Integral Representations in Weighted Bergman Spaces on the Tube Domains
aa r X i v : . [ m a t h . C V ] S e p Integral Representations in Weighted Bergman Spaceson the Tube Domains
Yun Huang ∗ , Guan-Tie Deng † , Tao Qian ‡ Herein, the Laplace transform representations for functions ofweighted holomorphic Bergman spaces on the tube domains aredeveloped. Then a weighted version of the edge-of-the-wedgetheorem is derived as a byproduct of the main results.
Key words : Weighted Bergman space, Tube domain, Laplacetransform, Integral representation, Regular cone
The classical Paley–Wiener theorem asserts that functions of the classical Hardy space H ( C + ) can be written as the Laplace transforms of L functions supported in the righthalf of the real axis, see [1]. This theorem has been extended to more general Hardy spaces,including the H p spaces cases (0 < p ≤ ∞ ), higher dimensional cases and weighted spaces,see [11, 13, 15, 14, 12, 9]. Integral representation theorems have also been investigated forBergman spaces. ∗ Department of Mathematics, Faculty of Science and Technology, University of Macau, Macao (Via HongKong). Email: [email protected]. † Corresponding author. School of Mathematical Sciences, Beijing Normal University, Beijing, 100875.Email:[email protected]. This work was partially supported by NSFC (Grant 11971042) and by SRFDP(Grant 20100003110004) ‡ Institute of Systems Engineering, Macau University of Science and Technology, Avenida Wai Long,Taipa, Macau). Email: [email protected]. The work is supported by the Macau Science and Technol-ogy foundation No.FDCT079/2016/A2 , FDCT0123/2018/A3, and the Multi-Year Research Grants of theUniversity of Macau No. MYRG2018-00168-FST.
1e first introduce some notations and definitions. Let B be a domain (open andconnected set) in R n and T B = R n + iB ⊂ C n be the tube over B . For any element z = ( z , z , . . . , z n ), z k = x k + iy k , by definition, z ∈ T B is and only if x = ( x , . . . , x n ) ∈ R n and y = ( y , . . . , y n ) ∈ B . The inner product of z, w ∈ C n is defined as z · w = z w + z w + · · · + z n w n . The associated Euclidean norm of z is | z | = √ z · ¯ z , where ¯ z = (¯ z , ¯ z , . . . , ¯ z n ).A nonempty subset Γ ⊂ R n is called an open cone if it satisfies (i) 0 / ∈ Γ, and (ii) αx + βy ∈ Γ for any x, y ∈ Γ and α, β >
0. The dual cone of Γ is defined as Γ ∗ = { y ∈ R n : y · x ≥ , for any x ∈ Γ } , which is clearly a closed convex cone with vertex at 0. We saythat the cone Γ is regular if the interior of its dual cone Γ ∗ is nonempty.For p + q = 1, define B p ( T B ) = ( F : F is holomorphic in T B and satisfies Z B (cid:18)Z R n | F ( x + iy ) | p dx (cid:19) q − dy < ∞ ) . Among the previous studies, Genchev showed that the function spaces B p (1 ≤ p ≤ A p ( T Γ ) = (cid:26) F : F is holomorphic on T Γ and satisfies Z T Γ | F ( x + iy ) | p dxdy < ∞ (cid:27) to obtain the corresponding integral representation results for A p ( T Γ ) in the range 1 ≤ p ≤ A p,s ( B, ψ ), ofwhich each is associated with a weight function of the form e − πψ ( y ) , where ψ ( y ) ∈ C ( B ) iscontinuous on B . The space A p,s ( B, ψ )(0 < p ≤ ∞ , < s ≤ ∞ ) is the collection of functions F ( z ) that are holomorphic in T B and satisfy k F k A p,s ( B,ψ ) = (cid:18)Z B (cid:18)Z R n | F ( x + iy ) e − πψ ( y ) | p dx (cid:19) s dy (cid:19) sp < ∞ , 0 < p, s < ∞ , k F k A p, ∞ ( B,ψ ) = sup ( e − πψ ( y ) (cid:18)Z R n | F ( x + iy ) | p dx (cid:19) p , y ∈ B ) < ∞ , 0 < p < ∞ , s = ∞ and k F k A ∞ , ∞ ( B,ψ ) = sup (cid:8) e − πψ ( y ) | F ( x + iy ) | , x ∈ R n , y ∈ B (cid:9) < ∞ , p = ∞ , s = ∞ . This paper is structured as follows. In §
2, we introduce our main work on the integralrepresentation for A p,s ( B, ψ ), which is separated into three cases, namely, A p,s ( B, ψ ) for2 ≤ p ≤ A p,s ( B, ψ ) for 0 < p < A p,s (Γ , ψ ) for p >
2, corresponding to Theorem 1, 2and 3 respectively. The proofs are given in §
3. Finally, some results, referring to Corollary2, Theorem 4 and Theorem 5, are derived as applications of the integral representationtheorems claimed in § In order to introduce our main results, we define the set U α ( B, ψ ) = (cid:26) t ∈ R n : Z B e − πα ( t · y + ψ ( y )) dy < ∞ (cid:27) (1)for α ∈ (0 , ∞ ) and U ∞ ( B, ψ ) = { t : inf y ∈ Γ ( y · t + ψ ( y )) > −∞} (2)for α = ∞ .The representation theorem for A p,s ( B, ψ ), where 1 ≤ p ≤ < s ≤ ∞ , is stated asfollows. Theorem . Assume that ≤ p ≤ , < s ≤ ∞ , then each F ( z ) ∈ A p,s ( B, ψ ) admits anintegral representation in the form F ( z ) = Z R n f ( t ) e πit · z dt, z ∈ T B , (3) in which, for p = 1 , f ( t ) ∈ C ( R n ) satisfies | f ( t ) | (cid:18)Z B e − sπ ( y · t + ψ ( y )) dy (cid:19) s ≤ k F k A ,s ( B,ψ ) (4) and, for < p ≤ and p + q = 1 , f ( t ) is a measurable function that satisfies Z B (cid:18)Z R n | f ( t ) e − π ( y · t + ψ ( y )) | q dt (cid:19) spq dy ! sp ≤ k F k A p,s ( B,ψ ) . (5) Moreover, f is supported in U s ( B, ψ ) for p = 1 and supported in U sp ( B, ψ ) for < p ≤ , < s ( p − ≤ . As given in the next theorem, integral representations in the form of Laplace transformare also available for 0 < p < < s ≤ ∞ .3 heorem . Assume that F ( z ) ∈ A p,s ( B, ψ ) , where < p < and < s ≤ ∞ . Then thereexists a continuous function f ( t ) such that f ( t ) e − πy · t ∈ L ( R n ) and (3) hold for y ∈ B . Considering the property of f ( t ) for the case of 0 < p <
1, we let B be a regular openconvex cone Γ and let ψ ∈ C (Γ) satisfy R ψ = lim y ∈ Γ ,y →∞ ψ ( y ) | y | < ∞ . (6)Then we obtain the following corollary. Corollary . Assume that Γ is a regular open convex cone and F ( z ) ∈ A p,s (Γ , ψ ) for < p < , < s ≤ ∞ , where ψ ∈ C (Γ) satisfies (6). Then there exists f ( t ) supported in Γ ∗ + D (0 , R ψ ) such that (3) holds and | f ( t ) | (cid:0)R Γ e − sπ ( y · t + R ψ | y | ) dy (cid:1) s is slowly increasing. Similarly, we establish an analogy for p > < s ≤ ∞ . Theorem . Assume that p > , < s ≤ ∞ , Γ is a regular open convex cone in R n and ψ ∈ C (Γ) satisfies (6). If F ( z ) ∈ A p,s (Γ , ψ ) satisfying lim y ∈ Γ ,y → Z R n | F ( x + iy ) | dx < ∞ , (7) then there exists f ( t ) ∈ L ( R n ) supported in U sp (Γ , ψ ) such that (3) holds for all z ∈ T Γ . The definition of A p,s ( B, ψ ) shows that A p,s ( B, ψ ) is a weighted Hardy space when s = ∞ and a weighted Bergman space when s = 1. Taking ψ ( y ) = 0, it becomes, for s = ∞ and s = 1, respectively, the classical Hardy space H p and the classical Bergman space A p .Therefore, our results herein can be regarded as generalizations of certain previously obtainedresults.For example, taking s = ∞ and B a regular open convex cone Γ, A p, ∞ ( B, ψ ) = H p (Γ , ψ )is the weighted Hardy spaces investigated in our previous paper [15]. Then Theorem 1, 2and 3 in [15] can be derived from our main work, including Theorem 1, 2, 3 and Corollary 1herein. For s = ∞ and ψ ( y ) = 0, letting B be some specific domains, some previous studiesfor the Hardy spaces, see [1, 13, 14, 12, 9], can be also derived from Theorem 1, 2, 3 andCorollary 1.On the other hand, letting s = 1, by using Theorem 1, 2, 3 and Corollary 1, we can obtainthe representation theorems for the standard Bergman spaces. Note that for s = 1, B = Γand ψ ( y ) = 0, we have A p,s ( B, ψ ) = A p ( T Γ ). We therefore conclude from Theorem 1 thatthe counterpart results of Theorem 1, 2 and 3 in [5] hold for the classical Bergman spaces4 p ( T Γ )(1 ≤ p ≤ ψ ( y ) = 0 and s = q −
1, where p + q = 1, then A p,s ( B, ψ ) = B p ( T B ). The integral representation theorems for those function spaces B p ( T B )(1 ≤ p ≤ s = 1, p = 2, ψ ( y ) = − α π log | y | and B a regular open convex cone Γ, Theorem 1 implies a higher dimensionalgeneralization of Theorem 1 of [10] in tube domains, which is established as Corollary 2 inthe sequel. This section is devoted to proving the results stated in § Proof of Theorem 1.
We first prove the case of p = 1. If F ( z ) ∈ A ,s ( B, ψ ), then F y ( x ) ∈ L ( R n ) as a function of x , and ˇ F y ( x ) as well, are both well defined. Next we prove thatˇ F y ( t ) e − πy · t is independent of y ∈ B . Without loss of generality, assume that a = ( a ′ , a n ), b = ( a ′ , b n ) ∈ B , and a + τ ( b − a ) ∈ B for 0 ≤ τ ≤
1, where a ′ = ( a , . . . , a n − ). The fact F y ( x ) ∈ L ( R n ) implies that Z ∞ Z Z R n − ( | F (( x ′ , x n ) + i ( a + τ ( b − a ))) | + | F (( x ′ , − x n ) + i ( a + τ ( b − a ))) | ) dx ′ dτ dx n < ∞ , which implieslim R →∞ Z Z R n − ( | F (( x ′ , R ) + i ( a + τ ( b − a ))) | + | F (( x ′ , − R ) + i ( a + τ ( b − a ))) | ) dx ′ dτ = 0 . Hence, we have | ˇ F b ( t ) e − πb · t − ˇ F a ( t ) e − πa · t | = (cid:12)(cid:12)(cid:12)(cid:12)Z R n (cid:0) F ( x + ib ) e πi ( x + ib ) · t − F ( x + ia ) e πi ( x + ia ) · t (cid:1) dx (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z R n Z ∂∂τ (cid:0) F ( x + i ( a + τ ( b − a ))) e πi ( x + i ( a + τ ( b − a ))) · t (cid:1) dτ dx (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z R n Z ∂∂y n (cid:16) F ( x + i (( y ′ , y n )) e πi ( x + i ( y ′ ,y n )) · t (cid:12)(cid:12) y n = a n + τ ( b n − a n ) ( b n − a n ) (cid:17) dτ dx (cid:12)(cid:12)(cid:12)(cid:12) = | b n − a n | (cid:12)(cid:12)(cid:12)(cid:12)Z R n Z i ∂∂x n (cid:0) F ( x + i ( a + τ ( b − a ))) e πi ( x + i ( a + τ ( b − a ))) · t (cid:1) dτ dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ | b n − a n | lim R →∞ Z Z R n − ( | F (( x ′ , R ) , ( a + τ ( b − a ))) | + | F (( x ′ , − R ) , ( a + τ ( b − a ))) | ) e − π | t | ( | a | + | b − a | ) dx ′ dτ = 0 . B is connected and open, by an iteration argument, we can show that ˇ F y ( t ) e − πy · t is independent of y ∈ B and we write it as g ( t ). Then g ( t ) = ˇ F y ( t ) e − πy · t holds for y ∈ B . Next, we show that g ( t ) e πy · t ∈ L ( R n ). Let us decompose R n into a finiteunion of non-overlapping polygonal cones, Γ , Γ , . . . , Γ N with their very vertexes at theorigin, i.e., R n = S Nk =1 Γ k . Then χ Γ k ( t ) g ( t ) e πy · t = χ Γ k ( t ) ˇ F y k ( t ) e − π ( y k − y ) · t . For any y ∈ B ,there exists δ such that D ( y , δ ) ⊂ B . Then for any y ∈ D ( y , δ ) and y k ∈ ( y + Γ k )satisfying δ ≤ | y k − y | < δ , we get ( y k − y ) · t = ( y k − y ) · t + ( y − y ) · t . Since y k − y , t ∈ Γ k , the angle between the segments O ( y k − y ) and Ot is less than, say π . Then( y k − y ) · t ≥ | y k − y |√ | t | − | y − y || t | ≥ ( √ − ) δ | t | ≥ δ | t | . Thus, it follows from H¨older’sinequality that Z Γ k | g ( t ) e πy · t | dt ≤ Z Γ k | ˇ F y k ( t ) e − π δ | t | | dt ≤ k F y k ( x ) k L ( R n ) Z Γ k e − π δ | t | dt < ∞ , which shows that g ( t ) e πy · t ∈ L (Γ k ). Hence g ( t ) e πy · t ∈ L ( R n ). Together with the relation g ( t ) = ˇ F y ( t ) e − πy · t for y ∈ B , there holds F ( z ) = R R n g ( t ) e − πiz · t for all y ∈ B . By letting f ( t ) = g ( − t ), we then obtain the desired formula (3) for p = 1 and z ∈ T B .Thus, f ( t ) e − πy · t ∈ L (Γ k ) implies thatsup t ∈ R n | f ( t ) | e − πy · t ≤ Z R n | F ( x + iy ) | dx | f ( t ) | e − πy · t e − πψ ( y ) ≤ Z R n | F ( x + iy ) | e − πψ ( y ) dx | f ( t ) | s Z B e − sπ ( y · t + ψ ( y )) dy ≤ Z B (cid:18)Z R n | F ( x + iy ) | e − πψ ( y ) dx (cid:19) s dy = k F k sA ,s ( B,ψ ) , (8)which implies (4). Next we prove supp f ⊂ U s ( B, ψ ). Suppose that t / ∈ U s ( B, ψ ), then (1)implies R B e − sπ ( y · t + ψ ( y )) dy = + ∞ for y ∈ B . It then follows from (8) that f ( t ) = 0, whichmeans the support of f , i.e., supp f ⊂ U s ( B, ψ ).Next we prove the case 1 < p ≤
2. Let B ⊆ B be a bounded connected open set,so there exists a positive constant R such that B ⊆ D (0 , R ). Assume that l ε ( z ) =(1 + ε ( z + · · · + z n )) N , where N is an integer satisfying N > n . Then for ε ≤ R , z = x + iy with | y | ≤ R , | l ε ( z ) | = | ((1 + ε ( z + · · · + z n )) ) N | = (cid:16)(cid:0) ε ( | x | − | y | ) (cid:1) + 4 ε ( x · y ) (cid:17) N ≥ (cid:0) ε ( | x | − | y | ) (cid:1) N ≥ (cid:18)
12 + ε | x | (cid:19) N | y | ≤ R , i.e., | l − ε ( z ) | ≤ ( + ε | x | ) N . For F y ( x ) = F ( x + iy ), set F ε,y ( x ) = F y ( x ) l − ε ( z ),then based on H¨older’s inequality, Z R n | F ε,y ( x ) | dx ≤ (cid:18)Z R n | F y ( x ) | p dx (cid:19) p (cid:18)Z R n (cid:12)(cid:12) l − ε ( x + iy ) (cid:12)(cid:12) q dx (cid:19) q < ∞ , where p + q = 1, which implies that F ε,y ( x ) ∈ L ( R n ). Then as in the proof for p = 1, g ε,y ( t ) = ˇ F ε,y ( t ) e − πy · t can be also proved to be independent of y ∈ B when 1 < p ≤
2. Put g ε,y ( t ) = g ε ( t ), then g ε ( t ) e πy · t = ˇ F ε,y ( t ) ∈ L ( R n ).On the other hand, it is obvious that F ε,y ( x ) → F y ( x ) pointwise as ε →
0. Now we provethat ˇ F y ( t ) e − πy · t is also independent of y ∈ B . Indeed, for a, b ∈ B and any compact subset K ⊂ R n , let R = max {| z | : z ∈ K } , (cid:18)Z K (cid:12)(cid:12) ˇ F a ( t ) e − πa · t − ˇ F b ( t ) e − πb · t (cid:12)(cid:12) q dt (cid:19) q ≤ (cid:18)Z K (cid:12)(cid:12) ˇ F a ( t ) e − πa · t − g ε ( t ) (cid:12)(cid:12) q dt (cid:19) q + (cid:18)Z K (cid:12)(cid:12) g ε ( t ) − ˇ F b ( t ) e − πb · t (cid:12)(cid:12) q dt (cid:19) q = (cid:18)Z K (cid:12)(cid:12) ˇ F a ( t ) e − πa · t − ˇ F ε,a ( t ) e − πa · t (cid:12)(cid:12) q dt (cid:19) q + (cid:18)Z K (cid:12)(cid:12) ˇ F ε,b ( t ) e − πb · t − ˇ F b ( t ) e − πb · t (cid:12)(cid:12) q dt (cid:19) q ≤ e πR R (cid:18)Z K (cid:12)(cid:12) ˇ F a ( t ) − ˇ F ε,a ( t ) (cid:12)(cid:12) q dt (cid:19) q + (cid:18)Z K (cid:12)(cid:12) ˇ F ε,b ( t ) − ˇ F b ( t ) (cid:12)(cid:12) q dt (cid:19) q ! ≤ e πR R (cid:18)Z R n | F a ( t ) − F ε,a ( t ) | p dt (cid:19) p + (cid:18)Z R n | F ε,b ( t ) − F b ( t ) | p dt (cid:19) p ! → , as ε →
0. Hence we obtain that ˇ F a ( t ) e − πa · t = ˇ F b ( t ) e − πb · t almost everywhere on R n andwrite it as g ( t ). Then we have g ( t ) = ˇ F y ( t ) e − πy · t .Next, we show that g ( t ) e πy · t ∈ L ( R n ). As in the proof for p = 1, let R n = S Nk =1 Γ k and D ( y , δ ) ⊂ B . Then for any y ∈ D ( y , δ ) and y k ∈ ( y + Γ k ) satisfying δ ≤ | y k − y | < δ ,we have ( y k − y ) · t ≥ | y k − y |√ | t | − | y − y || t | ≥ ( 34 √ −
14 ) δ | t | ≥ δ | t | for y k − y , t ∈ Γ k . Thus, from H¨older’s inequality Z Γ k | g ( t ) e πy · t | dt ≤ Z Γ k | ˇ F y k ( t ) e − π δ | t | | dt ≤ (cid:18)Z Γ k | ˇ F y k ( t ) | p dt (cid:19) p (cid:18)Z Γ k | e − qπ δ | t | | dt (cid:19) q < ∞ , which shows that g ( t ) e πy · t ∈ L (Γ k ) and the function G ( z ) defined by G ( z ) = Z R n g ( t ) e − πi ( x + iy ) · t dt
7s holomorphic in the tube domain T D ( y ,δ ) .Now we can prove that, for y ∈ B ,lim ε → Z R n g ε ( t ) e − πi ( x + iy ) · t dt = Z R n g ( t ) e − πi ( x + iy ) · t dt. In fact, if y ∈ B , (cid:12)(cid:12)(cid:12)(cid:12)Z R n ( g ε ( t ) − g ( t )) e − πi ( x + iy ) · t dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z R n (cid:12)(cid:12)(cid:0) ˇ F ε,y ( t ) e − πy · t − ˇ F y ( t ) e − πy · t (cid:1) e πiz · t (cid:12)(cid:12) dt = n X k =1 Z Γ k (cid:12)(cid:12)(cid:0) ˇ F ε,y k ( x ) − ˇ F y k ( x ) (cid:1) e − πi ( y k − y ) · t (cid:12)(cid:12) dt ≤ n X k =1 (cid:18)Z Γ k | ˇ F ε,y k ( x ) − ˇ F y k ( x ) | q dt (cid:19) q (cid:18)Z Γ k e − pπ δ | t | dt (cid:19) p ≤ C δ n X k =1 (cid:18)Z Γ k | F ε,y k ( x ) − F y k ( x ) | p dt (cid:19) p → ε →
0, where C pδ = R R n e − pπ δ | t | dt . It follows that lim ε → F ε ( z ) = G ( z ). Combining withlim ε → F ε ( z ) = F ( z ), we can state G ( z ) = F ( z ) for y ∈ B . Then there exists a measurablefunction g ( t ) such that F ( z ) = R R n g ( t ) e − πiz · t dt holds for y ∈ B . Since B is connected, wecan choose a sequence of bounded connected open set { B k } such that B ⊂ B ⊂ · · · and B = S ∞ k =0 B k . Together with the fact that g ( t ) = ˇ F y ( t ) e − πy · t is independent of y ∈ B k , thenˇ F y l ( t ) e − πy l · t = ˇ F y j ( t ) e − πy j · t = ˇ F y ( t ) e − πy · t for l = j , y l ∈ B l , y j ∈ B j and y ∈ B . Hence g ( t ) e πy · t = ˇ F y ( t ) holds for y ∈ B k , k = 0 , , , . . . . In other words, f ( z ) = R R n g ( t ) e − πiz · t dt holds for all y ∈ B . By letting f ( t ) = g ( − t ), we obtain the desired representation F ( z ) = R R n f ( t ) e πiz · t dt for y ∈ B when 1 < p ≤ p + q = 1, based on the Hausdorff-Young Inequality, (cid:18)Z R n | f ( t ) e − πy · t | q dt (cid:19) q ≤ (cid:18)Z R n | F ( x + iy ) | p dx (cid:19) p , (9)then (cid:18)Z R n | f ( t ) e − πy · t | q dt (cid:19) pq e − pπψ ( y ) dy ! s ≤ (cid:18)Z R n | F ( x + iy ) e − πψ ( y ) | p dx (cid:19) s . Performing integral about y ∈ B on both sides, we get Z B (cid:18)Z R n | f ( t ) e − πy · t | q dt (cid:19) pq e − pπψ ( y ) ! s dy ≤ Z B (cid:18)Z R n | F ( x + iy ) e − πψ ( y ) | p dx (cid:19) s dy Z B (cid:18)Z R n | f ( t ) e − πy · t | q dt (cid:19) pq e − pπψ ( y ) ! s dy ≤ k F k spA p,s ( B,ψ ) . (10)As a result, formulas (3) and (5) hold for 1 < p ≤
2. Now we prove that supp f ⊂ U sp ( B, ψ )when 0 < s ( p − ≤
1. For 0 < s ( p − ≤
1, we have qsp ≥
1. Then Minkowski’s inequalityand (10) imply that Z R n | f ( t ) | q (cid:18)Z B e − πps ( y · t + ψ ( y )) dy (cid:19) qps dt ≤ k F k qA p,s ( B,ψ ) < ∞ . (11)Consequently, It follows from (11) and (1) that f ( t ) = 0 for almost every t U sp ( B, ψ ).Therefore, supp f ⊂ U sp ( B, ψ ).In order to prove Theorem 2, we first introduce a lemma.
Lemma . Suppose that F ( z ) ∈ A p,s ( B, ψ ) , where < p < ∞ and < s ≤ ∞ , then for y ∈ B and positive constant δ such that D n ( y , δ ) ⊂ B , there exist constants N > and C n,N,p,s depending on n, N, p, s such that | F ( z ) | ≤ C n,N,p,s δ − np (1+ s ) e πψ δ ( y ) , (12) where ψ δ ( y ) = max { ψ ( η ) : | η − y | ≤ δ } .Proof. For y ∈ B , there exists δ > B δ = D ( y , δ ) ⊂ B . Then for F ( z ) = F ( x + iy ) ∈ A p,s ( B, ψ ), based on the subharmonic properties of | F ( z ) | t , we have | F ( z ) | t ≤ n δ n Z D n ( z,δ ) | F ( ξ + iη ) | t dξdη ≤ n δ n Z D n ( y ,δ ) (cid:18)Z D n ( x,δ ) | F ( ξ + iη ) | t dξ (cid:19) dη for y ∈ B δ , where Ω k is the volume of k -dimensional unit ball D k (0 ,
1) centered at the originwith radius 1, k = n, n . Let p = N = pt > max { , s } and p + q = 1. H¨older’s Inequalityimplies that | F ( z ) | t ≤ n δ n Z D n ( y ,δ ) (cid:18)Z D n ( x,δ ) | F ( ξ + iη ) | p dξ (cid:19) p dη (cid:18)Z D n ( x,δ ) q dξ (cid:19) q = ( δ n Ω n ) q δ n Ω n Z D n ( y ,δ ) (cid:18)Z D n ( x,δ ) | F ( ξ + iη ) | p dξ (cid:19) p dη. < s < ∞ , let p = sN . Then p >
1. Again, by H¨older’s Inequality, for p + q = 1, | F ( z ) | t ≤ ( δ n Ω n ) q δ n Ω n (cid:18)Z D n ( y ,δ ) (cid:18)Z D n ( x,δ ) | F ( ξ + iη ) | p dξ (cid:19) s dη (cid:19) p (cid:18)Z D n ( y ,δ ) q dη (cid:19) q ≤ ( δ n Ω n ) q + q δ n Ω n (cid:18)Z D n ( y ,δ ) (cid:18)Z D n ( x,δ ) | F ( ξ + iη ) e − πψ ( η ) | p dξ (cid:19) s e spπψ ( η ) dη (cid:19) p ≤ ( δ n Ω n ) − N (1+ s ) e spp πψ δ ( y ) δ n Ω n (cid:18)Z D n ( y ,δ ) (cid:18)Z D n ( x,δ ) | F ( ξ + iη ) e − πψ ( η ) | p dξ (cid:19) s dη (cid:19) p ≤ ( δ n Ω n ) − N (1+ s ) e spp πψ δ ( y ) δ n Ω n (cid:18)Z B (cid:18)Z R n | F ( ξ + iη ) e − πψ ( η ) | p dξ (cid:19) s dη (cid:19) p , where ψ δ ( y ) = max { ψ ( η ) : | η − y | ≤ δ } . Hence, | F ( z ) | ≤ δ − nN (1+ s ) Ω − N (1+ s ) n e spp πψ δ ( y ) Ω n ! t (cid:18)Z B (cid:18)Z R n | F ( ξ + iη ) e − πψ ( η ) | p dξ (cid:19) s dη (cid:19) tp ≤ Ω Np − p (1+ s ) n Ω Np n δ np (1+ s ) e sptp πψ δ ( y ) (cid:18)Z B (cid:18)Z R n | F ( ξ + iη ) e − πψ ( η ) | p dξ (cid:19) s dη (cid:19) sp sptp . Since sptp = 1, by letting C n,N,p,s = Ω Np − p (1+ 1 s ) n Ω Np n k F ( z ) k A p,s ( B,ψ ) , we obtain the desired inequal-ity | F ( z ) | ≤ C n,N,p,s δ − np (1+ s ) e πψ δ ( y ) . While s = ∞ , for p = sN = ∞ , we have | F ( z ) | t ≤ ( δ n Ω n ) − N δ n Ω n sup η ∈ D n ( y,δ ) (cid:12)(cid:12)(cid:12)(cid:12)Z D n ( x,δ ) | F ( ξ + iη ) | p dξ (cid:12)(cid:12)(cid:12)(cid:12) tp . Then | F ( z ) | ≤ ( δ n Ω n ) (2 − N ) Np ( δ n Ω n ) Np e πψ δ ( y ) sup η ∈ D n ( y,δ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18)Z D n ( x,δ ) | F ( ξ + iη ) | p dξ (cid:19) p e − πψ ( y ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = Ω Np − p n Ω Np n δ − np e πψ δ ( y ) k F ( z ) k A p, ∞ ( B,ψ ) . Obviously, the inequality (12) is also applicable in the case s = ∞ .Now we are ready to prove Theorem 2. Proof of Theorem 2.
For y ∈ B , there exists δ > B δ = D ( y , δ ) ⊂ B . Then for F ( z ) ∈ A p,s ( B, ψ ) and any y ∈ B δ , it follows from Lemma 1 that | F ( z ) | ≤ C n,N,p,s δ − np (1+ s ) e πψ δ ( y ) . Z R n | F ( z ) | dx = Z R n | F ( z ) | p +2 − p dx ≤ C − pn,N,p,s δ − n (2 − p ) p (1+ s ) e − p ) πψ δ ( y ) Z R n | F ( z ) | p dx. Therefore, Z R n | F ( z ) | e − πψ δ ( y ) dx ≤ C − pn,N,p,s δ − n (2 − p ) p (1+ s ) e − p ) πψ δ ( y ) Z R n | F ( z ) e − πψ ( y ) | p dxe pπψ ( y ) e − πψ δ ( y ) ≤ C − pn,N,p,s δ − n (2 − p ) p (1+ s ) e − p ) πψ δ ( y ) Z R n | F ( z ) e − πψ ( y ) | p dxe p − πψ δ ( y ) = C − pn,N,p,s δ − n (2 − p ) p (1+ s ) Z R n | F ( z ) e − πψ ( y ) | p dx. Taking integral with respect to y to both sides of the inequality, we have Z B δ (cid:18)Z R n | F ( z ) | e − πψ δ ( y ) dx (cid:19) s dy ≤ C (2 − p ) sn,N,p,s δ − n (2 − p )(1+ s ) p Z B δ (cid:18)Z R n | F ( z ) e − πψ ( y ) | p dx (cid:19) s dy, which concludes that F ∈ A ,s ( B δ , ψ δ ). Similarly, we can prove that Z B δ (cid:18)Z R n | F | e − πψ δ ( y ) dx (cid:19) s dy ≤ C (1 − p ) sn,N,p,s δ − n (1 − p )(1+ s ) p k F ( z ) k spA ,s ( B δ ,ψ ) . (13)Then F ( z ) ∈ A ,s ( B δ , ψ δ ).Following the proof of the case p = 1 in Theorem 1, there exists a continuous function f ( t ) such that F y ( x ) = R R n f ( t ) e πiz · t dt holds for y ∈ B δ and f ( t ) = ˆ F y ( t ) e πy · t is independentof y ∈ B . Together with the fact that f ( t ) e − πy · t ∈ L ( R n ) for all y ∈ B , we see that (3)holds for all y ∈ B . This completes the proof of Theorem 2.Before the proof of Corollary 1, we introduce the following lemma. Lemma . Assume that Γ is a regular open convex cone of R n . Let ψ ∈ C (Γ) satisfy (6),then U α ( ψ, Γ) ⊂ Γ ∗ + D (0 , R ψ ) , where U α ( ψ, Γ) is defined by (1) for < α < ∞ and by (2)for α = ∞ .Proof. For t / ∈ Γ ∗ + D (0 , R ψ ), there exist ε > ξ ∈ Γ ∗ such that d ( t , Γ ∗ ) = | ξ − t | ≥ R ψ + 3 ε and ξ · ( t − ξ ) = 0. Then for any ˜ t ∈ Γ ∗ ,(˜ t − t ) · ( ξ − t ) | ξ − t | ≥ | ξ − t | . Hence ˜ t · ( ξ − t ) = (˜ t − t + t − ξ + ξ ) · ( ξ − t ) ≥ | ξ − t | −| ξ − t | = 0, which means ξ − t ∈ Γ. Forany δ >
0, it follows from (6) that there exists ρ such that ψ ( y ) ≤ ( R ψ + δ ) | y | for | y | ≥ ρ . Let11 = ξ − t | ξ − t | ∈ Γ ∩ ∂D (0 , ε >
0, we can find an e ∈ Γ with | e | = 1 such that | e − e | < ε , which means there exists a positive constant δ < ε such that D ( e , δ ) ⊂ Γ.Thus, for any e ∈ D ( e , δ ) with | e | = 1, we have | e − e | ≤ | e − e | + | e − e | < ε . Choose ε satisfying 2 ε | t | ≤ ε and let Γ = { y = ρe : ρ > e ∈ D ( e , δ ) ∩ ∂D (0 , } ⊂ Γ.Then for any y ∈ Γ , − ρe · t = ρ ( − e + e − e ) · t ≥ ρ ( − ε | t | + | ξ − t | ) ≥ ρ ( R ψ + 2 ε ) and Z Γ e − πα ( t · y + ψ ( y )) dy ≥ Z Γ ∩{| y |≥ ρ } e − πα ( t · y +( R ψ + δ ) | y | ) dy ≥ Z ∞ ρ ρ n − dρ Z ∂D (0 , ∩ D ( e ,δ ) e παρ (2 ε − δ ) dσ ( ζ ) = + ∞ , which implies t / ∈ U α ( ψ, Γ). Therefore, U α ( ψ, Γ) ⊂ Γ ∗ + D (0 , R ψ ).Now we prove Corollary 1. Proof of Corollary 1.
For y ∈ Γ, there exists δ such that D ( y , δ ) ⊂ Γ. It follows fromTheorem 2 that there exists f ( t ) such that (3) holds for y ∈ D ( y , δ ). Since Γ is connected,(3) also holds for all y ∈ Γ. Applying the methods in the proof of Theorem 1 for p = 1,we obtain that such an f ( t ) is supported in U s (Γ , ψ δ ). Combing with Lemma 2, we havesupp f ⊂ U s (Γ , ψ δ ) ⊂ Γ ∗ + D (0 , R ψ δ ), where R ψ δ = lim y ∈ Γ ,y →∞ ψ δ ( y ) | y | . Since R ψ δ = R ψ for any y ∈ Γ, we see that U s (Γ , ψ δ ) is also a subset of Γ ∗ + D (0 , R ψ ). Hence,supp f ⊂ Γ ∗ + D (0 , R ψ ).Now we show that | f ( t ) | (cid:0)R Γ e − sπ ( y · t + R ψ | y | ) dy (cid:1) s is slowly increasing. For y , y ∈ Γ, y + y ∈ Γ, F y ( z ) = F ( x + i ( y + y )) ∈ A p,s (Γ , ψ ). As in Theorem 1, we have f ( t ) = g ( − t ) =ˇ F y + y ( − t ) e π ( y + y ) · t . Due to the relation R ψ = lim y ∈ B,y →∞ ψ ( y ) | y | , we have ψ δ ( y ) ≤ R ψ (1+ | y | + | y | ),where R ψ is a positive constant independent of y , y ∈ Γ. Then | f ( t ) | = | ˇ F y + y ( − t ) e π ( y + y ) · t | = (cid:12)(cid:12)(cid:12)(cid:12)Z R n F y + y ( x ) e − πix · t e − πψ δ ( y ) dx (cid:12)(cid:12)(cid:12)(cid:12) e π ( ψ δ ( y )+( y + y ) · t ) ≤ Z R n | F y ( z ) | e − πψ δ ( y ) dxe π ( R ψ (1+ | y | + | y | )+( y + y ) · t ) . Combining with (13), it follows that (cid:18)Z Γ | f ( t ) | s e − sπ ( y · t + R ψ | y | ) dy (cid:19) s ≤ (cid:18)Z Γ (cid:18)Z R n | F y ( z ) | e − πψ δ ( y ) dx (cid:19) s dy (cid:19) s e π ( R ψ (1+ | y | )+ y · t ) ≤ C − pn,N,p,s δ − n (1 − p )(1+ s ) sp k F y k pA ,s ( B,ψ ) e π ( R ψ (1+ | y | )+ y · t ) = C exp { J ( y , t ) } , C = C − pn,N,p,s k F y k pA ,s (Γ ,ψ ) and J ( y ) = − n (1 − p )(1+ s ) sp log δ + 2 π ( R ψ (1 + | y | ) + y · t ). Let J ( t ) = inf { J ( y , t ) : y ∈ Γ } , then | f ( t ) | (cid:18)Z Γ e − sπ ( y · t + R ψ | y | ) dy (cid:19) s ≤ C exp { J ( t ) } . Take y = ρv with ρ > v ∈ Γ with | v | = 1, then δ = d ( ρv, ∂ Γ) / ρε ,where ε = d ( v, ∂ Γ) /
2. Therefore, J ( t ) = inf ρ> (cid:26) − n (1 − p )(1 + s ) sp log( ερ ) + 2 πR ψ (1 + ρ ) + 2 πρ | t | (cid:27) , in which the infimum can be attained when ρ = n (1 − p )(1+ s )2 spπ ( R ψ + | t | ) . It follows that J ( t ) ≤ πR ψ + n (cid:18) p − (cid:19) (cid:18) s + 1 (cid:19) (cid:18) − log ε − log n (cid:18) p − (cid:19) (cid:18) s + 1 (cid:19) + log 2 π ( R ψ + | t | ) (cid:19) . Thus, there exists a positive constant M n,p,s,v such that | f ( t ) | (cid:18)Z Γ e − sπ ( y · t + R ψ | y | ) dy (cid:19) s ≤ Ce J ( t ) ≤ M n,p,s,v (1 + | t | ) n ( p − s +1) . The proof is complete.
Proof of Theorem 3.
We first prove the case when 2 < p < ∞ . Since Γ is a regular openconvex cone, intΓ = ∅ , where intΓ is denoted as the interior of Γ. Then for y ∈ Γ, wecan find a basis { e j } ⊂ intΓ ∗ such that y = P nj =1 e j y j and e j · y ≥
0. For ε >
0, let l ε ( z ) = (cid:16)Q nj =1 (1 − iεe j · z ) (cid:17) N with N > n (cid:16) − p (cid:17) and choose two positive constant A , B such that B | x | ≤ ε P nj =1 ( e j · x ) ≤ A | x | for all x ∈ R n . Thus, | l ε ( z ) | = n Y j =1 | − iεe j · z | ! N = n Y j =1 (cid:0) (1 + εe j · y ) + ε ( e · x ) (cid:1)! N ≥ n Y j =1 (cid:0) ε ( e j · x ) (cid:1)! N ≥ ε n X j =1 ( e j · x ) ! N ≥ (cid:0) ε B | x | (cid:1) N , i.e., | l − ε ( z ) | ≤ (1 + ε B | x | ) − N . For F ( x + iy ) ∈ A p,s (Γ , ψ ), F y ( x ) = F ( x + iy ) ∈ L p ( R n ) asa function of x . Let F ε ( z ) = F ε,y ( x ) = F y ( x ) l − ε ( z ), then F ε,y ( x ) ∈ L ( R n ) ∩ L ( R n ). Indeed,H¨older’s inequality implies that Z R n | F ε,y ( x ) | dx ≤ (cid:18)Z R n | F y ( x ) | p dx (cid:19) p (cid:18)Z R n | l − ε ( x + iy ) | q dx (cid:19) q ≤ C ,ε k F y k L p ( R n ) (14)13nd Z R n | F ε,y ( x ) | dx ≤ (cid:18)Z R n | F y ( x ) | p dx (cid:19) p (cid:18)Z R n | l − ε ( x + iy ) | pp − dx (cid:19) p − p ≤ C ,ε k F y k L p ( R n ) , where C ,ε = (cid:16)R R n dx (1+ ε B | x | ) qN (cid:17) q < ∞ , C ,ε = (cid:18)R R n dx (1+ ε B | x | ) pp − N (cid:19) p − p < ∞ .As the proof of p = 1 in Theorem 1, we can show g ε ( t ) e πy · t = ˇ F ε,y ( t ) ∈ L ( R n ). Thus, g ε ( t ) e πy · t = Z R n F ε,y ( x ) e πix · t dx, (15)then | g ε ( t ) | e πy · t ≤ R R n | F ε,y ( x ) | dx . Together with (14), there hold | g ε ( t ) | e π ( y · t − ψ ( y )) ≤ C ,ε (cid:18)Z R n | F ( x + iy ) e − πψ ( y ) | p dx (cid:19) p , Z Γ | g ε ( t ) | sp e spπ ( y · t − ψ ( y )) dy ≤ C ,ε Z Γ (cid:18)Z R n | F ( x + iy ) e − πψ ( y ) | p dx (cid:19) s dy, | g ε ( t ) | sp Z Γ e spπ ( y · t − ψ ( y )) dy ≤ C ,ε k F k spA p,s (Γ ,ψ ) . Now we prove that supp g ε ( t ) ⊂ − U ps (Γ , ψ ). Note that g ε ( t ) is continuous in R n . Then for t / ∈ − U ps (Γ , ψ ), formula (1) shows that R Γ e psπ ( y · t − ψ ( y )) dy = ∞ for y ∈ Γ. It follows fromthe above inequality that g ε ( t ) = 0 for t / ∈ − U ps (Γ , ψ ). As a result, supp g ε ( t ) ⊂ − U ps (Γ , ψ ).Since g ε ( t ) e πy · t ∈ L ( R n ), we can rewrite (15) as F ε,y ( x ) = Z R n g ε ( t ) e − πiz · t χ − U ps (Γ ,ψ ) ( t ) dt. (16)Plancherel’s Theorem implies that R R n | g ε ( t ) e πy · t | dt = R R n | F ε,y ( x ) | dx . Then based onFatou’s lemma, Z R n | g ε ( t ) | ≤ lim y ∈ Γ ,y → Z R n | F ( x + iy ) | dx < ∞ . Thus, there exist g ( t ) ∈ L ( R n ) and a sequence { ε k } tending to zero as k → ∞ such thatlim k →∞ R R n g ε k ( t ) h ( t ) dt = R R n g ( t ) h ( t ) dt for h ( t ) ∈ L . In fact, for t ∈ − U ps (Γ , ψ ), lemma 2implies that t ∈ − Γ ∗ k + D (0 , R ψ ). Then t can always be written as t + t with t ∈ − Γ ∗ k and | t | < R ψ . Hence, for y ∈ Γ, y · t = y · ( t + t ) ≤ −| t | k + | t || y | ≤ − ( | t | − | t | ) k + R ψ | t | ≤ ( R ψ − k ) | t | + R ψ k, implying that R R n | e πy · t χ − U ps ( B k ,ψ ) ( t ) | dt < ∞ . Therefore, on the right hand side of (16),lim k →∞ Z R n g ε k ( t ) e − πiz · t χ − U ps (Γ ,ψ ) ( t ) dt = Z R n g ( t ) e − πiz · t χ − U ps (Γ ,ψ ) ( t ) dt e πy · t χ − U ps (Γ ,ψ ) ( t ) ∈ L ( R n ). Whilst it is obvious that F ε ( z ) → F ( z ) when ε →
0. Sending k to ∞ on both sides of (16) and letting f ( t ) = g ( − t ), we obtain that f ∈ L ( R n ) and thesupport supp f is contained in U ps (Γ , ψ ), as well as the desired representation (3) holds forall z ∈ T Γ .We now prove the case when p = ∞ . For z = ( z , . . . , z n ) ∈ T Γ and ε >
0, we can alsoconstruct a function F ε,y ( x ) = F ε ( z ) = F y ( x ) l − ε ( z ), where l ε ( z ) = (cid:16)Q nj =1 (1 − iεe j · z ) (cid:17) N with N > n . Then Z R n | F ε,y ( x ) | dx ≤ sup x ∈ R n | F y ( x ) | Z R n | l − ε ( x + iy ) | dx ≤ e C ,ε k F y k L ∞ ( R n ) < ∞ (17)and Z R n | F ε,y ( x ) | dx ≤ sup x ∈ R n | F y ( x ) | Z R n | l − ε ( x + iy ) | dx ≤ e C ,ε k F y k L ∞ ( R n ) < ∞ , where e C ,ε = R R n dx (1+ ε B | x | ) N and e C ,ε = (cid:16)R R n dx (1+ ε B | x | ) N (cid:17) < ∞ . Hence F ε,y ∈ L ( R n ) ∩ L ( R n ). In this case, we also have g ε ( t ) e πy · t = ˇ F ε,y ( t ) ∈ L ( R n ). Then g ε ( t ) e πy · t = R R n F ε,y ( x ) e πix · t dx . Therefore, together with (17), | g ε ( t ) | e π ( y · t − ψ ( y )) ≤ e C ,ε sup x ∈ R n | F y ( x ) | e − πψ ( y ) , sup y ∈ Γ | g ε ( t ) | e π ( y · t − ψ ( y )) ≤ e C ,ε sup x ∈ R n ,y ∈ Γ | F ( x + iy ) | e − πψ ( y ) = e C ,ε k F k A ∞ , ∞ (Γ ,ψ ) < ∞ . Then we can similarly show that supp g ε ( t ) ⊂ − U ∞ (Γ , ψ ) ⊂ − Γ ∗ + D (0 , R ψ ). Applying thesame method for 2 < p < ∞ , we obtain the desired formula (3) holds for all z ∈ T Γ and thesupport supp f is contained in U ∞ (Γ , ψ ) ⊂ Γ ∗ + D (0 , R ψ ). In [10], denoting by A α ( C + ) a weighted Bergman space of functions holomorphic in C + satisfying k F k A α ( C + ) = R C + | F ( x + iy ) | y α dxdy < ∞ , and by L β ( R + ) the space of complex–valued measurable functions f on R + satisfying k f k L β ( R + ) = Γ( β )(4 π ) β R ∞ | f ( t ) | t − β dt < ∞ ,Duren stated an analogy of the Paley–Wiener theorem for Bergman space. Theorem A ([10]) For each α > −
1, the space A α ( C + ) is isometrically isomorphic underthe Fourier transform to the space L α +1 ( R + ). More precisely, F ∈ A α ( C + ) if and only if itis the Fourier transform F ( z ) = R ∞ f ( t ) e πiz · t dt of some function f ∈ L α +1 ( R + ), in whichcase k F k A α ( C + ) = k f k L α +1 ( R + ) . 15ased on Theorem 1, letting s = 1, p = 2, ψ ( y ) = − α π log | y | and B be a regular openconvex cone Γ, we establish Corollary 2, which can be regarded as a higher dimensional andtube domain generalization of Theorem A. Corollary . For each α > − , F ∈ A α ( T Γ ) if and only if there exists f ( t ) ∈ L α +1 (Γ ∗ ) such that F ( z ) = Z Γ ∗ f ( t ) e πiz · t dt holds for z ∈ T Γ and k F k A α ( T Γ ) = k f k L α +1 (Γ ∗ ) .Proof. By restricting the base B to be a regular open convex cone Γ and letting ψ ( y ) = ψ α ( y ) = − α π log | y | , F ∈ A α ( T Γ ) is also an element of A , (Γ , ψ α ). Applying Theorem 1 tosuch an F , we can show that there exists f ( t ) satisfying (5) such that F ( z ) = R R n f ( t ) e πiz · t dt and supp f ⊂ U (Γ , ψ α ). Based on (6), we have R ψ α = lim y ∈ Γ ,y →∞ ψ α ( y ) | y | = 0 . Thus, together with Lemma 2, the supporter of f ( t ) is contained in Γ ∗ and F ( z ) = R Γ ∗ f ( t ) e πiz · t dt .Moreover, R Γ R Γ ∗ | f ( t ) | e − π ( y · t + ψ α ( y )) dtdy ≤ k F k A , (Γ ,ψ α ) . Thus, Z Γ Z Γ ∗ | f ( t ) | e − π ( y · t + ψ α ( y )) dtdy = Z Γ ∗ Z Γ | f ( t ) | e − πy · t y α dydt = Z Γ ∗ | f ( t ) | Γ( α )(4 πt ) α +1 dt, which shows f ∈ L α +1 (Γ ∗ ). And Plancherel’s Theorem assures that k F k A α ( T Γ ) = k f k L α +1 (Γ ∗ ) .Conversely, note that F ( z ) = R Γ ∗ f ( t ) e πit · z dt . For f ( t ) ∈ L α +1 (Γ ∗ ), Plancherel’s theoremimplies that Z R n | F ( x + iy ) | dx = Z Γ ∗ e − πy · t | f ( t ) | dt, Z Γ Z R n | F ( x + iy ) | e − πψ α ( y ) dxdy = Z Γ Z Γ ∗ | f ( t ) | e − π ( y · t + ψ α ( y )) dtdy < ∞ , in which ψ α ( y ) = − α π log | y | . Hence, F ( z ) ∈ A , (Γ , ψ α ) = A α ( T Γ ). The proof is complete.By restricting the base B to be a regular open convex cone Γ, we establish the followingweighted version of the edge-of-the-wedge theorem (see [2]) as an application of Theorem 1. Theorem . Assume that Γ is a regular open convex cone in R n , ψ ∈ C (Γ) and ψ ∈ C ( − Γ) satisfy R ψ = lim y ∈ Γ ,y →∞ ψ ( y ) | y | < ∞ (18)16 nd R ψ = lim y ∈ Γ ,y →∞ ψ ( − y ) | y | < ∞ (19) respectively. If < p ≤ , < s ( p − ≤ , F ∈ A p,s (Γ , ψ ) and F ∈ A p,s ( − Γ , ψ ) ,satisfying lim y → Z R n | F ( x + iy ) − F ( x − iy ) | p dx = 0 , (20) then F and F can be analytically extended to each other and further form an entire function F . Furthermore, there exists a function f ∈ L ( R n ) supported in a bounded convex set K such that F ( z ) = R K f ( t ) e πit · z dt .Proof. Theorem 1 implies that there exists a function f j ( j = 1 ,
2) such that F j = Z R n f j ( t ) e πit · z dt holds, in which the supporter of f j is contained in U sp (( − j +1 Γ , ψ j ) for for 1 < p ≤
2. Basedon lemma 2, supp f j ⊂ ( − j +1 Γ ∗ + D (0 , R ψ j ). By the Hausdorff-Young inequality, (cid:18)Z R n | f ( t ) e πy · t − f ( t ) e − πy · t | q dt (cid:19) q ≤ (cid:18)Z R n | F ( x + iy ) − F ( x − iy ) | p dt (cid:19) p . Then it follows from Fatou’s lemma and (20) that k f − f k L q ( R n ) = 0. Thus, f = f almost everywhere on R n . Let f ( t ) = f ( t ) = f ( t ), and R = max { R ψ , R ψ } , then supp f ⊂ K ⊂ (Γ ∗ + D (0 , R )) T ( − Γ ∗ + D (0 , R )). Thus, K is a bounded convex set. Consequently, F ( z ) = R K e πiz · t f ( t ) dt is an entire function, where F ( z ) = F ( z ) for z ∈ T Γ and F ( z ) = F ( z )for z ∈ T − Γ .Similarly, we can prove the weighted version of the edge-of-the-wedge theorem for p > Theorem . Suppose that Γ is a regular open convex cone in R n , ψ ∈ C (Γ) and ψ ∈ C ( − Γ) satisfy (18) and (19) respectively. If F ∈ A p,s (Γ , ψ ) and F ∈ A p,s ( − Γ , ψ ) , where p > ,satisfying lim y ∈ Γ ,y → Z R n | F ( x + iy ) − F ( x − iy ) | dx = 0 , (21) then F and F can be analytically extended to each other and further form an entire function F . Furthermore, there exists a measurable function f ( t ) supported in a bounded convex set K such that F ( z ) = R K f ( t ) e πit · z dt .Proof. For F j ∈ A p,s (( − j +1 Γ , ψ j ) and p + q = 1, exists a measurable function f j such that F j = R R n f j ( t ) e πit · z dt and supp f j ⊂ U sp (( − j +1 Γ , ψ j ), where j = 1 ,
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