if,g)= [ f(z)7(7)da(z).

Pages 20
Views 54

Please download to get full document.

View again

of 20
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
transactions of the american mathematical society Volume 323, Number 2, February 1991 MÖBIUS INVARIANT HILBERT SPACES OF HOLOMORPHIC FUNCTIONS IN THE UNIT BALL OF C KEHE ZHU Abstract. We prove that there
transactions of the american mathematical society Volume 323, Number 2, February 1991 MÖBIUS INVARIANT HILBERT SPACES OF HOLOMORPHIC FUNCTIONS IN THE UNIT BALL OF C KEHE ZHU Abstract. We prove that there exists a unique Hubert space of holomorphic functions in the open unit ball of C whose (semi-) inner product is invariant under Möbius transformations. 1. Introduction Let Bn be the open unit ball in C and Aut(7?J be the Möbius group of biholomorphic mappings from Bn onto Bn. Let 77 be a Hubert space of holomorphic functions in Bn. In this paper, a Hubert space will be a linear space with a complete semi-inner product. Furthermore, we assume that 77 contains all polynomials and the polynomials are dense in 77. We say that 77 is Möbius invariant if f o tp g H and /o p\\ = \\f\\ whenever fi G H and tp G Aut(7?n). The main result of the paper is the following Theorem. There exists a unique Hubert space ofi holomorphic functions in Bn which is Möbius invariant. When n = 1, the above result was established in [1]. The unique Möbius invariant Hubert space in this case is the Dirichlet space 3 consisting of holomorphic functions / in the open unit disc D of C such that where da is the (normalized) in 3 is given by i /(z) 2^(2) +00, '0 area measure on D. The (semi-) inner product if,g)= [ f(z)7(7)da(z). Jb We remark that the condition /oç«= / for all / G 77 and tp G Aut(7?J is equivalent to the condition (/ o tp, g o tp) = (fi, g) for all /, g G 77 and tp G Aut(Bf. The organization of the paper is as follows: In the next section, we present some preliminary results which will be needed for the proof of the main theorem. In 3, we prove the uniqueness of invariant (semi-) inner products. In Received by the editors August 1, 1988 and, in revised form, February 9, Mathematics Subject Classification (1985 Revision). Primary 46E20; Secondary 30A10. Research supported in part by NSF Grant DMS American Mathematical Society /91 $1.00+ $.25 per page 824 KEHE ZHU 4 we prove that the inner product obtained in the uniqueness proof is indeed Möbius invariant. This will establish the existence. 5 and 6 discuss other possible ways of describing the invariant inner product and the invariant Hubert space. The author wishes to thank Don Marshall for useful discussions. 2. Preliminary results For any ordered «-tuple a = (ax,..., an) of nonnegative integers, we use the following abbreviated notations: \a\ = ax + + an, a! = af.---af., z =zf---zf, dhfi dlalf dza ' dzaf---dzan-' where z = (zx,..., zf gc and f(z) is holomorphic in Bn. Recall that the inner product in C is given by We simply write (z,w) = ^z k k k=\ \z\ = y/iz,z) = \/\zx\ \zn 2 for any z in C . Let f n be the group of unitary operators on the Hubert space C . It is clear that %n is a subgroup of Aut(7?n). In fact, í n is the isotropy subgroup of Aut(7?n) at 0 when we consider the natural action of Aut(7?n) on Bn. Therefore, for tp g Aut(ßn), we have tp G %n iff tp(0) = 0. For any a G Bn, define tpa g Aut(7?n) as follows: If a = 0, then pa(z) = -z. If a / 0, then Paiz) a-p z-\jl a -\a\2p^z l-iz,a) where Pa is the orthogonal projection from C onto the complex line [a] spanned in C by a, P^ is the orthogonal projection to the orthogonal compliment of [a] in C . It is easy to see that cpfo) = a, pa(a) = 0, and pa o tpa(z) = z. For these and other properties of pa, see [6, 7, 4]. Given tp G Aut(5n), let a = p~x(0) and U = p cpa, then U G Aut(Bn) and (7(0) = cp((pa(0)) = tp(a) = 0. Thus U is a unitary. Since tpa is involutive, we have tp = U otpa. This shows that Aut(7?J is generated by í n and {tpa: a G Bn}. The following lemma will be needed in the proof of the main theorem. MÖBIUS invariant HILBERT SPACES 825 Lemma 1. Aut(5n) is generated by %n and all tpa with a = (r, O,..., 0) and 0 r l. Proof. When a = (r, 0,..., 0), we simply write Pa = Pr- Since Aut(Bf) is generated by í n and { pa: a G Bn}, it suffices to show that each tpa (a G Bf) can be written as a product of unitaries and some tpr (0 r 1). Given a G Bn. If a = 0, then tpa is already a unitary. So we may assume a ^ 0. We show that there are U, V G %n such that pa = [j tp,, V. Suppose U G %An, then UPaU* is a one-dimensional projection in C . Since UPaU* (Ua) = UPa(a) = Ua, UP U* must be the projection onto the complex line [Ua], Thus UPaU* = PUa. It follows that UP^U* = Pya and UtpaU* = tpua. It is easy to see that there exists a unitary U G *2 n such that U(\a\, 0,..., 0) = a. The above argument now implies that pa = Utp,a, U*, completing the proof of Lemma 1. D Note that by the definition of tpa, we have - (z) = (izii. _ v^gzfi _^IZV VrK} ^1-rz,' 1-rz, ' 1-rz, ) for all zgbh and re[0, 1). In this paper, we'll be assuming that all Hubert spaces of holomorphic functions in Bn contain the polynomials and the polynomials are dense in them. It is natural to ask if this condition is always satisfied. For invariant function spaces, we have Lemma 2. Suppose H is a linear space of holomorphic functions in Bn with a complete semi-inner product which is invariant under Kut(Bf). Then H contains all the polynomials and the polynomials are dense in it iff 77 contains a nonconstant function and %n acts on 77 continuously. Proof. The only if part will follow from our main result. The if part has a proof similar to that of the one-dimensional case. See Proposition 2 in [2] and Lemma 3 in [8]. G Remark. We will not need Lemma 2 in the proof of the main theorem. Let dmn = dxx dyx... dxn dyn be the Euclidean measure in C = R2n. For any r 0, the volume of any Euclidean ball in C with radius r is nnr n/n\. Let dvn = ^dmn be the normalized volume measure on Bn so that B dvfz) = 1. When n = 1, we write da = dvx. Clearly, da = \dxdy = \rdrdq in polar coordinates. The following lemma will be needed in our later discussions. Lemma 3. If n 1 and f(z) = f(zx) only depends on z,, then f fi(z)dvn(z) = nf(l- JBn where D = Bx is the open unit disc in C. JD \zx\2) -xf(zx)da(zx), 826 KEHE ZHU Proof. f fi(z)dvn(z) = ^ f fi(z)dmn(z) JBn n JBn = fañ i fiizx)dxxdyx i dx2dy2--dxndyn \zf+- + \zn\ l-\zf '(1-lz I2) 1 m f f(z y-'u-w) i//(*i) ,»-1! ^1^1 = n f(l- Ja Zi Y-i/(2i)^Ä = n f(l-\zx\2) -xf(zx)da(zx) 3. The uniqueness Suppose 77 is a Hubert space of holomorphic functions in Bn with (semi-) inner product (, ). For any /, g G H, we write /w = Eva. g(z) = 2b/. a ß Since the polynomials are dense in 77, we have We compute the inner product (za, z^) in this section under the assumption that (, ) be Möbius invariant, that is, (/ o tp, g o tp) = (f, g) if fi, g g H and tp g Aat(B ). The main result of this section is Theorem A. If H is a Hubert space of holomorphic functions in Bn with a nonzero Möbius invariant (semi-) inner product (, ), then if', g) = EaAoriQ for some constant c 0 and all f(z) = J2a aaza, g(z) = J2a baza in 77. It follows from the above theorem that if 77 is an invariant Hubert space of holomorphic functions in Bn, then V a a ' ' ) Thus 77 is unique. Moreover, the above theorem also implies that the inner product in 77 is unique up to a positive multiple. The canonical inner product in 77 is MÖBIUS INVARIANT HILBERT SPACES 827 It is not clear at all that the above inner product is Möbius invariant. This will be proved in the next secton, thus establishing the existence of invariant Hubert spaces of holomorphic functions in Bn. In order to prove Theorem 4, it suffices to prove the following three equalities: (1) iz\zß) = 0 ifatß, (2) (z\za) = ^-(z\\z\a), (3) (zkx,zkx) = k(zx,zx). The constant c in Theorem 4 is then (z,, zx). The proof of ( 1) is almost trivial. Assume that a ^ ß, then there exists some 1 k n such that ak ± ßk. We may as well assume a, ^ ßx. Let U be the unitary operator on C defined by U(zx,z2,...,zf) = (zxe'e,z2,...,zf), where 6 is any real number. Since 77 is invariant under ÍAn, we must have (za, zß) = (zaou,zßou) = ei{a -ß )e(za, zß). Since 0 is arbitrary and a, ^ /?,, we must have (za, zß) = 0. The proof of (3) is similar to that of Theorem 1 in [1]. We reproduce it here for completeness. Recall that for r G [0, 1), f {2) _ (i^ 1 rzj _^S 1-rz, _^Vi 1 - rz. The invariance of (, ) gives (fi, fi) = (f pr, f cpr) for all f in H. f(z) = 1 - rz,, then by (1) and Therefore, /,/) = (!, l) + r2(z,,z,,,., r-zx I -rzx-r2 + rzx 1 - r2 loa z =1- r- =-!- =- 7 ^ ; 1-rz, 1-rz, 1 -rzx 4 I 1-r2 l-r2 (l,l) + r z,,z,) = 1 - rzx ' 1-rz, = (l-r2)2y.irkz\,fz[) c,j=0 /. = il-r) oo 2,2\-^ }^r 2k, k k, (zi'zi k=0 = (l,l) + r2(zx,zx)-2r2(l,l) oo k k, -, Jt-1 «c-1,., k-2 k-2,s 2k + Lsiiz\» zi)-2(zi. zi ) + (^i. zi ))' fe=2 Let 828 KEHE ZHU Since r g [O, 1) is arbitrary, we must have (1, 1) = O and (z\, z\) - 2(z\-x, z\~x) + (z\-2, z\-2) = 0, k 2. Now it is easy to show by induction that (z\, z\) = k(zx, zf for all k = 0, 1, 2,..., completing the proof of (3). In order to prove (2), we need the following Lemma 5. Suppose k 0 is an integer and Xa (\a\ = k) are complex numbers such that a =lt 1 for all t. H-+ L = 1 and t,. 0 (l j n), then k = I fior all \a\ = k. Proof. The assertion is clearly true for n = 1. We may as well assume that n 2. since for all r, H-h tn = 1, we have E^o «; = / ^ a 1 n for all tx tn = l and /7 0 (1 j n), where lal' an = r(a - 1), lal = fc. For a = k and?, -I-1- tn = 1, we can write \a\=k a 1 n = E^l-Cr,1(i-'1--'»-i)^r'- v' l«=ä = (!-*,-tn_xf x It follows that y a (_Í1_ I E v^' {n-i n-l) V1 fl ln-\. 'n-l v«!-l for all x 0 (1 j n 1). This implies that aa = 0 (\a\ = k). Hence Xa = 1 for all a = k, completing the proof of Lemma 5. o We can now prove the main equation (2). MÖBIUS INVARIANT HILBERT SPACES 829 Fix any k 1 and let U = («,-, ) x be any unitary matrix. We assume that U acts on C by the usual matrix multiplication Uz, where z G Cn is considered a column vector. It follows that zxou(z) = uxxzx uxnzn. Therefore, the invariance of (, ) implies that (z\, z\) = ((uxxzx + + uxnznf, («,,, z, + + uxnznf) Ep a! \ß\\ a, a _/J. _ß. a ßs E ^JWUl\'''UlnUU'''UlniZ Z) \a\=k\ß\=k ' H' /I l!\2 El a! \,.2a, i i2a, a a, a -* TT l nl 'Kl z z ) V Q- y Since U is unitary, we have w,, 2 H-h we have /i n\2 (zi'zi)=e (y-j 'i'-v z 'z) l«^2 = 1 Also U is arbitrary, thus for all tx + + tn = 1 and tj 0 (1 j n). Now if (z,, z\) = 0, then clearly all (za, z ) = 0 and so (f, f) = 0 for all f G H, contradicting the fact that (, ) is nonzero. Thus (z,, z, ) = 0 for all k 1. Now for any fixed k 1, let then Z_* ry! 1 «1 Q M-ft - for all?, H-hí = 1 and í 0 ( 1 / «). By Lemma 5, all Xa = 1, thus, q a, a., q UI, (z 'z ) = R!(zi 'zi This completes the proof of Theorem 4. D 4. The existence By Theorem 4 in the last section, if H is an invariant Hilbert space of holomorphic functions in Bn with semi-inner product (, ), then if,g) = cj^aabaßr\a\ for all f(z) = J2aaaza and g(z) = ^2abaza in 77. The purpose of this section is to show that if, g) = ^aj t-t.w\ is indeed a Möbius invariant inner product. 830 KEHE ZHU Theorem 6. Let 77 be the space of holomorphic functions f(z) = J2aaaza B n such that ElflJ O!lal +0 a ' '' Then 77 is a Möbius invariant Hubert space with (semi-) inner product V' & = EflAo[M a ' ' for all f(z) = EQ aaz, *(*) = EQ baza in 77. Proof. Clearly 77 is a nontrivial Hubert space and the polynomials are dense in 77. It only remains to show that its inner product is invariant under Aut(7?J. By Lemma 1, it suffices to prove the following identities: (1) ifi U,goU) = (f,g), (2) if Pr,g Pr) = ifi,g), for all /, g in 77 and U G %n, r G [0, 1). In order to prove (1), we recall the homogeneous expansion of holomorphic functions in Bn. Suppose f{z) = 2~ZQ aaza For any k 0, let /*(*)= E aaz \a\=k then fik is a homogeneous polynomial of degree k, and oo /(*) = /*««c=0 This is called the homogeneous expansion of /. It is easy to see that the homogeneous expansion of / is unique and invariant under linear transformations of C . Thus if A is a linear transformation on C , then ifioa)k=fkoa. In particular, for all U G 1 n, we have (fou)k=fikou. Equation ( 1 ) of this section will be a consequence of the following Lemma 7. For any f(z) = J2a aaza, g(z) = a baza in 77, we have m a ' ' fe=0 ' a where d V is the normalized volume measure on Bn. Proof. Since the volume measure is rotation invariant, we have / fk(z)gk(z)dv(z)= EöA/ \za\2dv(z). JBr \a\=k JBn MÖBIUS INVARIANT HILBERT SPACES 831 By 1.4.9(2) of [7], / Z dv(z) = -.- TTT. Therefore, / fk(z)gk(z)dv(z)=j2j^áyctaha. It follows that t^*l A:=0 and Lemma 7 is proved. AMftM «oo E(w + fc)!, v^ «a! r n!fc!,4- («+ )!** a «c=0 a =Jt v ' oo I = ^ y^ Tikab = y^-r frlala ë, Z f Z f ^1 a «Z f a!'! a a' fc=0 a =/c a ' ' D Now equation ( 1) of this section follows from the above lemma and the facts that homogeneous expansions are invariant under linear transformations and the volume measure is invariant under unitary transformations of C . In order to prove equation (2) of this section, take /, g G H and r G [0, 1). We have if Pr, g Pr) = \J2aaZa o g r,^bßzß o tp\ So it suffices to prove Note that a ß. a ß.. a ß. ( 0, OLÏ ß, (z otp z otp) = (z, z ) = l 1 RtM' a = ß. z-ocp(z)-.^hzii\ \..(_^HZ^ \l-rzj,, /. 1 - rzx I \ 1-rz, / When a ^ ß. for some 2 j n, then (zqopr, zß otpr) = 0 since no monomial in the Taylor expansion of za o cpr is equal to any monomial in the Taylar expansion of zß o tpr (just looking at the powers of z ). So we may as well assume that a = (a,, a2,..., af), ß = (ßx, a2,..., af). 832 KEHE ZHU We will prove in this case that Let N = a2-\-h a then I 0, a, ßx, (z otpr,zpotpr) = \ ' (z, z ), a, =/?,. Let then zaotpr(z) *,* JA ir-zx'.«, = (-lf(l-rz)2 (l-rz,)a +N^2 ^n z otpr(z) = (-ly(l-f) G(zx) = ir-z. il-rzx ir-z/ d -rzf)^ tf '-?* (l-rz,r ) = Y a,+n Z tckzl k=0 oo = E dkz\. fc=0 2 «Tïv 2 ' zn (z opr,z opr) = (l-r) }^ckdk(zxz22---zf,zxzf «c=0 We first settle the cases N = 0, 1 When N = 0, we have Ml-^E^^^(HiV), fe (A: + 7V)! «c=0 V) o^z^oeg^/^^ oo., _ / F'(zx)G'(zx)dA(: it=o /d 7d IA1 -^i A change of variable z, AAfAi. leads to r~zi 1 - rz. il4(z,). (zaotpr,zßotpf = i(z^)'(zß')'da(zx) Jd (0 ifax?ßx, I a. if a, = /?,, = (z, z ) for a2 = = an = 0. This is actually the one-dimensional result. See [1]. When N = 1, we have oo (za o tpr, zß o pr) = (1 - r2) Y, ck dk k=0 2% MÖBIUS INVARIANT HILBERT SPACES 833 By the change of variable for a2 -I-h an = 1 Now if TV 2, then r e -+ e , we get 1 re / «ß \ I f l-^2 r»(q,-/?,)/ 1 +r2-2r cos r 0 if a,,, 1 ifa,=^,, = (z, z ) fc!a,!-aj izaotpr,zßotpf = (l-r2)ny,ckdk J+NJik + N) By 1.4.9(2) of [7], (1 ff«2x-'-a Yc d klin-l)\ -il r (N-l)\ J * * ik + N - 1)! fc!(7v- 1)! rw= / lzil Hz), (^ + A^ where i V is the normalized volume measure on BN_,. Therefore, I I OO r. ia ß \ /i 2,./va2! -a! ^, /* fc,2.t,,, (z opr,z otpf = (l-r) 2_ 2^ckdk I z, dv(z) K ' k=0 JBN-l By Lemma 3, = {l-r2)nw^flb ^OWfdViz). [ F(zx)G(JfdV(z) = (N- 1) [ F(zx)G(zT)il - \zx\2f~2 da(z{) 7b _, 'N i Thus - '^«ÍKH 1 / r-z, (l-r2al- z,iy U-r^r V L4(z.) (l- z, 2)2- 834 KEHE ZHU Note that 2sN,,,,2s.N /.n ^ 1-/') il~\zl\ ) r~zi 1 - rz. and the measure da(zx )/( 1 - z, ) is invariant under Möbius transformations of D, thus a change of variable gives, a ß. a2!---a! f Q «2,jv íl4(z,) * *.*».}--rfr=3jr *.'V «-W ftt^7- Since the measure (l- z, if a, ^ /?,. Thus if a ^ ß. If a, = /?,, then a = ß and ) da(zx) i' is rotation invairant, we have zaxlzßfl-\zx\2f~2da(zx) = 0 / JO, a Ô,, a ß, (z otpr, z otpr) = (z,z) (' fr- Z Pr) = W=^lo^l{X lz. 2^_2^Z.) = a2!---a »! [\a (l-tf-2dt (N-2)\ J0 [l l) ai- Using the classical 5-function and T-function, we have fa,(l - tf~2 dt = B(ax + l,n-l) = Io Therefore, T(a, + 7V) aa - a\ a!, a a \ i* m* 'il /a a\ Z ^'Z 0^=(a1+A-l)! = Ñ!H = (Z 'Z)- This completes the rpoof of Theorem 6. o 5. Other descriptions of the invariant inner product (a,+7v-1)! The results in this section are somewhat negative. We point out many other ways of constructing the invariant inner product on the Dirichlet space 2 of the open unit disc D. Then we show one-by-one that these constructions fail in higher dimensions. The Dirichlet pairing if,g)3= i' fiz)tw)da(z) is clearly an invariant inner product on the Dirichlet space 2. Our problem here is to try to find a natural analog of this inner product in higher dimensions. As the first trial, one may be tempted to look at the pairing if,g)= f (Vf(z),Vg(z))c dv(z), JB MÖBIUS INVARIANT HILBERT SPACES 835 where V/(z) is the complex gradient of / at z. This is indeed an inner product, but it is easy to see that it is not invariant when n 2. Recall that the invariant Laplacian A of Bn is defined by Af(z)=A(fiotpz)(0), where A is the usual Laplacian in C . A is invariant in the sense that l(fotp)(z) = (lf)otp(z) for all (p G Aut(7in). On the unit disc D, A( / 2)(z) = 4(l- z 2)2 /(z) 2 for all holomorphic functions f. It follows that (fi,g)3= jl(fg)(z)k(z,z)da(z), where K(z, w) is Bergman kernel of D : Kiz,w)= l (1 - zw) The above inner product is invariant because A is invariant and the Bergman kernel is invariant. Naturally we look at the generalization to Bn : (fi,g)= [ l(fg)(z)k(z,z)dv(z), where K(z, w) is the Bergman kernel of Bn : K(z, w) =-i--t. (l-(z,w))n+x The pairing is clearly Möbius invariant since both A and the Bergman kernel are invairant. Unfortunately, when n 2, the only holomorphic functions / in B with n [ Ä(\fi\2)(z)K(z,z)dV(z) +oo are the constant functions. See [8]. Let P be the Bergman projection defined by Pf(z)= f K(z,w)fi(w)dV(w). JB Given a function g G L (Bn, dv), the Hankel operator 77 on the Bergman space L2a(Bn) = {fgl2(bn, dvf.fi is holomorphic} is defined by HJ=(I-P)(fig), 836 KEHE ZHU where 7 is the identity operator on L2(Bn, dv). Basic properties of Hankel operators can be found in [3-5]. Hankel operators depend on its symbol invariantly, that is, if tp G Aut(7?n), then Hgo9 = u9hgu where U is the unitary operator on L (Bn, dv) defined by U9fi(z) = J9(z)f( p(z)), JJz) is the complex Jacobian determinant of tp at z. It was shown in [3] that if / is holomorphic in D, then 77? is Hilbert-Schmidt iff / G 3. Moreover, (f,g)s=tr(h*jhl) for all /, g in 3A, where tr denotes the trace of an operator. Once again, this nice result does not generalize to higher dimensions. As shown in [8], when n 2, the only Hilbert-Schmidt Hankel operator 77? with / holomorphic in Bn is the zero operator (when / is a constant). The so-called Berezin transform is an invariant transform which has attracted much attention lately in function theory and operator theory [4, 5]. We have a brief discussion of it here because of its connection with invariant inner products. Given a function / in L (Bn, dv), let ^=L/wl%^dv{w) ze5- The function / is called the Berezin transform of / (see [4, 5]). Clearly, / = /, and the reproducing property of K(z, w) gives fi = f if fi is holomorphic. The invariance of the Bergman kernel K(z, w) implies the invariance of the Berezin transform namely, for all tp G Aut(Bf. Now consider the pairing and the space fotp(z) =f(tp(z)) (/,*}= f (Tgiz)-fiz)giz))K(z,z)dV(z) H = if: jb i\fi\\z) - /(z) 2)7C(z, z)dv(z) +oo, / holomorphic in Bn . Since the Berezin transform is invariant and the measure K(z, z)dv(z) is also invariant, 77^ is an invariant space. It is also easy to see that %An acts on 77n MÖBIUS INVARIANT HILBERT SPACES 837 continuously, thus by Lemma 2, Hn will be an invariant Hubert space if 77 contains a nonconstant function. When n = 1, we show that the function f(z) = z is in 77,. In fact, for f(z) = z in D, we have 1 log- 1-lzl \fi\2(z) - \f(z)\2 = i\fiotp2(w)-fi(z)\2da(w) (see [4]) z -w -f da(w) 1 - zw i I 2x2 l-\z\ ) It follows that for fi(z) = z in D, f(\fi\2(z)-\fi(z)\2)k(z, z)da(z) = f -L \z\ log - z da(z) +CO. By the uniqueness of invariant Hubert spaces, we must have 3 = 77, and where (/, g)s = c~ ifgi z)-f(z)g(z))k(z,z)da(z), [± log da(z) = 1. \z\ 1-Iz Once again th
Related Documents
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks

We need your sign to support Project to invent "SMART AND CONTROLLABLE REFLECTIVE BALLOONS" to cover the Sun and Save Our Earth.

More details...

Sign Now!

We are very appreciated for your Prompt Action!