. DERIVATIVES OF HARMONIC BERGMAN AND BLOCH FUNCTIONS ON THE BALL

Transcription

1 . DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI Abstract. On the setting of the unit ball of euclidean n-space, we investigate properties of derivatives of functions in the harmonic ergman space and the harmonic loch space. Our results are () size estimates of derivatives of the harmonic ergman kernel, (2) Gleason's problem, and (3) characterizations in terms of radial, tangential and ordinary derivative norms. In the course of proofs, some reproducing formulas are found and estimated.. Introduction For a xed positive integer n 2, let be the open unit ball in R n. The harmonic ergman space b p, p <, is the space of all harmonic functions f on such that jjfjj p = jfj p dv =p < where V is the volume measure on. The space b p is a closed subspace of L p = L p (; dv ) and thus a anach space. The harmonic loch space is the space of harmonic functions f on with the property that the function ( jxj 2 )jrf(x)j is bounded on. The space is also a anach space equipped with norm jjfjj = jf()j + sup( jxj 2 )jrf(x)j: x2 The harmonic little loch space is the space of harmonic functions f 2 with the additional property that ( jxj 2 )jrf(x)j is vanishing The main results of this paper are Theorems.,.2,.3 and.4 below. For a given multi-index = ( ; :::; n ) with each j a nonnegative integer, we use notations jj = + + n, x = x xn n = n j denotes the dierentiation with respect to j-th variable. We sometimes attach a variable subscript to indicate the specic variable with respect to which dierentiation is to be taken. Theorem.. Let R(x; y) denote the harmonic ergman kernel for. Given multi-indices and, there exists a positive constant C = C(; ) such that for all x; y 2. j@ x y R(x; y)j ( 2x y + jxj 2 jyj 2 ) (n+jj+jj)=2 It is the above estimate which enables us to obtain most of the following results. Key words and phrases. Harmonic ergman and loch functions, Gleason's problem, Derivative Norms. This research is partially supported by the Korea Research Foundation Grant (KRF DI5).

2 2 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI Theorem.2. For any positive integer m and multi-indices with jj = m, there exist linear operators A on harmonic functions in with the following property: () If f is harmonic on and all its partial derivatives up to order m vanish at, then f(x) = x A f(x) (x 2 ): jj=m (2) Each A is bounded on b p for p <. (3) Each A is bounded on and. We let Df(x) = x rf(x) denote the radial derivative of f. Also, we let T f denotes tangential derivatives of f (see Section 5). In terms of derivatives, the harmonic ergman spaces are characterized in the following way. Theorem.3. Let p < and m be a positive integer. Then, for a function f harmonic on, the following conditions are equivalent. () f 2 b p. (2) ( jxj 2 ) m D m f 2 L p. (3) ( jxj 2 ) m T f 2 L p for all with jj = m. (4) ( jxj 2 ) f 2 L p for all with jj = m. As a companion result for the harmonic loch space, we have the following. Here, C = C (). Theorem.4. Let m be a positive integer. Then, for a function f harmonic on, the following conditions are equivalent. () f 2 ( resp:). (2) ( jxj 2 ) m D m f 2 L (C resp:). (3) ( jxj 2 ) m T f 2 L (C resp:) for all with jj = m. (4) ( jxj 2 ) f 2 L (C resp:) for all with jj = m. It seems worth to compare and contrast Theorem.3 and Theorem.4 with known results for ergman spaces and loch spaces of holomorphic functions. Analogous results (only for ordinary derivative norms) for the holomorphic case are proved in [8]. A signicant dierence that we wish to point out here is that tangential and radial growth of harmonic ergman and loch functions are of the same order; this is not the case for holomorphic functions where additional smoothness occurs in the complex tangential directions (see Section 6.4 of [4] or [7]). In Section 2 we rst estimate sizes of derivatives of the harmonic ergman kernel. Then we obtain a couple of estimates which show how integrals of the harmonic ergman kernel behave near boundary. In Section 3 we prove a couple of reproducing formulas and estimate them as a preliminary step towards results in later sections. In Section 4 we give proofs of solutions to Gleason's problem. ased on the estimates in the previous section, the proofs are quite simple. Finally, in Section 5, we prove characterizations in terms of derivatives. The radial, tangential and ordinary derivatives are all considered. We introduce corresponding norms and show that all the norms are equivalent. Here, we employee a direct approach to estimate norms, which are quite dierent from previous ones used in some other settings. Constants. Throughout the paper we will use the same letter C to denote various constants which may change at each occurrence. For nonnegative quantities A and

3 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 3, we will often write A. if A is dominated by times some inessential positive constant. Also, we write A if A. and. A. 2. Kernel Estimates y the mean value property of harmonic functions, it is easily seen that point evaluations are continuous on b 2. Thus, to each x 2, there corresponds a unique R(x; ) 2 b 2 such that (2.) f(x) = f(y)r(x; y) dy for functions f 2 b 2 and thus for f 2 b, because b 2 is dense in b. Here and elsewhere, we let dy = dv (y). It is well known that the kernel function is real and hence the complex conjugation in the integral of (2.) can be removed. The explicit formula of the kernel function is also well known (see, for example, []): ( jxj 2 jyj 2 2 ) R(x; y) = nv () n n 4jxj 2 jyj 2 (x; y) (x; y) where Since jxkyj (x; y), it is clear that (x; y) = p 2x y + jxj 2 jyj 2 : (2.2) jr(x; y)j. n (x; y) for all x; y 2. The key step to our results is the optimal size estimates of derivatives of the reproducing kernel in terms of. Note that a standard argument using Cauchy's estimates cannot be directly applied. We rst prove a lemma. Lemma 2.. Given multi-indices,, and >, there exists a positive constant C = C(; ; ) such y (x; y) C +jj+jj (x; y) whenever x; y 2, jx yj ( jxj)=2 and jx yj ( jyj)=2. Proof. Fix x; y 2 such that jx yj ( jxj)=2 and jx yj ( jyj)=2. We let = (x; y). Note (2.3) Clearly we have (2.4) 2 = ( jxj 2 )( jyj 2 ) + jx yj 2 : j@ y 2 2 jj jj j.. for jj + jj 2. Also, note r x 2 = 2y + 2xjyj 2 and r y 2 = 2x + 2yjxj 2. Thus, jr x 2 j + jr y 2 j. jx yj + ( jxj) + ( jyj). so that (2.4) remains true for all and. First, we prove the lemma for = 2. We prove by induction on jj + jj. There is nothing to prove for jj + jj =. Let jj + jj = k and assume the lemma is

4 4 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI true for all, with j j + j j k. j j;x j;y Then, by (2.4) and induction hypothesis, we have j y j 2 x 2 y. j j j j 2 j j j j 2 j j j j 3 k where the sum is taken over all,,,,, such that j j+j j+j j = jj and j j + j j + j j = jj. This completes the proof for = 2. For general >, j y j 2 x y : Thus, using the result for = 2, one may complete the proof by induction as above. The following lemma is a consequence of the mean value property of harmonic functions and Cauchy's estimate. For details, see Corollary 8.2 of []. Lemma 2.2. Let p < and be a multi-index. Suppose f is harmonic on a proper open subset of R n. Then, we have j@ f(x)j p C d n+pjj jf(y)j p dy (x 2 ) where denotes the distance from x The constant C depends only on n, p, and. We are now ready to prove the following size estimates of derivatives of the reproducing kernel. Theorem 2.3. Given multi-indices and, there exists a positive constant C = C(; ) such that for all x; y 2. j@ x y R(x; y)j n+jj+jj (x; y) Proof. Note that jr(x; y)j 2 dy = R(x; x) ( jxj) n for all x 2. Thus, given a multi-index, we have j@ y R(x; y)j2. ( jxj) n ( jyj) n+2jj for all x; y 2 by Lemma 2.2. Given another multi-index, applying a standard argument using Cauchy's estimate, we conclude from the above (2.5) j@ y R(x; y)j. ( jxj) n=2+jj ( jyj) n=2+jj

5 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 5 for all x; y 2. Thus, for the case where jx yj < ( jxj)=2 or jx yj < ( jyj)=2, we have (x; y). jxj jyj by (2.3) and therefore obtain the desired estimate from (2.5). Now, assume jx yj ( jxj)=2 and jx yj ( jyj)=2. Note R(x; y) = C (x; y) (n+2) (x; y) + C 2 jxj 2 jyj 2 n (x; y) where (x; y) = ( jxj 2 jyj 2 ) 2. Let = (x; y) and = (x; y). Here, we will estimate the rst term. The estimate of the second term is similar and simpler. y Lemma 2. we have y ( n y x y y n 2 j j j j where the sum is taken over all,,, such that j j + j j = jj and j j + j j = jj. For j j + j j 2, we have (2.7) j@ y j.. 2 j j j j : Note r x = 4xjyj 2 ( jxj 2 jyj 2 ) and r y = 4yjxj 2 ( jxj 2 jyj 2 ). Thus, jr x j + jr y j. jxjjyj = jxj + jxj( jyj). jx yj so that (2.7) remains valid for all,. Therefore, by (2.6) and (2.7), we y ( n 2 ). n j j j j j j j j n jj jj ; which completes the proof. In conjunction with Theorem 2.3, the following two propositions describe integral behavior of derivatives of the reproducing kernel. Here and in the rest of the paper, we put (x) = jxj 2 (x 2 ) for simplicity. Also, we use the notation d for the surface area measure Proposition 2.4. Given c real, dene I c (x) = for x 2. Then the following hold. dy n+c (x; y) () For c <, I c is bounded on. (2) For c =, ( + log ) I is bounded on. (3) For c >, c I c is bounded on. Proof. Here, we give a proof for c (which are the cases we need later). The case c < is easily modied and left to the readers. Assume c. Recall that the integral of the Poisson kernel is constant. That jy jyj 2 d() = (@) jn

6 6 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI for all y 2. Hence, integrating in polar coordinates, we have so that I c (x) = = t n ( 2tx + t 2 jxj 2 ) (n+c)=2 t n d() jtx jn+c t n t I c (x) ( t 2 jxj 2 )( 2 jxj 2 d() c dt jtx jn t n = (@) ( t 2 jxj 2 )( tjxj) c dt. dt ( tjxj) +c for x 2. The rest of the proof is now straightforward. Proposition 2.5. Given s >, there exists a constant C = C(s) such that for all x; y 2. dt s (tx; y) C s (x; y) Proof. Fix x; y 2. Note that if x y or jxjjyj < =2, then all the terms are bounded above and bounded away from. Thus the estimate is trivial. So suppose x y > and jxjjyj =2 in the rest of the proof. Dene h(t) = (tx; y) for t and put t = x y=jxj 2 jyj 2 >. Then the function h attains its minimum at t with minimum value p (x y) 2 =jxj 2 jyj 2. Note that if t <, then (x y) 2 =jxj 2 jyj 2 < x y and so 2h 2 (t ) h 2 () 2( x y) ( 2x y + jxj 2 jyj 2 ) = jxj 2 jyj 2 : Thus, h(t ) h()= p 2. Now, since jxjjyj =2, we have and thus h 2 (t) = jxj 2 jyj 2 (t t ) 2 + h 2 (t ) & (t t ) 2 + h 2 () dt h s (t). h s () dt (t t ) 2 + h 2 () s=2 dt t 2 + h 2 () s=2 dt (t 2 + ) s=2 : Note that the last integral is nite for s >. Note also that if t, then it is elementary to see 2h 2 (t) jxj 2 jyj 2 (t ) 2 + h 2 () & (t ) 2 + h 2 () for all t 2 [; ], because jxjjyj =2. Hence, a similar argument yields the desired estimate.

7 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 7 3. Reproducing Formulas y the reproducing formula (2.), the orthogonal projection Q : L 2! b 2 is given by Qg(x) = g(y)r(x; y) dy (x 2 ) for g 2 L 2. Note that Q naturally extends to an operator from L into the space of all harmonic functions on. In this section, as a preliminary step towards our results in later sections, we rst prove that there are many other (nonorthogonal) projections which can be easily estimated by means of results in the previous section. Lemma 3.. Given a positive integer m, there are constants c j c m = ( )m+ m!, and c jk = c jk (m) such that m ' dv = c j j D j ' dv + m m c jk k= j+k D k ' dv: = c j (m) with whenever 2 b and ' is a function harmonic on an open set containing. Proof. Suppose ' and are functions harmonic on an open set containing. y Green's theorem, we have m ( 2m ') dv = [ ][ m ( 2m ')] dv D m ( 2m ') [D ][ m ( 2m ')] d: Clearly, the rst term of the right side of the above is. One can check that the remaining terms are also, because D m ( 2m ') and m ( 2m ') both vanish It follows that (3.) m ( 2m ') dv = : Note that, since harmonic polynomials are dense in b, this remains valid for general 2 b. We now calculate m ( 2m '). For an integer k 2, a straightforward calculation shows that r k (x) = 2kx k (x) and Hence, we obtain k = 2k(n + 2k 2) k + 4k(k ) k 2 : ( k ') = ( k )' + 2r k r' = 4k(k ) k 2 ' 4k k D' 2k(n + 2k 2) k ': Let k = 2m and apply the Laplacian m-times using this formula recursively. Since radial derivatives of harmonic functions are again harmonic, the result is m ( 2m ') = 4 m (2m)!' + m m m c j j D j ' + c jk j+k D k ': k= for some constants c j and c jk depending only on n and m. Note that c m = ( 4) m (2m)! m!. This, together with (3.), proves the lemma.

8 8 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI For a given positive integer m, let c j and c jk be the constants provided by Lemma 3. and dene an operator T m by m m m T m g = c j j D j g + c jk j+k D k g k= for g 2 C m (). For x 2, let R x = R(x; ), put R m (x; y) = T m R x (y), and consider an operator Q m dened by Q m g(x) = g(y)r m (x; y) dy (x 2 ) for g 2 L. Note that Q m is a linear operator taking L into the space of all harmonic functions on. It is well known that Q is bounded on L p if and only if < p <. The advantage of Q m is the boundedness on L p for all p <. Theorem 3.2. Let p < and m be a positive integer. Then Q m : L p! b p is bounded. Moreover, Q m f = f for f 2 b p. Proof. Given f 2 b p and x 2, apply Lemma 3. with = f and ' = R(x; ). Then, by the reproducing formula (2.), we obtain Q m f = f. Now, we show the L p -boundedness of Q m. Let 2 L p. Then jq P m j. m j= j j j where (x) = j (x) = j (y)jjr(x; y)j(y) dy j (y)jjd j y R(x; y)jj (y) dy (j = ; :::; m) for x 2. First, consider p =. y (2.2) and Proposition 2.4, we have (x) dx (y)j (y)j jr(x; y)j dx dy. j (y)j dy: A similar argument using Theorem 2.3 and Proposition 2.4 yields L -boundedness of each j. Now, assume < p <. y Theorem 2.3, we have j (x). j (y)j j (y) n+j (x; y) dy. j (y)j n (x; y) dy for all x 2 and j = ; ; :::; m. Hence, by the Schur test as in [5], each j is bounded on L p. As another consequence of Lemma 3., we have the following reproducing formula for harmonic loch functions. This reproducing formula will play an essential role in the estimates of harmonic loch functions in the next section. For m =, one can nd in [5] another proof by means of the extended Poisson kernel. Theorem 3.3. Let m be a positive integer. Then T m :! L is bounded. Moreover, T m takes into C and QT m f = f for f 2. It is well known (and not hard to prove) that Q : L! bounded and Q(C ). Note that the above theorem yields Q(L ) = and Q(C ) =, which are also well known.

9 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 9 Proof. Let f 2. For m =, we have T f = Df + (n=2 + )f. It is not hard to see that (3.2) jfj. jjfjj ( + log ): So, T f 2 L and jjt fjj. jjfjj. Now, given x 2, apply Lemma 3. with = R(x; ) and ' = f r, < r <, where f r is a dilate dened by f r (y) = f(ry) for y 2. Then, by the reproducing formula (2.), we obtain QT f r = f r. Note Df r = (Df) r! Df and jdf r j jjdfjj. Thus, Df r! Df in L by the Lebesgue dominated convergence theorem. Similarly, f r! f in L. Therefore, after taking the limit, we have QT f = f, or more explicitly, f(x) = T f(y)r(x; y) dy (x 2 ) and thus, dierentiating under the integral, we have f(x) = T f(y)@ x R(x; y) dy for every multi-index and x 2. Thus, for jj, we obtain by Theorem 2.3 and Proposition 2.4 jj (x)j@ f(x)j. jjfjj jj (x) dy n+jj (x; y). jjfjj for all x 2. It follows that jj k D k fjj. jjfjj for every positive integer k and therefore jjt m fjj. jjfjj. In particular, the integral QT m f is well dened. One can check D k f r = (D k f) r for each k. Thus, by the same limiting argument, we have QT m f = f for general m. Now, we prove T m ( ) C. Suppose f 2. Since f 2, we have Df 2 C. Also, f 2 C by (3.2). Thus, T f 2 C. Note that, for each < r < and with jj, we have by Proposition 2.4 and (3.3) jj (x)j@ f(x)j. jj (x) jyj>r jt f(y)j n+jj (x; y) dy + jj (x). sup jt f(y)j + jjfjj jj (x) jyj>r jyjr Now, take the limit jxj! with r xed and get jyjr dy n+jj (x; y) : lim sup jj (x)j@ f(x)j. sup jt f(y)j: jxj! jyj>r jt f(y)j n+jj (x; y) dy Since r is arbitrary and T f 2 C, we obtain f 2 C for jj and hence j D j f 2 C for each j. Consequently, since jt m fj. jfj + P m jj D j fj, we conclude T m f 2 C. 4. Gleason's Problem Let f 2 C (). Then, for x 2, we have (4.) f(x) f() = n x rf(tx) dt = x j A j f(x)

10 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI where A j f(x) j f(tx) dt: Note that if f is harmonic, then so is each A j f. Gleason's problem is to gure out whether the operators A j leave the spaces under consideration invariant. Recently the same problem has been considered on the setting of the upper half-space by Choe-Koo-Yi [2]. Their result reveals some pathology caused by unboundedness of the half-space. Also, see [8] for more references in this direction. For the ball, operators A j are expected to behave well, which turns out to be indeed the case. ased on all the estimates in the previous sections, proofs are also quite simple. Theorem 4.. Let p <. The operators A j are all bounded on b p. Proof. First, consider the case < p <. Let f 2 b p. Dierentiating under the integral sign of the reproducing formula (2.), we have and thus by Theorem 2.3 j@ j j f(x) = (x; j (x; y) dy dy. jf(y)j n+ (x; y) dy for x 2. It follows from Proposition 2.5 that dt ja j f(x)j. jf(y)j n+ (tx; y) dy. jf(y)j n (x; y) dy: for each j and x 2. Thus, by the Schur test as in [5], we get the boundedness of A j on b p. Next, consider the case p = and let f 2 b. This time we use the reproducing formula given by Theorem 3.2 with m =. Then, by a similar argument using Proposition 2.5 and Theorem 2.3, we obtain ja j f(x)j. (y)jf(y)j n (x; y) + n+ (x; y) for all j and x 2. Integrating both sides of the above, we obtain from Proposition 2.4 that ja j f(x)j dx. (y)jf(y)j n (x; y) + n+ dx dy (x; y). jf(y)j dy as desired. The analogous result is valid for the harmonic loch space. Theorem 4.2. The operators A j are all bounded on. In addition, Each A j takes into itself. Proof. Now suppose f 2 and let x 2. Then, by Theorem 3.3, we have f(x) = T f(y)r(x; y) dy: dy

11 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL Now, dierentiating under the integral, we have A j f(x) = T (tx; y) dy j for each j. Recall jjt fjj. jjfjj. It follows from Theorem 2.3 and Proposition 2.5 that jra j f(x)j. jjfjj. jjfjj. jjfjj ( jxj 2 ) : In other words, jja j fjj. jjfjj. For f 2, we have by Proposition 2.4 t dt n+2 (tx; y) dy dy n+ (x; y) jt jra j f(x)j. f(y)j jyj>r n+ (x; y) jyjr dy + jt f(y)j n+ (x; y) dy for < r <. Thus, the same argument as in the proof of Theorem 3.3 gives jra j fj 2 C so that A j f 2. Now, repeating the results of Theorem 4. and Theorem 4.2, we can prove Theorem.2. Proof of Theorem.2: For f harmonic on, a repetition of (4.) yields f(x) = f() + = f() + n n x j A j f(x) A j f()x j + n n k= x j x k A k A j f(x) = f() + rf() x + x j x k A k A j f(x) k= for x 2. The boundedness properties of operators A k A j follow from Theorem 4. and Theorem 4.2. For higher orders, one may repeat the same argument. n n 5. Derivative Norms In this section we prove the equivalence of various derivative norms. We will consider radial, tangential and ordinary derivative norms. For the half-space, such results are proved in [3]. For the holomorphic ergman spaces on the ball, such results (only for ordinary derivative norms) are proved in [8]. Our approach is direct and quite dierent from theirs. Since there is no smooth nonvanishing tangential vector eld n for n > 2, we dene tangential derivatives by means of a family of tangential vector elds generating all the tangent vectors. We dene T ij f of f 2 C () by T ij f(x) = (x j x i )f(x) (x 2 ) for i < j n. Note that tangential derivatives of harmonic functions are again harmonic. Given a nontrivial multi-index, we abuse the notation T = T i j T n i nj n for any choice of i ; :::; i n and j ; :::; j n.

12 2 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI Now, we introduce corresponding norms. For p < and positive integers m, put jjfjj p;m; = jf()j + jj m D m fjj p jjfjj p;m;2 = jf()j + jj m T fjj p jj=m jjfjj p;m;3 = j@ f()j + jj fjj p jj<m jj=m for functions f harmonic on. Our result is that all of these norms are equivalent. Estimates are somehow long and thus we proceed step by step through lemmas. Lemma 5.. Given an integer m, there exists a smooth dierential operator E m of order 2m with bounded coecients such that D 2m f = for functions f harmonic T 2 ij i<j Proof. Since D 2 = P x i x j + K, we have i<j A m f + E m f T 2 ij = i<j(x 2 j + x 2 i 2x i x j ) + K 2 = i6=j (x 2 j x i x j ) + K 2 = i;j (x 2 j x i x j ) + K 2 = jxj 2 D 2 + K + K 2 : Here, K and K 2 are rst order smooth dierential operators with bounded coecients. i<j T 2 ij A m = ( ) m D 2m + K 3 + K 4 for some K 3 of order 2m 2 and K 4 of order 2m. This implies the lemma. (5.) The following is an easy consequence of the mean value property of rf: sup jf(x) jxjr f()j p C jxjr+ jrf(x)j p dx for p <, >, < r <, and functions f harmonic on. The constant C is independent of f and r. The analogous inequalities for radial and tangential derivatives do not seem to be trivial. Recall that radial and tangential derivatives of harmonic functions are again harmonic. Proposition 5.2. Let p <, > and m be a positive integer. Then there exists a constant C = C(p; m; ) such that () sup jf(x) jxjr f()j p C jxjr+ jd m f(x)j p dx

13 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 3 (2) sup jf(x) jxjr f()j p C jj=m jxjr+ whenever < r < and f is harmonic on. Proof. Suppose < r < 2.2, jt f(x)j p dx and let f be a function harmonic on. y Lemma sup jf(x) f()j 2. n jf(x) f()j 2 dx jxjr jxjr+=2 n jxjr+=2. sup jdf(x)j 2 jxjr+=2 jdf(x)j 2 dx where the second inequality is easily veried by using the homogeneous expansion of f (see Theorem 5.3 of []). Thus, sup jf(x) f()j p. sup jdf(x)j p. jxjr jxjr+=2 jxjr+ jdf(x)j p dx This proves the lemma for m =. For general m, one can repeat the same process with harmonic functions D j f. Since D j f() = for j, we get (). We now prove (2). Let S r denote the sphere of radius r centered at the origin. Now, x x 2 S r, and pick a smooth curve : [; ]! S r such that () = x, (t ) = r for some t and j j = 2r. We claim (5.2) j(f ) (t)j. i<j jt ij f (t) j for all t 2 [; ]. Given t 2 [; ], let (t) = r where Since jj =, we may assume j j = p n. Let e j = e j j e, 2 j n, where e k denotes the unit vector in the positive direction of the k-th coordinate axis. Note that fe j g j2 is a basis for the tangent space T at. Following the Gram-Schmidt process, put a 2 = e 2 and dene (5.3) a j = e j j k=2 a k e j a k a k a k inductively for j = 3; :::; n. Then fa j g j2 is an orthogonal basis for T. What we need here is a uniform control of coecients. Since each a k in the sum of the above is spanned by e 2 ; :::; e k by construction, the j-th coordinate of a j comes only from e j and is equal to. Thus, ja j j = p n for each j. Now, writing (5.3) in the form a j = P j k=2 c jke k ; one may check inductively jc jk j < C for some C which depends only on n. Now, since (t) 2 T, we have (t) = n j=2 d j ja j j j n n c jk e k = k=2 k=2 j=k d j c jk ja j j e k for some constants d j such that P d 2 j = j (t)j 2 = (2r) 2. Since rf (t) e k = r T k f (t), it follows that n jrf (t) (t)j k=2 n j=k jd j jjc jk j rja j j jt k f (t) j. n k=2 jt k f (t) j;

14 4 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI which implies (5.2). Now, by (5.2), we have and therefore jf(x) f(r)j p jf(x) f()j t jf(x) j(f ) (t)j p dt. sup S r i<j f(r)j p d(). sup S r by subharmonicity. On the other hand, by Lemma 2.2 sup jt ij fj p. n jt ij f(y)j S r jyjr+ p dy jt ij fj p i<j jt ij fj p for all T ij. This proves the lemma for m = by the maximum principle. For general m, one can repeat the same process with harmonic functions T f. Since T f() = for jj, we obtain (2). Lemma 5.3. For p and r >, we have a p jh(t)j p t r p a dt jth (t)j p t r dt + jh(a)j p r + for C -functions h on (; a], a >. Proof. Suppose h is a C -function on (; a]. Then we have jh(t)j jh(a)j + a t jh (s)j ds for < t a. Thus, the inequality follows from Hardy's inequality (see, for example, [6]). We are now ready to prove norm equivalence for the harmonic ergman spaces. Theorem 5.4. Let p < and m be a positive integer. Then, there are positive constants C, C 2, C 3, C 4 such that jjfjj p C jjfjj p;m; C 2 jjfjj p;m;2 C 3 jjfjj p;m;3 C 4 jjfjj p for functions f harmonic on. In the proof below f is a given harmonic function on. Proof of jjfjj p;m;3. jjfjj p : Given x 2, take to be the ball x with center at x and radius ( jxj)=2. Note ( jxj)=2 jyj 3( jxj)=2 for y 2 x. Hence, for a multi-index and r real, we obtain by Lemma 2.2 (5.4) ( jxj) pjj+r j@ f(x)j p. ( jxj) n jf(y)j p ( jyj) r dy: x Inserting x = into the above, we have jj<m j@ f()j p. jfj p r dy:

15 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 5 Also, for jj = m, we obtain by (5.4) (5.5) jj=m j fj p r dv. =. ( jxj) n jf(y)j p r (y) x jf(y)j p r (y) dy dx jf(y)j p r (y) dy: x (y) dx dy ( jxj) n Consequently, taking r =, we have jjfjj p;m;3. jjfjj p. Proof of jjfjj p. jjfjj p;m; : For a given and r >, apply Lemma 5.3 to the function h(t) = f ( t) on (; =2]. What we get is and thus =2 =2 jf(t)j p ( t) r dt. jf(t)j p ( t 2 ) r t n dt. =2 =2 j rf(t)j p ( Integrating both sides of the above we have Note that jfj p r dv. jxj>=2 sup jf(x)j p. jf()j p + jxj=2 by Proposition 5.2. It follows that t) p+r dt + jf(=2)j p jdf(t)j p ( t 2 ) p+r dt + sup jf(x)j p : jx=2 jdfj p p+r dv + sup jf(x)j p : jxj>=2 jx=2 jxj3=4 jfj p r dv. jdf(x)j p dx. jf()j p + jdfj p p+r dv jdfj p p+r dv + jf()j p : Note Df() =. Thus, iterating the above with harmonic functions D j f, we have jfj p r dv. jd m fj p pm+r dv + jf()j p : So, taking r =, we have jjfjj p. jjfjj p;m;. Proof of jjfjj p;m;. jjfjj p;m;3 : This time we apply Lemma 5.3 on the interval (; ]. We obtain, for a given and r >, and thus jf(t)j p ( t) r dt. j rf(t)j p ( t) p+r dt + jf()j p jf(t)j p ( t 2 ) r t n dt. =2 jrf(t)j p ( t 2 ) p+r dt + sup jrf(x)j p + jf()j p : jxj=2

16 6 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI y using (5.) and proceeding in a way similar to the preceding proof, we obtain jfj p r dv. jrfj p p+r dv + jf()j p : Let j be an integer. Then, replacing f f, jj = j, and taking r = jp in the above, we have j fj p dv. j fj p dv + j@ f()j p jj=j jj=j+ jj=j and therefore jjfjj p;j;3. jjfjj p;j+;3. Consequently, jjfjj p;m;. P m jjfjj p;j;3. jjfjj p;m;3 as desired. Proof of jjfjj p jjfjj p;m;2 : So far, we've seen that jjfjj p jjfjj p;j; jjfjj p;j;3 for each j. Thus, it is clear that jjfjj p;m;2. P m jjfjj p;j;3 jjfjj p : For the other direction, let T ij be any tangential dierential operator and jj = 2m. Then, by (5.5) with r = p(2m ), we have j 2m T ij T fj p dv.. jrt fj p p(2m ) dv jt fj p p(2m ) dv: Accordingly, we have (5.6) jjfjj p;2m;2. jjfjj p;2m ;2 : Let E m be the dierential operator as in Lemma 5.. Then, we have (5.7) jjfjj p;2m;. jjfjj p;2m;2 + jj 2m E m fjj p by Lemma 5.. Let < a <. Then, by (5.5), we have jxj> a j 2m E m f(x)j p dx. a p Also, by Lemma 2.2 and Proposition 5.2, we have sup j 2m E m f(x)j C jxj a j 2m E m fj p dv. a p sup jxj a=2 for some constants C = C(a). Therefore, we have jf(x)j Cjjfjj p;2m;2 jfj p dv: (5.8) jj 2m E m fjj p C jjfjj p;2m;2 + C 2 ajjfjj p where C = C (a) and C 2 is independent of a. Therefore, since we already have jjfjj p jjfjj p;2m;, we have by (5.7) and (5.8) (5.9) jjfjj p C 3 jjfjj p;2m;2 + C 4 ajjfjj p where C 3 = C 3 (a) and C 4 is independent of a. Hence, taking a to be suciently small, we conclude from (5.6) and (5.9) that jjfjj p. jjfjj p;m;2.

17 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 7 We now turn to the norm equivalence for the harmonic loch spaces. For positive integers m, put jjfjj ;m; = jf()j + jj m D m fjj jjfjj ;m;2 = jf()j + jj m T fjj jj=m jjfjj ;m;3 = j@ f()j + jj fjj jj<m jj=m for functions f harmonic on. These norms also turn out to be equivalent. Theorem 5.5. Let m be a positive integer. Then, there are positive constants C, C 2, C 3, C 4 such that jjfjj C jjfjj ;m; C 2 jjfjj ;m;2 C 3 jjfjj ;m;3 C 4 jjfjj for functions f harmonic on. In the proof below f is a given harmonic function on. Proof of jjfjj jjfjj ;m; : The inequality jjfjj ;m;. jjfjj is implicit in the proof of Theorem 3.3. We now prove jjfjj. jjfjj ;m;. First, note by Theorem 3.3 jjfjj = jjqt m fjj. jjt m fjj. jjfjj + m jj j D j fjj : Hence, it is sucient to show that the rightmost side of the above is dominated by jjfjj ;m;. Let x 2 and jxj =2. We write x = jxj where First, note (5.) jf(=2)j. jf()j + sup jyj3=4 jdf(y)j by Proposition 5.2. Let j be an integer. Then, since we have by (5.) This yields which, in turn, yields jf(x) f(=2)j 2 jf(x)j. jf()j + jj j+ Dfjj ( + jxj jdf(t)j dt; =2 jxj ) dt ( t) j+ : sup j (x)jf(x)j. jf()j + jj j+ Dfjj ; jxj=2 jj j fjj. jf()j + jj j+ Dfjj : Taking j =, we obtain jjfjj. jf()j + jjdfjj. Also, for j < m, applying the above to D j f and iterating, we obtain jj j D j fjj. jj m D m fjj because D j f() =. We therefore conclude m jjfjj + jj j D j fjj. jjfjj ;m; which completes the proof.

18 8 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI Proof of jjfjj jjfjj ;m;3 : The inequality jjfjj ;m;3. jjfjj is implicit in the proof of Theorem 3.3. Next, we prove jjfjj. jjfjj ;m;3. For any multi-index with jj = j, we have (5.) and jxj j@ f(x)j j@ f()j + jxj jr@ f(tx=jxj)j dt dt ( t) j+. jjfjj ;j+;3 j (x) jr@ f(tx=jxj)j dt. jjfjj ;j+;3 jxj for all x 2. This yields jjfjj ;j;3. jjfjj ;j+;3 for each j. Hence, jjfjj jjfjj ;;3. jjfjj ;m;3 : as desired. Proof of jjfjj jjfjj ;m;2 : We already have jjfjj jjfjj ;j; jjfjj ;j;3 for each j. Thus, it is clear that jjfjj ;m;2. m jjfjj ;j;3 jjfjj : Imitating the proof of jjfjj p jjfjj p;m;2 of Theorem 5.4, we also have and jjfjj ;2m;2. jjfjj ;2m ;2 jjfjj jjfjj ;2m; C jjfjj ;2m;2 + C 2 ajjfjj where C = C (a) and C 2 is independent of a. Hence, taking a to be suciently small, we obtain jjfjj. jjfjj ;m;2. We now close the paper with the corresponding little-oh version. Theorem 5.6. Let m be a positive integer and f 2. Then, the following conditions are all equivalent. () f 2. (2) m D m f 2 C. (3) m T f 2 C for all with jj = m. (4) f 2 C for all with jj = m. Proof. The implication () =) (4) is implicit in the proof of Theorem 3.3. We show (4) =) (2). Assume f 2 C for jj = m. Consider such that jj = m. Then, by (5.), lim sup j m (x)@ f(x)j lim sup m (x) jxj! jxj! = lim sup jxj! m (x). sup m (y)jr@ f(y)j jyjr jxj jr@ f(tx=jxj)j dt jxj jr@ f(tx=jxj)j dt r

19 DERIVATIVES OF HARMONIC ERGMAN AND LOCH FUNCTIONS ON THE ALL 9 for every positive r <. Hence, taking the limit r!, we have f 2 C. Repeating the same, we obtain f 2 C for jj m and thus m D m f 2 C. We show (2) =) (). Assume m D m f 2 C. We rst show that j D j f 2 C for each j = ; :::; m. We only need consider the case j = m. So, assume m 2. Let x 2 and write x = jxj where Assume jxj > =2 and choose any r such that =2 < r < jxj. We have so that jxj jd m f(x) D m f(r)j r jd m f(t)j dt r! jxj. sup j m (y)d m dt f(y)j jyjr ( t) m. m+ (x) sup j m (y)d m f(y)j jyjr m (x)jd m f(x)j. m (x)jd m f(r)j + sup j m (y)d m f(y)j: jyjr Take the limit jxj! with r xed and get lim sup m (x)jd m f(x)j. sup j m (y)d m f(y)j: jxj! jyjr Since r > =2 is arbitrary and m D m f 2 C by assumption, the above yields m D m f 2 C, as desired. Now, since m m jt m fj. j j D j fj + j j D j fj j=. jfj + jjfjj + m. ( + log )jjfjj + j j D j fj m j j D j fj; we have T m f 2 C. We conclude f = QT m f 2 by Theorem 3.3, because Q maps C onto. Since we already have () () (4), the implication () =) (3) is clear. We now prove (3) =) (). Suppose (3) holds. y (5.4) (with p = jj = ; r = 2m ), we have 2m (x)jt ij T f(x)j. sup y2 x 2m (y)jt f(y)j for all x 2, T ij and with jj = 2m. This yields lim sup jxj! 2m (x) jj=2m jt f(x)j. lim sup jxj! 2m (x) jj=2m! jt f(x)j and thus, without loss of generality, we may assume 2m P jj=2m jt fj 2 C. Let E m be the dierential operator as in Lemma 5.. Since j 2m E m fj. jjfjj, we have 2m E m f 2 C and therefore 2m D 2m f 2 C by Lemma 5.. Since () () (2), it follows that f 2.

20 2 OO RIM CHOE, HYUNGWOON KOO, AND HEUNGSU YI References [] S. Axler, P. ourdon and W. Ramey, Harmonic function theory, Springer-Verlag, New York, 992. [2]. R. Choe, H. Koo, and H. Yi, Gleason's problem for harmonic ergman and loch functions on half-spaces, Integral Equations and Operator Theory, 36(2), 269{287. [3] W. Ramey and H. Yi, Harmonic ergman functions on half-spaces, Trans. Amer. Math. Soc. 348(996), 633{66. [4] W. Rudin, Function theory in the unit ball of C n, Springer-Verlag, New York, 98. [5] K. Stroetho, Harmonic ergman functions, Holomorphic spaces, MSRI Publications 33(998), 5{64. [6] E. Stein and G. Weiss, Fourier Analysis on Euclidean Spaces, Princeton University Press, Princeton, NJ, 97. [7] R. Timoney, loch functions in several complex variables I, ull. London Math. Soc. 2(98), 24{267. [8] K. hu, The ergman spaces, the loch space and Gleason's problem, Trans. Amer. Math. Soc. 39(988), 253{268. Department of Mathematics, Korea University, Seoul 36{7, Korea address: choebr@math.korea.ac.kr Department of Mathematics, Korea University, Seoul 36{7, Korea address: koohw@math.korea.ac.kr Department of Mathematics, Research Institute of asic Sciences, Kwangwoon University, Seoul 39{7, Korea address: hsyi@math.kwangwoon.ac.kr