HARDY SPACES AND PARTIAL DERIVATIVES OF CONJUGATE HARMONIC FUNCTIONS

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1 HARDY SPACES AND PARTIAL DERIVATIVES OF CONJUGATE HARMONIC FUNCTIONS ANATOLI RIABOGUINE AND DMITRY RYABOGIN Abstract. In this paper we give necessary and sufficient conditions for a harmonic vector and all its partial derivatives to belong to H p ( ) for all p >. We also obtain the Hardy-Littlewood-Sobolev imbedding-type result, formulated on the language of the conjugate harmonic functions.. Introduction and statements of main results Let f(z) be an analytic function in the unit disc D = {z : z < } and let () M p (r, f) = ( 2π 2π f(re iθ ) p dθ) /p, p >, r <. It is well-known (see, for example, [7], [8], and references therein) that f H p (D), p >, if M p (r, f) C <. It is also well-known that f H p (D) if and only if a subharmonic function f(z) p has a harmonic majorant. Using the Riemann conformal mapping theorem one can define the Hardy space on any domain U having more than one point in its boundary. An analytic function in a domain U belongs to H p (U), p >, if and only if a subharmonic function f(z) p has a harmonic majorant in U. In particular, this is true in the case of the upper half-plane domain U = R 2 = {z : Im z > }. On the other hand, if we want to define H p (R 2 ) in terms of the integral everages of type (), one has to take into account a weight appearing after a change of variables in the integral. This is one of the reasons for an appearence of another natural class of functions, h p (R 2 ), [6]. An analytic function f h p (R 2 ), p >, if M p (y, f) = ( f(x iy) p dx) /p C <. It is clear that not every bounded analytic function in the upper half-plane belongs to h p (R 2 ). Moreover, the above condition is only sufficient for the existence of a harmonic majorant for f(x iy) p, and analytic functions from h p (R 2 ) loose several important properties that analytic functions from H p (D) have. For example, if f H p (D), then f H q (D) for all < q < p. For f h p (R 2 ), the statement f H q (R 2 ), < q < p, is no longer true without additional assumptions on f. 99 Mathematics Subject Classification. Primary 3E25, secondary 42B25. Key words and phrases. Hardy spaces, subharmonic functions.

2 2 ANATOLI RIABOGUINE AND DMITRY RYABOGIN The reason for this is very simple, f is bounded in z r, < r <, and f H q ( z r ), for all q >, since M p (r, f) C(r ), < r r. This is why the class S p (R 2 ), p >, comes into play, [7], []. An analytic function f S p (R 2 ), p >, if for any y > there exists a constant C(y, f), such that M p (y, f) C(y ) y y. In particular, if C is independent of y, then f(x iy) h p (R 2 ). Now, if f S p (R 2 ), p >, and y, then f(x iy) converges to zero uniformly with respect to x, and f S q (R 2 ) q p. But again, the condition f S p (R 2 ) does not imply f S r (R 2 ), < r p. Consider, for example, f(x iy) = u(x, y) iv(x, y), where x u(x, y) = x 2 y 2, v(x, y) = y x 2 y 2. We have f S p (R 2 ), p >, but f / S q (R 2 ), < q. Thus, in the case of the upper half-plane, one can consider different problems concerning classes H p (R 2 ), h p (R 2 ), S p (R 2 ). In particular, we have h p (R 2 ) S p (R 2 ), and h p (R 2 ) Hp (R 2 ). In the case of the half-space, (, ), classes h p ( ), S p ( ), (all definitions are given in Section 2), were considered by Solomentsev [], and by Stein and Weiss, [3], [5]. One of the results, proved in [3] reads as follows: if F = (U(x, y), V (x, y), V 2 (x, y),..., V n (x, y)) is a harmonic vector, (x, y), then F p is subharmonic, provided p (n )/n. This leads to additional technical difficulties for p < (n )/n. Finally, Fefferman and Stein [4], introduced the classes of gradients H p ( ), p >, of k F, and showed that k F p is subharmonic provided p (n )/(n k ), k N. They also proved that H p ( ) = h p ( ) for p (n )/n. Nevertheless, Wolff [6] showed that H p ( ) h p ( ) for p < (n )/n. In this article we find new relations between the conjugate harmonic functions in and their partial derivatives. We study the following problem: what can we say about conjugate harmonic functions, provided we are given certain restrictions, imposed on partial derivatives of a harmonic vector F = (U, V, V 2,..., V n ). We give necessary and sufficient conditions for a harmonic vector and all its partial derivatives up to the order k to belong to H p ( ), p >. Our first results are Theorem. Let k N, p >. The harmonic vector F = (U, V,..., V n ) and all its partial derivatives of the order k belong to H p if and only if ( (2) ) M p (y, F ) C, 2) sup DnU(x, η) ) k C. η y Theorem 2. Let < p < q. The harmonic vector F = (U, V,..., V n ) and all its partial derivatives of the order k belong to H r, p r q, if and only if ( qdx (3) ) M p (y, F ) C, 2) sup DnU(x, η) ) k C. η y Conditions (2) allow to obtain the Hardy-Littlewood-Sobolev imbedding-type result, formulated on the language of the conjugate harmonic functions. Results of

3 H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 3 this type were obtained by Fefferman-Stein, [4], Flett [5], and others (see [2] and references therein). For us the case of small p > is of special interest. Theorem 3. Let k N, p >, and let F be as in Theorem. a) If < kp < n, then F H r r : p r np/(n kp). If kp n, then F H r r p. b) Let k >, m k, m N. If < (k m)p < n, then all partial derivatives of F of the order m belong to H r r : p r np/(n (k m)p). If (k m)p n, then all partial derivatives of F of the order m belong to H r r p. One of the methods of the proof of Theorems and 2 is the application of classes S p ( ) together with the Lagrange mean-value Theorem. We note that the first condition in (2) is natural not only because of the decomposition of the function into two parts, S p ( ) and H p ( ) (see Section 5). In fact, it (together with the second condition) implies M p (y y, F ) C(y ) y >. Moreover, the next result shows that it is necessary. Theorem 4. Let p >, and let F be a harmonic vector such that ) F x y, 2) M p(y, D n U) C, 3) D n U(x, y) C. Then a) F H r, r > np/(n p), provided < p < n. b) If p n, then there exist conjugate harmonic functions such that F / H r, r >. The paper is organized as follows. In section 2 we give all necessary definitions and auxiliary results used in the sequel. Section 3 is devoted to results needed for the proof of Theorem, and in section 4 we prove Theorem. In sections 5, 6 we prove Theorem 2. This result is used as a tool for the proof of Theorem 3. Auxiliary results for the proof of Theorem 3 are given in section 7. We prove Theorem 3 in section 8. The last section is devoted to the proof of Theorem 4. For convenience of the reader we split our proofs into elementary Lemmata. 2. Auxiliary results We begin with the definition of h p ( ). Let U(x, y) be a harmonic function in (, ). We say that the vector-function V (x, y) = (V (x, y),..., V n (x, y)) is the conjugate of U(x, y) in the sense of M. Riesz [2], [4], if V k (x, y), k =,..., n are harmonic functions, satisfying the generalized Cauchy-Riemann conditions: U n y V k =, x k k= V i x k = V k x i, U = V i, i k, k =,..., n. x i y If U(x, y) and V (x, y) are conjugate in in the above sense, then the vectorfunction F (x, y) = (U(x, y), V (x, y)) = (U(x, y), V (x, y),..., V n (x, y)) is called a harmonic vector.

4 4 ANATOLI RIABOGUINE AND DMITRY RYABOGIN Define M p (y) = M p (y, F ) = { F (x, y) p dx } /p, p >. Definition ([], [7]). We say that F (x, y) S p ( ), p > if for any y > there exists a constant C(y, F ), such that y y, M p (y, F ) C(y ). In particular, if C is independent of y, then F (x, y) h p ( ). Now we define the space H p ( ). We follow the work of Fefferman and Stein[4]. Let U(x, y) be a harmonic function in, and let U j j 2 j 3...j k denote a component of a symmetric tensor of rank k, j i n, i =,..., n. Suppose also that the trace of our tensor is zero, meaning n U jjj3...j k (x, y) =, j 3,..., j k. j= The tensor of rank k can be obtained from the above tensor of rank k by passing to its gradient: U j j 2...j k j k (x, y) = x jk (U j j 2 j 3...j k (x, y)), x = y, j k n. Definition 2 ([4]).We say that U(x, y) H p ( ), p >, if there exists a tensor of rank k of the above type with the properties: ( ) p/2dx U... (x, y) = U(x, y), sup U(j) 2 (x, y) <, (j) = (j,...j k ). It is well-known that the function (j) ( ) p/2 U(j) 2 (x, y) is subharmonic for p pk = (n k)/(n k ), see [3],[4],[4]. We will use the radial and nontangential maximal functions: Here (j) F (x) = sup F (x, y), N α (F )(x ) = sup F (x, y). (x,y) Γ α(x ) Γ α (x ) = {(x, y) : x x < αy}, α >, is an infinite cone with the vertex at x. It is well-known [4] that F (x, y) H p ( ) N α (F )(x) L p F (x) L p, p >. We also define the weak maximal function W F (x, y) = sup F (x, ζ), y >. ζ y The above expression is understood as folows: we fix x, and for fixed y we find the supremum over all ζ y. We will repeatedly use the following results.

5 H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 5 Lemma. ([4], p.73). Suppose u(x, y) is harmonic in, and for some p, < p <, sup u(x, y) p dx <, then (4) sup x u(x, y) Ay n/p, < y <. Theorem 5. ([5], p. 268). Let < p, k N, and let u : R be a harmonic function such that u(x, t) x y, K k,p t kp Dnu(x, k t) p dxdt < C. Then u(x, ) = lim t u(x, t) exists for almost all x, and for all t, u(x, t) p dx AC(k, n, p)k k,p. Theorem 6. ([5], p. 269). Let m N, p (n )/(m n ) (if n = we suppose p > ), and let u : R be harmonic. Then, for all t >, m u(x, t) p dx A(m, n, p)t mp 3t/2 ds u(x, t) p dx. t/2 Corollary. ([5], p. 27). Let p, m be as in Theorem 6, let b >, and let u : R be a harmonic function such that for all t > u(x, t) p dx Ct b. Then m u(x, t) p dx A(b, m, n, p)ct b mp, (t > ). In fact, the choice of p in Theorem 5 and Corollary may be independent of m (see Lemma 2). Theorem 7. []. Let p > and let F (x, y) = (U, V,..., V n ) be a harmonic vector satisfying (5) ) V i (x, y) x y, i =,..., n, 2) M p (y, U) C, 3) U C. Then F H r, r > p.

6 6 ANATOLI RIABOGUINE AND DMITRY RYABOGIN Notation. We denote by Di k f(x, y) the partial derivative of the function f of the order k with respect to x i, i =, 2,..., n. M(f)(x) denotes the usual Hardy- Littlewood maximal function of f(x). The notation f(x, y) x y means that f(x, y) converges to uniformly with respect to x, provided y, k f(x) = ( k f(x),..., k f(x) ). Everywhere below the constants A(k, n), C, K depend only on the x k x k n parameters pointed in parentheses, and may be different from time to time. 3. Auxiliary lemmata for the proof of Theorem. The next result shows that in Theorem 6 and Corollary the choice of p > may be independent on m N. We include it here for convenience of the reader. Lemma 2. Let p > and let F = (U, V,..., V n ) be such that V i x y, i =,..., n, M p (y, U) C. Then M p (y, k F ) ACy k, k N. Proof. By induction on k. Let k =. Fix p > and let l = inf{j N : p (n )/(j n )}. Let φ ij (x, y) be a coordinate of V i (x, y), j =,..., n, x n = y, i =,..., n, V = U. Since V i (x, y) x y, we may use the following relation (see [5] or [4]) φ ij (x, y) = (2l 2)! (2l 2)! We have φ ij (x, y) h ij (x, y), where h ij (x, y) (2l 2)! y (s y) 2l 2 D 2l n φ ij(x, s)ds = s 2l 2 D 2l n φ ij(x, s y)ds. s 2l 2 l D l nφ ij (x, s y) p ds. Theorem 3 of[5] implies (take w = l Dnφ l ij, a = 2l, A = A(l, n, p)), φ ij(x, y) p dx h ij(x, y) p dx A s (2l )p ds l Dn l φ ij(x, s y) p dx. Since Dnφ l ij (x, y) is the l th derivative of V i, we use Theorem 6 to get (6) l Dnφ l ij (x, y) p dx 2l F (x, y) p dx C y 2lp. This gives φ ij(x, y) p dx A(l, n, p)c s (2l )p (s y) 2lp ds = A(l, n, p)cy p,

7 H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 7 and the first induction step is proved. Assume that the statement is true for k. Then M p (y, k F ) ACy (k ). To prove it for k we define l as above and apply Corollary with b = k, m =, u = k F. Lemma 3. Let D i U x y, i =,..., n, and let (7) M p (y, D n U) Cy for some p >. Then ( (8) sup D n U(x, y y ) ) AC, y >, A = A(n, p, y ). Proof. Let p >. Then (see [2]) sup D n U(, y y ) p CM p (y y, D n U) AC, A = A(n, p, y ). Now let < p. Assume that (9) s p 2 U(x, s y ) p dxds < C(y ) <. Applying Lemma 2, Theorem 5 (with D n U instead of u and k = ) and the tensor representation of D n U from [4] we have D n U(x, y y ) H p. Thus, we prove (9). By Theorem 6, (7) yields () M p (y, 2 U) ACy 2. This gives (9). Indeed, the first integral in the right-hand side of (9) is finite, since On the other hand, s p ds M p (y y, 2 U) AC(y y ) 2 AC(y ), y >. 2 U(x, s y ) p dx AC s p (s y ) 2p ds AC(y ) <. Lemma 4. Let < p, and let F = (U, V,..., V n ) be a harmonic vector in. If M p (y, F ) C, then V i (and all its partial derivatives) x y, i =,..., n, V = U. Proof. Denote H = V i, and let B t (x, t ) be a ball of radius t, centered at (x, t ). Then Lemma gives H(x, t ) p A(p)t n H(z, s) p dzds B t(x z,s (t))

8 8 ANATOLI RIABOGUINE AND DMITRY RYABOGIN A(p)t n 2t ds z x <t ( 2A(p)t n sup ( H(z, s) p dz 2A(p)t n sup t Thus, sup x H(x, t ) ACt n/p. s ) H(z, t ) p dz ACt n. ) H(z, s) p dz Lemma 5. F = (U, V,..., V n ) H p if and only if V i (x, y) x y, i =,..., n and ( () sup U(x, η) ) C. η y Proof. The part only if is obvious. The part if follows by the reasons which are similar to those in Theorem 9 of [4] and the Fatou lemma. Lemma 6. Let F = (U, V,..., V n ) be a harmonic vector such that (2) ) M p (y, F ) C, 2) M p (y, D n U) Cy. Then (3) ( sup F (x, y y ) ) AC, y >, A = A(n, p, y ). Proof. Let H = V i, i =,,..., n, V = U. By the mean-value theorem, sup H(x, y y ) sup H(x, y y ) sup D n H(x, y y ). By virtue of Lemma 3 it is enough to show that ( (4) sup H(x, y y ) ) AC, y >, A = A(n, p, y ). Let p >. Then (4) follows from ) and the L p boundedness of the maximal function. Let < p. Then (4) is a consequence of Theorem 5 (take t = yy and repeat the arguments similar to those in Lemma 3). 4. Proof of Theorem. The part if is obvious. We show only if by proving subsequently that k F H p, k F H p,..., F H p. First of all, Lemmata 4, 5 give k F H p. Let us show that k F H p. It is enough to prove that DnU(x, k y) H p. By the mean-value theorem we have (5) sup D k n U(x, y) sup D k n U(x, y ) sup Dn k U(x, y),

9 H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 9 and it is enough to show that sup DnU(x, k y ) L p. In fact, we have some more, namely ( (6) sup DnU(x, k y y ) ) < C(y ) y >. To prove (6) we apply the mean-value theorem again, and write sup Dn k U(x, y y ) sup D k n U(x, y y ) sup Dn k U(x, y y ). Then k F H p implies ( sup DnU(x, k y y ) ) < C(y ). On the other hand, we have ( sup DnU(x, k y y ) ) < C(y ). This follows from Theorem 6, the estimate (7) j F (x, y y ) p dx AC(y y ) jp, j k, and the reasons which are similar to those in Lemma 3. Thus, we have (6) and k F H p. We repeat the argument and get i F H p, < i k. Now let i =. Then the first condition of our theorem, the fact that F H p, and the estimate U(x, y) U(x, y ) sup D n U(x, y) give F h p. Then we may apply Theorem 6 and Lemma 6 to obtain (3). To show that F H p it remains to use sup U(x, y) sup The proof of the theorem is complete. U(x, y ) sup D n U(x, y). 5. Auxiliary results for the proof of Theorem 2. Let V i be components of the harmonic vector F = (U, V,..., V n ). By the meanvalue theorem, (8) V i (x, y) = V i (x, y ) D n V i (x, y θ i ), < θ i (x, y) <. Lemma 7. We have (9) sup V i (x, y) sup V i (x, y ) sup D n V i (x, y θ i ), (2) sup D n V i (x, y θ i ) 2F (x),

10 ANATOLI RIABOGUINE AND DMITRY RYABOGIN where i =,..., n, V = U. Lemma 8. Let F = (U, V,..., V n ) H q and let θ i be as in (8), i =,..., n. Then ( qdx (2) sup D i U(x, y θ i ) ) AC. Proof. By (2) of Lemma 7, we have ( qdx (22) sup D n V i (x, y θ i ) ) AC. To get the desired result we apply the Cauchy-Riemann equations D i U(x, y) = D n V i (x, y). The next result uses the observation that θ i > on a set, controlled by estimate (4). Lemma 9. Let < p < q and let F (x, y) = (U(x, y), V (x, y),..., V n (x, y)) be a harmonic vector in. Then conditions (23) ) M p (y, F ) C, 2) M p (y, D n U) Cy, 3) F H q imply F h p. Proof. Let θ i be as in (8). Due to (8) and condition ) it is enough to show that D n V i (, y θ i (, y)) L p ( ). This will follow from the Cauchy-Riemann sup equations and (24) ( sup D i U(x, y θ i (x, y)) ) <. Thus, we prove (24). Condition 3) and Lemma imply AC (25) sup D i U(x, y θ i ) sup x x (y θ i ) AC(y α i(y)) n/q, n/q where α i (y) = inf x θ i(x, y). Define L i = {x AC : sup D i U(x, y θ i ) sup }. (y α i (y)) n/q Then x CL i (the complement of L i ) we have sup D i U(x, y θ i ) >. Then (26) CL i ( ) sup D i U(x, y θ i (x, y)) ( qdx sup D i U(x, y θ i (x, y)) ) AC due to condition 3) and Lemma 8. We estimate the integral in (24) over L i. Observe that for fixed x L i, sup D i U(x, y θ i (x, y)) sup D i U(x, y α i (y)),

11 L i H p SPACES OF CONJUGATE HARMONIC FUNCTIONS and α i inf α i(y) >. We put γ = min α i >, and take any < y γ. Then i=,...,n ( ) ( (27) sup D i U(x, y θ i (x, y)) sup U(x, y y ) ) AC. Indeed, condition 3) and Theorem 6 imply V i (x, y) x y. The same is true for all partial derivatives of V i, i =,..., n, V = U. Now the second inequality in (27) follows from 2) and Lemma 3. Taking into account (27), (26), we get (24). sup The next result uses the observation that under conditions (23) the supremum F (x, y) is reached at the boundary. Theorem 8. Let < p < q and let F = (U, V,..., V n ) be a harmonic vector in. Then F H r r : p r q if and only if conditions (23) are valid. Proof. The part only if follows from Theorem 6. We prove if. It is enough to show that F H p, or (sup F (, y) ) p L ( ). By the previous lemma F h p. Hence, applying the Fatou lemma we have ( ) (28) F (x, ) p dx = lim F (x, y) lim M p (y, F ) C. y y We claim that sup F (x, y) = F (x, ) and our lemma follows from (28). Since for fixed x the function W F (x, y) sup F (x, η) is nonincreasing in y, we have η y sup F (x, y) = sup sup F (x, η) = lim η y sup y η y F (x, η) = F (x, ). Even if sup F (x, y) = F (x, y ) for some y >, then W F (x, y) = W F (x, y ) for all y y, and we may put F (x, y ) = F (x, ). Theorem 9. Let < p < q, k N, and let F = (U, V,..., V n ) be a harmonic vector in. Then F H r r : p r q if and only if (29) ) M p (y, F ) C, 2) M p (y, D k nu) Cy k, 3) F H q. Proof. By Corollary, and the inverse statement, proved in [9], conditions (23), (29) are equivalent. Lemma. Let p > and let F = (U, V,..., V n ) be a harmonic vector in such that (3) ) M p (y, F ) C, 2) M p (y, U) C, 3) U(x, y) C. Then F H r, r p. Proof. By Theorem 7 we have F H r, r > p. Let r = p. By Lemma 4 and Theorem 6 we have M p (y, k U) Cy k, and we may use Theorem 9.

12 2 ANATOLI RIABOGUINE AND DMITRY RYABOGIN Lemma. Let < p < q and let F = (U, V,..., V n ) be a harmonic vector in such that ( qdx (3) ) M p (y, F ) C, 2) sup D n U(x, η) ) C. η y Then F H q. Proof. By the mean-value theorem, sup U(x, y) sup U(x, y ) sup U(x, y), and it is enough to show that (sup U(, y ) ) p L. To prove this, we apply Theorem 6, the mean-value theorem again, U(x, y ) U(x, y 2) sup D n U(x, y θ), and observe that conditions of the previous Lemma are satisfied with y instead of y. 6. Proof of Theorem 2. The proof is given in two lemmata presented below. Lemma 2. Let k N, < p < q and let F = (U, V,..., V n ) be a harmonic vector in. Then F and all its partial derivatives up to order k belong to H r, p r q, if and only if ( qdx (32) ) M p (y y, F ) C(y ) y >, 2) sup DnU(x, η) ) k C. η y Proof. The only if part is trivial. We prove if by induction. Let k =. We show at first that F H r, p r q. By Lemma we have F H q, and it is enough to show that F H p. To this end, we apply the mean-value theorem and repeate the proof of Lemma 9 begining with (24). As in Lemma 9 we define L i, and (26) follows from 2). The last estimate in (27) follows from Theorem 6 and Lemma. Conditions (3) are satisfied with U instead of F, D n U instead of U, and y y instead of y. To show that all partial derivatives of the first order belong to H r, p r q one has to proceed as above by changing U by 2 U, and Dn 2 U by D nu. Assume that the statement is true for k, and we have to prove it for k. By Theorem 6 we have ) with DnU k instead of F, and the result follows. Lemma 3. Conditions of the theorem are equivalent to conditions of the previous lemma. Proof. It is enough to prove that M p (y, F ) C and 2) of (32) imply ) of (32). This will follow from F H q. Since V i (x, y) x y, i =,..., n, it is enough to show that U H q. We will subsequently show that all DnU, k DnU, k 2..., U H q. In fact, we prove that DnU k H q. The proof of DnU, k 2..., U H q is similar.

13 H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 3 Observe that DnU k H q follows from DnU k h q. Indeed, let DnU k h q. By the mean-value theorem we have sup DnU(x, k y) sup D k nu(x, y ) sup DnU(x, k y), and we may apply Lemma to obtain ( sup DnU(x, k y ) ) <. Here we use Lemma with p = q, DnU(x, k y ) instead of U(x, y), k F instead of F. The assumption Dn k U hq and Lemma are used to satisfy the third condition of Lemma. The above inequality gives DnU k H q. Thus, it remains to prove that Dn k U(x, y) hq. By the mean-value theorem, DnU(x, k y) DnU(x, k y ) sup D k nu(x, y), and it is enough to prove that D k n U(x, y ) hq. Again, by the mean-value theorem, DnU(x, k y ) DnU(x, k y ) sup D k nu(x, y), but now we may use the assumption M p (y, F ) C to show that M r (y 2, DnU) k C, r p. To this end, we apply Theorem 6, Lemma, and take y instead of y. 7. Auxiliary results for the proof of Theorem 3 Lemma 4. Let p > and let F = (U, V,..., V n ) be a harmonic vector in such that (33) ) V j x y, 2) M p (y, U) C, 3) M p (y, D n U) C, j =,..., n. Then F H r r : p < r < np/(n p) provided < p < n, and F H r r > p, provided p n. Proof. Let < p < n. By ) it is enough to show that U H r r : p < r < np/(n p). Assume at first that < p < n, np/(n p). We have to show (see Theorem 5 and the definition of H p spaces from [4]) that (34) s r D nu(x, s) r dxds = This can be done by using inequalities proved in [4]: (35) M r (y, D n U) ACy n/pn/r, < C. (36) M r (y, F ) ACy n/pn/r, M r (y, U) ACy n/pn/r. The estimate in (35) follows from 3). The estimates in (36) follow from 2).

14 4 ANATOLI RIABOGUINE AND DMITRY RYABOGIN To estimate the first integral in the right-hand side of (34) we write s r ds D nu(x, s) r dx AC s r s nr p n ds AC, since r nr/p n > r < np/(n p). To estimate the second integral in the right-hand side of (34) we use (36), s r ds D nu(x, s) r dx AC s r nr r s p n ds AC, where r r nr/p n <. Assume now that < p < n, np/(n p) >. We show that F H r, < r < np/(n p). Here we use (37) U(x, y) U(x, ) The first inequality in (36) implies (38) M r (, F ) AC. D n U(x, η) dη. Since F (x, ) is bounded, (38) is true r > p, and it is enough to show that the second term in the right-hand side of (37) belongs to H r, < r < np/(n p). By Minkowsi s inequality we have ( ) /r D n U(x, y) r dxdy ( ) /rdy D nu(x, y) r dx y n/pn/r dy AC, n/p n/r >. If p <, np/(n p) >, then one has to split (p, np/(n p)) into two intervals (p, ], (, np/(n p)), and to repeat the previous parts of the proof. Finally, let p n. We take r > p n, i.e. r > and we may use Minkowski s inequality. Remark. The previous lemma is sharp in the case < p. There exists a harmonic vector F satisfying (33) such that F / H p, F / H np/(n p). Proof. We take for simplicity n =, p = /2, q = np/(n p) =. The multidimensional example can be constructed by using the Poisson kernel. Consider a harmonic vector x U(x, y) = x 2 y x ( 2 x 2 (y ), V (x, y) = y 2 x 2 y y ). 2 x 2 (y ) 2

15 Then F (x, y) = H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 5 x2 y 2 x 2 (y ) 2, First of all, F H r r : /2 < r <, (sup F (x, y) ) r dx A(r) On the other hand, it is obvious that F / H, and F / H /2, sup sup F (x, y) dx = F (x, y) dx = sup F (x, y) = dx x x 2 2 x x 2 dx ( x ( x )) r < C. dx x 2 /4 x 2 dx x =, dx x =. It remains to show that M /2 (y, U) C, M /2 (y, D n U) C. We have x(2y ) M /2 (y, U) = U(x, y) dx = 2 (x 2 y 2 )(x 2 (y ) 2 ) dx 2 2 xy (x 2 y 2 ) dx 2 2 dx x x2 C, 2 x ( x ). where in the first integral of the above inequality one has to make a substitution x/y = t. Now, 2xy D n U(x, y) = (x 2 y 2 ) 2x(y ) 2 (x 2 (y ) 2 ) 2 and the estimate M /2 (y, D n U) C is similar. Lemma 4 can be generalized to the case of partial derivatives of any order. Lemma 5. Let p > and let F = (U, V,..., V n ) be a harmonic vector in such that (39) ) V j x y, j =,..., n, 2) M p (y, U) C, 3) M p (y, D k nu) C for some k N. a) If < kp < n, then F H r r : p < r < np/(n kp). If kp n, then F H r r p. b) Let k >, m k, m N. If < (k m)p < n, then all partial derivatives of F of the order m belong to H r r : p < r < np/(n (k m)p). If (k m)p n, then all partial derivatives of F of the order m belong to H r r p.

16 6 ANATOLI RIABOGUINE AND DMITRY RYABOGIN Proof. The proof is a consequence of Lemma 4 and Theorem 2. For convenience of the reader we give it here. We begin with b), < (k m)p < n. Let m = k. Then 2) and Lemma 2 imply M p (y, k U) Cy (k ), and we may follow the proof of Lemma 4 with Dn k U instead of U. To estimate the integral, similar to (34), we use M r(y, k U) Cy (k ) n/pn/r instead of the second inequality in (36). We conclude that all partial derivatives of the order k belong to H r, p < r < np/(n p). Now let m = k 2. By the previous step, Dn k H np n p ɛ, where ɛ > is small enough. We repeat the proof of Lemma 4 with ɛ instead of p. We have all partial derivatives of the order k 2 in H r np, ɛ < r < n(np/(n p) ɛ). Letting n p n np/(n p)ɛ ɛ, we have np r < np. It remains to show that all partial derivatives of n p n 2p the order k 2 belong to H r, p < r < np. Since all partial derivatives of the order n p k belong to H r, p < r < np/(n p), we may use Theorem 2 with q = np ɛ, n p and k instead of k. Thus, we get b) for m = k 2, provided < (k m)p < n. The proof of the cases m = k 3, k 4,...,, < (k m)p < n, is similar. We prove b), the case (k m)p n. If m = k, then the result follows from Lemma 4 (repeat the proof with Dn k Let m = k 2. We write (37) with D k 2 np n p U instead of U). nu instead of U, and observe that all partial derivatives of the order k 2 belong to H r r p, provided p n. To get the case 2p n we apply Theorem 2. Thus, we get b) for m = k 2, provided (k m)p n. The proof of the cases m = k 3, k 4,...,, (k m)p n, is similar. The proof of b) is complete. We show a) by induction on k. For k =, a) is just Lemma 4. Assume that a) is true for k, < (k )p < n. To prove the statement for k, < kp < n, it is enough to show that F H r np, < r < np. To this end, we observe n (k )p n kp that b) and the induction hypothesis force the first derivative of F to be in H r, np np p < r <. Hence, we may apply Lemma 4 (with ɛ instead of p) n (k )p n (k )p to get F H r np, < r < np. n (k )p n kp Now, let a) be true for k, (k )p n. To prove the statement for k, kp n, we use Theorem Proof of Theorem 3 Now we prove Theorem 3. We prove a). By Theorem and Lemma 5, we have F H r, r : p r < np/(n kp). We have to prove that F H np/(n kp). We prove this in two steps. At first we assume that (4) I(a, np n kp ) y a F (x, y) np/(n kp) dxdy <, for some < a <, and prove that (4) implies F (x) L np/(n kp). Then we prove (4). Let (4) be true. Define E = {x : F (x) }, and let CE denote the complement of E. Since F H r, p r < np/(n kp), we have m(ce) <. If

17 F / L np/(n kp) (CE), then H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 7 I(a, np n kp ) y a dy F (x, y) np/(n kp) dx dy F (x, y) np/(n kp) dx. But the last integral may be arbitrary large, a contradiction. ( np It remains to prove (4). We split I(a, ) into two parts, n kp ) dy. Since M r (y, F ) C, p r < np/(n kp), we have M r (y, F ) ACy n/pn/r for r > p. Let r = np/(n kp), then nr/p n = nkp/(n kp) < np/(n kp). Chosing a (, ) such that a np/(n kp) <, we obtain y a dy F (x, y) np/(n kp) dx AC n 2 p y a y p(n kp) n dy <. To estimate y a dy F (x, y) np/(n kp) dx, we write y a dy F (x, y) np/(n kp) dx y a dy F (x, y) np/(n kp) dx. E CE The first integral is obviously finite. To get a bound for the second one we fix a : a np < as above, and choose r < np np such that ( r) n a >. n kp n kp n kp r Then we use sup F (x, y) ACy n/r to obtain x y a dy CE F (x, y) np/(n kp) dx AC y a ( np n kp r) n r dy (F (x)) r dx AC y a ( np n kp r) n r dy AC. Thus, we have a). We prove b). By Theorem and Lemma 5, all partial derivatives of F of the order m belong to H r, r : p r < np/(n (k m)p). We have to prove that all partial derivatives of F of the order m belong to H np/(n (k m)p). The proof of this is similar to the proof of a) with obvious changes, one has only take m F instead of F, and np/(n (k m)p) instead of np/(n kp). The proof of Theorem 3 is complete.

18 8 ANATOLI RIABOGUINE AND DMITRY RYABOGIN 9. Proof of Theorem 4. Lemma 6. Let α >, β >, p > and let W (x, y) in be such that W p be subharmonic and I(α, p, β) = s αpβ W p (x, s)ds <. Then the function W α (x, y) Γ(α) y > we have s α W (x, s y)ds is subharmonic, and y (W α (x, y)) p dx A(α, n, p, β, y )I(α, p, β). In particular W α S p ( ). Proof. The proof is based on the arguments, which are similar to those in [5], Lemma 6 and Theorem 3. We take U = W p (x, t), q = αp β in Lemma 6, [5]. Since W p is subharmonic, and I(α, p, β) is convergent, we may use the proof of Theorem 3, [5], to conclude that the integral defining W α (x, y) is convergent and W α (x, y) is subharmonic. There exist f and a constant A(n, p, β, α) such that f A(n, p, β, α)i(α, p, β), and we have W (x, sy) A(n, p, β, α)(sy) α (β)/p f /p (x), W α (x, y) A(n, p, β, α)y α (β)/p f /p (x), y y. The proof of the next lemma can be obtained by the reasons which are similar to those in []. Lemma 7. Let p >, α >, m N, and F be a harmonic vector such that F x y, I(α, m, p) t αmp Dn m U(x, t) p dxdt <. Then t α F (x, t) p dxdt A(n, p, α)ci(α, m, p). As a consequence of two previous lemmata we have Lemma 8. Let p >, α >, and let F be a harmonic vector such that t α F (x, t) p dxdt <. Then all tensor coordinates ( of the rank k of V i, i =,..., n, V = U, belong to p S p. In particular, sup F (, y y ) ) L y >.

19 H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 9 Now we prove Theorem 4. We start with a). Let np/(n p) < r <. We claim that ( ) (4) y r D n U(x, y) r dxdy = y r dy D nu(x, y) r dx <. The first integral in the right-hand side of the above equality is obviously finite for all r >. Since M r (y, D n U) ACy n/pn/r, r > p, the second integral is also finite. By Theorem 5 we have M r (y, U) C, r > np/(n p). The mean-value theorem, and condition 3) imply the boundedness of U. By Theorem 7, U H r, > r > np/(n p). Now, let r >, and let γ > be such that r γ nr/p n <, together with r nr/p n <. Then the integral y rγ D n U(x, y) r dxdy <, and Lemma 7 (with m =, α = γ ) gives y γ D n U(x, y) r dxdy <. r Since γ >, we apply Lemma 8 to obtain sup F (, y ) ) L, r > np/(n p). By Theorem 7, D n U H r, and the mean-value theorem implies U H r. We prove b). Let n = and consider U(x, y) = /2 log(x 2 (y 2) 2 ) /4 log 2 (x 2 (y 2) 2 ) arctan 2 (x/(y 2)), arctan(x/(y 2)) V (x, y) = /4 log 2 (x 2 (y 2) 2 ) arctan 2 (x/(y 2)). It is clear that U, V satisfy all conditions of the Theorem, nevertheless F / H r, provided r >. The proof of Theorem 4 is complete. Observe that all conditions of the Theorem are essential and inependent one on another. Indeed, let n =, p, U(x, y) = log(x 2 (y ) 2 ), ( V (x, y) = arctan x y. This functions satisfy all conditions of the theorem, but the first one, and F / H r, r >. Now, let n =, p = /2, x y U(x, y) = x 2 y2, V (x, y) = x 2 y 2.

20 2 ANATOLI RIABOGUINE AND DMITRY RYABOGIN We have ) and 2), but not 3), and F / H r, r >. Let n = 2, r = x 2 x 2 2 (y ) 2, F = (U, V, V 2 ), V = r, V 2 = x 2 r(r x ), U = y r(r x ). We have ) and 3), but not 2), and F / H r, r >. Finally, we note that one can not conclude that F H r, r = np/(n p). Indeed, let n =, p = /2, np/(n p) =. Then U = x x 2 (y ), V = y 2 x 2 (y ) 2 satisfy all conditions of the Theorem, but F H r, r >, r > np/(n p) and F / H. References [] A. A. Bonami, Integral inequalities for conjugate harmonic functions of several variables, Math. Sbornik., Vol (972), No. 2, p [2] D. L. Burkholder, R. V. Gandy and M. L. Silverstein, A maximal functions characterizations of the class H p, Trans. AMS., 57(97), [3] A. P. Calderón and A. Zygmund, On higher gradients of harmonic functions, Studia Math. 24 (964), No. 2, [4] C. Fefferman, E. M. Stein, H p spaces of several variables, Acta Math., 29, No 3, (972), [5] T. M. Flett, Inequalities for the p th mean values of harmonic and subharmonic functions with p, Proc. Lon. Math. Soc., Ser. 3, 2, No. 3, [6] V. I. Krylov, On functions regular in the half-plane, Math., Sb., (939), 6(48), pp [7] U. Kuran, Classes of subharmonic functions in (, ), Proc. Lond. Math. Soc., Ser 3., 6(966), No. 3, [8] I. Privalov, Subharmonic functions, Moscow, 937. [9] A. Riaboguine, Conjugate harmonic functions of the Hardy class. (Russian), Izv. Vyssh. Uchebn. Zaved. Mat. 99,, no. 9, 47-53; translation in Soviet Math. (Iz. VUZ) 35 (99), no. 9, [] A. Riaboguine, Boundary values of conjugate harmonic functions of several variables, (Russian), Izv. Vyssh. Uchebn. Zaved. Mat. (98), no. 2, [] E. D. Solomentsev, On classes of subharmonic functions in the half-space, Notes of Moscow State Univ.,, (958). [2] E. M. Stein, Singular integrals and differentiability properties of functions, Princeton University Press, Princeton NJ, 97. [3] E. M. Stein and G. Weiss, On the theory of harmonic functions of several variables, Acta Math., 3 (96), pp [4] E. M. Stein and G. Weiss, Generalization of the Caushy-Riemann equations and representation of the rotation group, Amer. J. Math.,9(968), [5] E. M. Stein and G. Weiss, An introduction to harmonic analysis on Euclidean spaces, Princeton University Press, Princeton NJ, 969. [6] T. Wolff, Counterexamples with harmonic gradient in R 3, Essays in honor of E. M. Stein, Princeton Mathematical Series, 42 (995), [7] A. Zygmund, Trigonometric series, 2nd ed., Cambridge University Press, Cambridge, 968.

21 H p SPACES OF CONJUGATE HARMONIC FUNCTIONS 2 Anatoli Riaboguine, Department of Mathematics, Ben Gurion University of the Negev, P.O.B. 653, Be er Sheva 845, Israel address: riabogin@cs.bgu.ac.il Dmitry Ryabogin, Department of Mathematics, Kansas State University, Manhattan, KS , USA address: ryabs@math.ksu.edu