Khavinson problem for hyperbolic harmonic mappings in Hardy space
aa r X i v : . [ m a t h . A P ] S e p KHAVINSON PROBLEM FOR HYPERBOLIC HARMONICMAPPINGS IN HARDY SPACE
JIAOLONG CHEN, DAVID KALAJ, AND PETAR MELENTIJEVI ´C
Abstract.
In this paper, we partly solve the generalized Khavinson conjecturein the setting of hyperbolic harmonic mappings in Hardy space. Assume that u = P Ω [ φ ] and φ ∈ L p ( ∂ Ω , R ), where p ∈ [1 , ∞ ], P Ω [ φ ] denotes the Poissonintegral of φ with respect to the hyperbolic Laplacian operator ∆ h in Ω, and Ωdenotes the unit ball B n or the half-space H n . For any x ∈ Ω and l ∈ S n − , let C Ω ,q ( x ) and C Ω ,q ( x ; l ) denote the optimal numbers for the gradient estimate |∇ u ( x ) | ≤ C Ω ,q ( x ) k φ k L p ( ∂ Ω , R ) and gradient estimate in the direction l |h∇ u ( x ) , l i| ≤ C Ω ,q ( x ; l ) k φ k L p ( ∂ Ω , R ) , respectively. Here q is the conjugate of p . If q = ∞ or q ∈ [ K − n − + 1 , K n − + 1] ∩ [1 , ∞ ) with K ∈ N = { , , , . . . } , then C B n ,q ( x ) = C B n ,q ( x ; ± x | x | ) for any x ∈ B n \{ } , and C H n ,q ( x ) = C H n ,q ( x ; ± e n ) for any x ∈ H n , where e n = (0 , . . . , , ∈ S n − . However, if q ∈ (1 , nn − ), then C B n ,q ( x ) = C B n ,q ( x ; t x ) for any x ∈ B n \{ } ,and C H n ,q ( x ) = C H n ,q ( x ; t e n ) for any x ∈ H n . Here t w denotes any unit vector in R n such that h t w , w i = 0 for w ∈ R n \ { } . Introduction
For n ≥
2, let R n denote the n -dimensional Euclidean space. For x = ( x , . . . , x n ) ∈ R n , sometimes we identify each point x with a column vector. For two column vec-tors x, y ∈ R n , we use h x, y i to denote the inner product of x and y . The ball { x ∈ R n : | x | < r } and the sphere { x ∈ R n : | x | = r } are denoted by B n ( r ) and S n − ( r ), respectively. In particular, let B n = B n (1), S n − = S n − (1), R = C and B = D . We also denote S n − = { x ∈ S n − : x n > } , S n − − = { x ∈ S n − : x n < } and H n = { x = ( x ′ , x n ) ∈ R n : x ′ ∈ R n − , x n > } . As is usual we identify R n − with R n − × { } . With this convention we have that ∂ H n = R n − .A mapping u ∈ C ( B n , R ) is said to be hyperbolic harmonic if u satisfies thehyperbolic Laplace equation∆ h u ( x ) = (1 − | x | ) ∆ u ( x ) + 2( n − − | x | ) n X i =1 x i ∂u∂x i ( x ) = 0 , Mathematics Subject Classification.
Primary 31B05; Secondary 42B30.
Key words and phrases.
Hyperbolic harmonic mappings, Hardy space, the generalized Khavin-son conjecture, estimates of the gradient. here ∆ denotes the usual Laplacian in R n . Meanwhile, a mapping u ∈ C ( H n , R )is said to be hyperbolic harmonic if u satisfies the hyperbolic Laplace equation∆ h u ( x ) = x n ∆ u ( x ) − ( n − x n ∂u∂x n ( x ) = 0 . For convenience, in the rest of this paper, we call ∆ h the hyperbolic Laplacian oper-ator .When n = 2, we easily see that hyperbolic harmonic mappings coincide withharmonic mappings. In this paper, we focus our investigations on the case when n ≥ Hardy space for hyperbolic harmonic mappings.
For p ∈ (0 , ∞ ], let L p ( S n − , R ) denote the space of Lebesgue measurable mappings from S n − into R satisfying k f k L p ( S n − , R ) < ∞ , where k f k L p ( S n − , R ) = (cid:18)Z S n − | f ( ξ ) | p dσ ( ξ ) (cid:19) p , if p ∈ (0 , ∞ ) , ess sup ξ ∈ S n − (cid:8) | f ( ξ ) | (cid:9) , if p = ∞ . Here and hereafter, dσ denotes the normalized surface measure on S n − so that σ ( S n − ) = 1. Similarly, let L p ( R n − , R ) denote the space of Lebesgue measurablemappings from R n − into R satisfying k f k L p ( R n − , R ) < ∞ , where k f k L p ( R n − , R ) = (cid:18)Z R n − | f ( x ) | p dV n − ( x ) (cid:19) p , if p ∈ (0 , ∞ ) , ess sup x ∈ R n − (cid:8) | f ( x ) | (cid:9) , if p = ∞ . Here and hereafter, dV n − denotes the Lebesgue volume measure in R n − .If φ ∈ L ( S n − , R ), we define the invariant Poisson integral or Poisson-Szeg¨ointegral of φ in B n by (cf. [19, Definition 5.3.2]) P B n [ φ ]( x ) = Z S n − P B n ( x, ζ ) φ ( ζ ) dσ ( ζ ) , where P B n ( x, ζ ) = (cid:18) − | x | | x − ζ | (cid:19) n − (1.1)is the Poisson-Szeg¨o kernel with respective to ∆ h satisfying Z S n − P B n ( x, ζ ) dσ ( ζ ) = 1(cf. [19, Lemma 5.3.1(c)]).If φ ∈ L ( R n − , R ), then the Poisson-Szeg¨o integral of φ is the function P H n [ φ ]defined by (cf. [19, Section 5.6]) P H n [ φ ]( x ) = Z R n − P H n ( x, y ′ ) φ ( y ′ ) dV n − ( y ′ ) , here x = ( x ′ , x n ) ∈ H n , x ′ , y ′ ∈ R n − and P H n ( x, y ′ ) = c n (cid:18) x n | x ′ − y ′ | + x n (cid:19) n − (1.2)is the Poisson-Szeg¨o kernel with respective to ∆ h satisfying Z R n − P H n ( x, y ′ ) dV n − ( y ′ ) = 1 . By calculations, we have c n = 2 n − Γ( n ) π n , (1.3)where Γ is the Gamma function.For p ∈ [1 , ∞ ], we use H p ( B n , R ) to denote the Hardy space consisting of hyper-bolic harmonic mappings of the form u = P B n [ φ ] with φ ∈ L p ( S n − , R ). Similarly, by H p ( H n , R ) we denote the Hardy space consisting of hyperbolic harmonic mappingsof the form u = P H n [ φ ] with φ ∈ L p ( R n − , R ) and p ∈ [1 , ∞ ].1.2. The Khavinson problem.
Let p ∈ [1 , ∞ ] and q be its conjugate. Assumethat u = P Ω [ φ ] and φ ∈ L p ( ∂ Ω , R ), where Ω denotes B n or H n . For fixed x ∈ Ω and l ∈ S n − , let C Ω ,q ( x ) and C Ω ,q ( x ; l ) denote the optimal numbers for the gradientestimate |∇ u ( x ) | ≤ C Ω ,q ( x ) k φ k L p ( ∂ Ω , R ) . (1.4)and gradient estimate in the direction l |h∇ u ( x ) , l i| ≤ C Ω ,q ( x ; l ) k φ k L p ( ∂ Ω , R ) . (1.5)respectively. Since |∇ u ( x ) | = sup l ∈ S n − |h∇ u ( x ) , l i| , we clearly have C Ω ,q ( x ) = sup l ∈ S n − C Ω ,q ( x ; l ) . (1.6)The generalized Khavinson conjecture states: Conjecture 1.1.
Let q ∈ [1 , ∞ ] and e n = (0 , . . . , , ∈ S n − . Then(1) for any x ∈ B n \{ } , we have C B n ,q ( x ) = C B n ,q ( x ; ± x | x | );(2) for any x ∈ H n , we have C H n ,q ( x ) = C H n ,q ( x ; ± e n ) . This conjecture actually dates back to 1992. Khavinson [8] obtained a sharppointwise estimate for the radial derivative of bounded harmonic functions in B .In a private conversation with Gresin and Maz’ya, he conjectured that the sameestimate holds for the norm of the gradient of bounded harmonic functions. Later,this conjecture was formulated by Kresin and Maz’ya in [10] for bounded harmonic unctions in B n . In the same paper, they obtained the sharp inequalities for the radialand tangential derivatives of such functions and solved the analogous problem forharmonic functions with the L p integrable boundary values for p = 1 and p = 2.Also, the same authors in [11] solved the half-space analog of this problem for p = 1, p = 2 and p = ∞ . Later, Kalaj and Markovi´c [6] established an analogous result forharmonic functions from D into C with L p integrable boundary values, p ≥ x ∈ B n and x is near the boundary. Kalaj [5]proved the conjecture in B , and Melentijevi´c [15] confirmed the conjecture in B .Very recently, Liu [12] showed that the conjecture is true in B n with n ≥
3. See[1, 7, 13, 17] and references therein for further discussions on the gradient estimatesfor analytic and harmonic functions.The main purpose of this paper is to consider the generalized Khavinson conjec-ture in the setting of hyperbolic harmonic mappings in Hardy space. For the casewhen q ∈ [1 , ∞ ), we have the following results. Theorem 1.1.
Let q ∈ (1 , nn − ) .(1) For any x ∈ B n \{ } and l ∈ S n − , C B n ,q (cid:0) x ; ± x | x | (cid:1) ≤ C B n ,q (cid:0) x ; l (cid:1) ≤ C B n ,q (cid:0) x ; t x (cid:1) = C B n ,q ( x ) , where t x is any unit vector in R n such that h t x , x | x | i = 0 .(2) For any x ∈ H n and l ∈ S n − , C H n ,q (cid:0) x ; ± e n (cid:1) ≤ C H n ,q (cid:0) x ; l (cid:1) ≤ C H n ,q (cid:0) x ; t e n (cid:1) = C H n ,q ( x ) . Theorem 1.2.
Let q ∈ [ K − n − + 1 , K n − + 1] ∩ [1 , ∞ ) , where K ∈ N = { , , , . . . } .(1) For any x ∈ B n \{ } and l ∈ S n − , C B n ,q (cid:0) x ; t x (cid:1) ≤ C B n ,q (cid:0) x ; l (cid:1) ≤ C B n ,q (cid:0) x ; ± x | x | (cid:1) = C B n ,q ( x ) . (2) For any x ∈ H n and l ∈ S n − , C H n ,q (cid:0) x ; t e n (cid:1) ≤ C H n ,q (cid:0) x ; l (cid:1) ≤ C H n ,q (cid:0) x ; ± e n (cid:1) = C H n ,q ( x ) . In the following, we give two special cases of Theorem 1.2.
Theorem 1.3. (1) For any x ∈ B n and l ∈ S n − , C B n , nn − ( x ) ≡ C B n , nn − ( x ; l ) ≡ n − − | x | ) n − n Γ( n )Γ( n − n − ) √ π Γ( n n − ) (1 + | x | ) ! n − n and C B n , ( x ) ≡ C B n , ( x ; l ) ≡ n − n ) √ π Γ( n +12 )(1 − | x | ) . (2) For any x ∈ H n and l ∈ S n − , C H n , nn − ( x ) ≡ C H n , nn − ( x ; l ) ≡ ( n − n )2 n +1 √ πx n − n n Γ( n − n − )Γ( n n − ) ! n − n nd C H n , ( x ; l ) ≡ C H n , ( x ) ≡ n ) √ π Γ( n − ) x n . For a ∈ R and k ∈ N , let ( a ) k denote the factorial function with ( a ) = 1 and( a ) k = a ( a + 1) . . . ( a + k − a is neither zero nor a negative integer, then (cf.[18, Page 23]) ( a ) k = Γ( a + k )Γ( a ) . (1.7)For r, s ∈ Z + = { , , , . . . } and x ∈ R , we define the generalized hypergeometricseries by r F s ( a , a , . . . , a r ; b , . . . , b s ; x ) = ∞ X k =0 ( a ) k ( a ) k . . . ( a r ) k k !( b ) k . . . ( b s ) k x k , (1.8)where a i , b j ∈ R (1 ≤ i ≤ r , 1 ≤ j ≤ s ) and b j is neither zero nor a negative integer(cf. [4, Chapter IV] or [18, Chapter 5]). If r = s + 1, then the series converges for | x | < | x | >
1. If r = s + 1 and P si =1 b i − P ri =1 a i >
0, then theseries is absolutely convergent on | x | = 1.Let Ω = B n or H n . Using an explicit formula for C Ω ,q (cid:0) x ; t x (cid:1) and C Ω ,q (cid:0) x ; x | x | (cid:1) , wecan reformulate Theorems 1.1 and 1.2 as follows, respectively. Theorem 1.4.
Let p ∈ ( n, ∞ ) and q be its conjugate.(1) If u = P B n [ φ ] and φ ∈ L p ( S n − , R ) , then for any x ∈ B n , we have the followingsharp inequality: |∇ u ( x ) | ≤ n − Γ( n )Γ( q +12 )(1 + | x | ) ( n − q − √ π Γ( q + n )(1 − | x | ) n ( q − ! q k φ k L p ( S n − , R ) × (cid:18) F (cid:18) ( n − − q )2 , − ( n − q − q + n | x | (1 + | x | ) (cid:19)(cid:19) q . (2) If u = P H n [ φ ] and φ ∈ L p ( R n − , R ) , then for any x ∈ H n , we have the followingsharp inequality: |∇ u ( x ) | ≤ ( n − n )2 q − q π nq − n +12 q x n ( q − q n k φ k L p ( R n − , R ) × Γ( q +12 )Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − , q + 12 ; 12 , q + n (cid:19)! q . Theorem 1.5.
Let p ∈ {∞} ∪ [ K + n − K , K + n − K − ] and q be its conjugate, where K ∈ Z + .
1) If u = P B n [ φ ] and φ ∈ L p ( S n − , R ) , then for any x ∈ B n , we have the followingsharp inequality: |∇ u ( x ) | ≤ n − Γ( n )Γ( q +12 )(1 + | x | ) ( n − q − √ π Γ( q + n )(1 − | x | ) n ( q − ! q k φ k L p ( S n − , R ) × (cid:18) F (cid:18) ( n − − q )2 , − ( n − q − , q + 12 ; 12 , q + n | x | (1 + | x | ) (cid:19)(cid:19) q . (2) If u = P H n [ φ ] and φ ∈ L p ( R n − , R ) , then for any x ∈ H n , we have the followingsharp inequality: |∇ u ( x ) | ≤ ( n − n )2 q − q π nq − n +12 q x n ( q − q n k φ k L p ( R n − , R ) × Γ( q +12 )Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − q + n (cid:19)! q . By using the expression of generalized hypergeometric series and letting q = 1, n +1 n − or nn − in Theorem 1.5, respectively, we easily arrive at the following threeresults, and we omit the proofs of them. Corollary 1.1. (1) If u = P B n [ φ ] and φ ∈ L ∞ ( S n − , R ) , then for any x ∈ B n , wehave the following sharp inequality: |∇ u ( x ) | ≤ n − n ) k φ k L ∞ ( S n − , R ) √ π (1 − | x | )Γ( n +12 ) . (2) If u = P H n [ φ ] and φ ∈ L ∞ ( R n − , R ) , then for any x ∈ H n , we have thefollowing sharp inequality: |∇ u ( x ) | ≤ n ) k φ k L ∞ ( R n − , R ) √ π Γ( n − ) x n . Corollary 1.2. (1) If u = P B n [ φ ] and φ ∈ L n +12 ( S n − , R ) , then for any x ∈ B n , wehave the following sharp inequality: |∇ u ( x ) | ≤ n − k φ k L n +12 ( S n − , R ) (1 − | x | ) n − n +1 Γ( n )Γ (cid:0) nn − (cid:1) √ π Γ (cid:0) n +12 n − (cid:1) (1 + | x | ) + 4Γ( n )Γ (cid:0) n − n − (cid:1) √ π Γ (cid:0) n +2 n − n − (cid:1) | x | ! n − n +1 . (2) If u = P H n [ φ ] and φ ∈ L n +12 ( S n − , R ) , then for any x ∈ B n , we have thefollowing sharp inequality: |∇ u ( x ) | ≤ ( n − n ) k φ k L n +12 ( R n − , R ) n +1 π n − n − x n − n +1 n Γ (cid:0) nn − (cid:1) Γ (cid:0) n +12 n − (cid:1) + Γ (cid:0) nn − (cid:1) (cid:0) n +2 n − n − (cid:1) ! n − n +1 . orollary 1.3. (1) If u = P B n [ φ ] and φ ∈ L n ( S n − , R ) , then for any x ∈ B n , wehave the following sharp inequality: |∇ u ( x ) | ≤ n − k φ k L n ( S n − , R ) (1 − | x | ) n − n Γ( n )Γ( n − n − ) √ π Γ( n n − ) (1 + | x | ) ! n − n . (2) If u = P H n [ φ ] and φ ∈ L n ( R n − , R ) , then for any x ∈ H n , we have thefollowing sharp inequality: |∇ u ( x ) | ≤ ( n − n ) k φ k L n ( R n − , R ) n +1 √ πx n − n n Γ( n − n − )Γ( n n − ) ! n − n . For the case when q = ∞ , we have the following result. Theorem 1.6. (1) For any x ∈ B n \{ } and l ∈ S n − , C B n , ∞ (cid:0) x ; l (cid:1) ≤ C B n , ∞ (cid:0) x ; ± x | x | (cid:1) = C B n , ∞ ( x ) = 2( n −
1) (1 + | x | ) n − (1 − | x | ) n . (2) For any x ∈ H n and l ∈ S n − , C H n , ∞ (cid:0) x ; l (cid:1) ≤ C H n , ∞ (cid:0) x ; ± e n (cid:1) = C H n , ∞ ( x ) = 2 n − ( n − n ) π n x nn . Using an explicit formula for C Ω , ∞ (cid:0) x ), we can reformulate Theorem 1.6 as follows,where Ω = B n or H n . Theorem 1.7. (1) If u = P B n [ φ ] and φ ∈ L ( S n − , R ) , then for any x ∈ B n , wehave the following sharp inequality: |∇ u ( x ) | ≤ n −
1) (1 + | x | ) n − (1 − | x | ) n k φ k L ( S n − , R ) . (2) If u = P H n [ φ ] and φ ∈ L ( R n − , R ) , then for any x ∈ H n , we have the followingsharp inequality: |∇ u ( x ) | ≤ n − ( n − n ) π n x nn k φ k L ( R n − , R ) . The rest of this paper is organized as follows. In Section 2, we establish somerepresentations for C B n ,q ( x ; l ) when q ∈ [1 , ∞ ). In Section 3, we prove Theorems1.1 ∼ C H n ,q ( x ; l )with q ∈ [1 , ∞ ). In Section 6, we present the proofs of Theorems 1.1 ∼ Case q ∈ [1 , ∞ ) and representations for C B n ,q ( x ; l )In this section, we first establish a general integral representation formula for thesharp quantity C B n ,q ( x ; l ) when q ∈ [1 , ∞ ). emma 2.1. For any q ∈ [1 , ∞ ) , x ∈ B n and l ∈ S n − , we have C B n ,q ( x ; l ) = 2( n − − | x | ) n ( q − q (cid:18)Z S n − | η − x | n − q − |h η, l i| q dσ ( η ) (cid:19) q . (2.1) Proof.
In order to calculate C B n ,q ( x ; l ), we let p be the conjugate of q with p ∈ (1 , ∞ ], φ ∈ L p ( S n − , R ) and u = P B n [ φ ] in B n . For any x ∈ B n and ζ ∈ S n − , it follows from(1.1) that ∇P B n ( x, ζ ) = − n − − | x | ) n − x | x − ζ | + (1 − | x | )( x − ζ ) | x − ζ | n (2.2)(cf. [2, Equation (2.25)]). Obviously, the mapping ( x, ζ )
7→ ∇P B n ( x, ζ ) is continuousin B n ( r ) × S n − , where r ∈ (0 , h∇ u ( x ) , l i = Z S n − h∇P B n ( x, ζ ) , l i φ ( ζ ) dσ ( ζ ) , (2.3)and so, |h∇ u ( x ) , l i| ≤ (cid:18)Z S n − |h∇P B n ( x, ζ ) , l i| q dσ ( ζ ) (cid:19) q k φ k L p ( S n − , R ) . (2.4)On the other hand, for every p ∈ (1 , ∞ ], x ∈ B n and l ∈ S n − , let φ l ( ζ ) = |h∇P B n ( x, ζ ) , l i| q/p sign h∇P B n ( x, ζ ) , l i on S n − and u l = P B n [ φ l ] in B n . Then we have |h∇ u l ( x ) , l i| = (cid:18)Z S n − |h∇P B n ( x, ζ ) , l i| q dσ ( ζ ) (cid:19) q k φ l k L p ( S n − , R ) . This, together with (1.5) and (2.4), implies that for any q ∈ [1 , ∞ ), x ∈ B n and l ∈ S n − , C B n ,q ( x ; l ) = (cid:18)Z S n − |h∇P B n ( x, ζ ) , l i| q dσ ( ζ ) (cid:19) q . (2.5)In the following, we calculate the integral above. For any η ∈ S n − and x ∈ B n ,let ζ = T x ( η ), where T x ( η ) = x − (1 − | x | ) η − x | η − x | . Then ζ = T x ( η ) is a transformation from S n − onto S n − (cf. [2, Section 2.5])satisfying x − ζ = (1 − | x | ) η − x | η − x | , | x − ζ | = 1 − | x | | η − x | (2.6)and dσ ( ζ ) = (1 − | x | ) n − | η − x | n − dσ ( η )(2.7) cf. [14, Page 250]). Combining (2.2) and (2.6), we get ∇P B n ( x, ζ ) = − n − η | η − x | n − (1 − | x | ) n , which, together with (2.7), yields that |h∇P B n ( x, ζ ) , l i| q dσ ( ζ ) = 2 q ( n − q | η − x | n − q − |h η, l i| q (1 − | x | ) n ( q − dσ ( η ) . Therefore, for any q ∈ [1 , ∞ ), x ∈ B n and l ∈ S n − , Z S n − |h∇P B n ( x, ζ ) , l i| q dσ ( ζ ) = 2 q ( n − q (1 − | x | ) n ( q − Z S n − | η − x | n − q − |h η, l i| q dσ ( η ) . From this and (2.5), we see that Lemma 2.1 is true. (cid:3)
Based on Lemma 2.1, we obtain the following two results. The first one is aboutthe symmetry property of the quantity C B n ,q ( x ; l ), which is useful. Lemma 2.2.
For any q ∈ [1 , ∞ ) , x ∈ B n , l ∈ S n − and unitary transformation A in R n , we have C B n ,q ( x ; l ) = C B n ,q ( Ax ; Al ) . Proof.
For any unitary transformation A , by replacing η with Aη in (2.1), we get C B n ,q ( Ax ; Al )= 2( n − − | x | ) n ( q − q (cid:18)Z S n − | Aη − Ax | n − q − |h Aη, Al i| q dσ ( Aη ) (cid:19) q = 2( n − − | x | ) n ( q − q (cid:18)Z S n − | η − x | n − q − |h η, l i| q dσ ( η ) (cid:19) q = C B n ,q ( x ; l ) , as required. (cid:3) For any q ∈ [1 , ∞ ), x ∈ B n and l ∈ S n − , let C B n ,q ( x ; l ) = Z S n − | η − x | n − q − |h η, l i| q dσ ( η ) . (2.8)Then we obtain the following result. Lemma 2.3.
For any q ∈ [1 , ∞ ) , α ∈ [0 , π ] and ρ ∈ [0 , , we have C B n ,q ( ρl α ; e n )= Γ( n )(1 + ρ ) ( n − q − √ π ∞ X k =0 k X j =0 ( ( n − − q )2 ) k ( − ( n − q − ) k Γ( k + q + n ) (cid:18) ρ ρ (cid:19) k × j Γ( j + q +12 )(1) k − j (2 j )! sin k − j α cos j α, where l α = sin α · e n − + cos α · e n and e n − = (0 , . . . , , , ∈ S n − . roof. For any q ∈ [1 , ∞ ), α ∈ [0 , π ] and ρ ∈ [0 , πn − C B n ,q ( ρl α ; e n )(2.9)= 2 πn − Z S n − | η n | q (1 + ρ − ρη n cos α − ρη n − sin α ) ( n − q − dσ ( η )= Z D | x | q (1 − x − y ) n − (1 + ρ − ρx cos α − ρy sin α ) ( n − q − dxdy = Z − | x | q Z √ − x −√ − x (1 − x − y ) n − (1 + ρ − ρx cos α − ρy sin α ) ( n − q − dydx, where η = ( η , . . . , η n ) ∈ S n − . Let y = t √ − x , where t ∈ ( − ,
1) and x ∈ ( − , Z √ − x −√ − x (1 − x − y ) n − (1 + ρ − ρx cos α − ρy sin α ) ( n − q − dy (2.10)= (1 − x ) n − Z − (1 − t ) n − (1 + ρ − ρx cos α − ρt √ − x sin α ) ( n − q − dt. Since for any m ≥ s ∈ ( − , − s ) m = ∞ X k =0 (cid:16) mk (cid:17) ( − s ) k = ∞ X k =0 ( − m ) k k ! s k (cf. [4, Page 85]). Then(1 + ρ − ρx cos α − ρt √ − x sin α ) ( n − q − = (1 + ρ ) ( n − q − (cid:18) − ρ ρ ( x cos α + t √ − x sin α ) (cid:19) ( n − q − = (1 + ρ ) ( n − q − ∞ X k =0 (cid:0) ( n − − q ) (cid:1) k k ! (cid:18) ρ ρ (cid:19) k ( x cos α + t √ − x sin α ) k = (1 + ρ ) ( n − q − ∞ X k =0 (cid:0) ( n − − q ) (cid:1) k k ! (cid:18) ρ ρ (cid:19) k × ∞ X j =0 ( − j ( − k ) j j ! x j t k − j (1 − x ) k − j sin k − j α cos j α. his, together with (2.9) and (2.10), shows that2 πn − C B n ,q ( ρl α ; e n )=(1 + ρ ) ( n − q − ∞ X k =0 ∞ X j =0 (cid:0) ( n − − q ) (cid:1) k k ! (cid:18) ρ ρ (cid:19) k ( − j ( − k ) j j ! × Z − | x | q x j (1 − x ) n − k − j dx Z − t k − j (1 − t ) n − dt · sin k − j α cos j α. It follows from the fact (cf. [18, Pages 18 and 19]) Z − | x | q x j (1 − x ) n − + k − j dx = Z x q − + j (1 − x ) n − + k − j dx = Γ( q +12 + j )Γ( n − + k − j )Γ( k + n + q )and Z − t k − j (1 − t ) n − dt = Z t k − j − (1 − t ) n − dt = Γ( k − j + )Γ( n − k − j + n − )that π Γ( n ) C B n ,q ( ρl α ; e n ) =(1 + ρ ) ( n − q − ∞ X k =0 ∞ X j =0 (cid:0) ( n − − q ) (cid:1) k Γ(2 k + 1) (cid:18) ρ ρ (cid:19) k ( − k ) j Γ(2 j + 1) × Γ( q +12 + j )Γ( k − j + )Γ( k + n + q ) · sin k − j α cos j α. Further, for any a ∈ R , b > k ∈ N , we have (cf. [18, Pages 23 and 24]).(2 a ) k = 2 k ( a ) k (cid:0) a + 12 (cid:1) k and √ π Γ(2 b ) = 2 b − Γ( b )Γ( b + 12 ) . (2.11)Therefore, C B n ,q ( ρl α ; e n ) =Γ( n ρ ) ( n − q − ∞ X k =0 k X j =0 (cid:0) ( n − − q )2 (cid:1) k (cid:0) − ( n − q − (cid:1) k Γ( k + )Γ( j + ) (cid:18) ρ ρ (cid:19) k × ( − k ) j ( − k ) j Γ( k − j + )Γ( j + 1)Γ( k + 1) Γ( q +12 + j )Γ( k + n + q ) sin k − j α cos j α. Note that( − k ) j Γ( k − j + 1) = ( − j Γ( k + 1) and ( 12 − k ) j Γ( k − j + 12 ) = ( − j Γ( k + 12 ) . his, together with (2.11), means C B n ,q ( ρl α ; e n )(2.12) = Γ( n ρ ) ( n − q − ∞ X k =0 k X j =0 (cid:0) ( n − − q )2 (cid:1) k (cid:0) − ( n − q − (cid:1) k Γ( j + )Γ( j + 1)Γ( k − j + 1) × Γ( q +12 + j )Γ( k + n + q ) (cid:18) ρ ρ (cid:19) k sin k − j α cos j α = Γ( n )(1 + ρ ) ( n − q − √ π ∞ X k =0 k X j =0 ( ( n − − q )2 ) k ( − ( n − q − ) k Γ( k + q + n ) (cid:18) ρ ρ (cid:19) k × j Γ( j + q +12 )(1) k − j (2 j )! sin k − j α cos j α. The proof of the lemma is complete. (cid:3) Proofs of Theorems 1.1 ∼ The aim of this section is to prove Theorems 1.1 ∼ Proofs of Theorems 1.2(1) and 1.5(1). (I) First, we prove Theorem 1.2(1).For any x ∈ B n and l ∈ S n − , in order to find the sharp quantity C B n ,q ( x ; l ), wechoose an unity transformation A in R n such that Ax = ρl α and Al = e n , where ρ = | x | ∈ [0 , l α = sin α · e n − + cos α · e n and α ∈ [0 , π ]. By Lemma 2.2, (2.1) and(2.8), we see that C B n ,q ( x ; l ) = C B n ,q ( Ax ; Al ) = C B n ,q ( ρl α ; e n )(3.1) = 2( n − − | x | ) n ( q − q C q B n ,q ( ρl α ; e n ) . Hence, to prove the theorem, it remains to calculate C B n ,q ( ρl α ; e n ). By Lemma 2.3,we obtain C B n ,q ( ρl α ; e n )= Γ( n )(1 + ρ ) ( n − q − √ π ∞ X k =0 k X j =0 ( ( n − − q )2 ) k ( − ( n − q − ) k Γ( k + q + n ) (cid:18) ρ ρ (cid:19) k × j Γ( j + q +12 )(1) k − j (2 j )! sin k − j α cos j α. In order to estimate the right side of the above equation, for any k ∈ N and j ∈ { , , . . . , k } , we let B q,k ( j ) = 4 j Γ( j + q +12 )(1) k − j (2 j )! . laim 3.1. For any k ∈ N and j ∈ { , , . . . , k } , (cid:18) kj (cid:19) Γ( q +12 ) k ! ≤ B q,k ( j ) ≤ k (cid:18) kj (cid:19) Γ( k + q +12 )(2 k )! . The first equality occurs if and only if j = 0 , and the second one occurs if and onlyif j = k . From the fact that (cid:18) kj (cid:19) = Γ( k + 1)Γ( j + 1)Γ( k − j + 1) , we obtain B n,k ( j ) (cid:30)(cid:18) kj (cid:19) = 4 j Γ( j + q +12 )Γ( j + 1)(2 j )!Γ( k + 1) . For k ∈ N and j ∈ { , , . . . , k } , set a q,k ( j ) = 4 j Γ( j + q +12 )Γ( j + 1)(2 j )!Γ( k + 1) . Since a q,k ( j + 1) a q,k ( j ) = 4( j + q +12 )( j + 1)(2 j + 2)(2 j + 1) = 2 j + q + 12 j + 1 > , we getΓ( q +12 ) k ! = a q,k (0) ≤ B n,k ( j ) (cid:30)(cid:18) kj (cid:19) = a q,k ( j ) ≤ a q,k ( k ) = 4 k Γ( k + q +12 )(2 k )! , which implies that Claim 3.1 is true.By the assumption in the theorem that q ∈ [ K − n − + 1 , K n − + 1] ∩ [1 , ∞ ) and K ∈ N , we see that for any k ≥ (cid:18) ( n − − q )2 (cid:19) k (cid:18) − ( n − q − (cid:19) k ≥ . Now, we infer from (1.7), (1.8), (2.11), (2.12) and Claim 3.1 that √ πC B n ,q ( ρl α ; e n )Γ( n )(1 + ρ ) ( n − q − (3.2) ≤ ∞ X k =0 k X j =0 ( ( n − − q )2 ) k ( − ( n − q − ) k Γ( k + q + n ) (cid:18) ρ ρ (cid:19) k k Γ( k + q +12 )(2 k )! × (cid:18) kj (cid:19) sin k − j α cos j α = ∞ X k =0 k ( ( n − − q )2 ) k ( − ( n − q − ) k Γ( k + q +12 )(2 k )!Γ( k + q + n ) (cid:18) ρ ρ (cid:19) k = Γ( q +12 )Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − , q + 12 ; 12 , q + n ρ (1 + ρ ) (cid:19) . he equality occurs if and only if α = 0 and α = π , which shows C B n ,q ( ρl α ; e n ) ≤ C B n ,q ( ± ρe n ; e n ) . (3.3)Similarly, we get from (2.12) and Claim 3.1 that √ πC B n ,q ( ρl α ; e n )Γ( n )(1 + ρ ) ( n − q − ≥ ∞ X k =0 k X j =0 ( ( n − − q )2 ) k ( − ( n − q − ) k Γ( k + q + n ) (cid:18) ρ ρ (cid:19) k Γ( q +12 ) k ! (cid:18) kj (cid:19) sin k − j α cos j α = ∞ X k =0 ( ( n − − q )2 ) k ( − ( n − q − ) k Γ( q +12 ) k !Γ( k + q + n ) (cid:18) ρ ρ (cid:19) k , which, together with (1.7) and (1.8), yields √ πC B n ,q ( ρl α ; e n )Γ( n )(1 + ρ ) ( n − q − (3.4) ≥ Γ( q +12 )Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − q + n ρ (1 + ρ ) (cid:19) . The equality occurs if and only if α = π , which implies C B n ,q ( ρl α ; e n ) ≥ C B n ,q ( ρe n − ; e n ) = C B n ,q ( ρe n ; e n − ) . (3.5)Then for any x ∈ B n \{ } , l ∈ S n − , and t x ∈ S n − with h t x , x | x | i = 0, by (1.6),(3.1), (3.3), (3.5) and Lemma 2.2, we see that C B n ,q (cid:0) x ; t x (cid:1) = C B n ,q ( | x | e n ; e n − ) ≤ C B n ,q (cid:0) x ; l (cid:1) = C B n ,q ( | x | l α ; e n ) ≤ C B n ,q ( | x | e n ; ± e n ) = C B n ,q (cid:0) x ; ± x | x | (cid:1) = C B n ,q ( x ) . In particular, for x ∈ B n \{ } , we have C B n ,q (cid:0) x ; t x (cid:1) < C B n ,q (cid:0) x ; ± x | x | (cid:1) . The proof of Theorem 1.2(1) is complete.(II) By (1.4), (1.6), (3.1) and (3.2), we see that Theorem 1.5(1) is true. (cid:3)
Proof of Theorem 1.3(1).
For any x ∈ B n and l ∈ S n − , by (3.1) and letting q = 1 in (3.2) and (3.4), respectively, we get C B n , ( x ) ≡ C B n , ( x ; l ) ≡ n − n ) √ π Γ( n +12 )(1 − | x | ) . Similarly, by (3.1) and letting q = nn − in (3.2) and (3.4), respectively, we obtain C B n , nn − ( x ) ≡ C B n , nn − ( x ; l ) ≡ n − − | x | ) n − n Γ( n )Γ( n − n − ) √ π Γ( n n − ) (1 + | x | ) ! n − n . Hence, Theorem 1.3(1) follows. (cid:3) .3. Proofs of Theorems 1.1(1) and 1.4(1). (I) First, we prove Theorem 1.1(1).Since q ∈ (1 , nn − ), we see that for any k ∈ Z + , (cid:18) ( n − − q )2 (cid:19) k (cid:18) − ( n − q − (cid:19) k < . For any ρ ∈ [0 ,
1) and α ∈ [0 , π ], similar arguments as in the proofs of (3.2) and(3.4) guarantee that √ πC B n ,q ( ρl α ; e n )Γ( n )(1 + ρ ) ( n − q − (3.6) ≥ Γ( q +12 )Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − , q + 12 ; 12 , q + n ρ (1 + ρ ) (cid:19) and √ πC B n ,q ( ρl α ; e n )Γ( n )(1 + ρ ) ( n − q − (3.7) ≤ Γ( q +12 )Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − q + n ρ (1 + ρ ) (cid:19) . The equality holds in (3.6) for α = 0 and α = π . The equality holds in (3.7) for α = π . Therefore, C B n ,q ( ρl α ; e n ) ≥ C B n ,q ( ρe n ; ± e n ) and C B n ,q ( ρl α ; e n ) ≤ C B n ,q ( ρe n ; e n − ) . (3.8)Then for any x ∈ B n \{ } , l ∈ S n − and t x ∈ S n − with h t x , x | x | i = 0, we obtainfrom (1.6), (3.1), (3.8) and Lemma 2.2 that C B n ,q (cid:0) x ; ± x | x | (cid:1) = C B n ,q (cid:0) | x | e n ; ± e n (cid:1) ≤ C B n ,q (cid:0) x ; l (cid:1) = C B n ,q ( | x | l α ; e n ) ≤ C B n ,q ( | x | e n ; e n − ) = C B n ,q (cid:0) x ; t x (cid:1) = C B n ,q ( x ) . In particular, for any x ∈ B n \{ } , we have C B n ,q (cid:0) x ; ± x | x | (cid:1) < C B n ,q (cid:0) x ; t x (cid:1) . The proof of Theorem 1.1(1) is complete.(II) By (1.4), (1.6), (3.1) and (3.7), we see that Theorem 1.4(1) is true. (cid:3) Proofs of Theorems 1.6 and 1.7 for the unit ball case
The aim of this section is to prove Theorems 1.6(1) and 1.7(1). First, we establisha representation for C B n , ∞ ( x ; l ). Lemma 4.1.
For any x ∈ B n and l ∈ S n − , we have C B n , ∞ ( x ; l ) = sup ζ ∈ S n − |h∇P B n ( x, ζ ) , l i| . roof. Let u = P B n [ φ ] in B n , where φ ∈ L ( S n − , R ). For any x ∈ B n and l ∈ S n − ,it follows from (2.3) that |h∇ u ( x ) , l i| ≤ sup ζ ∈ S n − |h∇P B n ( x, ζ ) , l i| · k φ k L ( S n − , R ) , which means C B n , ∞ ( x ; l ) ≤ sup ζ ∈ S n − |h∇P B n ( x, ζ ) , l i| . (4.1)Next, we show the sharpness of (4.1). For any x ∈ B n and l ∈ S n − , by (2.2), wehave h∇P B n ( x, ζ ) , l i = − n − − | x | ) n − (cid:10) x | x − ζ | + (1 − | x | )( x − ζ ) , l (cid:11) | x − ζ | n . (4.2)Obviously, the mapping ( x, l, ζ )
7→ h∇P B n ( x, ζ ) , l i is continuous in B n × S n − × S n − .Then for any x ∈ B n and l ∈ S n − , there exists ζ ∗ = ζ ∗ ( x, l ) ∈ S n − such thatmax ζ ∈ S n − h∇P B n ( x, ζ ) , l i = h∇P B n ( x, ζ ∗ ) , l i , (4.3)where ζ ∗ ( x, l ) means that the point ζ ∗ depends only on x and l .For i ∈ Z + , ζ ∈ S n − and x ∈ B n , we let φ i ( ζ ) = χ Ω i ( ζ ) || χ Ω i || L ( S n − , R ) and u i ( x ) = P B n [ φ i ]( x ) , where Ω i = { ζ ∈ S n − : | ζ − ζ ∗ | ≤ i } and χ is the indicator function. Obviously, forany i ∈ Z + , || φ i || L ( S n − , R ) = 1 and h∇ u i ( x ) , l i = Z S n − h∇P B n ( x, ζ ) , l i χ Ω i ( ζ ) || χ Ω i || L ( S n − , R ) dσ ( ζ ) . (4.4) Claim 4.1.
For any x ∈ B n and l ∈ S n − , lim i →∞ Z S n − h∇P B n ( x, ζ ) , l i · χ Ω i ( ζ ) || χ Ω i || L ( S n − , R ) dσ ( ζ ) = max ζ ∈ S n − h∇P B n ( x, ζ ) , l i . By the definition of χ Ω i and the continuity of the mapping ( x, l, ζ )
7→ h∇P B n ( x, ζ ) , l i ,we obtain lim i →∞ (cid:0) h∇P B n ( x, ζ ) , l i − h∇P B n ( x, ζ ∗ ) , l i (cid:1) · χ Ω i ( ζ ) = 0 . Then for any ε >
0, there exists a positive integer m = m ( ε ) such that for any i ≥ m , (cid:12)(cid:12) h∇P B n ( x, ζ ) , l i − h∇P B n ( x, ζ ∗ ) , l i (cid:12)(cid:12) · χ Ω i ( ζ ) < ε. Since R S n − χ Ω i ( ζ ) || χ Ω i || L S n − , R ) dσ ( ζ ) = 1, then for any i ≥ m , (cid:12)(cid:12)(cid:12)(cid:12)Z S n − h∇P B n ( x, ζ ) , l i χ Ω i ( ζ ) || χ Ω i || L ( S n − , R ) dσ ( ζ ) − h∇P B n ( x, ζ ∗ ) , l i (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z S n − (cid:12)(cid:12)(cid:0) h∇P B n ( x, ζ ) , l i − h∇P B n ( x, ζ ∗ ) , l i (cid:1) · χ Ω i ( ζ ) (cid:12)(cid:12) · χ Ω i ( ζ ) || χ Ω i || L ( S n − , R ) dσ ( ζ ) ≤ ε, which, together with (4.3), yields that Claim 4.1 is true. ow, it follows from (4.4) and Claim 4.1 thatlim i →∞ h∇ u i ( x ) , l i = max ζ ∈ S n − h∇P B n ( x, ζ ) , l i · lim i →∞ k φ i k L ( S n − , R ) . This, together with (4.1), shows that for any x ∈ B n and l ∈ S n − , C B n , ∞ ( x ; l ) = max ζ ∈ S n − h∇P B n ( x, ζ ) , l i , which is what we need. (cid:3) For any α ∈ [0 , π ] and ρ ∈ [0 , l α = sin α · e n − + cos α · e n and C B n , ∞ ( ρe n ; l α ) = max ζ ∈ S n − (cid:12)(cid:12) − ρ ρ sin α · ζ n − + ( ζ n − ρ ρ ) cos α (cid:12)(cid:12) | ρe n − ζ | n , (4.5)where ζ = ( ζ , ζ , . . . , ζ n ) ∈ S n − . Then, we obtain the following result. Lemma 4.2.
For any α ∈ [0 , π ] and ρ ∈ [0 , , we have C B n , ∞ ( ρe n ; l α ) = 2( n − ρ )(1 − ρ ) n − C ∞ ( ρe n ; l α ) . (4.6) Proof.
For any x ∈ B n and l ∈ S n − , by (4.2), we havesup ζ ∈ S n − |h∇P B n ( x, ζ ) , l i| (4.7) = 2( n − − | x | ) n − max ζ ∈ S n − (cid:12)(cid:12)(cid:10) x | x − ζ | + (1 − | x | )( x − ζ ) , l (cid:11)(cid:12)(cid:12) | x − ζ | n . Let x = ρe n and l = l α , where ρ ∈ [0 , ζ ∈ S n − (cid:12)(cid:12)(cid:10) x | x − ζ | + (1 − | x | )( x − ζ ) , l (cid:11)(cid:12)(cid:12) | x − ζ | n = (1 + ρ ) max ζ ∈ S n − (cid:12)(cid:12) − ρ ρ sin α · ζ n − + ( ζ n − ρ ρ ) cos α (cid:12)(cid:12) | ρe n − ζ | n . This, together with Lemma 4.1, (4.5) and (4.7), implies that (4.6) holds true. (cid:3)
Lemma 4.3.
For any x ∈ B n , l ∈ S n − and unitary transformation A in R n , wehave C B n , ∞ ( x ; l ) = C B n , ∞ ( Ax ; Al ) . Proof.
For any x ∈ B n , l ∈ S n − and unitary transformation A in R n , it follows fromLemma 4.1 and (4.7) that C B n , ∞ ( Ax ; Al )= 2( n − − | x | ) n − max ζ ∈ S n − (cid:12)(cid:12)(cid:10) Ax | Ax − ζ | + (1 − | x | )( Ax − ζ ) , Al (cid:11)(cid:12)(cid:12) | Ax − ζ | n . et ξ = A − ζ . By the fact | Aξ − Ax | = | ξ − x | and h Ax, Al i = h x, l i , we find C B n , ∞ ( Ax ; Al ) = 2( n − − | x | ) n − max ξ ∈ S n − (cid:12)(cid:12)(cid:10) Ax | x − ξ | + (1 − | x | )( Ax − Aξ ) , Al (cid:11)(cid:12)(cid:12) | x − ξ | n = 2( n − − | x | ) n − max ξ ∈ S n − (cid:12)(cid:12)(cid:10) x | x − ξ | + (1 − | x | )( x − ξ ) , l (cid:11)(cid:12)(cid:12) | x − ξ | n = C B n , ∞ ( x ; l ) , as required. (cid:3) Proofs of Theorems 1.6(1) and 1.7(1). (I) First, we prove Theorem 1.6(1).For any x ∈ B n and l ∈ S n − , we choose an unity transformation A in R n such that Ax = ρe n and Al = l α , where ρ = | x | ∈ [0 , l α = sin α · e n − + cos α · e n and α ∈ [0 , π ]. For any x ∈ B n , by (1.6), Lemmas 4.2 and 4.3, we get C B n , ∞ ( x ) = sup l ∈ S n − C B n , ∞ ( x ; l ) = sup l ∈ S n − C B n , ∞ ( Ax ; Al ) = sup α ∈ [0 ,π ] C B n , ∞ ( ρe n ; l α )(4.8) = 2( n − ρ )(1 − ρ ) n − sup α ∈ [0 ,π ] C B n , ∞ ( ρe n ; l α ) . Hence, to prove the theorem, it remains to estimate the quantity sup α ∈ [0 ,π ] C B n , ∞ ( ρe n ; l α ). Claim 4.2.
For any α ∈ [0 , π ] and ρ ∈ [0 , , sup α ∈ [0 ,π ] C B n , ∞ ( ρe n ; l α ) = C B n , ∞ ( ρe n ; ± e n ) = 1(1 + ρ )(1 − ρ ) n − . Using (4.5) and spherical coordinate transformation (cf. [2, Equation (2.2)]), wefind C B n , ∞ ( ρe n ; l α ) = max β ∈ [0 ,π ] max γ ∈ [0 ,π ] (cid:12)(cid:12) − ρ ρ sin α sin β cos γ + (cos β − ρ ρ ) cos α (cid:12)(cid:12) (1 + ρ − ρ cos β ) n . Since the maximum in γ is attained either at γ = 0 or γ = π , we get C B n , ∞ ( ρe n ; l α ) = max β ∈ [0 , π ] (cid:12)(cid:12) − ρ ρ sin α sin β + (cos β − ρ ρ ) cos α (cid:12)(cid:12) (1 + ρ − ρ cos β ) n . (4.9)Therefore,sup α ∈ [0 ,π ] C B n , ∞ ( ρe n ; l α ) ≥ C B n , ∞ ( ρe n ; ± e n ) = max β ∈ [0 , π ] (cid:12)(cid:12) cos β − ρ ρ (cid:12)(cid:12) (1 + ρ − ρ cos β ) n (4.10) ≥ (cid:12)(cid:12) − ρ ρ (cid:12)(cid:12) (1 − ρ ) n = 1(1 + ρ )(1 − ρ ) n − . On the other hand, by (4.9) and CauchySchwarz inequality, we have C B n , ∞ ( ρe n ; l α ) ≤ − ρ ) n − max β ∈ [0 , π ] (cid:0) − ρ ρ sin β ) + (cos β − ρ ρ ) (1 + ρ − ρ cos β ) ! . ince (cid:0) − ρ ρ sin β ) + (cos β − ρ ρ ) (1 + ρ − ρ cos β ) = 1(1 + ρ ) , we have C B n , ∞ ( ρe n ; l α ) ≤ ρ )(1 − ρ ) n − . This, together with (4.10), impliessup α ∈ [0 ,π ] C B n , ∞ ( ρe n ; l α ) = C B n , ∞ ( ρe n ; ± e n ) = 1(1 + ρ )(1 − ρ ) n − , and so, the claim is true.By (4.8), Claim 4.2 and Lemma 4.3, we see that for any x ∈ B n \{ } and l ∈ S n − , C B n , ∞ ( x ; l ) ≤ C B n , ∞ ( x ; ± x | x | ) = C B n , ∞ ( x ) = 2( n −
1) (1 + | x | ) n − (1 − | x | ) n . (4.11)The proof of Theorem 1.6(1) is complete.(II) By (1.4) and (4.11), we see that Theorem 1.7(1) is true. (cid:3) Integral representations for C H n ,q ( x ; l ) with q ∈ [1 , ∞ )For w = ( w , w ) ∈ H , the gradient estimate |∇ U ( w ) | ≤ πw sup x ∈ H | U ( x ) | is sharp if U is a bounded harmonic function in H . Using the conformal transfor-mation from D onto H given by w = i (1+ z )1 − z , one easily transfers the above inequalityinto the following pointwise optimal estimate: |∇ U ( z ) | ≤ π (1 − | z | ) sup x ∈ B | U ( x ) | , where this time U is a bounded harmonic function in D . For the above inequalities,we refer to [3, 9, 14].Assume that u = P H n [ φ ] with φ ∈ L p ( R n − , R ) and T : B n → H n is the M¨obiustransform given by T ( w ) = − e n + 2( w + e n ) | w + e n | . It follows from [19, Exercise 3.5.17 and Theorem 5.3.5] that P H n [ φ ]( x ) = P B n [ φ ] (cid:0) T ( w ) (cid:1) = P B n [ φ ◦ T ]( w ) , where x = T ( w ) ∈ H n . However, it seems that this transformation does not provideany fruitful connection between the two sharp constants C B n ,q ( x ; l ) and C H n ,q ( x ; l ).The main purpose of this section is to establish some general representations forthe sharp quantity C H n ,q ( x ; l ) when q ∈ [1 , ∞ ). Before the proofs, for convenience,we use x ′ to denote the point ( x , . . . , x n − ) ∈ R n − , where x = ( x , . . . , x n ) ∈ R n . emma 5.1. For x = ( x ′ , x n ) ∈ H n and y = ( y ′ , ∈ R n , let ψ ( y ) = y − x | y − x | ∈ S n − − .Then we have dS n − (cid:0) ψ ( y ) (cid:1) = x n | y − x | n dV n − ( y ′ ) , where dS n − is the ( n − -dimensional Lebesgue surface measure.Proof. Let x = ( x ′ , x n ) ∈ H n , y = ( y ′ , ∈ R n and ψ ( y ) = y − x | y − x | ∈ S n − − . Then forany 1 ≤ i ≤ n − ≤ j ≤ n , ∂∂y i ψ j ( y ) = δ i,j | y − x | − ( y i − x i )( y j − x j ) | y − x | and ∂∂y n ψ j ( y ) = 0 , where ψ = ( ψ , . . . , ψ n ) and δ i,j = (cid:26) , if i = j ,0 , if i = j . For any w ∈ R n \{ } , defineΦ( w ) = w w w · · · w w w . . . ...... . . . . . . 0 w n w · · · w n w n − and Ψ( w ) = 1 | w | | w | − w w w · · · w w n − w w | w | − w · · · w w n − w n − w w n − w · · · | w | − w n −
00 0 · · · . Making elementary calculations, we obtain the Jacobian matrix Dψ ( y ) = (cid:0) ∇ ψ ( y ) · · · ∇ ψ n ( y ) (cid:1) T = | y − x | · · ·
00 . . . . . . ...... . . . | y − x | · · · − Φ( y − x ) | y − x | and (cid:0) Dψ ( y ) (cid:1) T Dψ ( y ) = Ψ( y − x ) , where T is the transpose and ∇ ψ i are understood as column vectors. Since theeigenvalues of (cid:0) Dψ ( y ) (cid:1) T Dψ ( y ) are λ = 0 , λ = · · · λ n − = 1 | y − x | and λ n = ( y n − x n ) | y − x | , we see that dS n − ( ψ ( y )) = | y n − x n || y − x | n dV n − ( y ′ ) = x n | y − x | n dV n − ( y ′ ) . The proof of the lemma is complete. (cid:3) ased on Lemma 5.1, we get the following integral representation of C H n ,q ( x, l ),where q ∈ [1 , ∞ ). Lemma 5.2.
For q ∈ [1 , ∞ ) , x ∈ H n and l ∈ S n − , we have C H n ,q ( x ; l ) = 2 n − ( n − n ) π n x n ( q − q n Z S n − (cid:12)(cid:12)(cid:10) e n − h ξ, e n i ξ, l (cid:11)(cid:12)(cid:12) q · h ξ, e n i n − q − n dS n − ( ξ ) ! q . Proof.
Assume that p is the conjugate of q with p ∈ (1 , ∞ ], φ ∈ L p ( R n − , R ) and u = P H n [ φ ] in H n . Let x = ( x ′ , x n ) ∈ H n and y = ( y ′ , ∈ R n . For any i ∈{ , , . . . , n − } , by (1.2), we obtain that ∂∂x i u ( x ) = − n − c n Z R n − x n − n ( x i − y i )( | x ′ − y ′ | + x n ) n φ ( y ′ ) dV n − ( y ′ )= 2( n − c n Z R n − x n − n ( y i − x i ) | y − x | n φ ( y ′ ) dV n − ( y ′ )and ∂∂x n u ( x ) = ( n − c n Z R n − (cid:18) x n − n | y − x | n − + 2 x n − n ( y n − x n ) | y − x | n (cid:19) φ ( y ′ ) dV n − ( y ′ ) , where c n is the constant from (1.3). Therefore, for any x ∈ H n and l ∈ S n − , ∇ u ( x ) = Z R n − ∇P H n ( x, y ′ ) φ ( y ′ ) dV n − ( y ′ )(5.1)and C H n ,q ( x ; l ) ≤ (cid:18)Z R n − |h∇P H n ( x, y ′ ) , l i| q dV n − ( y ′ ) (cid:19) q , (5.2)where ∇P H n ( x, y ′ ) = ( n − c n (cid:18) x n − n e n | y − x | n − + 2 x n − n ( y − x ) | y − x | n (cid:19) . (5.3)In order to calculate the quantity C H n ,q ( x ; l ), we first estimate the right-hand sideof (5.2). Since x n > y n = 0, then x = y . Let e xy = y − x | y − x | . By calculations, we deduce2 x n − n ( y − x ) | y − x | n = 2 x n − n | y − x | n − · x n ( y − x ) | y − x | = − x n − n | y − x | n − (cid:28) y − x | y − x | , e n (cid:29) y − x | y − x | = − x n − n | y − x | n − h e xy , e n i e xy , which implies ∇P H n ( x, y ′ ) = ( n − c n x n − n e n − h e xy , e n i e xy | y − x | n − , (5.4) nd so, (cid:18)Z R n − |h∇P H n ( x, y ′ ) , l i| q dV n − ( y ′ ) (cid:19) q (5.5) = ( n − c n Z R n − x ( n − qn (cid:12)(cid:12)(cid:10) e n − h e xy , e n i e xy , l (cid:11)(cid:12)(cid:12) q | y − x | n − q dV n − ( y ′ ) ! q . In view of Lemma 5.1, we get x ( n − qn | y − x | n − q dV n − ( y ′ ) = x n (1 − q ) − n (cid:18) x n | y − x | (cid:19) n − q − n x n | y − x | n dV n − ( y ′ )(5.6) = x n (1 − q ) − n h ξ, − e n i n − q − n dS n − ( ξ ) . Combing (5.6) and (5.6) yields that (cid:18)Z R n − |h∇P H n ( x, y ′ ) , l i| q dV n − ( y ′ ) (cid:19) q (5.7) = ( n − c n x n ( q − q n Z S n − − (cid:12)(cid:12)(cid:10) e n − h ξ, e n i ξ, l (cid:11)(cid:12)(cid:12) q · h ξ, − e n i n − q − n dS n − ( ξ ) ! q = ( n − c n x n ( q − q n Z S n − (cid:12)(cid:12)(cid:10) e n − h ξ, e n i ξ, l (cid:11)(cid:12)(cid:12) q · h ξ, e n i n − q − n dS n − ( ξ ) ! q , where ξ = y − x | y − x | .Next we show the sharpness of (5.2). Fix l ∈ S n − and w ∈ H n with w =( w , . . . , w n ). For any y ′ ∈ R n − and x ∈ H n , we define φ l ( y ′ ) = |h∇P H n ( w, y ′ ) , l i| qp · sign h∇P H n ( w, y ′ ) , l i and u l ( x ) = P H n [ φ l ]( x ). The similar arguments as above show that k φ l k L p ( R n − , R ) = Z R n − |∇P H n ( w, y ′ ) | q dV n − ( y ′ )= ( n − c n w n ( q − q n Z S n − (cid:12)(cid:12) e n − h ξ, e n i ξ (cid:12)(cid:12) q · h ξ, e n i n − q − n dS n − ( ξ ) ! q < ∞ . Moreover, (5.1) yields that h∇ u l ( w ) , l i = Z R n − |h∇P H n ( w, y ′ ) , l i| q dV n − ( y ′ )= (cid:18)Z R n − |h∇P H n ( w, y ′ ) , l i| q dV n − ( y ′ ) (cid:19) q k φ l k L p ( R n − , R ) . hen the arbitrary of w and l shows the sharpness of (5.2). This, together with(5.7), implies C H n ,q ( x ; l ) = (cid:18)Z R n − |h∇P H n ( x, y ′ ) , l i| q dV n − ( y ′ ) (cid:19) q = ( n − c n x n ( q − q n Z S n − (cid:12)(cid:12)(cid:10) e n − h ξ, e n i ξ, l (cid:11)(cid:12)(cid:12) q · h ξ, e n i n − q − n dS n − ( ξ ) ! q , as required. (cid:3) For q ∈ [1 , ∞ ), x ∈ H n and l ∈ S n − , let C H n ,q ( x ; l ) = Z S n − (cid:12)(cid:12)(cid:10) e n − h ξ, e n i ξ, l (cid:11)(cid:12)(cid:12) q · h ξ, e n i n − q − n dS n − ( ξ ) . (5.8)Thus, in order to estimate C H n ,q ( x ; l ), we only need to calculate C H n ,q ( x ; l ). Byspherical coordinate transformation, we can reformulate C H n ,q ( x ; l ) as follows. Lemma 5.3.
For q ∈ [1 , ∞ ) , x ∈ H n and l ∈ S n − , we have C H n ,q ( x ; l ) = 12 ( n − q Z S n − |h η, l i| q (cid:0) h η, e n i (cid:1) ( n − q − dS n − ( η ) . Proof.
Let ξ = ( ξ , . . . , ξ n ) ∈ S n − and l = ( l , . . . , l n ) ∈ S n − . By (5.8), we get C H n ,q ( x ; l ) = Z S n − (cid:12)(cid:12)(cid:12) ξ n n − X k =1 ξ k l k + (2 ξ n − l n (cid:12)(cid:12)(cid:12) q ξ n − q − nn dS n − ( ξ ) . Let ξ n = cos θ ,ξ n − = sin θ cos θ , ... ξ = sin θ sin θ . . . sin θ n − cos θ n − ,ξ = sin θ sin θ . . . sin θ n − sin θ n − , where θ ∈ [0 , π ), θ , . . . , θ n − ∈ [0 , π ] and θ n − ∈ [0 , π ]. Then we obtain from [2,Section 2.2] that C H n ,q ( x ; l ) = Z π sin n − θ dθ Z π sin n − θ dθ · · · Z π sin θ n − dθ n − × Z π h (2 θ , . . . , θ n − , l ) | cos θ | n − q − n dθ n − , where h (2 θ , . . . , θ n − , l )= (cid:12)(cid:12) sin 2 θ sin θ · · · sin θ n − sin θ n − l + sin 2 θ sin θ · · · sin θ n − cos θ n − l + · · · + sin 2 θ cos θ l n − + cos 2 θ l n (cid:12)(cid:12) q . et ϑ = 2 θ and ϑ i = θ i for i ∈ { , . . . , n − } . Therefore, C H n ,q ( x ; l ) = 12 Z π sin n − ϑ dϑ Z π sin n − ϑ dϑ · · · Z π sin ϑ n − dϑ n − × Z π h ( ϑ , . . . , ϑ n − , l ) (cid:0) cos ϑ (cid:1) n − q − n dϑ n − = Z π sin n − ϑ dϑ Z π sin n − ϑ dϑ · · · Z π sin ϑ n − dϑ n − × Z π h ( ϑ , . . . , ϑ n − , l ) (cid:0) cos ϑ (cid:1) n − q − n − dϑ n − . Set η n = cos ϑ ,η n − = sin ϑ cos ϑ , ... η = sin ϑ sin ϑ . . . sin ϑ n − cos ϑ n − ,η = sin ϑ sin ϑ . . . sin ϑ n − sin ϑ n − . Then η = ( η , . . . , η n ) ∈ S n − \{− e n } , | cos ϑ | = r η n h ( ϑ , . . . , ϑ n − , l ) = |h η, l i| q . Hence, C H n ,q ( x ; l ) = 12 ( n − q Z S n − |h η, l i| q (1 + η n ) ( n − q − dS n − ( η ) . The proof of the lemma is complete. (cid:3) Proofs of Theorems 1.1 ∼ The aim of this section is to prove Theorems 1.1 ∼ Proofs of Theorems 1.6(2) and 1.7(2). (I) First, we prove Theorem 1.6(2).For any x ∈ H n and l ∈ S n − , we infer from (1.5), (1.6) and (5.1) that C H n , ∞ ( x ; l ) ≤ sup y ′ ∈ R n − |h∇P H n ( x, y ′ ) , l i| . (6.1)Let e xy = y − x | y − x | , where y = ( y ′ , ∈ R n . Then (5.4) and the fact x n | y − x | = h e xy , − e n i yield that |h∇P H n ( x, y ′ ) , l i| = ( n − c n x n − n (cid:12)(cid:12)(cid:10) e n − h e xy , e n i e xy , l (cid:11)(cid:12)(cid:12) | y − x | n − (6.2) = ( n − c n x nn (cid:12)(cid:12)(cid:10) e n − h e xy , e n i e xy , l (cid:11)(cid:12)(cid:12) · h e xy , − e n i n − , here c n is the constant from (1.3). Note that e xy ∈ S n − − . Therefore,sup y ′ ∈ R n − |h∇P H n ( x, y ′ ) , l i| = ( n − c n x nn sup ξ ∈ S n − − (cid:12)(cid:12)(cid:10) e n − h ξ, e n i ξ, l (cid:11)(cid:12)(cid:12) · h ξ, − e n i n − (6.3) = ( n − c n x nn sup ξ ∈ S n − (cid:12)(cid:12)(cid:10) e n − h ξ, e n i ξ, l (cid:11)(cid:12)(cid:12) · h ξ, e n i n − . Taking into account the equality (cid:12)(cid:12) e n − h ξ, e n i ξ (cid:12)(cid:12) = 1 givessup y ′ ∈ R n − |h∇P H n ( x, y ′ ) , l i| ≤ ( n − c n x nn sup ξ ∈ S n − (cid:12)(cid:12) e n − h ξ, e n i ξ (cid:12)(cid:12) · h ξ, e n i n − (6.4) = ( n − c n x nn sup ξ ∈ S n − h ξ, e n i n − = ( n − c n x nn , which, together with (1.3) and (6.1), implies C H n , ∞ ( x ) = sup l ∈ S n − C H n , ∞ ( x ; l ) ≤ ( n − c n x nn = 2 n − ( n − n ) π n x nn . (6.5)In the following, we show that the constant in (6.5) is sharp. For each x ∈ H n ,by (6.2) ∼ (6.4), we see thatsup l ∈ S n − sup y ′ ∈ R n − |h∇P H n ( x, y ′ ) , l i| = |h∇P H n ( x, x ′ ) , ± e n i| = 2 n − ( n − n ) π n x nn . (6.6)Fix w = ( w ′ , w n ) ∈ H n . For any j ∈ Z + , we define φ j ( y ′ ) = χ Ω j ( y ′ ) || χ Ω j || L ( H n , R ) in R n − and u j ( x ) = P H n [ φ j ]( x ) in H n , where Ω j = { y ′ ∈ R n − : | y ′ − w ′ | ≤ j } .Then for j ∈ Z + , x ∈ H n and l ∈ S n − , || φ j || L ( H n , R ) = 1 and h∇ u j ( x ) , l i = Z R n − h∇P H n ( x, y ′ ) , l i χ Ω j ( y ′ ) || χ Ω j || L ( H n , R ) dV n − ( y ′ ) . (6.7)For y ′ , w ′ ∈ R n − and x ∈ H n , using (5.3) and the definition of χ Ω j , we find (cid:12)(cid:12) h∇P H n ( x, y ′ ) − ∇P H n ( x, w ′ ) , l i (cid:12)(cid:12) · χ Ω j ( y ′ ) ≤ ( n − c n x n − n (cid:12)(cid:12) | y − x | n − − | w − x | n − (cid:12)(cid:12) | y − x | n − | w − x | n − + 2( n − c n x n − n (cid:12)(cid:12)(cid:12)(cid:12) y − x | y − x | n − w − x | w − x | n (cid:12)(cid:12)(cid:12)(cid:12) , where w = ( w ′ , ∈ R n and y = ( y ′ , ∈ R n . Clearly, (cid:12)(cid:12) | y − x | n − − | w − x | n − (cid:12)(cid:12) | y − x | n − | w − x | n − ≤ | y − w | x n − n n − X k =0 | y − x | k | w − x | n − − k ≤ (2 n − | y ′ − w ′ | x n − n ( | w − x | + | y ′ − w ′ | ) n − nd (cid:12)(cid:12)(cid:12)(cid:12) y − x | y − x | n − w − x | w − x | n (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) ( y − x )( | w − x | n − | y − x | n ) x nn + ( y − w ) | y − x | n x nn (cid:12)(cid:12)(cid:12)(cid:12) ≤ | y − x | · | y − w | x nn n − X k =0 | y − x | k | w − x | n − − k + | y − w | · | y − x | n x nn ≤ (2 n + 1) | y ′ − w ′ | x nn ( | w − x | + | y ′ − w ′ | ) n . Therefore, lim j →∞ (cid:12)(cid:12) h∇P H n ( x, y ′ ) − ∇P H n ( x, w ′ ) , l i (cid:12)(cid:12) · χ Ω j ( y ′ ) = 0 , which means for any ε >
0, there exists a positive integer m = m ( ε, x, w ) suchthat for any j ≥ m and y ′ ∈ R n − , (cid:12)(cid:12) h∇P H n ( x, y ′ ) − ∇P H n ( x, w ′ ) , l i (cid:12)(cid:12) · χ Ω j ( y ′ ) < ε. Since R R n − χ Ω j ( y ′ ) || χ Ω j || L H n, R ) dV n − ( y ′ ) = 1, then for any j ≥ m , (cid:12)(cid:12)(cid:12)(cid:12)Z R n − h∇P H n ( x, y ′ ) , l i χ Ω j ( y ′ ) || χ Ω j || L ( H n , R ) dV n − ( y ′ ) − h∇P H n ( x, w ′ ) , l i (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z R n − (cid:12)(cid:12) h∇P H n ( x, y ′ ) − ∇P H n ( x, w ′ ) , l i · χ Ω j ( y ′ ) (cid:12)(cid:12) · χ Ω j ( y ′ ) || χ Ω j || L ( H n , R ) dV n − ( y ′ ) ≤ ε. Combining this with (6.7), we concludelim j →∞ h∇ u j ( x ) , l i = lim j →∞ Z R n − h∇P H n ( x, y ′ ) , l i χ Ω j ( y ′ ) || χ Ω j || L ( H n , R ) dV n − ( y ′ ) = h∇P H n ( x, w ′ ) , l i . Replacing x by w and l by ± e n in the above equalities, we obtain from (6.6) thatlim j →∞ h∇ u j ( w ) , ± e n i = h∇P H n ( w, w ′ ) , ± e n i = 2 n − ( n − n ) π n x nn lim j →∞ || φ j || L ( H n , R ) . Hence, the sharpness of (6.5) follows and we get C H n , ∞ ( x ) = sup l ∈ S n − C H n , ∞ ( x ; l ) = C H n , ∞ ( x ; ± e n ) = 2 n − ( n − n ) π n x nn . (6.8)The proof of Theorem 1.6(2) is complete.(II) By (1.4) and (6.8), we see that Theorem 1.7(2) is true. (cid:3) .2. Proofs of Theorems 1.1 ∼ For l ∈ S n − , choosean unitary transformation A such that Ae n = l α and Al = e n , where l α = sin α · e n − + cos α · e n and α ∈ [0 , π ]. It follows from Lemma 5.3 that C H n ,q ( x ; l ) = 12 ( n − q Z S n − |h Aη, Al i| q (cid:0) h Ae n , Aη i (cid:1) ( n − q − dS n − ( Aη )(6.9) = 12 ( n − q Z S n − |h ζ , e n i| q (cid:0) h l α , ζ i (cid:1) ( n − q − dS n − ( ζ )= 12 ( n − q − Z S n − |h ζ , e n i| q | ζ − l α | n − q − dS n − ( ζ ) . (I) First, we prove Theorems 1.2(2) and 1.5(2). Assume that q ∈ [ K − n − + 1 , K n − +1] ∩ [1 , ∞ ), where K ∈ N . By (2.8), (3.2) and (3.4), we know that for any ρ ∈ [0 , √ π Γ( q + n )Γ( q +12 )Γ( n )(1 + ρ ) ( n − q − Z S n − |h ζ , e n i| q | ζ − ρl α | n − q − dσ ( ζ )(6.10) ≤ F (cid:18) ( n − − q )2 , − ( n − q − , q + 12 ; 12 , q + n ρ (1 + ρ ) (cid:19) and √ π Γ( q + n )Γ( q +12 )Γ( n )(1 + ρ ) ( n − q − Z S n − |h ζ , e n i| q | ζ − ρl α | n − q − dσ ( ζ )(6.11) ≥ F (cid:18) ( n − − q )2 , − ( n − q − q + n ρ (1 + ρ ) (cid:19) . Observe that12 + q + n − ( n − − q )2 − − ( n − q − − q + 12 > q + n − ( n − − q )2 − − ( n − q − > . Then we deduce from [18, Chapter 5] that the above two series are absolutely con-vergent when ρ ≤
1. Let ρ = 1. It follows from (6.9) ∼ (6.11) that C H n ,q ( x ; l )(6.12) ≤ Γ( q +12 )Γ( n ) ω n − √ π q ( n − Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − , q + 12 ; 12 , q + n (cid:19) and C H n ,q ( x ; l )(6.13) ≥ Γ( q +12 )Γ( n ) ω n − √ π q ( n − Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − q + n (cid:19) , where ω n − denotes the ( n − S n − . The equalityholds in (6.10) for l α = ± e n , which means the equality holds in (6.12) when l = ± e n .Further, the equality holds in (6.11) for l α = ± e n − , which means the equality holds n (6.13) when l = t e n , where t e n ∈ S n − and h t e n , e n i = 0. These, together withLemma 5.2 and (5.8), yield that for any x ∈ H n and l ∈ S n − , C H n ,q ( x ; t e n ) ≤ C H n ,q ( x ; l ) ≤ C H n ,q ( x ; ± e n ) = C H n ,q ( x ) . (6.14)Hence, Theorem 1.2(2) holds true and Theorem 1.5(2) follows from (1.4), (5.8),(6.12) and Lemma 5.2.(II) Next, we show Theorems 1.1(2) and 1.4(2). Assume that q ∈ (1 , nn − ). By(2.8), (3.6), (3.7) and (6.9), we arrive at the following two sharp estimates: C H n ,q ( x ; l ) ≥ Γ( q +12 )Γ( n ) ω n − √ π q ( n − Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − , q + 12 ; 12 , q + n (cid:19) , and C H n ,q ( x ; l )(6.15) ≤ Γ( q +12 )Γ( n ) ω n − √ π q ( n − Γ( q + n ) F (cid:18) ( n − − q )2 , − ( n − q − q + n (cid:19) . Further, for any x ∈ H n and l ∈ S n − , we obtain from (2.8), (3.8), (6.9) and Lemma5.2 that C H n ,q ( x ; e n ) ≤ C H n ,q ( x ; l ) ≤ C H n ,q ( x ; t e n ) = C H n ,q ( x ) . Therefore, Theorem 1.1(2) holds true and Theorem 1.4(2) follows from Lemma 5.2,(5.8) and (6.15).(III) Now, we present the proof of Theorem 1.4(2). For any l ∈ S n − and x ∈ H n ,by letting q = nn − in (6.12) and (6.13), we have C H n , nn − ( x ; l ) ≡ π n Γ( n − n − )2 n − Γ( n n − ) and C H n , nn − ( x ) ≡ C H n , nn − ( x ; l ) . (6.16)Similarly, for any l ∈ S n − and x ∈ H n , by letting q = 1 in (6.12) and (6.13), we get C H n , ( x ; l ) ≡ − n π n − Γ( n ) and C H n , ( x ) ≡ C H n , ( x ; l ) . (6.17)Therefore, Theorem 1.4(2) follows from Lemma 5.2, (5.8), (6.16) and (6.17). (cid:3) Remark . In the case q = 2, we can find a very explicit sharp point estimate forthe half-space case. For any x ∈ H n and l ∈ S n − , it follows from (5.8), (6.9), (6.14),Lemma 5.2 and [15, Lemma 1] (or [2, Theorem G]) that C H n , ( x ; l ) ≤ C H n , ( x ; ± e n ) = 12 n − Z S n − |h η, e n i| | − h η, e n i| n − dS n − ( η )= π n − n − Γ( n − ) Z − (1 − t ) n − t (1 − t ) n − dt. et t = cos 2 s , where t ∈ [0 , π ]. Then C H n , ( x ; ± e n ) = π n − n − Γ( n − ) Z π (sin 2 s ) n − (cos 2 s ) (sin s ) n − ds. Note that for any i, j ≥ Z π sin i s cos j sds = Γ( i +12 )Γ( j +12 )2Γ( i + j + 1)(cf. [18, Page 19]). Therefore, Z π (sin 2 s ) n − (cos 2 s ) (sin s ) n − ds = 2 n − Γ( n +32 )Γ( n − )Γ(2 n ) . Hence, we obtain from Lemma 5.2 and (5.8) that C H n , ( x ; l ) ≤ C H n , ( x ) = C H n , ( x ; ± e n ) = 2 n − ( n − n ) π n x n +12 n π n − Γ( n +32 )Γ( n − )Γ( n − )Γ(2 n ) ! and |∇ u ( x ) | ≤ n − ( n − n ) k φ k L ( R n , R ) π n x n +12 n π n − Γ( n +32 )Γ( n − )Γ( n − )Γ(2 n ) ! . Funding.
The first author is partially supported by NSFS of China (No. 11571216,11671127 and 11801166), NSF of Hunan Province (No. 2018JJ3327), China Schol-arship Council and the construct program of the key discipline in Hunan Province.The third author is partially supported by MPNTR grant 174017, Serbia.
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Jiaolong Chen, Key Laboratory of High Performance Computing and Stochas-tic Information Processing (HPCSIP) (Ministry of Education of China), School ofMathematics and Statistics, Hunan Normal University, Changsha, Hunan 410081,People’s Repulic of China
E-mail address : [email protected] University of Montenegro, Faculty of Natural Sciences and Mathematics, Cetinjskiput b.b. 81000 Podgorica, Montenegro
E-mail address : [email protected] Matematiˇcki fakultet, University of Belgrade, Serbia
E-mail address : [email protected]@matf.bg.ac.rs