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

0.32 MB·

by  Tim Ferguson

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 stud-
ied 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 0<c<1. Combing this inequality with (14) and (15) we get
(−1)nddtnn(tαe−tα +K2e−tα) 1 .
( 1)n dn e ctα ≤ cK2(1 c)
− dtn − −
This proves (iii) since e ctα and e tα are completely monotone.
− −
For (iv), let f(t) = max{cKK11,cK2(11 c)}e−cstα,g(t) = stαe−stα. By (i), (ii)
−
and (iii) we have that both f +g,f g are completely monotone in t. Recall
−
6
that complete monotonicity is preserved by multiplication. Note that
1
fj+1+gj+1 = [(fj gj)(f g)+(fj +gj)(f +g)]
2 − −
1
fj+1 gj+1 = [(fj gj)(f +g)+(fj +gj)(f g)].
− 2 − −
We get, by induction, that(max{cKK11,cK2(11 c)α})je−jcstα−sjtjαe−jstα is com-
−
pletely monotone for any s>0, 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>0,0<c<1,j N, we have
∈
cK1φs,α(λ) φcs,α(λ) (20)
≤
0 ∂λ(λ1+αK1φs,α(λ)), (21)
≤
j
jK j
sj∂jφ (λ) max 1, φ , (22)
| s s,α | ≤ cK1 cK2(1 c)α cs,α
(cid:18) (cid:26) − (cid:27)(cid:19)
10
s∂ φ (λ) φ (λ), (23)
s s,α αs,α
| | ≤ 1 α
(cid:18) − (cid:19)
j
10j
sj∂jφ (λ) φ (λ). (24)
| s s,α | ≤ 1 α αs,α
(cid:18) − (cid:19)
Proof. These are direct consequences of Theorem 1, the identity (3), and the
Hausdorff-Bernstein-Widder Theorem because K i , except that (21) re-
i ≤ 1 α
quiresa little morecalculation. To prove(21),note tha−t(5)andTheorem1(ii)
imply that
φ (λ)
∂s ss,αK1 =−s−K1−1(K1φs,α(λ)−s∂sφs,α(λ))≤0.
Since φs,α(λ)=s−α1φ1,α(s−α1λ), we get
1 1
+K1 s−α1−K1−1φ1,α(s−α1λ) s−α1−1λs−α1−K1(∂λφ1,α)(s−α1λ) 0.
− α − α ≤
(cid:18) (cid:19)
7
That is
1 1 1
(1+K1α)φ1,α(s−αλ)+λs−α(∂λφ1,α)(s−αλ) 0.
≥
Therefore
(1+K α)φ (λ)+λ∂ φ (λ) 0,
1 s,α λ s,α
≥
since ∂λφs,α(λ)=s−α2∂λφs,α(s−α1λ). This is (21).
Lemma 5. For any s>0,0<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 <N <
∈
, one can easily see that u(α) is continuous in α for α (0,1). We conclude
∞that u(α) is bounded on [ 1 ,1] for any n N. Note that∈(17) also implies that
2n 2 ∈
∞
φ (λ) lnλdλ
1,αβ
| |
Z0
∞ ∞
= φ (λ) lnλdλφ (s)ds
s,α 1,β
| |
Z0 Z0
= ∞ ∞φ1,α(v) ln(sα1v)dvφ1,β(s)ds
| |
Z0 Z0
∞ ∞ 1
φ (v)( lns lnv )dvφ (s)ds (27)
1,α 1,β
≥ ± α| |−| |
Z0 Z0
∞ ∞ 1
φ (v)( lns + lnv )dvφ (s)ds (28)
1,α 1,β
≤ α| | | |
(cid:18) Z0 Z0 (cid:19)
8
Our change in the order of integration is justified because all the terms are
positive. Note 0∞φt,α(s)ds=1 for any t,α. (27) and (28) imply that
R 1 1
u(α) u(β) u(αβ) u(α)+ u(β) (29)
| − α |≤ ≤ α
We then obtain (25). (26) follows from (25).
Remark(BellPolynomials). WedefinethecompleteBellpolynomialB (x ,...,x )
n 1 n
by its generating function
∞ uj ∞ un
exp x = B (x ,...,x )
j n 1 n
 j! n!
j=1 n=0
X X
 
From this, we get the formula
dn ∞ uj
B (x ,...,x )= exp x
n 1 n dun  j j!
j=1 (cid:12)u=0
X (cid:12)
 (cid:12)
Now, for s>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 posi-
tive 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