GAUSSIAN INTEGRAL MEANS OF ENTIRE FUNCTIONS CHUNJIEWANGANDJIEXIAO 3 1 0 ABSTRACT. Foranentirefunctionf : C Candatriple(p,α,r) 2 7→ ∈ (0, ) ( , ) (0, ], the Gaussian integral means of f (with n resp∞ect×toth−e∞are∞ame×asure∞dA)isdefinedby a J M (f,r)= e−α|z|2dA(z) −1 f(z)pe−α|z|2dA(z). p,α 3 | | 1 Viaderivingam(cid:16)aZx|izm|<urmprinciplefor(cid:17)M Z|(zf|<,rr),weestablishnotonly p,α ] Fock-SobolevtraceinequalitiesassociatedwithMp,p/2(zmf(z), )(as V m = 0,1,2,...), but also convexitiesof r lnM (zm,r) an∞d r p,α C M2,α<0(f,r)inlnrwith0<r< . 7→ 7→ ∞ . h t a m 1. INTRODUCTION [ 3 Let dA be the Euclidean area measure on the finite complex plane C. v Suppose α is real and 0 < p < . For any entire function f : C C, we 9 ∞ 7→ consideritsGaussianintegralmeans 4 3 f(z) pe−α|z|2dA(z) 0 M (f,r) = |z|<r| | r (0, ). 1. p,α e−α|z|2dA(z) ∀ ∈ ∞ R 0 |z|<r 3 Uponwriting R 1 : M(r) = 2π f(reiθ) pdθ; v 0 | | i v(r) = re−αr2; X  R r i = √ 1 theimaginaryunit, a − − weget   r d v(r) M(r) M(s) v(s)ds M (f,r) = 0 − 0, dr p,α 2π rv(s)ds 2 ≥ R (cid:0) (cid:1) 0 and hence the function r Mp,α(f,r(cid:0))Ris strictly(cid:1)increasing on (0, ) un- less f is constant. Conseq7→uently, letting r 0 and r in M ∞(f,r) p,α → → ∞ 2000MathematicsSubjectClassification. Primary30C80,30H20,52A38,53C43. Key words and phrases. Maximum principle, trace inequality, logarithmic convexity, Fock-Sobolevspace. ThisworkwasinpartsupportedbyNSERCofCanadaandcompletedduringthefirst- namedauthor’svisit(2012.9-12)toMemorialUniversity. 1 2 CHUNJIEWANGANDJIEXIAO respectively,wefind thefollowingmaximumprincipleforr (0, ): ∈ ∞ f(0) p = M (f,0) M (f,r) p,α p,α | | ≤ f(z) pe−α|z|2dA(z) M (f, ) = C| | ≤ p,α ∞ e−α|z|2dA(z) R C withequalityifand onlyiff isaconstant.R Besides the above maximum principle we are here motived mainly by [15,6,7,13,12,14,2]totakeafurtherlookattheGaussianintegralmeans M (f,r) from two perspectives. The first is to treat the last inequality as p,α a space embedding: if dµ (z) = 1 dA(z) (with 1 being the character- r |z|<r E isticfunctionofE C)then ⊂ f(z)e−|z2|2 pdµr(z) e−p|z2|2 dA(z) Mp,p/2(f, ). | | ≤ ∞ ZC (cid:18)Z|z|<r (cid:19) Such an interpretation leads to characterizing a given nonnegative Borel measureµ on Csuch thatthefollowingFock-Sobolevtraceinequality f Lq(C,µ) k k 1 f(z)e−|z2|2 qdµ(z) q ≡ | | C (cid:18)Z (cid:19) 1 . M (zmf(z), ) p p,p/2 ∞ (cid:16) (cid:17) 1 zmf(z)e−|z2|2 pdA(z) p ≈ | | C (cid:18)Z (cid:19) f Fp,m ≡ k k holds for all holomorphic functions f : C C in p,m. In the above and 7→ F below: 0 < p,q < ; • X . Y (i.e∞. Y & X) means that there is a constant c > 0 such that X• cY -moreover- X Y isequivalenttoX . Y . X; ≤ ≈ m isnonnegativeinteger; • p = p,0 and p,m stand for the so-called Fock space and Fock- • F F F Sobolev space of order m 1 respectively. Interestingly, for an entire functionf : C C onehas≥(cf. [2]): 7→ f p,m f(0) + + f(m−1)(0) + f(m) < . Fp,0 ∈ F ⇐⇒ | | ··· | | k k ∞ B(a,r) = z C : z a < r is the Euclidean disk centered at a• Cwithradi{us r∈> 0. | − | } ∈ GAUSSIANINTEGRALMEANSOFENTIREFUNCTIONS 3 As stated in Theorem 3 of Section 2, the above-required measure is fully determinedby sup µ(B(a,r)) < as 0 < p q < ; a∈C (1+|a|)qm ∞ ≤ ∞ p  µ(B(a,r)) p−q dA(a) < as 0 < q < p < .  C (1+|a|)qm ∞ ∞ (cid:16) (cid:17) R Asaparticularlyinterestingandnaturalby-productofthischaracterization,  we can also use the Taylor expansion of an entire function at the origin to gettheoptimalGaussianPoincare´ inequality(see[8, (1.6)]aswellas[5, p. 115] and [16, Theorem 1] for the endpoint case corresponding to f 1,1 ∈ F withf(0) = 0) f(z)e−|z2|2 2dA(z) π f(0) 2 f′(z)e−|z2|2 2dA(z) f 2,1 | | − | | ≤ | | ∀ ∈ F C C Z Z which,plustheforegoingmaximum-principle-basedestimate(cf. [2, (1)]) f′(z) e−|z2|2 (2π)−1 f′(z)e−|z2|2 dA(z) f 1,1, | | ≤ | | ∀ ∈ F C Z derivesthefollowingGaussianisoperimetric-Sobolevinequalityf 1,1: ∈ F 2 f(z)e−|z2|2 2dA(z) π f(0) 2 (2π)−1 f′(z)e−|z2|2 dA(z) | | − | | ≤ | | C C Z (cid:18)Z (cid:19) whosesharp formis 2 f(z)e−|z2|2 2dA(z) π f(0) 2 (4π)−1 f′(z)e−|z2|2 dA(z) | | − | | ≤ | | C C Z (cid:18)Z (cid:19) since this inequality can be proved valid for the entire functions f(z) = zk with k = 1,2,3,... through a direct computation with the polar coordinate system,themathematicalinductionand theinequalityforthegammafunc- tionΓ( )below: · Γ k+1 k +1 2 . Γ k ≤ 2 (cid:0) 2 (cid:1) r The second is to decide: w(cid:0)hen(cid:1) lnr lnM (zk,r) is convex for r p,α 7→ ∈ (0, ),namely,whentheGaussianHadamardThreeCircleTheorembelow ∞ r r r ln 2 lnM (zk,r) ln 2 lnM (zk,r )+ ln lnM (zk,r ) p,α p,α 1 p,α 2 r ≤ r r 1 1 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) holds for 0 < r r r < . The expected result is presented in The- 1 2 ≤ ≤ ∞ orem 7 of Section 3, saying that for a nonnegativeinteger k and a positive 4 CHUNJIEWANGANDJIEXIAO numberp, lnr lnM (zk,r)is concaveas r (0, )under0 < α < ; p,α 7→ ∈ ∞ ∞ c (0, ) lnr lnM (zk,r)isconvexand concave  p,α ∃ ∈ ∞ ∋ 7→  as r (0,c]andr [c, )respectivelyunder < α 0. ∈ ∈ ∞ −∞ ≤ As a consequence, we have that if < α, p < 0 then the function   −∞ − lnr lnM (zk,r) is convex as r (0, (2+pk)/( 2α)] and hence p,α 7→ ∈ − the function lnr lnM (f,r) is convex as r (0, 1/( α)] for any entirefunctionf :7→C C2.,αIn otherwords, p ∈ − 7→ p r r r ln 2 lnM (f,r) ln 2 lnM (f,r )+ ln lnM (f,r ) 2,α 2,α 1 2,α 2 r ≤ r r 1 1 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) when 0 < r r r < 1/( α). However, as proved in Remark 9 via 1 2 ≤ ≤ − consideringtheentirefunction1+z, thelastconvexitycannotbeextended p to(0, ). ∞ 2. TRACE INEQUALITIES FOR FOCK-SOBOLEV SPACES Weneedtwolemmas. Thefirstlemmacomesfrom[2]and[18,17,4,11]. Lemma 1. Let p,σ,a,t,λ (0, ). ∈ ∞ (i) If m is a nonnegative integer, p (z) is the Taylor polynomial of ez of m orderm 1 (with theconventionthatp = 0), andb > (mp+2),then 0 − − ezw pm(zw) pe−a|w|2 w bdA(w) . z bep4a2|z|2 z σ. | − | | | | | ∀ | | ≥ C Z Furthermore,thislastinequalityholdsalsoforall z C when b pm. (ii)If f : C C isan entirefunction,then ∈ ≤ 7→ p p f(z)e−λ2|z|2 . f(w)e−λ2|w|2 dA(w) z C. ∀ ∈ (cid:12) (cid:12) ZB(z,t)(cid:12) (cid:12) (iii) The(cid:12)re exists a p(cid:12)ositive cons(cid:12)tant r such(cid:12)that for any 0 < r < r , the (cid:12) (cid:12) (cid:12) 0 (cid:12) 0 Fockspace p exactlyconsistsofallfunctionsf = c k , where F w∈rZ2 w w k (z) = exp(zw¯ w 2/2); P w −| | c : w rZ2 lp;  w { ∈ } ∈  1 k{cw}klp = w∈rZ2|cw|p p; Z2 = n+im : n,m = 0, 1, 2,... ; { (cid:0)P (cid:1)± ± } rZ2 = r(n+im) : n,m = 0, 1, 2,... .   { ± ± }  Moreover    f inf c f p, Fp w lp k k ≈ k{ }k ∀ ∈ F wheretheinfimumistakenoverallsequences c givingrisetotheabove w { } decomposition. GAUSSIANINTEGRALMEANSOFENTIREFUNCTIONS 5 The second lemmais the so-called Khinchine’s inequality, which can be found,forexample,in[7]. Lemma 2. Suppose p (0, ) and c C. For the integer part [t] of j ∈ ∞ ∈ t (0, )let ∈ ∞ 1, 0 t [t] < 1/2 r (t) = ≤ − 0 1, 1/2 t [t] < 1 (cid:26) − ≤ − and r (t) = r (2jt) j = 1,2, . j 0 ∀ ··· Then m 21 1 m p p1 c 2 ≈ c r (t) dt . j j j | | Xj=1 ! Z0 (cid:12)(cid:12)Xj=1 (cid:12)(cid:12) ! (cid:12) (cid:12) As the main result of this section, t(cid:12)he forthcom(cid:12) ing family of analytic- (cid:12) (cid:12) geometrictrace inequalitiesfortheFock-Sobolevspacesisanaturalgener- alizationoftheso-calleddiagonalCarlesonmeasuresfortheFock-Sobolev spaces in[2]. Theorem 3. Let m be a nonnegative integer, r (0, ), and µ be a non- negativeBorel measureon C. ∈ ∞ (i)If0 < p q < , then ≤ ∞ f . f f p,m Lq(C,µ) Fp,m k k k k ∀ ∈ F when andonlywhen µ(B(a,r)) sup < . a∈C (1+ a )mq ∞ | | Equivalently,a µ(B(a,r))(1+ a )−qm is ofclassL∞(C). 7→ | | (ii)If 0 < q < p < , then ∞ f . f f p,m Lq(C,µ) p,m k k k k ∀ ∈ F when andonlywhen p µ(B(a,r)) p−q < where s (0, ). (1+ a )mq ∞ ∈ ∞ a∈sZ2(cid:18) | | (cid:19) X Equivalently,a µ(B(a,r))(1+ a )−qm is ofclassLp/(p−q)(C). 7→ | | Proof. (i) Suppose 0 < p q < . The following argument is similar to ≤ ∞ thatofTheorem10 in[2]. Assumefirstlythat f . f holdsforallf p,m. Taking Lq(C,µ) Fp,m f = 1showsthatµ(Kk) .k 1foranykcokmpactset K C. ∈ F Fix anya C and let ⊂ ∈ f(z) = (eza p (za))/zm m − 6 CHUNJIEWANGANDJIEXIAO inthelastassumption. ThenLemma1(i)implies eza p (za) q − m e−12|z|2 dµ(z) . (ep2|a|2)pq = eq2|a|2. zm C Z (cid:12) (cid:12) (cid:12) (cid:12) Inparticular, (cid:12) (cid:12) (cid:12) (cid:12) eza p (za) q − m e−12|z|2 dµ(z) . eq2|a|2. zm ZB(a,r)(cid:12) (cid:12) (cid:12) (cid:12) If a > 2r,then z m(cid:12)iscomparableto(1+(cid:12) a )m forB(a,r). So | | | | (cid:12) (cid:12)| | eza q 1 e−zapm(za) qe−q2|z|2dµ(z) . (1+ a )mqe2q|a|2 | | | − | | | ZB(a,r) holdsforall a > 2r. Notethat | | lim inf 1 e−zap (za) = 1. m |a|→∞z∈B(a,r)| − | Thus eza qe−2q|z|2dµ(z) . (1+ a )mqeq2|a|2 | | | | ZB(a,r) holdsforthesufficientlylarge a . But thislastinequalityisclearly truefor | | smaller a as well. So wehave | | eza qe−q2|z|2dµ(z) . (1+ a )mqe2q|a|2 a C. | | | | ∀ ∈ ZB(a,r) Completing a square in the exponent, we can rewrite the inequality above as e−q2|z−a|2dµ(z) . (1+ a )mq | | ZB(a,r) therebydeducing µ(B(a,r)) . (1+ a )mqeq2r2 a C. | | ∀ ∈ Conversely,assumethat µ(B(a,r)) . (1+ a )mq a C. | | ∀ ∈ Weproceed toestimatetheintegral kfkqLq(C,µ) = |f(z)e−21|z|2|qdµ(z) C Z of any given function f p,m. For any positive number s let Q denote s thefollowingsquareinC∈wFithvertices 0,s,si,and s+si: Q = z = x+iy : 0 < x s & 0 < y s . s { ≤ ≤ } Itis clearthat C = (Q +a) a∈sZ2 s ∪ GAUSSIANINTEGRALMEANSOFENTIREFUNCTIONS 7 isadecompositionofCintodisjointsquares ofsidelengths. Thus kfkqLq(C,µ) = |f(z)e−12|z|2|qdµ(z). a∈sZ2ZQs+a X Fix positivenumberss and tsuch thatt+√s = r. By Lemma1(ii) f(z)e−21|z|2 p . f(w)e−21|w|2 pdA(w) | | | | ZB(z,t) 1 . wmf(w)e−21|w|2 pdA(w) (1+ z )mp | | | | ZB(z,t) holds for all z C. Now if z Q +a, where a sZ2 implies B(z,t) s ∈ ∈ ∈ ⊂ B(a,r)bythetriangleinequality,andhence1+ z 1+ a . Consequently, | | ≈ | | 1 f(z)e−21|z|2 p . wmf(w)e−12|w|2 pdA(w). | | (1+ a )mp | | | | ZB(a,r) Thisamountsto q f(z)e−12|z|2 q . 1 wmf(w)e−21|w|2 pdA(w) p . | | (1+ a )mq | | | | (cid:18)ZB(a,r) (cid:19) Therefore, q kfkqLq(C,µ) . (µ1(B+(aa,)rm))q |wmf(w)e−21|w|2|pdA(w) p . a∈sZ2 | | (cid:18)ZB(a,r) (cid:19) X Combiningthislast estimatewiththepreviousassumptiononµ and p q, ≤ weobtain q kfkqLq(C,µ) . |wmf(w)e−21|w|2|pdA(w) p a∈sZ2(cid:18)ZB(a,r) (cid:19) X q p . wmf(w)e−21|w|2 pdA(w) . | | a∈sZ2ZB(a,r) ! X NotethatthereexistsapositiveintegerN suchthateachpointinCbelongs toat mostN ofthedisksB(a,r), wherea sZ2. So, onegets ∈ q kfkqLq(C,µ) . |wmf(w)e−12|w|2|pdA(w) p ≈ kfkqFp,m, C (cid:18)Z (cid:19) asdesired. (ii)Suppose0 < q < p < . Thefollowingproofis inspiredby[13]. ∞ 8 CHUNJIEWANGANDJIEXIAO First assume that f . f holds for all f p,m. For any Lq(C,µ) Fp,m k k k k ∈ F c lp, wemay choose r (t) as in Lemma2, therebygetting j j { } ∈ { } c r (t) lp & c r (t) = c . j j j j lp j lp { } ∈ k{ }k k{ }k Thenby Lemma1 (iii)weknowthat ∞ c r (t)k (z) zmf(z) j j aj ≡ j=1 X is in p with norm f inf c . Here a is the sequence of Fp,m j lp j F k k ≈ k{ }k { } allcomplexnumbersofsZ2 and k (z) = eza−1|a|2. In particular, a 2 ∞ f(z) = c r (t)k (z)z−m. j j aj j=1 X Accordingtotheassumptionwehave ∞ k (z) q c r (t) aj dµ(z) = f q . f q , j j zme|z|2/2 k kLq(C,µ) k kFp,m C Z (cid:12)Xj=1 (cid:12) (cid:12) (cid:12) whencegett(cid:12)ingby Lemma2, (cid:12) ∞ q |cjkaj(z)e−21|z|2|2|z|−2m 2dµ(z) . kfkqFp,m. C Z (cid:16)Xj=1 (cid:17) Also, note that if a > 2r then z m is comparable to (1 + a )m for z | | | | | | ∈ B(a,r). So ∞ q |cjkaj(z)e−12|z|2|2|z|−2m 2dµ(z) C Z (cid:16)Xj=1 (cid:17) ∞ ∞ q = cje−21|z−aj|2 2 z −2m 2dµ(z) | | | | Xl=1 ZQs+al (cid:16)Xj=1 (cid:17) ∞ cl q z −mqe−2q|z−al|2dµ(z) ≥ | | | | l=1 ZQs+al X ∞ & cj q z −mqe−2q|z−aj|2dµ(z) | | | | j=1 ZB(aj,r) X ∞ µ(B(a ,r)) & c q j . | j| (1+ a )mq j j=1 | | X GAUSSIANINTEGRALMEANSOFENTIREFUNCTIONS 9 So, acombinationofthepreviously-establishedinequalitiesgives ∞ µ(B(a ,r)) c q j . c q = c q . | j| (1+ a )mq k{| j|}klp k{| j| }klp/q j j=1 | | X Sincep/(p q)istheconjugatenumberofp/q,anapplicationoftheRiesz − representationtheoremyields µ(B(a ,r)) j p lp−q. (1+ a )mq ∈ (cid:26) | j| (cid:27) Conversely, assume that the last statement holds. Note that the first part oftheargument fortheabove(i)tellsthat q kfkqLq(C,µ) . (µ1(B+(aa,)rm))q |wmf(w)e−21|w|2|pdA(w) p a∈sZ2 | | (cid:18)ZB(a,r) (cid:19) X holdsforallf p,m. ApplyingHo¨lder’sinequalitytothelastsummation ∈ F weobtain p−q p µ(B(a,r)) p−q p f q . k kLq(C,µ) (1+ a )mq ! a∈sZ2(cid:18) | | (cid:19) X q p wmf(w)e−21|w|2 pdA(w) . × | | a∈sZ2ZB(a,r) ! X Onceagain,noticethatthereexistsapositiveintegerN suchthateachpoint inC belongstoat mostN ofthedisksB(a,r), wherea sZ2. So, ∈ p−q p µ(B(a,r)) p−q p f q . f q . k kLq(C,µ) (1+ a )mq k kFp,m ! a∈sZ2(cid:18) | | (cid:19) X Thiscompletestheargument. (cid:3) The following extends [1, Theorem 5] (cf. [10, Theorem 1]), and [3, Theorem1], respectively. Corollary 4. Let φ : C C be an entire function. For p (0, ) and 7→ ∈ ∞ a nonnegative integer m define two linear operators acting on an entire functionf : C C: 7→ C f(z) = f φ(z) z C; φ ◦ ∀ ∈ T f(z) = zf(w)φ′(w)dw z C. ( φ 0 ∀ ∈ R 10 CHUNJIEWANGANDJIEXIAO (i)ThecompositionoperatorC : p,m q existsasaboundedoperator φ F 7→ F ifandonlyif sup Rφ−1(B(a,r))e−q|z|2/2dA(z) < when 0 < p q < ; a∈C (1+|a|)mq ∞ ≤ ∞  Rφ−1(B(a,r))e−q|z|2/2dA(z) p/(p−q)dA(a) < when 0 < q < p < .  C (1+|a|)mq ∞ ∞ (cid:16) (cid:17) R (ii) The Riemann-Stieltjes integral operator T : p,m q exists as a φ F 7→ F boundedoperatorifandonlyif R |φ′(z)| qdA(z) sup B(a,r) (1+|z|) < when 0 < p q < ; a∈C (1+|a|)mq ∞ ≤ ∞  R |φ′(cid:0)(z)| qd(cid:1)A(z) p/(p−q)  B(a,r) (1+|z|) dA(a) < when 0 < q < p < .  C (1+|a|)mq ∞ ∞ (cid:0) (cid:1) (cid:16) (cid:17) ProRof. (i)ForanyBorelsetE Cletφ−1(E)bethepre-imageofE under ⊂ φand q z 2 µ(E) = exp | | dA(z). − 2 Zφ−1(E) (cid:16) (cid:17) Then kCφfkqLq(C,µ) = |f(z)e−|z2|2|qdµ(z) ∀ f ∈ Fp,m. C Z An application of Theorem 3 with the above formula gives the desired re- sult. (ii) According to [3, Proposition 1], an entire function f : C C be- 7→ longsto q ifand onlyif F f′(z) e−|z2|2 q | | dA(z) < . 1+ z ∞ C Z (cid:16) | | (cid:17) So, T f q isequivalentto φ ∈ F f(z)φ′(z) e−|z2|2 q | | dA(z) < . 1+ z ∞ C Z (cid:16) | | (cid:17) Now,choosing φ′(z) q dµ(z) = | | dA(z) 1+ z (cid:16) | |(cid:17) inTheorem 3, weget theboundednessresultforT . (cid:3) φ