used in [Axler et al. 992] to calculate the ergman kernel R, but instead use Green s identity to relate R to the extended Poisson kernel P. uppose u is harmonic on and fix y. Let v also be harmonic on, and define the function w by w(x) = ( x 2 )v(x) = x 2 v(x) v(x). Observe that the function w is harmonic on, for w(x) = ( x 2 ) v(x) + 2 x 2 v(x) + x 2 v(x) = 2nv(x) + 4x v(x) = 2nv(x) + 4Rv(x). (2 2) Note that w(x) = 2v(x) x+( x 2 ) v(x), so that D n w(x) = w(x) x/ x = 2v(x) x + ( x 2 )D n v(x). In particular, D n w 2v on. ince u on and w on, it follows from ( ) (applied to u and w) that u w dv = ud n w ds = 2 uv ds = 2nV () uv dσ. (2 3) It is clear that our choice for v should be the extended Poisson kernel: v(x) = P (x, y) for x. Then u w dv = 2nV () u(ζ) P (ζ, y) dσ(ζ) = 2nV ()u(y). We conclude that the harmonic function w/(2nv ()) is the reproducing kernel at y. Using (2 2) we obtain the following formula for the ergman kernel R(x, y) = y elementary calculus, we get the formula R(x, y) = nv () (np (x, y) + 2x xp (x, y)). (2 4) nv ()( 2x y + x 2 y 2 ) n/2 ( n( x 2 y 2 ) 2 2x y + x 2 y 2 4 x 2 y 2 (2 5) In the next section we will need an estimate on R(x, y). y Cauchy-chwarz x y x y. It follows that ( x y ) 2 = + x 2 y 2 2 x y 2x y+ x 2 y 2, and thus ( x 2 y 2 ) 2 = ( + x y ) 2 ( x y ) 2 4( 2x y + x 2 y 2 ). Therefore we have R(x, y) ). 4. (2 6) V ()( 2x y + x 2 y 2 ) n/2
6 56 KAREL TROETHOFF 3. L p -oundedness of the ergman Projection and Duality Let Q denote the orthogonal projection of L 2 () onto b 2 (). If f L 2 () and y, then Q[f](y) = Qf, R y = f, R y and we have the following formula: Q[f](y) = R(x, y) f(x) dv (x). (3 ) For fixed y the function R(, y) is bounded, so that we can use the formula above for Q[f] to extend the domain of Q to L p (), where p <. Our goal is to prove the following theorem. Theorem 3.. Let < p <. Then Q maps L p () boundedly onto b p (). Proof. We will use the so-called chur Test. pecifically, we will show the existence of a positive function h and a constant C such that h(x) q R(x, y) dv (x) Ch(y) q for all y, and (3 2) h(y) p R(x, y) dv (y) Ch(x) p for all x, (3 3) where q denotes the conjugate index of p, that is, q = p/(p ). That this will imply the result is then proved as follows. Given f L p (, dv ), applying Hölder s inequality and (3 2) we have f(x) Q[f](y) h(x) R(x, y) dv (x) h(x) ( f(x) p h(x) p ( C /q h(y) ) /p ( R(x, y) dv (x) f(x) p h(x) p ) /p R(x, y) dv (x). h(x) q R(x, y) dv (x) ) /q Thus, applying Fubini s theorem to reverse the order of integration, and using (3 3) we obtain ( Q[f](y) p dv (y) C p/q h(y) p f(x) p ) h(x) p R(x, y) dv (x) dv (y) = C p/q f(x) p ( ) h(x) p h(y) p R(x, y) dv (y) dv (x) C p/q f(x) p h(x) p (Ch(x)p ) dv (x) = C p f(x) p dv (x), proving the theorem. We claim that the function h(x) = ( x 2 ) /(pq) works, that is, satisfies (3 2) and (3 3). y symmetry in p and q, it will suffice to find a constant C p for which ( x 2 ) /p R(x, y) dv (y) C p ( y 2 ) /p for all y. (3 4)
7 HARMONIC ERGMAN PACE 57 Fix y \{}. For < r < and ζ it follows from (2 6) that R(rζ, y) 4 V () ( ry ζ + r 2 y 2 ) = 4 n/2 V () ζ ry n. Using ( 5) and ( 4) we have Now y 2 R(x, y) dv (x) = nv () ( x 2 ) /p 4n 2n = 2n ( t) /p ( t y 2 ) dt Also (recall that /p = /q) r n ( r 2 ) /p r n ( r 2 ) /p 2r( r 2 ) /p r 2 y 2 dr ( t) /p ( t y 2 ) dt. y 2 y 2 ( t) /p ( t y 2 ) dt ( y 2 ) Addition yields R(rζ, y) dσ(ζ) dr dσ(ζ) dr ζ ry n ( t) /p dt p( y 2 ) /p. y 2 ( t) /p dt = ( y 2 ) q( y 2 ) /p = q( y 2 ) /p. ( t) /p ( t y 2 ) dt (p + q)( y 2 ) /p. Thus (3 4) is proved with C p = 2n(p + q). This concludes the proof of the L p -boundedness of Q. In fact, from the chur Test, the estimates above and the observation that C q = C p, we obtain the following bound on the norm of Q as an operator from L p () onto b p (): Q 2np 2 /(p ). Remark. The norm estimate given above is far from being sharp; it can be improved by estimating the integrals using binomial series, as in [Forelli and Rudin 974/75; Rudin 98]. Remark 2. As we will see in the next section, Theorem 3. does not hold for p = or p =.
8 58 KAREL TROETHOFF Duality. It is a consequence of the L p -boundedness of the ergman projection that the spaces b p () and b q () are dual to each other: if v b q (), then the function ϕ defined by ϕ(u) = u, v (u b p ()) defines a bounded linear functional on b p (), and every bounded linear functional on b p () is of the form above. To prove the latter statement, suppose that ϕ b p (). y the Hahn- anach theorem, ϕ extends to a bounded linear functional ψ on L p (). There exists a g L q () such that ψ(f) = f, g for all f L p (). In particular, if u b p (), then ϕ(u) = u, g. Note that v = Q[g] Q(L q ()) = b q (). Using Fubini s theorem to reverse the order of integration it can then be shown that u, v = u, g, and we obtain that ϕ(u) = u, v, for all u b p (). 4. The loch pace and the Dual of b () A harmonic function u on is said to be a loch function if u = sup( x 2 ) u(x) <. x The harmonic loch space is the set of all harmonic loch functions on. We will show that is the dual of the ergman space b (). We will first prove that = Q[L ()]. The hard part of the proof will be to show that each function u is of the form u = Q[g], where g is a bounded function on. If u, then it follows from the inequality Ru(x) u(x) that the function ( x 2 )Ru is bounded on. Thus we will try to relate Q[( x 2 )Ru] to u. We will first find an expression for Q[( x 2 )u]. Combining (2 4) and (3 ) we have nv ()Q[( x 2 )u](y) = 2 ( x 2 )u, RP y + n ( x 2 )u, P y. To rewrite ( x 2 )u, RP y we use the same idea as the derivation of formula (2 4): assuming u to be harmonic on an open set containing, apply identity ( ) with u and v(x) = ( x 2 ) 2 P (x, y). The outward normal derivative D n v is on ; thus ( ) gives us ( u(x) x ( x 2 ) 2 P (x, y) ) dv (x) =. (4 ) It is easily verified that x ( ( x 2 ) 2 P (x, y) ) = 8 x 2 P (x, y)+4n( x 2 )P (x, y)+8( x 2 )R x P (x, y). Thus (4 ) shows that x 2 u(x)p (x, y) dv (x) = 2 n u(x)( x 2 )P (x, y) dv (x) + u(x)( x 2 ) R x P (x, y) dv (x),
9 HARMONIC ERGMAN PACE 59 and using (3 ) and (2 4) we can write the equation above as x 2 u(x)p (x, y) dv (x) = 2 nv ()Q[( x 2 )u](y). (4 2) It follows from ( ) that u(rζ)d nv(rζ) dσ(ζ) = D nu(rζ)v(rζ) dσ(ζ) for harmonic functions u and v on and < r <. Multiplication by nv ()r n+ and integration over r yields the formula x 2 u(x)rv(x) dv (x) = x 2 Ru(x)v(x) dv (x). (4 3) Applying (4 2) to the function Ru, and making use of (4 3), we also have x 2 u(x)r x P (x, y) dv (x) = 2 nv ()Q[( x 2 )Ru](y). (4 4) Combining (4 2), (4 4) and (2 4) we arrive at Q[ x 2 u](y) = 2 nq[( x 2 )u](y) + Q[( x 2 )Ru](y), and, writing u = Q[( x 2 )u] + Q[ x 2 u] we obtain the formula in the following theorem. Theorem 4.. If u, then u = Q[( x 2 )Ru + ( 2 n + )( x 2 )u]. (4 5) Proof. We have shown the stated result for u harmonic on an open set containing. To get the result for general u, let < r < and consider the dilate u r of u, defined by u r (x) = u(rx), x. ince u r is harmonic on the set {x R n : x < /r}, equation (4 5) holds for u r. It is easily seen that (Ru) r = Ru r. We leave it as an exercise to show that ( x 2 )Ru r ( x 2 )Ru and ( x 2 )u r ( x 2 )u in L 2 () as r. The boundedness of Q and the continuity of point evaluation at y imply that Q[( x 2 )Ru r ](y) Q[( x 2 )Ru](y) and Q[( x 2 )u r ](y) Q[( x 2 )u](y) as r, and since u r (y) u(y) formula (4 5) follows. Corollary 4.2. = Q[L ()]. Proof. If g L (), and u = Q[g], then we claim that u. Differentiating u(x) = g(y)r(x, y) dv (y) we obtain D j u(x) = g(y) R(x, y) dv (y), x j and consequently u(x) g x R(x, y) dv (y). (4 6)
10 6 KAREL TROETHOFF Using (2 5) we have ( ) n(y y 2 x) n( x 2 y 2 ) 2 x R(x, y) = ( 2x y + x 2 y 2 ) +n/2 ( 2x y + x 2 y 2 ) 4 x 2 y 2 ( 4n( x 2 y 2 ) y 2 x + ( 2x y + x 2 y 2 ) n/2 ( 2x y + x 2 y 2 ) + + 2n( x 2 y 2 ) 2 (y x y 2 ) ( 2x y + x 2 y 2 ) 2 8 y 2 x Noting that y y 2 x 2 = y 2 2 y 2 x y + y 4 x 2 = y 2 ( 2x y + y 2 x 2 ), we see that y y 2 x ( 2x y + y 2 x 2 ) /2. Recalling that also x 2 y 2 ( 2x y + y 2 x 2 ) /2 we obtain x R(x, y) C ( 2x y + x 2 y 2) (n+)/2. (4 7) ). Thus x R(x, y) dv (x) CnV () r n dσ(ζ) dr. ζ ry n+ Now, ζ ry n+ dσ(ζ) = r 2 y 2 r 2 y 2 ( r y ) P (ry, ζ) dσ(ζ) ζ ry P (ry, ζ) dσ(ζ) ( r y ) 2. Multiply by nv ()r n and integrate with respect to r to obtain ( 2x y + x 2 y 2 ) dv (x) nv () (+n)/2 ( y ). (4 8) Using that /( y ) = ( + y )/( y 2 ) 2/( y 2 ), we conclude from (4 6), (4 7) and (4 8) that ( y 2 ) u(y) 2nV ()C g for all y, establishing our claim that u. This proves the inclusion Q[L ()]. The other inclusion Q[L ()] follows from Theorem 4., for if u, the function g = ( x 2 )Ru + ( 2 n + )( x 2 )u is bounded on. This corollary can be used to show that Theorem 3. does not hold for p = or p =. It is not difficult to construct unbounded harmonic loch functions (the function u(x) = log(( x ) 2 + x 2 2 ) provides an example), so Corollary 4.2 shows that the operator Q is not L -bounded. y duality it follows that Q is not L -bounded either.
11 HARMONIC ERGMAN PACE 6 The dual of b (). We will show how Corollary 4.2 can be used to prove that is the dual of b (). Our first concern is to define a pairing : given a function v we could try to define ϕ by ϕ(u) = u, v (u ), but we have to be careful here: the function v need not be bounded, so there is no guarantee that u v is integrable for all u b () (in fact, if v is unbounded, some standard functional analysis can be used to prove that there must exist a function u b () for which u v is not integrable). This problem is easily overcome by using dilates. If u is a harmonic function on and < r <, the dilate u r, defined by u r (x) = u(rx), x, is bounded on. If u C( ), then u is uniformly continuous on, and it follows that u r u uniformly on as r. Using the fact that C( ) is dense in L () it is easy to verify that if u b (), then u r u in b () as r. We are now ready to extend the usual pairing. Given u b () and v we claim that lim r u r, v exists. To prove this, write v = Q[g], where g L (). ince u r b 2 () we have u r, v = u r, Q[g] = u r, g. The inequality u r, v u s, v = u r u s, g u r u s g together with the fact that u r u s as r, s (because u r u s u r u + u u s ) implies that lim r u r, v exists. It is clear that the function ϕ defined by ϕ(u) = lim r u r, v for u b () is a linear functional on b (). Using u r, v u r g we see that ϕ(u) u g, so ϕ b (). We claim that every element of b () is of the form above. For let ϕ b (). y the Hahn-anach theorem, ϕ extends to a bounded linear functional ψ on L (, dv ). There exists a g L () such that ψ(f) = f, g for all f L (). In particular, ϕ(u) = u, g for all u b (). Now, if u b (), then u r u in b () as r, and thus ϕ(u) = lim r ϕ(u r ). ince u r b 2 () we have u r, g = Q[u r ], g = u r, Q[g]. Thus, if we set v = Q[g], then ϕ(u) = lim r u r, v, as was to be proved. The predual of b (). In this subsection we will identify the predual of b () under the above pairing. We define the space, called the harmonic little loch space, to be the set of all harmonic functions u on for which ( x 2 ) u(x) as x. Clearly,. We leave it as an exercise for the reader to show that if u, then u r u as r (the converse is also true). Now suppose that u and v b (). We claim that lim r u r, v exists. To show this, we use Corollary 4.2 (or rather its proof) to conclude that u r, v u, v r u r u, v + u, v v r C u r u v + C u v v r, and the statement follows. o if v b (), then ϕ(u) = lim r u r, v defines a bounded linear functional on.
12 62 KAREL TROETHOFF Our claim is that all bounded linear functionals on arise this way. uppose ϕ is a bounded linear functional on. Then ψ(g) = ϕ(q[g]) defines a bounded linear functional on C( ): ψ(g) = ϕ(q[g]) ϕ g 2 ϕ g. y the Riesz representation theorem ψ(g) = g dµ, for all g C( ), where µ is a finite complex orel measure on. If u, then g(x) = ( x 2 ){Ru(x) + ( 2 n + )u(x)} is in C( ), and, using Theorem 4., we have ϕ(u) = ψ(g) = g(y) dµ(y). Define v(x) = ( y 2 ){R x R(x, y) + ( 2n + )R(x, y)} d µ(y) for x. The function v is clearly harmonic in. We claim that in fact v b (). To show this, use Fubini s theorem to get v(x) dv (x) ( y 2 ) { R x R(x, y) + ( 2n + ) R(x, y) } dv (x) d µ (y) In the proof of Corollary 4.2 we have seen that there exists a positive constant C such that R xr(x, y) dv (x) C/( y 2 ). Also, using (2 6) and ( 5), we have R(x, y) dv (x) 4n r n dσ(ζ) dr ζ ry n Thus, v(x) dv (x) ( y 2 ) = 4n 4n C r n r 2 y 2 dr r n y 2 = 4 y 2. d µ (y) = C y 2 d µ (y) = C µ <, and our claim that v b () is proved. Now, assuming u to be harmonic on, applying Fubini s theorem we have u, v = ( y 2 ) u(x){r x R(x, y) + ( 2n + )R(x, y)} dv (x) dµ(y). imilarly to (4 3) we have u(x)r xr(x, y) dv (x) = Ru(x)R(x, y) dv (x) = Ru(y), thus u(x){r x R(x, y) + ( 2 n + )R(x, y)} dv (x) = Ru(y) + ( 2 n + )u(y) and we obtain u, v = ( y 2 ) { Ru(y) + ( 2 n + )u(y)} dµ(y) = g(y) dµ(y) = ϕ(u). Hence ϕ(u r ) = u r, v, and since u r u in and ϕ is continuous on we have ϕ(u) = lim r u r, v. Note that this pairing coincides with the one we saw earlier: if u and v b (), then lim r u r, v = lim r u, v r.
13 HARMONIC ERGMAN PACE 63 Acknowledgments This article was written while visiting the Free University, Amsterdam, The Netherlands, on sabbatical leave; I thank the University of Montana for awarding me a sabbatical and the Mathematics Department of the Free University for its hospitality and support. References [Axler 988]. Axler, ergman spaces and their operators, pp. 5 in urveys of some recent results in operator theory, vol., edited by J.. Conway and.. Morrel, Pitman Res. Notes Math. er. 7, Longman ci. Tech., Harlow, 988. [Axler et al. 992]. Axler, P. ourdon, and W. Ramey, Harmonic function theory, Graduate Texts in Mathematics 37, pringer, New York, 992. [Coifman and Rochberg 98] R. R. Coifman and R. Rochberg, Representation theorems for holomorphic and harmonic functions in L p, pp. 66 in Representation theorems for Hardy spaces, Astérisque 77, oc. Math. France, Paris, 98. [Forelli and Rudin 974/75] F. Forelli and W. Rudin, Projections on spaces of holomorphic functions in balls, Indiana Univ. Math. J. 24 (974/75), [Rudin 98] W. Rudin, Function theory in the unit ball of C n, Grundlehren der Mathematischen Wissenschaften 24, pringer, New York, 98. Karel troethoff Department of Mathematical ciences University of Montana Missoula, MT United tates ma kms@selway.umt.edu
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