H -calculus for semigroup generators on BMO ∞ 7 1 0 Tim Ferguson Tao Mei∗ Brian Simanek 2 r March 14, 2017 a M 1 Abstract 1 We prove that the negative generator L of a semigroup of positive A] contractions onL∞ hasboundedH∞(Sη)-calculuson BMO(√L) forany angle η > π/2, provided L satisfies Bakry-E´mery’s Γ2 0 criterion. F ≥ Our arguments only rely on the properties of the underlying semigroup . h and works well in the noncommutative setting. A key ingredient of our t argument is a quasi monotone property for the subordinated semigroup a m Tt,α=e−tLα,0<α<1, that is provedin thefirst part of this article. [ Introduction 4 v 3 Let∆= ∂2bethenegativeLaplacianoperatoronRn. TheassociatedPoisson 2 semigrou−poxfoperatorsP =e t√∆,t 0hasmanynicepropertiesthatmakeit 6 t − ≥ a very useful tool in the classicalanalysis. In particular,the Poissonsemigroup 6 0 hasaquasimonotonepropertythatthereexistconstantscr,j suchthat,forany . nonnegative function f L1(Rn, 1 dx), 1 ∈ 1+x2 | | 0 7 tj∂jP f c P f, (1) 1 | t t |≤ r,j rt : for any 0 < r < 1,j = 0,1,2,.... As a first result of this article, we show v i that the quasi monotone property (1) extends to all subordinated semigroups X T =e tLα for all 0<α<1 if L generatesa semigroupof positive preserving t,α − r operatorsonaBanachlatticeX. Thecaseof0<α 1 iseasyandispreviously a ≤ 2 known because of a precise subordination formula (see e.g. [24, 20]). Functional calculus is a theory of studying functions of operators. The so- called H -calculus is a generalizationof the Riesz-Dunford analytic functional ∞ calculusanddefines Φ(L)via aCauchy-typeintegralforan(unbounded) secto- rial operator L and a function Φ that is bounded and holomorphic in a sector S ofthecomplexplane. LissaidtohavetheboundedH -calculuspropertyif η ∞ the so-defined Φ(L) extends to bounded operators on X and Φ(L) c Φ k k≤ k k∞ for all such Φ’s. The theory of bounded H -calculus has developed rapidly ∞ in the last thirty years with many applications and interactions with harmonic ∗ResearchpartiallysupportedbytheNSFgrantDMS-1632435. 1 analysis, Banach space theory, and the theory of evolution equations, starting with A. McIntosh’s seminal work in 1986 ([23], [14],[22],[31]). It is a majortaskin the study inthe bounded H -calculustheory to deter- ∞ minewhichoperatorshavesuchastrongproperty. Cowling,Duong,andHieber &Pru¨ss([7,11,16])provethattheinfinitesimalgeneratorofasemigroupofpos- itive contractions on Lp,1 < p < always has the bounded H (S )-calculus ∞ η on Lp for any η > π. When the s∞emigroup is symmetric, the angle can be re- 2 duced to η >ω = π π by interpolation. It is not surprising that this result p |2 − p| fails for L in general. One may want to seek a BMO-type space that could ∞ be an appropriate alternative for p = case. The main theorem of this arti- ∞ cle states that the negative generator L of a semigroup of positive contractions on L always has bounded H -calculus on the space BMO(√L), provided L ∞ ∞ satisfies Bakry-E´mery’sΓ2 criterion. The classical BMO norm of a function f L1(Rn, 1 dx) can be defined ∈ 1+x2 as | | 1 2 2 f =sup e t√∆ f e t√∆f . (2) BMO spaces aksskoBcMiatOe(d√∆w)ith st>em0(cid:13)(cid:13)(cid:13)(cid:13)igr−oup g(cid:12)(cid:12)(cid:12)en−erat−ors ha(cid:12)(cid:12)(cid:12)ve(cid:13)(cid:13)(cid:13)(cid:13)Lb∞een intensively studied recently (e.g. [12] and the after-works). When a cubic-BMO is available, one can often compare it with the semigroup BMO and they are equivalent in many cases. In this article, we follow the approach from [20, 24] and consider the BMO(√L)-(semi)norm defined similarly to (2), merely replacing ∆ with the studied semigroup generator L. The corresponding space BMO(√L) inter- polates well with Lp-spaces when the semigroup is symmetric Markovian (see Lemma 11). Undertheassumptionsofourmaintheorem,wealsostudysemigroup-BMO spacesBMO(Lα),0<α<1andprovethatthey areallequivalent. We further prove that the imaginary power Lis is bounded on the associated semigroup- BMO space BMO(Lα) with a bound . (1+|s|)|32|exp(||π2s||) (see (71),(72)). This complements Cowling’s Lp-estimate (see [7, Corollary 1]) and fixes a mis- take in [20] (see the Remark in Section 3). Therelatedtopicsandestimatesonsemigroupgeneratorshavebeenstudied withgeomtric/metricassumptionsontheunderlyingmeasurespace. Thisarticle isfromafunctionalanalysispointofviewandtriestoobtainageneralresultby abstractarguments. CowlingandHieber/Pru¨ss’smethodfortheirH -calculus ∞ results on Lp is based on the transference techniques of Coifman and Weiss, which does not work for non-UMD Banach spaces, such as BMO. Our method istoconsiderthefractionalpowerofthegeneratortotaketheadvantagesofthe quasi-monotoneproperty(1). Ourargumentworkswellforthenoncommutative case,thatis,forLthatgeneratesasemigroupofcompletelypositivecontractions on a semifinite von Neumann algebra. We analyzea few examples to illustrate our results atthe end ofthe article. We use c for an absolute constant which may differs from line to line. 2 1 The complete monotonicity of a difference of exponential power functions A nonnegative C -function f(t) on (0, ) is completely monotone if ∞ ∞ ( 1)k∂kf(t) 0 − t ≥ for all t. Easy examples are f(t) = e λt for any λ > 0. It is well-known − that completely monontonicity is preserved by addition, multiplication, and taking pointwise limits. So the Laplace transform of a positive Borel measures on [0, ), which is an average of e λt in λ, is completely monotone. The − ∞ Hausdorff-Bernstein-WidderTheoremsaysthatthe reverseis alsotrue;namely thatafunctioniscompletelymonotoneifandonlyifitistheLaplacetransform ofapositiveBorelmeasureson[0, ). Inparticular,g (t)=e stα iscompletely s − ∞ monotoneandistheLaplacetransformofapositiveintegrableC functionφ ∞ s,α on (0, ) for all s>0,0<α<1. ∞ e stα = ∞e λtφ (λ)dλ= ∞e sα1λtφ (λ)dλ. (3) − − s,α − 1,α Z0 Z0 The function φ is uniquely determined by the inverse Laplace transform s,α φs,α(λ)=s−α1φ1,α(s−α1λ)= −1(e−szα)(λ)= 1 σ+i∞ezλe−szαdz, (4) L 2πi Zσ−i∞ for σ >0,λ>0. The derivative ∂ φ is again an integrable function (see e.g. s s,α [32, page 263]), and tαe stα = ∞e λt∂ φ (λ)dλ. (5) − − s s,α − Z0 The properties of φ are important in the study of the fractional powers of s,α semigroup generators. The goal of this section is to prove a few pointwise inequalities for φ , s,α which will be used in the next section. For that purpose, we first prove the complete monotonicity of severalvariants of e stα. − For k,n N,1 k n, let a(n) be the real coefficients in the expansion ∈ ≤ ≤ k dn n dtne−tα =(−1)n ak(n)t−n+kαe−tα. k=1 X It is easy to see that dn n e ctα =( 1)n cka(n)t n+kαe ctα. dtn − − k − − k=1 X (n) We define a = 0 if k > n or k 0. The proof of the following lemma is k ≤ simple and elementary. We leave it for the reader to verify. 3 Lemma 1. The a(n)’s satisfy the relation k a(n+1) =(n kα)a(n)+αa(n) (6) k − k k 1 − for all k Z,n N. ∈ ∈ Lemma 2. Let K ,i = 1,2 be the first integer such that K α. Then, for all j Z,n N, wei have K+i ≥ ∈ ∈ a(n) (j+1)a(n) 0 if k K (7) k+j − k+j+1 ≥ ≥ 1 (n) (n) (n) (n) (j+1)(a (j+2)a ) a (j+1)a if k K (8) k+j+1− k+j+2 ≤ k+j − k+j+1 ≥ 2 Proof. We only need to prove the case j 0. Let D be the right derivative for ≥ discrete functions: Df =f(j+1) f(j). It is easy to see that the productrule holds D(jf)(j)=jD f(j)+f(j+−1). Fix k Z. Let j ∈ f (j)=a(n) j! (9) n k+j for j 0, where we use the convention that 0!=1. By (6), we have ≥ f (j)=(n (k+j)α)f (j)+αjf (j 1), n+1 n n − − forallj 1andf (0)=(n kα)f (0)+αa(n) . Takingthediscretederivative ≥ n+1 − n k 1 on both sides, we get − Df (j) = (n (k+j)α)Df (j) αf (j+1)+αjDf (j 1)+αf (j) n+1 n n n n − − − = (n (k+j+1)α)Df (j)+αjDf (j 1). (10) n n − − for j 1 and Df (0)=(n (k+1)α)Df (0) αa(n) . By induction, we get ≥ n+1 − n − k 1 − Dif (j)=(n (k+j+i)α)Dif (j)+αjDif (j 1). (11) n+1 n n − − for all i 1,j 1 and Dif (0)=(n (k+i)α)Dif (0)+( 1)iαa(n) . ≥ ≥ n+1 − n − k 1 Let k = K in (9). Note that the condition Df (j) 0 trivial−ly holds 1 n ≤ (j) for n K + j because a = 0 for i > j. In particular, Df (j) 0 for ≤ 1 i n ≤ all j 0,n = K . We apply induction on n. Assume Df (j) 0 holds for 1 n ≥ ≤ all j 0. The equality (10) implies that Df (j) 0 for all j 0 satisfying n+1 n (≥K +j+1)α,whichholdsifn+1 K +j+1si≤nce n α.≥Ontheother ≥ 1 ≥ 1 n+1 ≥ hand, if n+1 K +j we have Df (j) 0 trivially. So Df (j) 0 for 1 n+1 n+1 all j 0. Ther≤efore, Df (j) 0 andequival≤ently (7) holds for all n N≤,j 0. n ≥ ≤ ∈ ≥ The argument for (8) is similar. Let k =K in (9). Note that D2f (j) 0 2 n ≥ isequivalentto(8)forj 0,whichtriviallyholdsforn K +j sinceK K 2 2 1 ≥ ≤ ≥ and a(n) (j +1)a(n) 0. In particular, (8) holds for n = K ,j 0. K2+j − K2+j+1 ≥ 2 ≥ Assume that (8) holds for n = m,j 0. We consider the case n = m+1. ≥ If n = m+1 K +j, (8) holds trivially. Otherwise, m+1 K +j +1 2 2 ≤ ≥ and by applying (11) we see that D2f 0. By induction, (8) holds for all n+1 n N,j 0. ≥ ∈ ≥ 4 Remark. The argument of the previous lemma shows that ( 1)iDif (j) 0 n for all n N,j 0 if we choose k so that k α. − ≥ ∈ ≥ k+i ≤ For a fixed K K , let 1 ≥ n Fn(x)=x−K aj(n)xj = ∞ aK(n+)jxj. (12) j=1 j= X X−∞ and for a fixed K K , let 2 ≥ n Gn(x)=x−K (a(jn)−jaj(n+)1)xj = ∞ (aK(n+)j −jaK(n+)j+1)xj. (13) j=0 j= X X−∞ Lemma 3. Let f(x) = F (x), or G (x) for the given suitable K. We have n n (f(x)e x) 0 and f(x+rx) erxf(x) for all r,x>0. − ′ ≤ ≤ Proof. It is easy to see that f(x) f (x) 0 for x > 0 by Lemma 2. So ′ − ≥ (f(x)e x) =(f f)e x 0 and hence f(x+rx) erxf(x) for r >0. − ′ ′ − − ≤ ≤ We now come to the main result of this section. Theorem 1. Let 0<α, c<1, and s 0 be fixed. Then ≥ (i) e−cstα cK1e−stα is completely monotone in t. − (ii) K e stα +stαe stα is completely monotone in t. 1 − − (iii) 1 e cstα stαe stα is completely monotone in t. cK2(1 c) − − − − (iv) (max jK1, j )je cstα sjtjαe stα are completely monotone in t {cK1 cK2(1 c)} − ± − for any j N. − ∈ Proof. By dilation, we may assume s = 1. We prove (i) first. Let x = tα and F be as in 12, n dn n e tα =( 1)nt n a(n)xke x =( 1)nt n+Kαe xF (x) dtn − − − k − − − − n k=1 X and dn n dtne−ctα =(−1)nt−n ckak(n)xke−xe−rx =(−1)nt−n+KαcKe−cxFn(cx). (14) k=1 X Applying Lemma 2 and Lemma 3 to F gives us n dn e ctα dtn − cK, dn e tα ≥ dtn − for any K K . This implies (i) since e tα is completely monotone for any 1 − ≥ 0<α 1. ≤ 5 We now prove (ii). Let g(s,t)=e−stαs−K1. Then ∂sg(s,t), is the limit of − the family of functions 1 sK1+1(c 1)(e−stα −c−K1e−cstα) − as c 1, which are completely monotone in t by (i). So → K1e−stα +stαe−stα = sK1+1∂sg(s,t) − is completely monotone in t. For (iii), we denote by f(n)(t)=∂nf(t) and, for K K K , write t ≥ 2 ≥ 1 1 (tαe−tα)(n)+K(e−tα)(n) = [t(e−tα)′](n)+K(e−tα)(n) −α 1 = [t(e−tα)(n+1)+n(e−tα)(n)]+K(e−tα)(n) −α = (−1)αnt−n " ∞ (ak(n+1)−(n−Kα)ak(n))tkαe−tα# k=1 X = (−1)nt−n" ∞ (ak(n−)1−(k−K)ak(n))tkαe−tα# k=1 X = ( 1)nt n+Kα ∞ (a(n) ka(n) )tkαe tα − − " K+k−1− K+k − # k= X−∞ = ( 1)nt−n+Kαxe−xGn(x) (15) − withx=tαandG (x)definedasin13,whichdependsonK. Lemma3saysthat n G (x)e xdeceasesinxifK K andnotethatG (x)e x = (F (x)e x) 0. n − 2 n − n − ′ ≥ − ≥ We have 1 x xG (x)e x G (s)e sds n − n − ≤ (1 c) − Zcx 1 x = (Fn(s)e−s)′ds (1 c) − − Zcx 1 F (cx)e cx, n − ≤ (1 c) − for 00, which implies (iv). We will apply Theorem 1 to pointwise estimates of φ (λ). Let us first list s,α a few basic properties of φ . s,α Lemma 4. For any s>0,0<α,β <1, we have 1 s2 3 φs,12(λ) = 2√πse−4λλ−2. (16) ∞ φ (λ) = φ (λ)φ (s)ds. (17) 1,αβ s,α 1,β Z0 1 1 φs,α(λ) = s−αφ1,α(s−αλ), (18) αs∂ φ (λ) = φ (λ)+λ∂ φ (λ). (19) s s,α s,α λ s,α − Proof. (16) is well-known(see e.g. [32], page268). (17), (18) canbe easilyseen from (3) and (4). (18) implies (19). Corollary 1. For all λ,s>0,00,0<β <α<1, we have that ∞ 1 c ln s−αu φs,α(u)du< . (25) β Z0 (cid:12) (cid:16) (cid:17)(cid:12) ∞ ∞ u (cid:12) (cid:12) c ln (cid:12) φ (u)φ (cid:12) (v)dudv < . (26) v s,α s,α β2 Z0 Z0 (cid:12) (cid:16) (cid:17)(cid:12) (cid:12) (cid:12) Proof. Sinceφs,α(u)=s−(cid:12)α1φ1,α((cid:12)s−α1u),thelefthandsideof(25)isindependent of s. We only need to prove the case s = 1. For α = 1, we can verify directly 2 from (16) that (25) holds. Denote by u(α) the left hand side of (25). We then get u(1) < . Using (17), we get u( 1 ) < . Now, for α > 1 , we use (17) 2 ∞ 2n ∞ 2n again and get ∞ φ (λ) = φ (λ)φ (s)ds 1,21n Z0 s,α 1,α21n 1 φ (λ)φ (s)ds ≥ Z0 s,α 1,α21n 1 (by (20)) ≥ φ1,α(λ)Z0 sK1(α)φ1,α21n(s)ds c φ (λ). α 1,α ≥ We conclude that u(α)< for all 0<α<1. Since φ (λ) is continuous as a 1,α ∞ function in α and this continuity is uniform for λ [δ,N] for any 0<δ 0, let (cid:12) dj x = s tα = s(α) tα j, (30) j − dtj − j − where (α) denotes the falling factorial. Then j ∞ x uj = stα ∞ (α)j u j =stα stα 1+ u α =stα s(t+u)α j j! − j! t − t − Xj=1 Xj=1 (cid:16) (cid:17) (cid:16) (cid:17) Applying Theorem 1 part (i), we see that for all n N, c (0,1), and t >0 it ∈ ∈ holds that dn e sc(t+u)α dun − (cid:12)u=0 cK1, (cid:12) ≥ dn e s(t+u)α(cid:12) dun − (cid:12) (cid:12)u=0 (cid:12) where K1 is as in Lemma 2. We can rew(cid:12)rite this inequality as (cid:12) dn esctα sc(t+u)α dun − e(1−c)stα (cid:12)(cid:12)u=0 ≥cK1. dn estα s(t+u)α (cid:12) dun − (cid:12) (cid:12)u=0 (cid:12) We conclude that if we define x by (30), then(cid:12) j (cid:12) B (cx ,...,cx ) e(1−c)stα n 1 n cK1 (31) B (x ,...,x ) ≥ n 1 n foralln N, c (0,1),andt>0. Allofthesecalculationsareeasilyreversible, ∈ ∈ and we conclude that (31) is actually equivalent to part (i) of Theorem 1. 9 2 Positive semigroups and BMO Let (M,σ,µ) be a sigma-finite measure space. Let L1(M) be the space of all complex valued integrable functions and L (M) be the space of all complex ∞ valued measurable and essentially bounded functions on M. Denote by f the ∗ pointwise complex conjugate of a function f on M and by f,g the duality h i bracket fg . ∗ DefinitiRon 1. A map T from L (M) to L (M) is called positive if Tf 0 ∞ ∞ ≥ for f 0. If T is positive on L (M), then T id is positive on matrix valued ∞ functi≥on spaces L (M) M for all n N, i.e⊗. T is completely positive. ∞ n ⊗ ∈ A positive map T commutes with complex conjugation, i.e. T(f )=T(f) . ∗ ∗ For two positive maps S,T, we will write S T if S T is positive. ≥ − We will need the following Kadison-Schwarz inequality for completely positive maps T, T(f)2 T(1) T(f 2), f L (M). (32) L∞ ∞ | | ≤k k | | ∈ 2.1 Postive semigroups Wewillconsiderasemigroup(T ) ofpositive,weak*-continuouscontractions t t 0 ≥ on L with the weak* continuity at t = 0+. That is a family of positive, ∞ weak*-continuous contractions T ,t 0 on L such that T T =T , T =id t ∞ s t s+t 0 ≥ and T (f),g f,g as t 0+ for any f L ,g L1. t ∞ h i→h i → ∈ ∈ Such a semigroup (T ) always admits an infinitesimal negative generator y L = limy→0 id−yTy which has a weak*-dense domain D(L) ⊂ L∞. We will write T = e yL. These definitions and facts extend to the noncommutative y − setting. Namely, given a semifinite von Neumann algebra and a normal M semifinite faithful trace τ, we let L ( ) = and L1( ) be the completion ∞ M M M of f : f L1 =τ f < . Here g =(g∗g)21 and g∗ denotes the adjoint { ∈M k k | | ∞} | | operatorsofg andwe set f,g =τ(fg ). We saya mapT on is completely ∗ h i M positive if (T id)(f) 0 for any f 0,f M . We say f weak* n λ ⊗ ≥ ≥ ∈ M⊗ converges to f if lim f ,g = f,g for all g L1( ) (see [21] for details). λ λ The so-called subohrdinaitedhsemiigroups T ∈ =eMyLα,0<α<1 are defined y,α − as ∞ ∞ T f = T fφ (u)du= T fφ (u)du, (33) t,α u t,α 1 1,α tαu Z0 Z0 with φ given in Section 1. The generator Lα is given by t,α Lα(f)=Γ( α)−1 ∞(Tt id)(f)t−1−αdt, (34) − − Z0 for f D(L). There are other (equivalent) formulations for Lα. The formula ∈ (34) is due to Balakrishnan (see [5] and [32, page 260]). For T = e tzid with t − Re(z) 0, Lα =zα with a chosen principal value so that Re(zα) 0. ≥ ≥ 10

$H^\infty$-calculus for semigroup generators on BMO

Similar $H^\infty$-calculus for semigroup generators on BMO