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On L^p-solutions to the Laplace equation and zeros of holomorphic functions

Bruna, Joaquim,Ortega Cerdà, Joaquim

Abstract

The problem we solve in this paper is to characterize, in a smooth domain $\Omega$ in $\Bbb R^n$ and for $1\le p\le\infty$, those positive Borel measures on $\Omega$ for which there exists a subharmonic function $u\in L^p(\Omega)$ such that $\Delta u=\mu$. The motivation for this question is mainly for $n=2$, in which case it is related with problems about distributions of zeros of holomorphic functions: If $\{a_n\}^{\infty}_{n=1}$ is a sequence in $\Omega\subset\Bbb C$ with no accumulation points in a simply connected domain $\Omega$, and $\mu=2\pi\sum_n\delta_{a_n}$, then all solutions $u$ of $\Delta u=\mu$ are of the form $u=\log |f|$, with $f$ holomorphic vanishing exactly on the poits $a_n$. Thus our results give the characterization of the zero sequences of holomorphic functions with $\log |f|\in L^p(\Omega)$. A related class had been considered by Beller.

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ON L p -SOLUTIONS OF THE LAPLACE EQUATION AND ZEROS OF HOLOMORPHIC FUNCTIONS Joaquim Bruna and Joaquim Ortega-Cerd  a 1. Introduction The problem we solve in this pap er is to characterize, in a smo oth domain  in R n and for 1  p  1 , those p ositive Borel measures on  for which there exists a subharmonic function u 2 L p () such that  u =  . The motivation for this question is mainly for n =2, in whichcase it is related with problems ab out distributions of zeros of holomorphic functions: if f a n g 1 n =1 is a sequence in   C with no accumulation points in a simply connected domain , and  = 2  X n  a n , then all solutions u of  u =  are of the form u = log j f j ,with f holomorphic vanishing exactly on the points a n .Thus our results give the characterization of the zero sequences of holomorphic functions with log j f j2 L p (). A related class had been considered by Beller (Bel1]). In section 2 we consider rst  = R n . In this case, it is easily seen lo oking at the behavior at 1 that the problem only makes sense for n  3and p > n= ( n ; 2), and that it is equivalent to the description of the measures  on R n having a Newtonian potential in L p ( R n ). This was achieved byMuckenhoupt and Wheeden, and we givean alternative pro of as well. The results and the metho ds of pro ofs serve as a guideline for the b ounded case. In section 3 we solve the problem for a b ounded smo oth domain  in R n .Among other characterizations we nd that  u =  has a solution in L p () if and only if the fractional maximal function   ( x )= sup r   ( x ) r 2 ; n  ( B ( x r )) belongs to L p (). Here  ( x )is the distance from x to @ and  is a xed constant, 0 <  < 1. The case p = 1 was previously known (Hei], DH]) in dimension 2, and amounts to say that Z   ( x ) 2 d ( x ) < + 1 : Supp orted by DGICYT grant PB92-0804-C02-02 and by CIRIT grantGRQ94-2014. 1 2 J. BRUNA AND J. ORTEGA-CERD  A We also consider at the end the case p = 1 . In section 4 we lo ok closely to the case  = D , the unit disc in C . We nd here a linear solution op erator u = K   ] of  u =  with u 2 L p (), and relate the problem with others, suchasTo eplitz op erators, Carleson measures for Bergman spaces, and the bilaplacian. In section 5 we particularize to the motivating question, the zeros of holomorphic functions. It turns out that the condition characterizing the zero sequences f a n g of functions with log j f j2 L p ( D ) dep ends b oth on the radii and the angular disp osition of the points a n , as in many other classes of functions. We nd adyadic expression of it and also we nd the best condition on the sequence of radii r n = j a n j . A probabilistic version is also considered. As a nal remark, we point out that we have b ound ourselves to the situation describ ed |subharmonic functions and p ositive measures| in view of the applications we had in mind, zeros of holomorphic functions. It is quite probable that our results extend to measures or distributions, and also replacing the laplacian by a general strongly elliptic op erator, but we have not tried to do so. Another generalization, replacing the Laplace equation  u =  by the equation i@  @u =  in C n and with applications to zero varieties of holomorphic functions of several variables, app ears in Ort]. 2. Subharmonic functions in L p ( R n ) In this section we shall characterize the laplacians of subharmonic functions in L p ( R n ). For general facts on subharmonic functions that we will use, see for instance Hay]. First note that if h is harmonic in L p ( R n ), the mean-value prop erty implies that h  0. Hence all subharmonic functions u in L p ( R n ) are completely determined by its laplacian  = u , a p ositive Borel measure in R n . The sub-mean value prop erty implies that any such u is non-p ositive, hence we can consider in each ball B (0 R )the Green decomp osition u ( x )= h R ( x )+ Z j y j R G R ( x y ) d ( y )  j x j <R: Here h R is the least harmonic ma jorant of u in B (0 R ) and G R ( x y ) is the Green function for the ball B (0 R ), given by G R ( x y )= 8 > > < > > : c n  j x ; y j 2 ; n ; R n ; 2 j y j 2 ; n j x ; y 0 j 2 ; n  n> 2 c 2 log j x ; y 0 jj y j R j x ; y j  n =2 with y 0 = yR 2 j y j ; 2 . Clearly u ( x )  h R ( x )  h R 0 ( x )  0 if R  R 0 .The limit h ( x ) = lim R !1 h R ( x ), the least harmonic ma jorant of u in R n , is then in L p ( R n ), hence ON L p SOLUTIONS TO THE LAPLACE EQUATION 3 is identically zero. Thus we conclude that u ( x )= lim R !1 Z j y j R G R ( x y ) d ( y ) : It is immediate to check that for n> 2, lim R !1 G R ( x y )= c n j x ; y j 2 ; n , while this limit is innite for n =2. By the monotone convergence theorem, this means that there are no non-zero subharmonic functions in L p ( R 2 ) and that for n> 2 they are exactly those Newtonian p otentials u ( x )= c n Z R n j x ; y j 2 ; n d ( y ) b elonging to L p ( R n ). Now note that unless  is identically zero, j u ( x ) j dominates j x j 2 ; n for large j x j and so p must b e greater than n= ( n ; 2). In conclusion, wehave seen that the question only makes sense for n> 2 and p>n= ( n ; 2) and that it is equivalent to the description of the p ositive measures  for which the Riesz p otential of order 2 is in L p ( R n ). We use the notation T   ( x )= Z R n j x ; y j  ; n d ( y ) for the Riesz potential of order  , 0 <  < n . We consider the maximal fractional function of  , dened by    ( x )=sup t> 0  ( B ( x t )) j B ( x t ) j 1 ;  n = sup t> 0 t  ; n  ( B ( x t )) : Note that    is pointwise dominated by T   .Next theorem was proved by Muckenhoupt and Wheeden in MW] for absolutely continuous measures. We will give an alternative pro of for it contains some concepts that will be used later. Theorem 2.1. Let  be a positive measure in R n , n> 2 , 1 <p< 1 and  such that 0 <<n . Then T   2 L p ( R n ) if and only if    2 L p ( R n ) . Proof. Write T   in the form T   ( x )=( n ;  ) Z 1 0 t  ; n ; 1  ( B ( x t )) dt: Consider in R n +1 + , with co ordinates ( x t ), x 2 R n , t> 0, the pro duct measure d ~  ( x t )= d ( x )  t  ; 1 dt . Then we can write T   ( x )=( n ;  ) Z ;( x ) t ; n d ~  ( y t ) def = A (~  )( x ) 4 J. BRUNA AND J. ORTEGA-CERD  A where ;( x )is the cone f ( y t ) : j x ; y j < t g .By CMS, x 6], A (~  ) b elongs to L p ( R n ), 1 <p , simultaneously with the function C (~  )( x ) = sup  ~  ( Q ( B )) j B j  x 2 B  : Here Q ( B ) denotes the Carleson window over the ball B .(We p oint out that this denition of A , C do es not coincide with that of CMS]). In our case ~  ( Q ( B )) '  ( B ) j B j =n , whence C (~  )( x )     ( x ).  3. Subharmonic functions in L p () In this section we will characterize, for asmo oth b ounded domain  in R n , n  2, the p ositive (Borel) measures  on  such that the Poisson equation   =  has a (subharmonic) solution u 2 L p (), 1  p 1 . 3.1. For x 2 , we denote by  ( x ) the distance from x to the b oundary of . For a p ositive measure  on  and a xed  , 0 <  < 1, we consider lo cal versions of the fractional maximal function and Newtonian potentials in the previous section. These are    ( x )= sup r   ( x )  ( B ( x r )) r n ; 2 and T   ( x )= Z B ( x ( x )) j x ; y j 2 ; n d ( y ) B ( x r ) b eing the ball of radius r centered at x . Theorem 3.1. The fol lowing are equivalent for a positive measure  on  and 1  p< 1 : (a) There is a subharmonic function u 2 L p () such that  u =  . (b) T   2 L p () for some 0 << 1 . (c)    2 L p () for some 0 << 1 . In case p =1 ,an equivalent condition is (d) R   ( x ) 2 d ( x ) < + 1 . In particular, conditions (b), (c) do not dep end on  . The theorem should be compared with Theorem 2.1. 3.2. It is trivial that (b) implies (c), b ecause     T   . An application of Fubini's theorem shows that up to constants the integral in (d) is the L 1 -norm of T   ,using that  ( x ) '  ( y ) whenever j x ; y j  ( x ). Obviously    ( x ) &  ( B ( x  ( x ))  ( x ) 2 ; n : and so    2 L 1 () implies again byFubini's theorem that (d) holds. Therefore (b), (c) and (d) are all equivalent for p =1. We will now prove that    2 L p (), 1 <p< 1 implies T   2 L p () for some  . The rst step is alo cal version of Theorem 2.1. ON L p SOLUTIONS TO THE LAPLACE EQUATION 5 Lemma 3.2. For a positive measure  on B (0  8 " ) and 1 <p< 1 (1) Z B (0 " )  Z j x ; y j " j x ; y j 2 ; n d ( y ) ! p dA ( x )  C p Z B (0  3 " )  sup r  5 "  ( B ( x r )) r n ; 2  p dA ( x ) : Proof. We use the same methodasinTheorem 2.1. The inner integral of the left-hand side equals ( n ; 2) Z " 0 t 1 ; n  ( B ( x t )) dt: We consider in the half-space R n +1 + with co ordinates ( x t ) the pro duct measure d ~  ( x t )= 1I B (0  2 " ) ( x ) d ( x )  1I (0 " ) ( t ) tdt , so that the previous integral equals Z ;( x ) t ; n d ~  ( x t )= A (~  )( x )  x 2 B (0 " ) : By the result of CMS] already used in Theorem 2.1, the left-hand side of (1) is b ounded by a constant times the L p -norm of C (~  )in R n . Now, for j x j 3 " C (~  )( x ) ' sup r  5 "  ( B ( x r )) r n ; 2  hence the integral of C (~  ) p over j x j 3 " is already b ounded by the right hand side of (1). For j x j > 3 " , C (~  )( x ) .  ( B (0  2 " )) " 2 j x j n and hence Z j x j 3 " ( C (~  )( x )) p dA ( x ) . " n ; p ( n ; 2)  ( B (0  2 " )) p  whichis also b ounded by the right hand side.  Assume  < 1 = 8 and consider  covered by the family of balls f B ( x   ( x )) x 2  g . By Besicovitch covering lemma (see Fed 2.8.14] for instance), there is a nite set of families B 1 ::: B N of balls in the covering whose total union is also and such that each family consists of disjoint balls. Then, writing B x for B ( x   ( x )), B 0 x for B ( x 3  ( x )), Z  T p   ( x ) dA ( x )  N X i =1 X B x 2 B i Z B x T p   ( x ) dA ( x ) which by Lemma 3.2 is dominated by N X i =1 X B x 2 B i Z B 0 x   p 5  ( x ) dA ( x ) : 6 J. BRUNA AND J. ORTEGA-CERD  A We claim now that, for each family B i , the balls B 0 x cover each p oint of  at most a nite numb er (dep ending only on n ) of times. Indeed, if y 2 B 0 x then m ; 1  ( y )   ( x )  m ( y ) for some constant m = m (  ), whence the balls B x 2 B i , such that y 2 B 0 x have volume '  ( y ) n ,are disjoint and are all of them in the ball B ( y (3  + m )  ( y )), and the claim follows. It is then clear that the last expression is b ounded by C ( p  ) Z    p 5  ( x ) dA ( x ) : Therefore,    2 L p () implies T   2 L p () for  small enough. 3.3. We will prove now that (a) implies (b). For r <  ( x ) consider the Green decomp osition of u in B ( x r ) 1  n r n ; 1 Z j x ; y j = r u ( y ) d ( y ) ; u ( x )= Z j x ; y j <r G r ( x y ) d ( y ) : Here  n is the surface measure of the unit sphere and G r ( x y )is Green's function for the ball B ( x r ). Multiply the ab ove by nr n ; 1 , integrate between 0 and R   ( x ), and divide by R n . The result is, for n  3 n  n R n Z j x ; y j R u ( y ) dA ( y ) ; u ( x )= = 1 ( n ; 2)  n Z j x ; y j R j x ; y j 2 ; n  1 ; j x ; y j n R n  1 ; n 2  ; n 2 j x ; y j n ; 2 R n ; 2  d ( y ) and for n =2 1 R 2 Z j x ; y j R u ( y ) dA ( y ) ; u ( x )= = 1 2  Z j x ; y j R  log R j x ; y j + 1 2  j x ; y j 2 R 2 ; 1  d ( y ) : The functions 1+ ; n 2 ; 1  x n ; n 2 x n ; 2 and log 1 x + 1 2 ( x 2 ; 1) are positivedecreasing functions in (0  1). Therefore for each  < 1 there is c (  )such that 1 j B ( x R ) j Z B ( xR ) u ( y ) dA ( y ) ; u ( x )  c (  ) Z j x ; y j R j x ; y j 2 ; n d ( y ) : Writing  ( x )= R with R = p  ( x ), it follows that for (a) ) (b) it is enough to prove that for a xed  < 1, u 2 L p () implies that the average of u over B x = B ( x   ( x )) belongs to L p () as well. To see this we use Holder's inequality Z      1  ( x ) n Z B x u ( y ) dA ( y )     p dA ( x )  Z  1  ( x ) n Z B x j u ( y ) j p dA ( y ) dA ( x ) : If y 2 B x , then  ( x ) '  ( y ) and Fubini theorem nishes the pro of that (a) implies (b). ON L p SOLUTIONS TO THE LAPLACE EQUATION 7 3.4. Next weprove that (d) implies (a) in case p =1. More precisely wewanttoshow that whenever  is a p ositive measure such that Z   ( x ) 2 d ( x )= M < + 1 then there is u 2 L 1 () such that  u =  and k u k 1  CM , where C is a constant dep ending only on . The dual statement of this is contained in next theorem Theorem 3.3. For any test function  in  , j ( x ) j C ( x ) 2 k  k 1  x 2  : These are indeed equivalent statements. Assume Theorem 3.3 holds and let  be as ab ove. Dene a linear functional  as follows () = Z  ( x ) d ( x ) is well dened, and by Theorem 3.3 j () j CM k  k 1 : By Hahn-Banach theorem,  extends to a continuous functional on the space C c () of continuous functions with compact supp ort, with norm  CM . By Riesz's representation theorem, we conclude that there is a measure  on , with total variation k  k CM such that () = Z  ( x ) d ( x ) : This means that   =  in the weak sense, and then  must be in fact absolutely continuous d = udx , with k u k 1  CM and  u =  . It is immediate to check that this argument can be reversed. Proof of Theorem 3.3. Let G ( x y ) denote the fundamental solution for  in R n , G ( x y )= c n j x ; y j 2 ; n if n  3, G ( x y )= c 2 log j x ; y j in R 2 , so that ( x )= Z  G ( x y )( y ) dA ( y ) : Since j ( x ) j k  k 1 Z  j G ( x y ) j dA ( y )  C k  k 1 it is enough to prove the inequality for points x close to @ . Let U be a tubular neighborhood of @ , that is, there is awell-dened pro jection p : U ! @  of class C 1 , p ( x )is the unique closest p oint to x , and the line between x and p ( x )is orthogonal to @ at p ( x ). Let z =2 p ( x ) ; x be the symmetric point of x with resp ect @ . 8 J. BRUNA AND J. ORTEGA-CERD  A The idea is that in the integral representation of  ab ove, one can subtract to G ( x y ) any harmonic function h in  b ecause these annihilate ( y ), by Green's theorem. Fixed x , we take as h ( y ) the second-order Taylor p olynomial of G (  y ) at z , evaluated at x , h ( y )= G ( z y )+2  ( x ) @ @n G ( z y )+ 2  ( x ) 2 @ 2 @n 2 G ( z y ) : Here n is the unit direction from z to x .Thus we have j ( x ) jk  k 1 Z  j G ( x y ) ; h ( y ) j dA ( y ) If j x ; y j 3  ( x ), by Taylor's formula, j G ( x y ) ; h ( y ) j .  ( x ) 3 j x ; y j n +1 and hence the contribution of  n B ( x 3  ( x )) in the last integral is dominated by  ( x ) 2 . If j x ; y j 3  ( x ), we write j G ( x y ) ; h ( y ) j j G ( x y ) ; G ( z y ) j +3  ( x ) jr z G ( z y ) j +5  ( x ) 2 jr 2 z G ( z y ) j : This is b ounded by c j x ; y j 2 ; n if n> 2, and by c +log 3  ( x ) j x ; y j if n =2. In b oth cases the integral over B ( x 3  ( x )) is estimated by  ( x ) 2 .  The same pro of will go through if for any p oint x close to the b oundary,we can nd apoint z outside the op en set , such that the distance of z to the b oundary of  is comparable to j z ; x j .This holds, for instance, if  satises a uniform exterior cone condition. H. Shapiro Sha], has let us known that it is possible to get the same result under very mild regularity assumptions on the b oundary of . 3.5. To complete the pro of of Theorem 3.1, we nally prove that (b) implies (a) when p > 1. Let  be a test function supp orted in j y j < 1, equal to one on j y j < 1 = 2, 0    1 with 0 < < 1 = 3, 0 < < small enough to be dened later, we dene an \approximate solution" by K (  )( x )= Z R n   x ; y  ( x )  K ( x y ) d ( y ) : Here K ( x y ) is the fundamental solution for  in R n , K ( x y )= G ( x y )= c n j x ; y j 2 ; n if n> 2 if n =2 we need to mo dify it by a function H ( x y ) K ( x y )= G ( x y ) ; H ( x y )= c 2 log ( x ; y ) ; H ( x y ) where H ( x y )is harmonic in x , equal to G ( x y 0 )for y close to @ , y 0 being the symmetric of y with resp ect to @ , as in the previous paragraph. Another choice, in all ON L p SOLUTIONS TO THE LAPLACE EQUATION 9 dimensions, is K ( x y ) = G  ( x y ) the Green's function for , if  is smo oth enough. In any event,  K (  )=  + R (  ) with R (  )( x )= Z R n  2 r x   x ; y  ( x )  r x K ( x y )+  x   x ; y  ( x )  K ( x y )  d ( y ) : We claim now that under the assumption (b), both K (  )and  2 R (  )are in L p (). This is clear if n> 2, b ecause both are pointwise estimated by T   .If n =2, b oth are pointwise estimated by  ( B ( x   ( x ))) + Z B ( x  ( x )) log  ( x ) j x ; y j d ( y ) and so it is enough to prove that the function dened by this last integral, call it ( x ), is in L p (). By Holder's inequality and Fubini's theorem Z  ( x ) p dA ( x )  Z   Z E  log  ( x ) j x ; y j  p  ( B ( x   ( x ))) p ; 1 dA ( x )  d ( y ) : The inner integral is over the set E = f x : y 2 B ( x   ( x )) g .With " = 2  1 ;  one has B ( x   ( x ))  B ( y " ( y )) if x 2 E ,and a computation in p olar co ordinates as in the previous section gives Z B ( y" ( y ))  log " ( y ) j x ; y j  p dA ( x ) '  ( y ) 2 : Therefore Z  ( x ) p dA ( x ) . Z   ( B ( y " ( y ))) p ; 1  ( y ) 2 d ( y ) : Now we can just rep eat the same argument: this last integral is comparable to Z   ( B ( y " ( y ))) p ; 1 ( Z B ( y " ( y )) dA ( x ) ) d ( y )= = Z   Z F  ( B ( y " ( y )) p ; 1 d ( y )  dA ( x )  the set F b eing now f y : x 2 B ( y " ( y )) g .Cho osing " suchthat 2 " 1 ; " =  , B ( y " ( y ))  B ( x  ( x )) for y 2 F and hence the ab ove is b ounded by Z   ( B ( x  ( x ))) p dA ( x )= Z  T   ( x ) p dA ( x ) < + 1 : It is then enough to prove that given a function f (= K (  )) with  2 f 2 L p () there is u 2 L p () such that  u = f in the distribution sense, with Z  j u j p dA  M Z   2 p j f j p dA the constant M dep ending only on . In the same way as b efore it is enough to prove the dual statement that follows: 16 J. BRUNA AND J. ORTEGA-CERD  A part of a subharmonic function is controlled by the p ositive part in L 1 .By the same reason log + j f j can be replaced by j log j f jj in the denition of W p ( D ). The connections with the problem  u =  is well known and classic: if Z = f a n g is a sequence with no accumulation p oints in D and  = 2  X n  a n ,the fact that an holomorphic function f vanishes exactly at Z is equivalent to say that  log j f j =  . 5.1. For a sequence Z = f a j g in D with no accumulation p oints in D , we call n jk the number of p oints of Z in the dyadic region T kj ,and N k = 2 k X j =1 n jk the numb er of p oints of Z in the annulus 1 ; 2 ; k < j z j 1 ; 2 ; k ; 1 ,counting multiplicities. Theorem 5.1. Z is the zero sequence of a function f 2A p ( D ) if and only if (6) 1 X k =0 2 ; 2 k 2 k X j =1 n p kj < + 1 and it is the zero sequence of f 2W p ( D ) if and only if (7) 1 X k =0 2 ; ( p +1) k N p k < + 1 : Proof. The rst part is the restatement of Theorem 3.1 in subsection 4.4. We prove now the necessity of condition (7). Let f 2 W p ( D ) and assume without loss of generality that f (0) = 1. Set  = 2  X n  a n , N ( r )=  fj z j r g . By Jensen's formula, for r close to 1 N ( r )(1 ; r ) ' Z r 0 log r j  j d (  )= Z 2  0 log j f ( re i ) j d and hence, Z 1 0 N ( r ) p (1 ; r ) p dr < + 1 whichis easily seen to be equivalent to (7). For the su#ciency, we will prove that the solution u = K   ] of  u =  , considered in the previous section, satises Z 1 0  Z 2  0 u + ( re it ) dt  p dr < + 1 : ON L p SOLUTIONS TO THE LAPLACE EQUATION 17 Note that the estimate (5) of the kernel K ( z  )of K implies that K   ]is a subharmonic function in L 1 lo c ( D ) whenever  satises Z D (1 ;j  j ) 2 d (  ) < + 1  or X k 2 ; 2 k N k < + 1  and that (7) implies this (as corresp onds to the inclusion W p ( D )  W 1 ( D ) = A 1 ( D )). Moreover (4) implies u + ( z ) . Z D (1 ;j  j 2 ) 2 j 1 ; z j 2 d (  ) b ecause log s +1 ; s  0 for s  1. Then Z 2  0 u + ( re it ) dt . Z D (1 ;j  j 2 ) 2 (1 ; r j  j ) d (  ) : The right-hand side is comparable to Z j  j r (1 ;j  j ) d (  )+ 1 1 ; r Z r< j  j < 1 (1 ;j  j ) 2 d (  ) : In terms of the distribution function N ( r ), we nally obtain Z 2  0 u + ( re it ) dt . Z r 0 N ( t ) dt + N ( r )(1 ; r )+ 1 1 ; r Z 1 r N ( t )(1 ; t ) dt and the result follows using Hardy's and Holder's inequalities.  The problem of describing the zero sequences of B p ( D ) remains op en (Bel1], Bel2], BH]). This is a harder problem than the ones we deal with, mainly b ecause it is a nonlinear problem. Note the di$erentcharacter of conditions (6) and (7). While (7) only dep ends on the radii of the p oints of Z , (6) also dep ends on their angular distribution. We can see directly that (6) is stronger than (7), because by Jensen's inequality for convex functions, N p k (2 ; k ) p ; 1 =2 k 0 @ 2 ; k 2 k X j =1 n kj 1 A p  2 k X j =1 n p kj : The only way to have equality ab ove is that all n kj are equal, that is, the p oints in Z are uniformly distributed in the angle. Only in this case conditions (6), (7) are the same. In particular, (7) is the necessary and su#cient condition on a sequence of radii r n for the existence of f 2A p ( D ) (or f 2B p ( D )) with zeros a n , r n = j a n j .Next we will make this even more precise. 18 J. BRUNA AND J. ORTEGA-CERD  A 5.2. Let f r n g be a xed increasing sequence of radii. Let  n ( w )be a sequence of indep endent random variables, all uniformly distributed in 0  2  ]. We consider the random sequence f a n g = Z ( w )with a n ( w ) = r n e i n ( w ) and ask ab out the probability that Z ( w )be a zero sequence of A p ( D ) or B p ( D ). Questions of this typ e have been considered in other classes of functions in Co c], Bom], Leb], Rud], Hor], Mas] and BH]. Theorem 5.2. If the sequence f r n g satises (7), then Z ( w ) is the sequenceofzeros of a function f 2A p ( D ) with probability 1. Proof. It is enough to prove that the random variable X ( w )= 1 X k =0 2 k X j =1 n kj ( w ) p 2 ; 2 k has nite mean, by Theorem 5.1. Here n kj is the random variable that counts the number of points a n = r n e i n ( w ) in T kj ,which has binomial law B ( N k  2 ; k ) for each j =1 :::  2 k . Then E ( n p kj ) . p X l =1 (2 ; k N k ) l and E ( X ) . 1 X k =0 2 k X j =1 p X l =1 (2 ; k ) l +1 N l k ' ' p X l =1 Z 1 0 N ( r ) l (1 ; r ) l dr . Z 1 0 N ( r ) p (1 ; r ) p dr < + 1 :  5.3. Finally wemake some remarks ab out prop erties of the zero sequences of functions in A p ( D ) and B p ( D ). It is immediate from the characterization in Theorem 5.1 that those of A p ( D ) are hyperbolical ly stable ,in the sense that whenever f a n g is such a sequence, and     a n ; b n 1 ; a n b n     < 0 < < 1 then f b n g is also a zero sequence of A p ( D ). It is also clear that a subsequence of a zero sequence of A p ( D ) is also a zero sequence of A p ( D ). Even though the characterization of those of B p ( D ) is unknown, this later prop erty can still b e proved using an idea from Lue3]. Theorem 5.3. Every subsequence of a zero sequence of B p ( D ) is also a zero sequence of B p ( D ) . Proof. We will show that if  is a positive Borel measure on D for which there is u , u + 2 L p ( D ), with  u =  , and 0     , then there is v suchthat  v =  ON L p SOLUTIONS TO THE LAPLACE EQUATION 19 and v + 2 L p ( D ). Recall from subsection 4.1 the decomp osition v = P  v ]+ K  v ], in particular v (0) = 1  Z D v (  ) dA (  )+ 1 2  Z D f log j  j 2 +1 ;j  j 2 g  v (  ) dA (  ) : Fixed z 2 D , we apply this to v = u  ' z , where ' z is the automorphism of D permuting z with 0, and change variables in the integrals. The result is u ( z )= Q  u ]( z )+ 1 2  Z D ( log      ; z 1 ; z     2 + (1 ;j  j 2 )(1 ;j z j 2 ) j 1 ; z j 2 ) d (  ) with Q  u ]( z )= 1  Z D  1 ;j z j 2 j 1 ; z j 2  2 u (  ) dA (  ) : Subtracting from u = P  u ]+ K  u ], and using (4), wesee that Q  u ]( z )= P  u ]( z )+ Z D j z j 2  1 ;j  j 2 j 1 ; z j  2 d (  ) : We takeas solution of  v =  the function v = K   ]+ P  u ]. From (4), v ( z ) . Z D j z j 2  1 ;j  j 2 j 1 ; z j  2 d (  )+ P  u ] and therefore v ( z ) . Q  u ]( z ). Then v + ( z ) . 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Ortega-Cerda: Dept. Matema  tica Aplicada I, Universitat Polit  ecnica de Catalunya, Diagonal 647, 08028 Barcelona, Spain E-mail address : [email protected]