INTEGRAL OPERATORS INDUCED BY THE FOCK KERNEL

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1 INTEGRAL OPERATORS INDUED BY THE FOK KERNEL MILUTIN DOSTANIĆ AND KEHE ZHU ABSTRAT We sudy he L boundedness and find he norm of a lass of inegral oeraors indued by he reroduing kernel of Fok saes over 1 INTRODUTION Our analysis will ake lae in he n-dimensional omlex Eulidean sae For any wo oins z (z 1,, z n ) and w (w 1,, w n ) in we wrie z, w z 1 w z n w n, and z z z n 2 For any > we onsider he Gaussian robabiliy measure ( ) n dv (z) e z 2 dv(z) π on, where dv is ordinary Lebesgue volume measure on Le H( ) denoe he sae of all enire funions on We hen define F L (, dv ) H( ) for < < These saes are ofen alled Fok saes, or Segal- Bargman saes, over See [1][2][3] [7][9][11][16][18] For > and > we are going o wrie [ ] 1 f, f(z) dv (z), and f, g f(z) g(z) dv (z) Dae: Deember 2, 26 2 Mahemais Subje lassifiaion 32A36 and 32A15 Key words and hrases Fok saes, Gaussian measure, inegral oeraors Dosanić is arialy suored by MNZZS Gran N o ON1441 Zhu is arially suored by he US Naional Siene Foundaion 1

2 2 MILUTIN DOSTANIĆ AND KEHE ZHU I is well known ha eah Fok sae F is a losed linear subsae of L (, dv ) In ariular, in he Hilber sae seing of L 2, here exiss a unique orhogonal rojeion P : L 2 (, dv ) F 2 Furhermore, his rojeion oinides wih he resriion of he following inegral oeraor o L 2 (, dv ): S f(z) e z,w f(w) dv (w) (1) The inegral kernel above, K (z, w) e z,w, is he reroduing kernel of F 2 The urose of his aer is o sudy he aion of he oeraor S on he saes L (, dv s ), where s > We also onsider he losely relaed inegral oeraor T f(z) K (z, w) f(w) dv (w), or more exliily, T f(z) e z,w f(w) dv (w) (2) The main resul of he aer is he following Main Theorem Suose >, s >, and 1 Then he following ondiions are equivalen (a) T is bounded on L (, dv s ) (b) S is bounded on L (, dv s ) () 2s Furhermore, he norms of T and S on L (, dv s ) saisfy whenever 2s S T 2 n The equivalene of (a), (b), and () is no new; i is imlii in [11] for examle So our main onribuion here is he ideniy T 2 n The aurae alulaion of he norm of an inegral oeraor is an ineresing bu ofen diffiul roblem We menion a few suessful examles in he lieraure: he norm of he auhy rojeion on L of he uni irle is deermined in [13], he norm of he auhy rojeion on L saes of more general domains is esimaed in [5], an asymoi formula for he norm of

3 INTEGRAL OPERATORS 3 he Bergman rojeion on L saes of he uni ball is given in [21], and he norm of he Berezin ransform on he uni disk is alulaed in [6] As a onsequene of he heorem above, we see ha he densely defined oeraor P : L (, dv ) L (, dv ) is unbounded for any 2 This is in shar onras o he heory of Hardy saes and he heory of Bergman saes For examle, if P is he Bergman rojeion for he oen uni ball B n, ha is, if P is he orhogonal rojeion P : L 2 (B n, dv) L 2 (B n, dv) H(B n ), where H(B n ) is he sae of holomorhi funions in B n, hen P : L (B n, dv) L (B n, dv) is bounded for every > 1 A similar resul holds for he auhy-szëgo rojeion in he heory of Hardy saes See [14] and [2] A more general lass of inegral oeraors indued by he Bergman kernel on he uni ball B n have been sudied in [8][12][19] We wish o hank Peer Duren and James Tung for bringing o our aenion he referenes [11] and [15] 2 PRELIMINARIES For an n-ule m (m 1,, m n ) of nonnegaive inegers we are going o wrie m m m n, m! m 1! m n! If z, we also wrie z m z m 1 1 z mn n When he dimension n is 1, we use da insead of dv, and da insead of dv Thus for > and z, we have da (z) π e z 2 da(z), where da is ordinary area measure on he omlex lane Lemma 1 Le m (m 1,, m n ) be an n-ule of nonnegaive inegers For any > and > we have n z m Γ((m k /2) + 1) dv (z) n m k/2 In ariular, k1 z m 2 dv (z) m! m

4 4 MILUTIN DOSTANIĆ AND KEHE ZHU Proof We evaluae he inegral in olar oordinaes n z m dv (z) z k m k da (z k ) k1 n z k m k e z k 2 da(z k ) π k1 n 2 r mk+1 e r2 dr k1 n k1 n k1 n k1 1 m k/2 r m k/2 e r dr Γ((m k /2) + 1) m k/2 r m k/2 e r dr The seond inegral is obviously a seial ase of he firs one Reall ha he resriion of he oeraor S o L 2 (, dv ) is nohing bu he orhogonal rojeion ono F 2 onsequenly, we have he following reroduing formula Lemma 2 If f is in F 2, hen S f f, ha is, f(a) e a,z f(z) dv (z) for all a A seial ase of he reroduing formula above is he following: K (a, a) K (a, z) 2 dv (z), a (3) As an aliaion of his ideniy, we obain he following fundamenal inegrals for owers of kernel funions in Fok saes Lemma 3 Suose > and s is real Then for all a e s z,a dv (z) e s2 a 2 /4

5 Proof I follows from (3) ha INTEGRAL OPERATORS 5 e s z,a dv (z) e sa/2,z 2 dv (z) n K (sa/2, z) 2 dv (z) K (sa/2, sa/2) This roves he desired ideniy e s2 a 2 /4 We need wo well-known resuls from he heory of inegral oeraors The firs one onerns he adjoin of a bounded inegral oeraor Lemma 4 Suose 1 < and 1/ + 1/q 1 If an inegral oeraor T f(x) K(x, y)f(y) dµ(y) is bounded on L (X, dµ), hen is adjoin is he inegral oeraor given by T f(x) Proof See [1] for examle X T : L q (X, dµ) L q (X, dµ) X K(y, x)f(y) dµ(y) The seond resul is a useful rierion for he boundedness of inegral oeraors on L saes, usually referred o as Shur s es Lemma 5 Suose H(x, y) is a osiive kernel and T f(x) H(x, y)f(y) dµ(y) X is he assoiaed inegral oeraor Le 1 < < wih 1/ + 1/q 1 If here exiss a osiive funion h(x) and osiive onsans 1 and 2 suh ha H(x, y)h(y) q dµ(y) 1 h(x) q, x X, and X X H(x, y)h(x) dµ(x) 2 h(y), y X, hen he oeraor T is bounded on L (X, dµ) Moreover, he norm of T on L (X, dµ) does no exeed 1/q 1 1/ 2 Proof See [2] for examle

6 6 MILUTIN DOSTANIĆ AND KEHE ZHU 3 INTEGRAL OPERATORS INDUED BY THE FOK KERNEL For any s > we rewrie he inegral oeraors S and T defined in (1) and (2) as follows ( ) n z,w +s w 2 w 2 S f(z) e f(w) dv s (w), s and T f(z) ( ) n z,w +s w 2 w 2 e f(w) dv s (w) s I follows from Lemma 4 ha he adjoin of S and T wih rese o he inegral airing f, g s f(z)g(z) dv s (z) is given reseively by ( ) n S f(z) e (s ) z 2 e z,w f(w) dv s (w), (4) s and T f(z) ( ) n e (s ) z 2 e z,w f(w) dv s (w) (5) s We firs rove several neessary ondiions for he oeraor S o be bounded on L (, dv s ) Lemma 6 Suose < <, >, and s > If S is bounded on L (, dv s ), hen 2s Proof onsider funions of he following form: f x,k (z) e x z 2 z k 1, z, where x > and k is a osiive ineger We firs use Lemma 1 o alulae he norm of f x,k in L (, dv s ) ( s ) n f x,k dv s z 1 k e (x+s) z 2 dv(z) π n ( ) n n s x + s ( s x + s z1 k dv x+s (z) ) n n Γ((k/2) + 1) (x + s) k/2

7 INTEGRAL OPERATORS 7 We hen alulae he losed form of S (f x,k ) using he reroduing formula from Lemma 2 ( ) n S (f x,k )(z) e z,w w π 1e k (+x) w 2 dv(w) ( ) n n e (+x) z/(+x),w w1 k dv +x (w) + x ( ) n n ( ) k z1 + x + x ( ) n+k z k + x 1 We nex alulae he norm of S (f x,k ) in L (, dv s ) wih he hel of Lemma 1 again ( ) (n+k) S (f x,k ) dv s z 1 k dv s (z) + x n ( ) n (n+k) Γ((k/2) + 1) + x s k/2 Now if he inegral oeraor S is bounded on L (, dv s ), hen here exiss a osiive onsan (indeenden of x and k) suh ha ( ) (n+k) ( ) n Γ((k/2) + 1) s Γ((k/2) + 1) + x s k/2 x + s (x + s), k/2 or ( ) (n+k) ( ) n+(k/2) s + x s + x Fix any x > and look a wha haens in he above inequaliy when k We dedue ha ( ) 2 s + x s + x ross mulily he wo sides of he inequaliy above and simlify The resul is 2 2s + sx Le x Then 2 2s, or 2s This omlees he roof of he lemma Lemma 7 Suose 1 < < and S is bounded on L (, dv s ) Then > s

8 8 MILUTIN DOSTANIĆ AND KEHE ZHU Proof If > 1 and S is bounded on L (, dv s ), hen S is bounded on L q (, dv s ), where 1/ + 1/q 1 Alying he formula for S from (4) o he onsan funion f 1 shows ha he funion e (s ) z 2 is in L q (, dv s ) From his we dedue ha q(s ) < s, whih is easily seen o be equivalen o s < Lemma 8 If S is bounded on L 1 (, dv s ), hen 2s Proof Fix any a and onsider he funion f a (z) e z,a e z,a, z n Obviously, f a 1 for every a On he oher hand, i follows from (4) and Lemma 3 ha ( ) n S (f a )(a) e (s ) a 2 e w,a dv s (w) s ( ) n n e (s ) a 2 e 2 a 2 /(4s) s Sine S ha is bounded on L ( ), here exiss a osiive onsan suh ( ) n e (s ) a 2 e 2 a 2 /(4s) S (f a ) f a s for all a This learly imlies ha whih is equivalen o Therefore, we have 2s s + 2 4s, (2s ) 2 Lemma 9 Suose 1 < 2 and S is bounded on L (, dv s ) Then 2s Proof One again, we onsider funions of he form f x,k (z) e x z 2 z k 1, z,

9 INTEGRAL OPERATORS 9 where x > and k is a osiive ineger I follows from (4) and Lemma 2 ha ( ) n S (f x,k )(z) e (s ) z 2 e z,w w π 1e k (s+x) w 2 dv(w) ( ) n n e (s ) z 2 e (s+x) z/(s+x),w w1 k dv s+x (w) s + x ( ) n n ( ) k z1 e (s ) z 2 s + x s + x ( ) n+k e (s ) z 2 z k s + x 1 Suose 1 < 2 and 1/ + 1/q 1 If he oeraor S is bounded on L (, dv s ), hen he oeraor S is bounded on L q (, dv s ) So here exiss a osiive onsan, indeenden of x and k, suh ha S (f x,k ) q dv s f x,k q dv s I follows from he roof of Lemma 1 ha ) n Γ((qk/2) + 1) (qx + s) qk/2 ( s f x,k q dv s qx + s n On he oher hand, i follows from Lemma 7 and is roof ha s q(s ) >, so he inegral I S (f x,k ) q dv s an be evaluaed wih he hel of Lemma 1 as follows ( ) q(n+k) ( s ) n I z 1 qk e (s q(s )) z 2 dv(z) s + x π ( ) n q(n+k) ( ) n s z 1 qk dv s q(s ) (z) s + x s q(s ) ( ) n q(n+k) ( ) n s Γ((qk/2) + 1) s + x s q(s ) (s q(s )) qk/2 Therefore, ( ) q(n+k) ( ) n s Γ((qk/2) + 1) s + x s q(s ) (s q(s )) qk/2 is less han or equal o ( ) n s Γ((qk/2) + 1) qx + s (qx + s), qk/2

10 1 MILUTIN DOSTANIĆ AND KEHE ZHU whih easily redues o ( ) q(n+k) ( ) n+(qk/2) s q(s ) s + x s + qx One again, fix x > and le k We find ou ha ( ) 2 s q(s ) s + x s + qx Using he relaion 1/ + 1/q 1, we an hange he righ-hand side above o s ( 1)s + x I follows ha whih an be wrien as 2 ( 1)s + 2 x ( s)(s 2 + 2sx + x 2 ), ( s)x 2 + [2s( s) 2 ]x + s 2 ( s) 2 ( 1)s Le q(x) denoe he quadrai funion on he lef-hand side of he above inequaliy Sine s > by Lemma 7, he funion q(x) aains is minimum value a x 2 2s( s) 2( s) Sine 2, he numeraor above is greaer han or equal o 2 2s + s 2 ( s) 2 I follows ha x and so h(x) h(x ) for all real x (no jus nonnegaive x) From his we dedue ha he disriminan of h(x) anno be osiive Therefore, [2s( s) 2 ] 2 4( s)[s 2 ( s) 2 ( 1)s] Elemenary alulaions reveal ha he above inequaliy is equivalen o Therefore, 2s ( 2s) 2 Lemma 1 Suose 2 < < and S is bounded on L (, dv s ) Then 2s Proof If S is bounded on L (, dv s ), hen S is bounded on L q (, dv s ), where 1 < q < 2 and 1/ + 1/q 1 I follows from (4) ha here exiss a osiive onsan, indeenden of f, suh ha [ e(s ) z 2 f(w)e ( s) w 2] q dv (w) dv s (z) e z,w

11 is less han or equal o f(w) q dv s (w), where f is any funion in L q (, dv s ) Le INTEGRAL OPERATORS 11 f(z) g(z)e (s ) z 2, where g L q (, dv s q(s ) ) (reall from Lemma 7 ha s q(s ) > ) We obain anoher osiive onsan (indeenden of g) suh ha S g q dv s q(s ) g q dv s q(s ) for all g L q (, dv s q(s ) ) Therefore, he oeraor S is bounded on L q (, dv s q(s ) ) Sine 1 < q < 2, i follows from Lemma 9 ha q 2[s q(s )] I is easy o hek ha his is equivalen o 2s We now omlee he roof of he firs ar of he main heorem As was oined ou in he inroduion, his ar of he heorem is known before We inluded a full roof here for wo uroses Firs, his gives a differen and self-onained aroah Seond, as a by-rodu of his differen aroah, we are going o obain he inequaliy T 2 n, whih is one half of he ideniy T 2 n Theorem 11 Suose >, s >, and 1 Then he following ondiions are equivalen (a) The oeraor T is bounded on L (, dv s ) (b) The oeraor S is bounded on L (, dv s ) () The weigh arameers saisfy 2s Proof When 1, ha (b) imlies () follows from Lemma 8, ha () imlies (a) follows from Fubini s heorem and Lemma 3, and ha (a) imlies (b) is obvious When 1 < <, ha (b) imlies () follows from Lemmas 9 and 1, and ha (a) imlies (b) is sill obvious So we assume 1 < < and roeed o show ha ondiion () imlies (a) We do his wih he hel of Shur s es (Lemma 5) Le 1/ + 1/q 1 and onsider he osiive funion h(z) e λ z 2, z, where λ is a onsan o be seified laer Reall ha T f(z) H(z, w)f(w) dv s (w),

12 12 MILUTIN DOSTANIĆ AND KEHE ZHU where ( ) n H(z, w) e z,w e (s ) w 2 s is a osiive kernel We firs onsider he inegrals I(z) H(z, w)h(w) q dv s (w), z If λ saisfies > qλ, (6) hen i follows from Lemma 3 ha ( ) n I(z) e z,w e ( qλ) w 2 dv(w) π ( ) n n e z,w dv qλ (w) qλ ( ) n n e 2 z 2 /4( qλ) qλ If we hoose λ so ha 2 4( qλ) qλ, (7) hen we obain ( ) n H(z, w)h(w) q dv s (w) h(z) q (8) qλ n for all z We now onsider he inegrals J(w) H(z, w)h(z) dv s (z), w If λ saisfies s λ >, (9) hen i follows from Lemma 3 ha ( ) n J(w) e z,w e (s ) w 2 h(z) dv s (z) s ( ) n n e (s ) w 2 e z,w e (s λ) z 2 dv(z) π ( ) n n e (s ) w 2 e 2 w 2 /4(s λ) s λ ( ) n e [(s )+2 /4(s λ)] w 2 s λ

13 INTEGRAL OPERATORS 13 If we hoose λ so ha s + 2 4(s λ) λ, (1) hen we obain ( ) n H(z, w)h(z) dv s (z) h(w) (11) s λ n for all w In view of Shur s es and he esimaes in (8) and (11), we onlude ha he oeraor T would be bounded on L (, dv s ) rovided ha we ould hoose a real λ o saisfy ondiions (6), (7), (9), and (1) simulaneously Under our assumion ha 2s i is easy o verify ha ondiion (7) is he same as ondiion (1) In fa, we an exliily solve for qλ and λ in (7) and (1), reeively, o obain qλ 2, 2s λ 2 The relaions 2s and 1/+1/q 1 learly imly ha he wo resuling λ s above are onsisen, namely, λ 2q 2s 2 (12) Also, i is easy o see ha he above hoie of λ saisfies boh (6) and (9) This omlees he roof of he heorem Theorem 12 If 1 < and 2s, hen S f dv s T f dv s 2 n f dv s for all f L (, dv s ) Proof Wih he hoie of λ in (12), he onsans in (8) and (11) boh redue o 2 n Therefore, Shur s es ells us ha, in he ase when 1 < <, he norm of T on L (, dv s ) does no exeed 2 n When 1, he desired esimae follows from Fubini s heorem and Lemma 3 Theorem 12 above an be saed as S T 2 n, wih S and T onsidered as oeraors on L (, dv s ) We now roeed o he roof of he inequaliy T 2 n Several lemmas are needed for his esimae Lemma 13 For > and 1 we have [ ( uv lim h (uv) 1 u + v 4 ex h + 2 hv ) ( dv] du 2 2π)

14 14 MILUTIN DOSTANIĆ AND KEHE ZHU Proof We begin wih he inner inegral I(u) (uv) 1 4 ex ( uv u + v 2 hv ) dv Le a h and hange variables aording o v 2 Then Wrie I(u) 2u 1 4 e u 2 a 2 + ( u a ex( a 2 + u ) d u ) 2 u + u 4a, make a hange of variables aording o x ( u/), and simlify he resul We obain I(u) 2u 1 4 e uh u x + e ax2 dx I is hen lear ha we an rewrie I(u) as follows where and ϕ 2 (u) 2 e I(u) ϕ 1 (u) + ϕ 2 (u), ϕ 1 (u) 2 + e uh e ax2 dx, u u 1 4 uh + u e ax2 ( x + u ) u dx For he funion ϕ 2 we raionalize he numeraor in is inegrand o obain ϕ 2 (u) 2 u u 1 2 e uh e uh + u + x e ax2 u dx x e ax2 dx A simle alulaion of he las inegral above hen gives ϕ 2 (u) 4 u 1 2 e uh (13) Similarly, we have ϕ 1 (u) 2 e uh e ax2 dx 2π a uh e (14)

15 and We now use he above esimaes o show ha lim h h + INTEGRAL OPERATORS 15 ϕ 1 (u) 1 ϕ 2 (u) du, (15) lim h + ϕ 2 (u) du (16) In fa, aording o (13) and (14), ( ) 1 2π hϕ 1 (u) 1 4 ϕ 2 (u) h u 1 2 e uh, a from whih we derive ha h ϕ 1 (u) 1 ϕ 2 (u) du ( ) 1 2π h 4 u 1 2 e uh du a ( ) 1 2π h + 4 ω 1 2 e ω dω a Sine a 1 as h 2 +, we obain (15) On he oher hand, i follows from (13) ha ( ) 4 ϕ 2 (u) u 2 e uh, so h ϕ 2 (u) du ( 4 ) h h u 2 e uh du a 1 h /2 ω 2 e ω dω Le h + and use he fa ha a 1 as h 2 + We obain (16) By he hange of variables s uh/, we have ( ) ( 2 + h ϕ 1 (u) du h e uh e dx) ax2 du h ( 2+1 a + ( ) e s h u e ax2 s h dx) ds Le h +, noie ha a 1 as h 2 +, and use Lebesgue s dominaed onvergene heorem We ge [ + ] ( lim h ϕ h + 1 du 2 e s e x2 2 dx ds 2 2π) (17)

16 16 MILUTIN DOSTANIĆ AND KEHE ZHU If 1, i is easy o see ha he funion g(z) 1 + z z 1 + z 1 is oninuous and bounded on Relaing z by z/w, we see ha z + w z ( z 1 w + w ) for all z and w, where is a osiive onsan ha only deends on This along wih (15) and (16) shows ha lim h h + (ϕ 1 (u) + ϕ 2 (u)) ϕ 1 (u) du ombining his wih (17), we onlude ha lim h h + This roves he desired resul Lemma 14 Le (ϕ 1 (u) + ϕ 2 (u)) du K(x, y) and define an inegral oeraor by Af(x) n 2n x n y n 4 n (n!) 2 A : L (, ) L (, ) ( 2 2π ) e 2 (x+y) K(x, y)f(y) dy, where 1 and is any fixed osiive onsan Then he norm of A on L (, ) saisfies A 2 Proof I follows from he asymoi behavior of he Bessel funion J (x) (see age 199 of [17] for examle) ha (z/2) 2k (k!) 2 ez 2πz as z Thus k n u n (n!) 2 u e2 1 4π u 4 as u Fix an arbirary η > and hoose some u > suh ha u n 2 e u 2 > (1 η) 1 (n!) 4π u 4 for every u u n (18)

17 Le δ 2 u / I follows from (18) ha INTEGRAL OPERATORS 17 e xy K(x, y) > (1 η) (19) (xy) 1 4 2π for all x δ and all y δ Fix a osiive number ε and le ( f ε (x) ex εx ) Then and so [ f ε On he oher hand, i follows from (19) ha ( ) 1 Af ε Af ε (x) dx δ [ δ (1 η) 2π ( dx ] 1/ f ε (x) dx ε 1, A ε 1 Afε (2) δ ) ] 1 e 2 (x+y) K(x, y)e εy dy [ ( dx ombining his wih (2), we obain ( (1 η) A [ε dx 2π δ δ δ δ e 2 (x+y)+ xy εy (xy) 1 4 e 2 (x+y)+ xy εy (xy) 1 4 dy dy ) ] 1 ) ] 1 Afer he hange of variables x u and y v we obain [ A 1 η ( ε ) e 1 2 (u+v)+ uv εv ] 1 du dv (21) 2π (uv) 1 4 δ δ Le δ and ε/ h learly, h + when ε + Le ε + in (21) and aly Lemma 13 We obain A 1 η 2π 2 2π 2(1 η) Sine η > is arbirary, we obain A 2, and he roof of Lemma 14 is omlee We an now rove he main resul of he aer

18 18 MILUTIN DOSTANIĆ AND KEHE ZHU Theorem 15 If 1 < and 2s, hen he norm of T on L (, dv s ) is given by T 2 n Proof In view of Theorem 12 i is enough for us o rove he inequaliy T 2 n Reall ha when n 1, we use he noaion da s insead of dv s For f L (, da s ) we onsider Then Φ L (, dv s ) and we have T T Φ Φ Φ(z 1,, z n ) f(z 1 ) f(z n ) e z ζ f(ζ) da (ζ) da s (z) f(ζ) da s (ζ) When f runs over all uni veors in L (, da s ), he suremum of he quoien inside he brakes above is exaly he h ower of he norm of he oeraor T on L (, da s ) So we only need o rove he inequaliy T 2 n for n 1 Now we assume n 1, 1, and le be he inegral oeraor defined by T : L (, da s ) L (, da s ) T f(z) e z ζ f(ζ) da (ζ) To obain a lower esimae of he norm of T on L (, da s ), we aly T o a family of seial funions More seifially, we onsider funions of he form f(z) G( z 2 )e z 2 /2, z, where G is any uni veor in L (, ) I follows from olar oordinaes and he assumion 2s ha f f da s s G(x) dx s (22) n

19 On he oher hand, T f(z) INTEGRAL OPERATORS 19 e zw G( w 2 )e w 2 /2 da (w) e zw/2 2 G( w 2 )e w 2 /2 da (w) (zw/2) n 2 G( w 2 )e w 2 /2 da (w) n! n 2n z 2n w 2n G( w 2 )e w 2 /2 da 4 n (n!) 2 (w) n 2n z 2n 4 n (n!) y n G(y)e y/2 dy 2 n e y/2 K( z 2, y)g(y) dy e z 2 /2 AG( z 2 ), where he kernel K and he oeraor A are from Lemma 14 Using olar oordinaes and he assumion 2s one more ime, we obain T f T f(z) da s (z) s AG(x) dx (23) By (22) and (23) we mus have T T f f AG(x) dx (24) Take he suremum over G and aly Lemma 14 The resul is T A 2 This omlees he roof of he heorem We onlude he aer wih wo orollaries orollary 16 For any s > and 1 he Fok sae Fs is a omlemened subsae of L (, dv s ), ha is, here exiss a losed subsae Xs of L (, dv s ) suh ha L (, dv s ) F s X s, where denoes he dire sum of wo subsaes Proof hoose > suh ha 2s Then by Theorem 11, he oeraor S is a bounded rojeion from L (, dv s ) ono Fs This shows ha Fs is omlemened in L (, dv s )

20 2 MILUTIN DOSTANIĆ AND KEHE ZHU The following resul is obviously a generalizaion of Theorem 11, bu i is also a dire onsequene of Theorem 11 orollary 17 Suose a >, b >, s >, and 1 Then he following ondiions are equivalen (a) The inegral oeraor T a,b f(z) is bounded on L (, dv s ) (b) The inegral oeraor S a,b f(z) a z 2 +(a+b) z,w b w 2 e a z 2 +(a+b) z,w b w 2 e is bounded on L (, dv s ) () The arameers saisfy (a + b) 2(s + a) f(w) dv(w) f(w) dv(w) Proof The boundedness of S a,b on L (, dv s ) is equivalen o he exisene of a osiive onsan, indeenden of f, suh ha f(w) dv(w) dv s+a (z) e (a+b) z,w b w 2 is less han or equal o f(z)e a z 2 dv s+a (z) Relaing f(z) by f(z)e a z 2, we see ha he above ondiion is equivalen o e (a+b) z,w f(w) dv a+b (w) dv s+a (z) f dv s+a This is learly equivalen o he boundedness of S a+b on L (, dv s+a ), whih, aording o Theorem 11, is equivalen o (a + b) 2(s + a) Therefore, ondiions (b) and () are equivalen The equivalene of (a) and () is roved in exaly he same way REFERENES [1] V Bargmann, On a Hilber sae of analyi funions and an assoiaed inegral ransform, omm Pure Al Mah 14 (1961), [2] Berger and L oburn, Toeliz oeraors on he Segal-Bargmann sae, Trans Amer Mah So 31 (1987), [3] Berger and L oburn, Hea flow and Berezin-Toeliz esimaes, Amer J Mah 116 (1994), [4] Berger, L oburn, and K Zhu, Toeliz oeraors and funion heory in n- dimensions, Sringer Leure Noes in Mahemais 1256 (1987), 28-35

21 INTEGRAL OPERATORS 21 [5] M Dosanić, Norm esimae of he auhy ransform on L (Ω), Inegral Equaions and Oeraor Theory 52 (25), [6] M Dosanić, Norm of he Berezin ransform on L saes, o aear in J d Analyse Mah [7] G Folland, Harmoni Analysis in Phase Sae, Ann Mah Sudies 122, Prineon Universiy Press, 1989 [8] F Forelli and W Rudin, Projeions on saes of holomorhi funions in balls, Indiana Univ Mah J 24 (1974), [9] V Guillemin, Toeliz oeraors in n dimensions, Inegral Equaions and Oeraor Theory 7 (1984), [1] P Halmos and V Sunder, Bounded Inegral Oeraors on L 2 Saes, Sringer-Verlag, Berlin, 1978 [11] S Janson, J Peere, and R Rohberg, Hankel forms and he Fok sae, Revisa Ma Ibero-Amer 3 (1987), [12] O Kures and K Zhu, A lass of inegral oeraors on he uni ball of, o aear in Inegral Equaions and Oeraor Theory [13] SK Pihorides, On he bes values of he onsans in he heorems of M Riesz, Zygmund, and Kolmogorov, Sudia Mah 44 (1972), [14] W Rudin, Funion Theory in he Uni Ball of, Sringer-Verlag, New York, 198 [15] P Sjögren, Un onre-exemle our le noyau reroduisan de la mesure gaussienne dans le lan omlexe, Seminaire Paul Krée (Equaions aux dérivées arielles en dimension infinie), Paris, 1975/1976 [16] Y- Tung, Fok saes, PhD disseraion a he Universiy of Mihigan, 25 [17] GN Wason, A Treaise of he Theory of Bessel Funions, seond ediion, ambridge Universiy Press, 1995 [18] J Xia and D Zheng, Sandard deviaion and Shaen lass Hankel oeraors on he Segal-Bargmann sae, Indiana Univ Mah J 53 (24), [19] K Zhu, A Forelli-Rudin ye heorem, omlex Variables 16 (1991), [2] K Zhu, Saes of Holomorhi Funions in he Uni Ball, Sringer-Verlag, New York, 24 [21] K Zhu, A shar norm esimae of he Bergman rojeion on L saes, onemorary Mah 44 (26), MATEMATIČKI FAKULTET, STUDENTSKI TRG 16, 11 BEOGRAD, SERBIA address: domi@mafbgayu DEPARTMENT OF MATHEMATIS, SUNY, ALBANY, NY 12222, USA address: kzhu@mahalbanyedu