Semigroup Forum  Springer-Verlag New York Inc.  Version of July 20, 1997

On Perturbations of Differentiable Semigroups

Bogdan D. Doytchinov, William J. Hrusa and Stephen J. Watson

Communicated by Jerome A. Goldstein

Abstract

Let X be a Banach space and let A be the infinitesimal generator of a differentiable semigroup {T (t), t ≥ 0}, i.e. a C_{0} - semigroup such that t ↦ T (t)x is differentiable on (0, ∞) for every x ∈ X. Let B be a bounded linear operator on X and let {S(t), t ≥ 0} be the semigroup generated by A + B. Renardy recently gave an example which shows that {S(t), t ≥ 0} need not be differentiable. In this paper we give a condition on the growth of ∥T′(t)∥ as t ↓ 0 which is sufficient to ensure that {S(t), t ≥ 0} is differentiable. Moreover, we use Renardy’s example to study the optimality of our growth condition. Our results can be summarized roughly as follows:

(i) If lim sup_{t→0^{+}} \frac{t log∥T′(t)∥}{log( \frac{1}{t})} = 0 then \{S(t), t ≥ 0\} is differentiable.

(ii) If 0 < L = lim sup_{t→0^{+}} \frac{t log∥T′(t)∥}{log(\frac{1}{t})} < ∞ then t ↦ S(t) is differentiable

on (L, ∞) in the uniform operator topology, but need not be differentiable near zero.

(iii) For each function α : (0, 1) → (0, ∞) with α(t)/ log(\frac{1}{t}) → ∞ as t ↓ 0, Renardy’s example can be adjusted so that lim sup_{t→0^{+}} \frac{t log∥T′(t)∥}{α(t)} = 0 and t ↦ S(t) is nowhere differentiable on (0, ∞).

We also show that if lim sup_{t→0^{+}} t^{p} ∥T′(t)∥ < ∞ for a given p ∈ [1, ∞), then lim sup_{t→0^{+}} t^{p} ∥S′(t)∥ < ∞; it was known previously that if lim sup_{t→0^{+}} t^{p} ∥T′(t)∥ < ∞ then {S(t), t ≥ 0} is differentiable and lim sup_{t→0^{+}} t^{2p−1} ∥S′(t)∥ < ∞.

1. Introduction and Statement of Results

Let (X, ∥·∥) be a complex Banach space and let {T (t), t ≥ 0} be a linear C_{0} - semigroup on X with infinitesimal generator A. We say that {T (t), t ≥ 0} is differentiable if the mapping t ↦ T (t)x is differentiable on (0, ∞) for every x ∈ X; we say that {T (t), t ≥ 0} is eventually differentiable if there exists t_{0} ∈ [0, ∞) such that t ↦ T (t)x is differentiable on (t_{0}, ∞) for every x ∈ X. It is well known that {T (t), t ≥ 0} is differentiable if and only if T (t) maps X into D(A), the domain of A, for every t > 0. Moreover, if {T (t), t ≥ 0} is differentiable then t ↦ T (t) is of class C^{∞} on (0, ∞) in the uniform operator topology, T (t) maps X into ∩_{n=1}^{∞}D(A^{n}) for every t > 0, and

T^{(n)}(t) = A^{n}T (t) = (AT (\frac{t}{n}))^{n} = (T′(\frac{t}{n}))^{n}

2 DOYTCHINOV, HRUSA and WATSON

for every n = 1, 2, 3, ... and every t > 0. We note also that if {T (t), t ≥ 0} is differentiable and A is unbounded then ∥T′(t)∥ = ∥AT (t)∥ → ∞ as t ↓ 0; there is no upper limit to the rate of growth of ∥AT (t)∥ as t ↓ 0. We refer to [4] and Section 2.4 of [6] for more information on the basic properties of differentiable and eventually differentiable semigroups.

Let B be a bounded linear operator on X. Then A + B generates a C_{0} - semigroup {S(t), t ≥ 0}. It is of interest to know which properties of {T (t), t ≥ 0} are inherited by {S(t), t ≥ 0}. In 1953, Phillips [7] gave an example of a C_{0} - semigroup {T (t), t ≥ 0} and a special choice of B such that T (t) = 0 for all t ≥ 1 and yet there is no t_{0} ∈ [0, ∞) at which the mapping t ↦ S(t) is continuous in the uniform operator topology. Such a semigroup {S(t), t ≥ 0} cannot be eventually differentiable; moreover, there can be no t_{0} ∈ [0, ∞) such that S(t_{0}) is compact.

Roughly speaking, Phillips’ example shows that one should not expect eventual regularity of {T (t), t ≥ 0} to be inherited by {S(t), t ≥ 0}. On the other hand it is known that if {T (t), t ≥ 0} is an analytic semigroup, or if T (t) is compact for every t > 0, or if t ↦ T (t) is continuous on (0, ∞) in the uniform topology, then {S(t), t ≥ 0} must also have the corresponding property ([2], [5], [7]). In 1968, Pazy [5] showed that a certain decay condition on the resolvent of A (that was already known to imply differentiability of {T (t), t ≥ 0}) is sufficient to ensure differentiability of {S(t), t ≥ 0} for every bounded linear B. Pazy also raised the question of whether or not differentiability of {T (t), t ≥ 0} alone is enough to imply differentiability of {S(t), t ≥ 0}. Renardy [8] recently gave a negative answer to this question by constructing a counterexample on l^{2}.

In this paper we show that if {T (t), t ≥ 0} is differentiable and ∥AT (t)∥ obeys a certain growth condition as t ↓ 0 then {S(t), t ≥ 0} is differentiable for every bounded linear B. We show further that no weaker growth condition is sufficient to ensure the same conclusion, and we discuss in detail the special case when ∥AT (t)∥ grows exactly at a certain critical level. In addition, we make an observation concerning a special class of differentiable semigroups discussed by Crandall & Pazy [1]. This class is described in the following definition.

Definition 1. Let {T (t), t ≥ 0} be a C_{0} - semigroup with infinitesimal generator A and let p ∈ [1, ∞) be given. We say that {T (t), t ≥ 0} is of class D^{p} if it is differentiable and

lim sup_{t→0^{+}} t^{p} ∥AT (t)∥ < ∞.

We note that D^{1} is precisely the class of analytic semigroups.

The remark below summarizes several earlier results that are relevant to the present discussion. We write R(λ; A) = (λI − A)^{−1} for all λ ∈ ρ(A), the resolvent set of A.

Remark 1. Let {T (t), t ≥ 0} be a C_{0} - semigroup with infinitesimal generator A and choose M ≥ 1, ω ∈ R such that ∥T (t)∥ ≤ Me^{ωt} for all t ≥ 0. Let B be a bounded linear operator on X and let {S(t), t ≥ 0} be the C_{0} - semigroup generated by A + B.

a) Pazy [5] showed that if

lim_{|τ|→∞} log |τ| · ∥R(μ + iτ; A)∥ = 0 for some real μ ≥ ω  (1)

then {S(t), t ≥ 0} is differentiable. [ It was shown earlier by Hille [3] that (1) implies differentiability of {T (t), t ≥ 0}.]

b) Yosida [9] has shown that if {T (t), t ≥ 0} is differentiable and

lim sup_{t→0^{+}} t log ∥AT (t)∥ = 0  (2)

then (1) holds. It is incorrectly asserted in [9] that (1) implies (2); a correction is made in [10]. In fact, (1) implies no limitation on the growth of ∥AT (t)∥ as t ↓ 0; this can be seen by considering diagonal operators on l^{2}.

c) Crandall & Pazy [1] showed that if {T (t), t ≥ 0} is of class D^{p} for a given p ∈ [1, ∞) then {S(t), t ≥ 0} is of class D^{2p−1}.

Our first theorem gives a sufficient condition for differentiability of the perturbed semigroup and improves the result quoted in Remark 1(c).

Theorem 1. Let {T (t), t ≥ 0} be a differentiable semigroup with infinitesimal generator A. Let B be a bounded linear operator on X and let {S(t), t ≥ 0} be the C_{0} - semigroup generated by A + B. If

0 ≤ L = lim sup_{t→0^{+}} \frac{t log ∥AT (t)∥}{log(\frac{1}{t})} < ∞  (3)

then the mapping t ↦ S(t) is of class C^{1} on (L, ∞) in the uniform operator topology. In pariticular, if L = 0 then {S(t), t ≥ 0} is differentiable. Further, if {T (t), t ≥ 0} is of class D^{p} for a given p ∈ [1, ∞) then {S(t), t ≥ 0} is also of class D^{p}.

Remark 2. We note that the lim sup in (3) is automatically nonnegative except in the trivial case when A is the zero operator.

Remark 3. There are differentiable semigroups {T (t), t ≥ 0} which satisfy

lim sup_{t→0^{+}} \frac{t log ∥AT (t)∥}{log(\frac{1}{t})} = 0

but do not satisfy (1). Such a semigroup can be produced by putting p(n) = e^{\sqrt{n}} and taking m(n) to be the integer part of \sqrt{n} in Renardy’s example.

Our next result says that for any rate of growth of ∥AT (t)∥ that violates (3) there is an example in which the perturbed semigroup fails to be eventually differentiable.

Theorem 2. Let α : (0, 1) → (0, ∞) be a given function satisfying

\frac{α(t)}{log(\frac{1}{t})} → ∞ as t → 0^{+}.

Then there exists a differentiable semigroup {T (t), t ≥ 0} on l^{2} with infinitesimal generator A and a bounded linear operator B on l^{2} such that

lim sup_{t→0^{+}} \frac{t log ∥AT (t)∥}{α(t)} = 0

and the semigroup generated by A + B fails to be eventually differentiable (and hence fails to be differentiable).

Our last theorem shows that there are examples in which (3) holds with arbitrarily small L > 0 and {S(t), t ≥ 0} fails to be differentiable.

Theorem 3. Let L^{∗} ∈ (0, ∞) be given. Then there exists a differentiable semigroup {T (t), t ≥ 0} on l^{2} with infinitesimal generator A and a bounded linear operator B on l^{2} such that

0 < lim sup_{t→0^{+}} \frac{t log ∥AT (t)∥}{log(\frac{1}{t})} ≤ L^{∗}

and the semigroup {S(t), t ≥ 0} generated by A+B fails to be differentiable. (Note that {S(t), t ≥ 0} is eventually differentiable by Theorem 1.)

The paper is organized as follows. In Section 2, we describe Renardy’s example and make some observations that will be used to prove Theorems 2 and 3. In particular we give an estimate for ∥AT (t)∥ as t ↓ 0. This estimate is based on an inequality for matrix exponentials that is recorded in the Appendix. Section 3 contains proofs of Theorems 1 through 3. The proof of Theorem 1 is based on a representation formula of Phillips which expresses the perturbed semigroup as an infinite series. We obtain estimaties that are sufficient to justify termwise differentiation of the series. The proofs of Theorems 2 and 3 are based on Renardy’s example.

2. Renardy’s Example

In this section we describe Renardy’s example [8] and make several observations that will be used to prove Theorems 2 and 3. In particular, we give an estimate for ∥AT (t)∥ . Throughout this section X = l^{2}, the space of all (complex) square - summable sequences, and we denote the perturbing operator by C rather than B for consistency of notation with [8].

The operator A is block diagonal. The n^{th} block is of size m(n) × m(n) and is given by

A_{n} = ( \begin{matrix}−n + ip(n) & 0 & ⋯ & 0 & 0 \\ n & −n + ip(n) & ⋯ & 0 & 0 \\ 0 & n & ⋯ & 0 & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & ⋯ & −n + ip(n) & 0 \\ 0 & 0 & ⋯ & n & −n + ip(n) \end{matrix} ) .

Here p(n) is real and positive and it is assumed that

p(n + 1) − p(n) > 2n + 2 for all sufficiently large n,  (4)
lim_{n→∞} \frac{n}{log p(n)} = ∞, lim_{n→∞} \frac{2^{m(n)}m(n)}{np(n)} = 0.  (5)

The domain of A is the set of all u ∈ l^{2} such that Au ∈ l^{2}. Observe that the spectrum of A consists precisely of the eigenvalues −n+ip(n). Renardy showed that under the above assumptions A generates a differentiable semigroup {T (t), t ≥ 0}.

The operator C is also block diagonal. The n^{th} block C_{n} is again of size m(n) × m(n) and is constructed as follows. With

μ(n) = n^{−1/m(n)},

and

x^{n} = n^{−1}(μ(n)^{−1}, μ(n)^{−2}, ...μ(n)^{−m(n)}),

C_{n}z is defined to be zero when z is orthogonal to x^{n} and

C_{n}x^{n} = (−1, 0, 0, ..., 0).

Then C is a bounded operator; in fact, ∥C∥ ≤ 1. It is important to note that each of the numbers ip(n) − n(1 − μ(n)), n = 1, 2, ... is an eigenvalue of A + C.

Renardy showed that if

lim_{n→∞} \frac{n(1 − n^{−1/m(n)})}{log p(n)} = 0  (6)

then the semigroup {S(t), t ≥ 0} generated by A + C is not differentiable. His argument actually yields a slightly more refined conclusion: If

lim sup_{n→∞} \frac{n(1 − n^{−1/m(n)})}{log p(n)} < ∞  (7)

then {S(t), t ≥ 0} is not differentiable and if (6) holds then {S(t), t ≥ 0} fails to be eventually differentiable. (To see that (6) implies {S(t), t ≥ 0} cannot be eventually differentiable one simply uses Theorem 2.4.7 of [6] in place of Theorem 2.4.8 of [6]. Moreover if (77) holds then {S(t), t ≥ 0} still violates the conditions of Theorem 2.4.8 of [6].)

The results of [8] are summarized in the following theorem; here N = {1, 2, 3, ...}.

Theorem 4. (Renardy). Let m : N → N and p : N → (0, ∞) be given and assume that (4) and (5) hold. Let A and C be as constructed above. Then A generates a differentiable semigroup {T (t), t ≥ 0} on l^{2} and C is a bounded linear operator on l^{2}. Let {S(t), t ≥ 0} be the semigroup generated by A + C. If (6) holds then {S(t), t ≥ 0} fails to be eventually differentiable and if (7) holds then {S(t), t ≥ 0} fails to be differentiable.

In order to prove Theorems 2 and 3 we need a good estimate for ∥AT (t)∥ as t ↓ 0. To this end we note that the norm of a block-diagonal operator is the supremum of the norms of the individual blocks. Let us fix n ∈ N. For each m(n) × m(n) matrix B we write

|||B||| = sup{|Bx|_{2} : x ∈ C^{m(n)}, |x|_{2} = 1}  (8)

where | · |_{2} is the Euclidean norm on C^{m(n)}. Noting that A_{n} is the sum of a diagonal and a nilpotent matrix we can express e^{tA_{n}} in the form

e^{tA_{n}} = e^{(−n+ip(n))t}P_{n}(nt)

where P_{n} is an m(n) × m(n) matrix-valued function whose entries are polynomials of degree ≤ m(n) − 1. It follows immediately from (8) and the special form of A_{n} that

|||A_{n}e^{tA_{n}}||| ≤ (2n + p(n))e^{−nt}|||P_{n}(nt)|||.  (9)

By examining the form of P_{n}, one can show that

|||P_{n}(nt)||| ≤ m(n)2^{m(n)}e^{\frac{nt}{2}} .  (10)

Details are given in the Appendix. By combining (9) and (10) we obtain the following estimate for ∥AT (t)∥ .

Lemma 1. Let m : N → N and p : N → (0, ∞) be given and assume that (4) and (5) hold. Let A be as constructed above and let {T (t), t ≥ 0} be the semigroup generated by A. Then, for each t > 0 we have

∥AT (t)∥ ≤ sup_{n∈N} m(n)2^{m(n)}(2n + p(n))e^{\frac{−nt}{2}} .

3. Proofs of Theorems

Proof of Theorem 1. We use a representation formula of Phillips [7] to express S(t) in the form

S(t) = ∑_{n=0}^{∞} V_{n}(t),  (11)

where the V_{n} are given inductively by

V_{0}(t) = T (t), V_{n+1}(t) = ∫_{0}^{t} T (s)BV_{n}(t − s)ds.

We choose M ≥ 1 and ω ∈ R such that

∥T (t)∥ ≤ Me^{ωt} for all t ≥ 0.

It is straightforward to verify that

∥V_{n}(t)∥ ≤ M^{n+1} ∥B∥^{n} e^{ωt} \frac{t^{n}}{n!}

for all n ∈ N∪{0} and all t ≥ 0. We shall use induction and the assumption that {T (t), t ≥ 0} is differentiable to show that each V_{n} is of class C^{1} on (0, ∞) in the uniform operator topology and that

∥V′_{n}(t)∥ ≤ M^{n+1} ∥B∥^{n} e^{ωt} \frac{t^{n−1}}{(n − 1)!} +M^{2n} ∥B∥^{n} ∥AT (\frac{t}{n + 1})∥ e^{\frac{n}{n+1} ωt} \frac{t^{n}}{n!} (12)

for all t > 0 and all n ∈ N. Since V_{0}(t) = T (t), it follows immediately that V_{0} is of class C^{1} (in fact, of class C^{∞}) on (0, ∞) in the uniform operator topology. The proof that V_{1} is of class C^{1} and that (12) holds for n = 1 is very similar to the induction step below; the details are therefore omitted. Let us assume now that V_{n} is of class C^{1} on (0, ∞) and that (12) holds for a given n ∈ N. We observe that

V_{n+1}(t) = ∫0^{\frac{t}{n+2}} T (s)BV_{n}(t − s)ds + ∫ t_{\frac{t}{n+2}} T (s)BV_{n}(t − s)ds

= ∫0^{\frac{t}{n+2}} T (s)BV_{n}(t − s)ds + ∫_{0}^{\frac{n+1}{n+2} t} T (t − s)BV_{n}(s)ds.  (13)

Formal differentiation of (13) gives

V′_{n+1}(t) = T (\frac{t}{n + 2})BV_{n}(\frac{n + 1}{n + 2}t)

+ ∫0^{\frac{t}{n+2}} T (s)BV′_{n}(t − s)ds + ∫_{0}^{\frac{n+1}{n+2} t} T′(t − s)BV_{n}(s)ds.

The induction hypothesis justifies the formal differentiation and yields the estimate

∥V′_{n+1}(t)∥ ≤ Me^{ω \frac{t}{n+2}} ∥B∥ M^{n+1} ∥B∥^{n} e^{ω \frac{n+1}{n+2} t} \frac{1}{n!}(\frac{n + 1}{n + 2}t)^{n}  (14)

+ ∫0^{\frac{t}{n+2}} Me^{ωs} ∥B∥ M^{n+1} ∥B∥^{n} e^{ω(t−s)} \frac{(t − s)^{n−1}}{(n − 1)!} ds

+ ∫0^{\frac{t}{n+2}} Me^{ωs} ∥B∥ M^{2n} ∥B∥^{n} ∥AT (\frac{t − s}{n + 1})∥ e^{\frac{n}{n+1} ω(t−s)} \frac{(t − s)^{n}}{n!} ds

+ ∫_{0}^{\frac{n+1}{n+2} t} ∥AT (t − s)∥ · ∥B∥ M^{n+1} ∥B∥^{n} e^{ωs} \frac{s^{n}}{n!} ds

for all t > 0. We observe now that for s ∈ [0, \frac{t}{n+2}] we have

∥AT (\frac{t − s}{n + 1})∥ = ∥AT (\frac{t}{n + 2})T (\frac{t}{(n + 1)(n + 2)} − \frac{s}{n + 1})∥

≤ ∥AT (\frac{t}{n + 2})∥ Me^{\frac{ω}{n+1} ( \frac{t}{n+2} −s)}  (15)

and for s ∈ [0, \frac{n+1}{n+2}t] we have

∥AT (t − s)∥ = ∥AT (\frac{t}{n + 2})T (\frac{n + 1}{n + 2}t − s)∥

≤ ∥AT (\frac{t}{n + 2})∥ Me^{ω( \frac{n+1}{n+2} t−s)}.  (16)

Substitution of (15) and (16) into (14) gives, after simplification,

∥V′_{n+1}(t)∥ ≤ M^{n+2} ∥B∥^{n+1} e^{ωt} \frac{t^{n}}{n!}

+M^{2n+2} ∥B∥^{n+1} ∥AT (\frac{t}{n + 2})∥ e^{\frac{n+1}{n+2} ωt} \frac{1}{n!} (∫0^{\frac{t}{n+2}} (t − s)^{n}ds + ∫_{0}^{\frac{n+1}{n+2} t} s^{n}ds)

which yields the desired result.

We can now use Weierstrass’ theorem to justify termwise differentiation of (11). Let t_{1} > t_{0} > L be given and notice that

∥AT (\frac{t}{n})∥ = ∥AT (\frac{t_{0}}{n})T (\frac{t − t_{0}}{n})∥ ≤ ∥AT (\frac{t_{0}}{n})∥ Me^{\frac{ω}{n} (t_{1}−t_{0})}  (17)

for all n ∈ N and all t ∈ [t_{0}, t_{1}]. Combining (12) and (17) we find that

∥V′_{n}(t)∥ ≤ a_{n}

for all n ∈ N and all t ∈ [t_{0}, t_{1}], where

a_{n} = M^{n+1} ∥B∥^{n} e^{ωt_{1}} \frac{t_{1}^{n−1}}{(n − 1)!} + ∥AT (\frac{t_{0}}{n + 1})∥ M^{2n+1} ∥B∥^{n} e^{ωt_{1}} \frac{t_{1}^{n}}{n!}.

For convergence of the series ∑_{n=1}^{∞} a_{n}, it suffices to have

lim sup_{n→∞}(\frac{1}{n!} ∥A(\frac{t_{0}}{n + 1})∥)^{1/n} = 0,

or, equivalently,

lim_{n→∞} \frac{1}{n} log(\frac{1}{n!} ∥AT (\frac{t_{0}}{n + 1})∥) = −∞.  (18)

Now, by (3),

\frac{1}{n} log ∥AT (\frac{t_{0}}{n + 1})∥ ≤ \frac{L}{t_{0}} log n + o(log n)

whereas, by Stirling’s formula

\frac{1}{n} log(n!) = log n + o(log n),

and consequently (18) holds since t_{0} > L. We may therefore differentiate (11) termwise to conclude that t ↦ S(t) is of class C^{1} on [t_{0}, t_{1}] in the uniform operator topology, for every [t_{0}, t_{1}] ⊂ (L, ∞).

Finally, let p ∈ [1, ∞) be given and assume that {T (t), t ≥ 0} is of class D^{p}. Then we may choose C > 0 such that

∥AT (t)∥ ≤ Ct^{−p} for all t ∈ (0, 1].

Observe that

∥(A + B)S(t)∥ = ∥S′(t)∥ ≤ ∑_{n=0}^{∞} ∥V′_{n}(t)∥

≤ ∑_{n=1}^{∞} M^{n+1} ∥B∥^{n} e^{ωt} \frac{t^{n−1}}{(n − 1)!} + ∑_{n=0}^{∞} ∥AT (\frac{t}{n + 1})∥ M^{2n} ∥B∥^{n} e^{\frac{n}{n+1} ωt} \frac{t^{n}}{n!}

≤ M^{2} ∥B∥ e^{(ω+M∥B∥)} + Ct^{−p} ∑_{n=0}^{∞} \frac{(n + 1)^{p}}{n!} M^{2n} ∥B∥^{n} e^{ω}

for all t ∈ (0, 1]. It follows that {S(t), t ≥ 0} is of class D^{p}.

Proof of Theorem 2. Let α : (0, 1) → (0, ∞) be a given function satisfying

\frac{α(t)}{log(\frac{1}{t})} → ∞ as t → 0^{+}.

We choose a continuous decreasing function β : (0, ∞) → (0, ∞) such that

\frac{α(t)}{β(t)} → ∞, \frac{β(t)}{log(\frac{1}{t})} → ∞ as t → 0^{+}  (19)

and define f : (0, ∞) → (0, ∞) by

f(t) = \frac{2β(t)}{t^{2}} for all t ∈ (0, ∞).

Notice that f is continuous and strictly decreasing; moreover, the range of f is (0, ∞). In addition, f^{−1} is strictly decreasing, x ↦ xf^{−1}(x) is strictly increasing, and xf^{−1}(x) → ∞ as x → ∞.

We put

m(n) = [[\frac{1}{4}nf^{−1}(n)]], p(n) = exp(\frac{1}{4}nf^{−1}(n))  (20)

for all n ∈ N. Here [[z]] denotes the integer part of z. It is straightforward to verify that (4), (5) and (6) are satisfied. Let {T (t), t ≥ 0} be as in Renardy’s example. (See Section 2.) In view of Theorem 4, it remains only to show that

lim sup_{t→0^{+}} \frac{t log ∥AT (t)∥}{α(t)} = 0.  (21)

By Lemma 1, we have

log ∥AT (t)∥ ≤ sup_{n∈N}(log(m(n)2^{m(n)}(2n + p(n))) − \frac{nt}{2})  (22)

for all t > 0. Using (20) and the properties of f it is straightforward to verify the following two claims.

Claim 1. For every N ∈ N, there exists δ > 0 such that

sup_{n∈N}(log(m(n)2^{m(n)}(2n+p(n)))− \frac{nt}{2}) = sup_{n≥N}(log(m(n)2^{m(n)}(2n+p(n)))− \frac{nt}{2})

for all t ∈ (0, δ).

Claim 2. There exists N ∈ N such that

log(m(n)2^{m(n)}(2n + p(n))) ≤ 2 log p(n)

for all n ∈ N with n ≥ N.

By combining Claims 1 and 2 and using (20) and (22) we may choose δ > 0 such that δ < f^{−1}(1) and

log ∥AT (t)∥ ≤ \frac{1}{2} sup_{n∈N}(nf^{−1}(n) − nt) for all t ∈ (0, δ).

Since δ < f^{−1}(1) we have

sup_{n∈N}(nf^{−1}(n) − nt) > 0 for all t ∈ (0, δ).

On the other hand

xf^{−1}(x) − xt ≤ 0 for all x ≥ f(t)

and consequently

log ∥AT (t)∥ ≤ \frac{1}{2} max \{xf^{−1}(x) − xt : x ∈ [1, f(t)]\}

≤ \frac{1}{2} max \{xf^{−1}(x) : x ∈ [1, f(t)]\}  (23)

for all t ∈ (0, δ). (Notice that f(t) > 1 for all t ∈ (0, δ).) Since x ↦ xf^{−1}(x) is increasing and f(t)f^{−1}(f(t)) = 2β(t)/t, we conclude from (23) that

log ∥AT (t)∥ ≤ \frac{β(t)}{t} for all t ∈ (0, δ).  (24)

It follows immediately from (19) and (24) that (21) holds.

Proof of Theorem 3. Let L^{∗} ∈ (0, ∞) be given and choose a continuous, strictly decreasing function f : (0, ∞) → (0, ∞) such that

f(t) = \frac{2L^{∗} log(\frac{1}{t})}{t^{2}} for all t ∈ (0, \frac{1}{2}]

and f(t) → 0 as t → ∞. We put

m(n) = [[\frac{1}{4}nf^{−1}(n)]], p(n) = exp(\frac{1}{4}nf^{−1}(n))

for all n ∈ N. It is straightforward to verify that (4) and (5) hold and that

lim_{n→∞} \frac{n(1 − n^{−1/m(n)})}{log p(n)} = \frac{16}{L^{∗}} .

Let {T (t), t ≥ 0} and {S(t), t ≥ 0} be as in Renardy’s example. Arguing as in the proof of Theorem 2, we find that

lim sup_{t→0^{+}} \frac{t log ∥AT (t)∥}{log(\frac{1}{t})} ≤ L^{∗}.  (25)

The desired result now follows from Theorem 4. (Note that Theorem 1 implies that the lim sup in (25) is > 0.)

Appendix. An Inequality for Matrix Exponentials

Let m ∈ N be given. For each m × m (complex) matrix B with components B_{ij} we write

∥B∥_{1} = ∑_{i,j=1}^{m} |B_{ij}|,

|||B||| = sup{|Bx|_{2} : x ∈ C^{m}, |x|_{2} = 1},

where | · |_{2} is Euclidean norm on C^{m}. We note that

|||B||| ≤ m ∥B∥_{1} .

Let a, b ∈ C be given and define the m × m matrix A by

A = ( \begin{matrix}a & 0 & ⋯ & 0 & 0 \\ b & a & ⋯ & 0 & 0 \\ 0 & b & ⋯ & 0 & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & ⋯ & a & 0 \\ 0 & 0 & ⋯ & b & a \end{matrix} ) .

Lemma 2. For each t ≥ 0 we have

|||Ae^{tA}||| ≤ m2^{m}(|a| + |b|)e^{(|b|+2Re(a)) \frac{t}{2}} .  (26)

Proof. Observe first that

e^{tA} = e^{at} ( \begin{matrix}1 & 0 & ⋯ & 0 & 0 \\ bt & 1 & ⋯ & 0 & 0 \\ \frac{(bt)^{2}}{2!} & bt & ⋯ & ⋯ & ⋯ \\ ⋯ & ⋯ & ⋯ & 1 & 0 \\ \frac{(bt)^{m−1}}{(m−1)!} & \frac{(bt)^{m−2}}{(m−2)!} & ⋯ & bt & 1 \end{matrix} )

and that

|||Ae^{tA}||| ≤ m ∥Ae^{tA}∥_{1} ≤ m(|a| + |b|) ∥e^{tA}∥_{1} .  (27)

Notice that for x ≥ 0 we have

∑_{j=0}^{i} \frac{x^{j}}{j!} = ∑_{j=0}^{i} 2^{j} \frac{(\frac{x}{2})^{j}}{j!} ≤ 2^{i}e^{\frac{x}{2}} ,

which leads to the estimate

∥e^{tA}∥_{1} = e^{tRe(a)} ∑_{i=0}^{m−1} ∑_{j=0}^{i} \frac{(|b|t)^{j}}{j!} ≤ e^{tRe(a)} ∑_{i=0}^{m−1} 2^{i}e^{\frac{t}{2} |b|} = 2^{m}e^{(|b|+2Re(a)) \frac{t}{2}} .

(28)

Combining (27) and (28) we obtain (26).

Acknowledgement

We are indebted to Raul Kangro for several valuable discussions.

References

[1] Crandall, M.G. and A. Pazy, On the differentiability of weak solutions of a differential equation in Banach space, J. Math. Mech. 18 (1969), 1007–1016.

[2] Hille, E., Representation of one-parameter semi-groups of linear transformations, Proc. Nat. Acad. Sci. 28 (1942), 175–178.

[3] Hille, E., “Functional Analysis and Semi-Groups,” Amer. Math. Soc. Colloq. 31, New York, 1948.

[4] Hille, E., On the differentiability of semi-groups of operators, Acta Sci. Math. (Szeged) 12B (1950), 19–24.

[5] Pazy, A., On the differentiability and compactness of semigroups of linear operators, J. Math. Mech. 17 (1968), 1131–1141.

[6] Pazy, A., “Semigroups of Linear Operators and Applications to Partial Differential Equations,” Springer, New York, 1983.

[7] Phillips, R.S., Perturbation theory for semi-groups of linear operators, Trans. Amer. Math. Soc. 74 (1953), 199–221.

[8] Renardy, M.On the stability of differentiability of semigroups, Semigroup Forum (to appear).

[9] Yosida, K., On the differentiablity of semi-groups of linear operators, Proc. Japan Acad.34 (1958), 337–340.

[10]  Yosida, K., Math. Reviews 20 (1959), #5435.

Department of Mathematics Carnegie Mellon University Pittsburgh, PA 15213