REMARKS ON `1 AND `∞ −MAXIMAL REGULARITY FOR POWER-BOUNDED OPERATORS N.J. KALTON AND P. PORTAL Abstract. We discuss `p −maximal regularity of power-bounded operators and relate the discrete to the continuous time problem for analytic semigroups. We give a complete characterization of operators with `1 − and `∞ −maximal regularity. We also introduce an unconditional form of Ritt’s condition for power-bounded operators, which plays the role of the existence of an H ∞ −calculus, and give a complete characterization of this condition in the case of Banach spaces which are L1 −spaces, C(K)−spaces or Hilbert spaces. 2000 Mathematcis Subject Classification. 47A30, 47A60, 47D06, 39A12.

1. Introduction Let T be a power-bounded operator on a Banach space X. In [4] and [5], Blunck studied `p −maximal regularity for the discrete equation un = T un−1 + xn

n≥1

where u0 = 0. See §2 for precise definitions. Blunck studied the case 1 < p < ∞. The cases p = 1 and p = ∞ are studied in [17] where some discrete analogues of the results of Baillon [2] and Guerre-Delabri`ere [9] are given. However these analogues are not completely satisfying and, moreover, the proofs of Theorems 4.4 and 4.5 are rather confused. In this paper we improve these results and also give a complete description of operators T with `1 − or `∞ −regularity. We then point out that one can obtain the analogous and apparently new descriptions for closed operators A such that −A generates a bounded analytic semigroup and have either L1 − or L∞ −maximal regularity. We relate our results to classical result of Da Prato and Grisvard on L∞ −maximal regularity in real interpolation spaces. In the final section we introduce and study an unconditional form of Ritt’s condition for power-bounded operators. This is analogous to McIntosh’s definition of an H ∞ −calculus for sectorial operators [14]. We show that on an L1 −space the unconditional Ritt condition is equivalent to `1 −maximal regularity and dually on a C(K)−space it is equivalent to `∞ −maximal regularity. These results use Grothendieck’s theorem. Finally we give a discrete analogue The first author acknowledges support from NSF grant DMS-0244515. 1

2

N.J. KALTON AND P. PORTAL

of a result of Auscher, McIntosh and Nahmod [1] characterizing the unconditional Ritt condition on Hilbert spaces. 2. Preliminaries Suppose −A is the generator of a bounded analytic semigroup on a (complex) Banach space X. We shall say that A has Lp −maximal regularity if the solution of the abstract Cauchy problem u0 + Au = f (t)

0 ≤ t < ∞,

u(0) = 0 given by Z

t

u(t) =

e−(t−s)A f (s)ds

0 0

has the property that u ∈ Lp (R+ , X) whenever f ∈ Lp (R+ , X). This is equivalent to the requirement that there is a constant C such that

p 1/p Z ∞ Z t

Ae−(t−s)A f (s) ds dt ≤ Ckf kp .

0

0

(Note we do not require that u ∈ Lp (R+ , X) which is sometimes additionally required.) Similarly suppose T is a bounded operator. We say that T satisfies Ritt’s condition (or generates a discrete analytic semigroup) [18] if there is a constant C so that k(1 − λ)R(λ, T )k ≤ C

(2.1)

|λ| ≥ 1.

The following characterization of operators satisfying Ritt’s condition is due to Nagy and Zemanek [15] and Lyubich [13]. Theorem 2.1. T satisfies Ritt’s condition (2.1) if and only if T is powerbounded and sup kn(T n−1 − T n )k < ∞.

(2.2)

n≥1

Note for future reference that (2.2) implies sup kn2 (I − T )2 T n−1 k < ∞.

(2.3)

n≥1

More generally if C = sup kn(T n−1 − T n )k then (2.4)

knr (I − T )r T n−1 k ≤ Crr

n = 1, 2, . . . , r = 1, 2 . . . .

REMARKS ON `1 AND `∞ −MAXIMAL REGULARITY FOR POWER-BOUNDED OPERATORS 3

We say that T has `p -maximal regularity if the solution of the difference equation un = T un−1 + xn n = 1, 2 . . . , u0 = 0 ∞ has the property that (un − un−1 )∞ n=1 ∈ `p (X) whenever (xn )n=1 ∈ `p (X). This is equivalent to the requirement that there exists a constant C such that

p !1/p !1/p ∞ X n ∞

X X

(2.5) ≤C T n−k (T − I)xk kxk kp .

n=1 k=1

k=1

This definition was suggested and investigated by Blunck [4] and [5]. It was shown by Blunck [4] that a necessary condition for T to have `p -maximal regularity for some p is that T satisfies Ritt’s condition (2.1). There is a simple connection between these problems: Proposition 2.2. In order that A has Lp −maximal regularity it is necessary and sufficient that the operators Th = e−hA has `p −maximal regularity uniformly (i.e. with uniform constants) for 0 < h < ∞. Proof. Suppose 0 < h < ∞ and suppose (xn )∞ n=1 ∈ `p (X). Let Z t F (t) = Ae−(t−s)A f (s)ds 0

where f=

∞ X

xk χ(k−1)h,kh) .

k=1

Similarly let vn =

n X

Thn−k (Th − I)xk

n = 1, 2, . . . .

k=1

Then F (nh) = −vn and more generally F ((n − 1)h + τ ) = −e−τ A vn−1 + (I − e−τ A )xn

0 < τ < h.

It follows that kF ((n − 1)h + τ )k ≤ M kvn−1 k + (M + 1)kxn k

0 0, so that (3.11)

∞ X

k(e−khA − e−(k−1)hA )xk ≤ Ckxk,

h > 0, x ∈ X.

k=0

Hence letting h → 0 we have (3.10). (ii) ⇒ (i). Equation (3.10) trivially implies (3.11).



Theorem 3.7. Let −A be the generator of a bounded analytic semigroup. The following conditions on A are equivalent: (i) A has L∞ −maximal regularity. (ii) There is a constant C so that (3.12)

kxk ≤ C sup ktAe−tA xk + lim sup ke−tA xk t>0

x ∈ X.

t→∞

Remark. If A is has dense range then we have limt→∞ e−tA x = 0 for every x ∈ X and we can drop the last term. Proof. (i) ⇒ (ii) is very similar to the preceding theorem case (i). (ii) ⇒ (i) Observe that if f ∈ L∞ (R+ , X) with kf k∞ ≤ 1, then

Z t

Z t

2 −(t+τ −s)A −(t−s)A

τ A e )f (s)ds k Ae f (s)dsk ≤ C sup

τ >0 0 0 Z t τ ≤ C1 sup ds 2 τ >0 0 (t + τ − s) ≤ C2 , where C1 , C2 are suitable constants depending on A.



At this point let us remark that it is now easy to recover the results of Da Prato and Grisvard [7] on L∞ −maximal regularity in real interpolation spaces. Da Prato and Grisvard consider maximal regularity on a finite interval which is equivalent to maximal regularity on the infinite interval for s + A for some s > 0. Thus it is enough to consider the case of an invertible operator. Let us consider the real interpolation space (X, Dom(A))(θ,∞) for 0 < θ < 1 which is defined by the norm kxk(θ,∞) = sup t−θ K(t, x) t>0

10

N.J. KALTON AND P. PORTAL

where K(t, x) = K(t, x; X, Dom(A)) = inf{kyk + tkAzk : y + z = x}. The space (X, Dom(A))(θ,∞) can be given several equivalent norms, see [7] and [12]; we will need one of these which we now describe for completeness. If if x = y + z we have ktAe−tA xk ≤ ktAe−tA kkyk + ke−tA kktAzk so that ktAe−tA xk ≤ CK(t, x). On the other hand K(t, x) ≤ kx − e−tA xk + ktAe−tA xk Z t kAe−sA xkds + ktAe−tA xk. ≤ 0

If t ≥ 1 then K(t, x) ≤ kxk while if 0 < t < 1 we have K(t, x) ≤ 2θ−1 tθ sup s1−θ kAe−sA xk. 00

Hence for a suitable constant sup t2−θ kA2 e−tA xk ≤ C sup kAe−tA xk0 . t>0

t>0

REMARKS ON `1 AND `∞ −MAXIMAL REGULARITY FOR POWER-BOUNDED OPERATORS 11

Now −tA

kAe

Z

∞

kA2 e−sA xkds ≤ (1 − θ)−1 tθ−1 sup t2−θ kA2 e−tA xk

xk ≤

t>0

t

so that kxk0 ≤ C(1 − θ)−1 sup kAe−tA xk0 . t>0

 4. The unconditional Ritt condition In this section we study the discrete analogue of the H ∞ −calculus for sectorial operators which was introduced by McIntosh [14]. Before proceeding we develop some basic ideas which will be useful later. Let assume that T satisfies the Ritt condition. For any m ≥ 0 we consider the operator Vm defined by ∞ X ck (T km − T (k+1)m ) (4.1) Vm = k=0

where ck =

(2k)! 22k (k!)2

.

√ Note that there is a constant M so that |ck | ≤ M/ k for k ≥ 1 so that if follows from (2.2) that there series in (4.1) converges absolutely. Of course V0 = 0. Lemma 4.1. For m ≥ 1 we have Vm2 = I − T m . Proof. Consider the function ∞ X Fm (t) = ck (tkm T km − t(k+1)m T (k+1)m )

0 ≤ t ≤ 1.

k=0

Since tkm T km −t(k+1)m T (k+1)m = tkm (T km −T (k+1)m )+(tkm −t(k+1)m )T (k+1)m )

0≤t≤1

it follows that the series on the right converges uniformly to Fm (t) for 0 ≤ t ≤ 1. If 0 < t < 1 we have ∞ X m m Fm (t) = (I − t T ) ck tkm T km k=0

and as −1/2

(1 − z)

=

∞ X k=0

ck z k

|z| < 1

12

N.J. KALTON AND P. PORTAL

we deduce that Fm (t)2 = I − tm T m

0 < t < 1.

Letting t → 1, and using uniform convergence, we deduce that Vm2 = I − T m .  Lemma 4.2. Suppose T satisfies the Ritt condition and define ρk (x) =

max

k(T u − T v )xk

x∈X

k = 0, 1, 2, . . .

kT n − T n+1 xk

x ∈ X,

k = 0, 1, 2, . . . .

2k −1≤u≤2k+1 −1 2k −1≤v≤2k+2 −1

and σk (x) =

max

2k −1≤n 1

k(I − T )R(λ, T )k ≤ C|λ|−1

|λ| > 1.

i.e. Hence k(1 − λ)R(λ, T )k ≤ 1 + C|λ|−1

|λ| > 1. 

∞ Now let (rk )∞ k=0 and (mk )k=0 be any pair of sequences of integers such that

(4.6)

0 ≤ mk < 2k+3 − 1,

2k − 1 ≤ rk < 2k+1 − 1

k = 0, 1, . . . .

Lemma 4.4. Suppose T satisfies the unconditional Ritt condition. Then there ∞ is a constant C such that for any pair of sequences (rk )∞ k=0 and (mk )k=0 satisfying (4.6) we have k

(4.7)

N X

ak T rk Vmk k ≤ C max |ak |. 0≤k≤N

k=0

Proof. Suppose max0≤k≤N |ak | ≤ 1 and x ∈ X and x∗ ∈ X ∗ with kxk = kx∗ k = 1. We have N N X X rk ∗ |x∗ (T rk Vmk x)| |x ( ak T Vmk x)| ≤ k=0

k=0

≤

N X ∞ X

cl |x∗ ((T rk +mk l x − T rk +mk (l+1) )x)|

k=0 l=0

=

∞ X

dj |x∗ ((T j − T j+1 )x)|,

l=1

where dj =

X rk ≤j

c[j−rk /mk ] .

14

N.J. KALTON AND P. PORTAL

Assume 2s − 1 ≤ j < 2s+1 − 1. Then rk ≤ j implies that k ≤ s. If k ≤ s − 2 then [j − rk /mk ] > 2s−1 /mk ≥ 2s−4−k . Thus we have dj ≤ 2 + M

s−2 X

2(k−s+4)/2 .

k=0

Thus we get an estimate dj ≤ C1 M where C1 is an absolute constant. Hence N X |x∗ ( ak T rk Vmk x)| ≤ C0 C1 M k=0

where C0 is the constant in (4.4) and thus k

N X

ak T rk Vmk k ≤ C0 C1 M.

k=0

 The following result is the discrete analogue of a similar result for sectorial operators with an H ∞ −calculus proved in [11]. We recall that a Banach space X is called a GT-space (for Grothendieck theorem) if every bounded operator T : X → `2 is absolutely summing. See [16]. Theorem 4.5. Let X be a GT-space (e.g. X = L1 , `1 or X = L1 /H1 ). Let T : X → X be any operator. Then T has `1 −maximal regularity if and only if T satisfies the unconditional Ritt condition. Proof. Let assume that T satisfies the unconditional Ritt condition (4.4). Let C1 be the constant in Lemma 4.4 (4.7). ∞ Suppose (uk )∞ k=0 and (vk )k=0 are two sequences of natural numbers such that 2k − 1 ≤ uk ≤ 2k+1 − 1 and 2k − 1 ≤ vk ≤ 2k+2 − 1. For k ≥ 0 we write uk+1 = rk +sk where 0 ≤ rk −sk ≤ 1 and mk = vk −uk . Thus 2k +1 ≤ rk , sk ≤ 2k+1 −1 ∞ and 0 ≤ mk < 2k+3 i.e. (rk )∞ k=0 and (mk )k=0 satisfy (4.6). At this point we introduce the hypothesis that X is a GT-space which means that there is a constant K so that for any operator T : X → `2 we have π1 (T ) ≤ KkT k where π1 (T ) is the usual absolutely summing norm. For any P 2 x∗0 , x∗1 , . . . , x∗N ∈ X ∗ with kx∗k k ≤ 1 and any a0 , . . . , aN ∈ C with N k=0 |ak | ≤ 1 +1 consider the operator S : X → `N defined by 2 Sx = (ak x∗k (x))N k=0 . Then kSk ≤ 1 and so π1 (S) ≤ K. Hence for any x ∈ X we have N X k=0

kST rk Vmk xk ≤ KC1 kxk.

REMARKS ON `1 AND `∞ −MAXIMAL REGULARITY FOR POWER-BOUNDED OPERATORS 15

In particular N X

|ak ||x∗k (T rk Vmk x)| ≤ KC1 kxk.

k=0 ∗ N Since this is true for all such choices of (ak )N k=0 and (xk )k=0 we have N X ( kT rk Vmk xk2 )1/2 ≤ KC1 kxk

x ∈ X.

k=0

Now for any fixed x∗0 , . . . , x∗N ∈ X ∗ with kx∗k k ≤ 1 consider the operator +1 R : X → `N defined by 2 Rx = (x∗k (T sk Vmk x))N k=0 and observe that π1 (R) ≤ KkRk ≤ K 2 C1 . It thus follows that N X

kRT rk Vmk xk ≤ K 2 C12 kxk

x∈X

k=0

and as before N X

|x∗k (T rk +sk Vm2 k x)| ≤ K 2 C12 kxk

x ∈ X.

kT uk (I − T mk )xk ≤ K 2 C12 kxk

x ∈ X.

k=0

Again this implies N X k=0 ∞ k k+1 We conclude that if (uk )∞ and uk ≤ vk ≥ k=0 and (vk )k=0 satisfy 2 ≤ uk < 2 k+2 2 then ∞ X k(T uk − T vk )xk ≤ C2 kxk x∈X k=1 2

C12 .

where C2 = K Thus we have an estimate ∞ X ρk (x) ≤ C2 kxk

x∈X

k=1

which implies by Lemma 4.2 an estimate ∞ X k=0

2k σk (x) ≤ C3 kxk

x∈X

16

N.J. KALTON AND P. PORTAL

and hence that

∞ X

kT k−1 x − T k xk ≤ C3 kxk.

k=1

The result now follows by Theorem 3.1. The converse direction is trivial.



The dual result is now easy: Theorem 4.6. Let X be a Banach space such that X ∗ is a GT-space (e.g. Xis a C(K) space or the disc algebra). If T is a power-bounded operator on X then T has `∞ −maximal regularity if and only if T satisfies the unconditional Ritt condition. Finally we establish a corresponding result for Hilbert spaces. The continuous analogue which we discuss later is due to McIntosh [14]. See also further discussion in [6] and [1] Theorem 4.7. Let T be a power-bounded operator on a Hilbert space H. Then T satisfies the unconditional Ritt condition if and only if there is a constant C such that (4.8) ! 21 ∞ X + lim sup kT n xk ≤ Ckxk x ∈ H. C −1 kxk ≤ kkT k−1 x − T k xk2 n→∞

k=1

Proof. Suppose T satisfies the unconditional Ritt with constant C0 . Pcondition k−1 (T x − T k x) is weakly We first observe that for every x ∈ H the series ∞ k=1 unconditionally Cauchy and hence unconditionally convergent to some P x where P is a projection whose kernel is the eigenspace {x : T x = x}. We may therefore easily reduce to the case when limn→∞ T n x = 0 for every x ∈ H. ∞ k k+1 Then for any pair of sequences (uk )∞ −1 k=0 , (vk )k=0 with 2 − 1 ≤ uk ≤ 2 k k+2 and 2 − 1 ≤ vk ≤ 2 − 1 we have an estimate N X k k (T uk − T vk )k ≤ 2C0 k = ±1, k = 1, 2, . . . , N k=0

and it follows from the generalized Parallelogram Law that for any x ∈ X we have !1/2 ∞ X ≤ 2C0 kxk x ∈ H. (4.9) k(T uk − T vk )xk2 k=0

Thus (4.10)

∞ X k=0

!1/2 ρk (x)2

≤ 2C0 kxk

x ∈ H.

REMARKS ON `1 AND `∞ −MAXIMAL REGULARITY FOR POWER-BOUNDED OPERATORS 17

Now by Lemma 4.2 we can deduce an estimate !1/2 ∞ X kkT k−1 x − T k xk2 ≤ C1 kxk

x∈H

k=0

for a suitable constant C1 . Thus the right-hand side of (4.8) follows. Notice we also have the same inequality for the adjoint T ∗ i.e. ∞ X

!1/2 kk(T ∗ )k−1 x − (T ∗ )k x∗ k2

≤ C1 kx∗ k

x∗ ∈ H

k=0

We now turn to the left-hand estimate. If x ∈ H, pick x∗ ∈ H ∗ with kx∗ k = 1 and x∗ (x) = kxk. Then, since we assume limn→∞ kT n xk = 0, kxk = x∗ (x) ∞ X ≤ |x∗ (T k−1 (I − T )x| k=1

≤ 4C1

∞ X

!1/2 kkT k−1 x − T k xk2

k=0

by an application of (3.7) combined with (4.9) and (4.10). We now turn to the converse. Assuming (4.8) let us first show that T satisfies the Ritt condition. Note that we have limn→∞ kT n−1 (I − T )xk = 0 for every x. Therefore knT n−1 (I − T )xk2 ≤ C 2

∞ X

n2 kkT n+k−2 (I − T )2 xk2

k=1

≤ C2

∞ X

k 3 kT k−1 (I − T )2 xk2

k=1

≤ 6C

2

= 6C 2 ≤ 6C

∞ X k X k=1 j=1 ∞ X ∞ X

j(k + 1 − j)kT k−1 (I − T )2 xk2 jkkT j+k−2 (I − T )2 xk2

j=1 k=1 ∞ X 4 j−1

kT

j=1

≤ 6C 6 kxk2 ,

(I − T )xk2

18

N.J. KALTON AND P. PORTAL

so that kT n−1 (I − T )k ≤

√

6C 3 /n.

Thus T satisfies the Ritt condition (2.1). suppose that (xk )∞ k=1 is any finitely nonzero sequence in H. Let y = PNow ∞ 1/2 k−1 k k (T − T )x . Note that limn→∞ T n y = 0. Then k k=1 kj 1/2 (T j−1 − T j )yk ≤

∞ X

(jk)1/2 kT j+k−2 (I − T )2 xk k

k=1

≤ C0 ≤ C0

∞ X (jk)1/2 kxk k 2 (j + k) k=1 ∞ X k=1

1 kxk k. j+k

The matrix ajk = (1/(j + k))j,k defines a bounded operator on `2 by Hilbert’s inequality. Thus for a suitable constant C1 we have: ! 21 ! 12 ∞ ∞ X X jk(T j−1 − T j )yk2 ≤ C1 kxk k2 j=1

k=1

P 2 1/2 . for a suitable constant C1 . We conclude from (4.8) that kyk ≤ C1 ( ∞ k=1 kxk k ) ∗ ∗ Now suppose x ∈ H . For N ∈ N pick x1 , . . . , xN with kxk k = 1 and x∗ (T k−1 (I − T )x) = k(T ∗ )k−1 (I − T ∗ )x∗ k. Then for any sequence of scalars a1 , . . . , aN we have N X

ak k

1/2

∗ k−1

k(T )

∗

∗

∗

(I−T )x k ≤ kx kk

k=1

N X

ak k

1/2

T

k−1

N X (I−T )xk k ≤ C1 kx k( |ak |2 )1/2 .

k=1

∗

k=1

Hence ∞ X

! 12 kk(T ∗ )k−1 (I − T ∗ )x∗ k2

≤ C1 kx∗ k.

k=1

At this point we can appeal to (3.7) of Lemma 3.2. We have ∞ X

|x∗ (T k−1 (I − T )x)| ≤ C2 kxkkx∗ k

x ∈ H, x∗ ∈ H ∗ .

k=1

This implies the unconditional Ritt condition.



Now suppose again that −A is the generator of a bounded analytic semigroup on a Banach space X, and suppose also for convenience that A has

REMARKS ON `1 AND `∞ −MAXIMAL REGULARITY FOR POWER-BOUNDED OPERATORS 19

dense domain and range (i.e. A is sectorial). Then limt→∞ ke−tA xk = 0 for x ∈ X. The continuous version of the unconditional Ritt condition is Z ∞ |x∗ (Ae−tA x)|dt ≤ Ckxkkx∗ k, x ∈ X, x∗ ∈ X ∗ . (4.11) 0

Let us call this the continuous unconditional Ritt condition. If (4.11) holds then e−tA uniformly satisfies the unconditional Ritt condition. We recall that A has an H ∞ (Σψ )−calculus where Σψ = {z : | arg z| < ψ} if f (A) is a bounded operator for every f ∈ H ∞ (Σψ ); see [6] for more details. Let ω(A) be the infimum of all φ so that we have the resolvent estimates kλR(λ, A)k ≤ C

| arg z| ≥ φ

and let ωH (A) be the infimum of all φ so A has an H ∞ (Σφ )−calculus. It easy to show that if ωH (A) < π/2 then A satisfies the continuous unconditional Ritt condition. Conversely, it follows from Theorem 4.5 of [6] that if A satisfies the continuous unconditional Ritt condition, then A has an H ∞ (Σψ )-calculus as long as ψ > π/2; thus ωH (A) ≤ π/2. If X is a Hilbert space, then results of McIntosh [14] imply that ωH (A) = ω(A) < π/2. One cannot apply this argument for an arbitrary Banach space [10]. Thus it is open whether the continuous Ritt condition is equivalent to ωH (A) < π/2. It is easy as in Theorem 3.6 to prove continuous versions of Theorems 4.5, 4.6 and 4.7: Theorem 4.8. Let A be the generator of a bounded analytic semigroup with dense domain and range. Then: (i) If X is a GT-space then A satisfies the continuous unconditional Ritt condition if and only if there is a constant C so that Z ∞ −1 kAe−tA xkdt ≤ Ckxk x ∈ X. C kxk ≤ 0

(ii) If X ∗ is a GT-space then A satisfies the continuous unconditional Ritt condition if and only if there is a constant C so that C −1 kxk ≤ sup t||Ae−tA xk ≤ Ckxk

x ∈ X.

t>0

(iii) If X is a Hilbert space then A satisfies the continuous unconditional Ritt condition if and only if there is a constant C so that Z ∞  21 −1 −tA 2 ≤ Ckxk x ∈ X. C kxk ≤ tkAe xk dt 0

20

N.J. KALTON AND P. PORTAL

In view of our remarks above (iii) is simply a special case of the result of McIntosh [14] on the equivalence of the H ∞ −calculus with certain quadratic estimates. Similarly (i) is a close relative with Proposition 7.2 of [11].

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[2]

[3] [4] [5] [6] [7]

[8] [9] [10] [11] [12]

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REMARKS ON `1 AND `∞ −MAXIMAL REGULARITY FOR POWER-BOUNDED OPERATORS 21

Department of mathematics, University of Missouri-Columbia, Columbia, MO 65211, USA E-mail address: [email protected] Centre for Mathematics and its Applications, Australian National University, Canberra, ACT 0200 Australia E-mail address: [email protected]