On an extension of Kato-Voigt perturbation theorem for

substochastic semigroups and its application

Lecture given at the ICMAA 2000, National Sun Yat-sen University, Taiwan

J. Banasiak ^{∗}

School of Mathematical and Statistical Sciences, University of Natal, Durban 4041 South Africa

Abstract The aim of this paper is to discuss some recent developments of the perturbation method introduced first by Kato for the Kolmogoroff equation and later extended by Voigt and Arlotti to deal with a range of problems related to the solvability of the so-called Master Equation. In the application part, the paper contains a detailed analysis of the fragmentation equation.

Key words: perturbation theory for semigroups, substochastic semigroups, fragmentation model AMS subject classification: 47D06, 34G10.

1 Introduction

In many branches of the applied sciences we are interested in the evolution of the density function (t, ξ) → u(t, ξ) where t is the time and ξ ∈ Ω is an element of some state space - it could be the velocity or energy in the kinetic theory, the number of bacteria in a sample in the population theory, mass in the coagulationfragmentation theory. The function u is then interpreted as the probability (density) of finding an individual which at the time t has the property ξ. If we assume that the probability at time t is completely determined by the probability at an earlier time t_{0} (that is we are dealing with a Markov process), then the time evolution of u is described by the master equation (e.g. [11])

∗This work was supported by the University of Natal Research Fund

∂_{t}u(t, ξ) = ∫_{Ω} [k(ξ′, ξ)u(t, ξ′) − k(ξ, ξ′)u(t, ξ)]dμ_{ξ′} = ∫_{Ω} k(ξ′, ξ)u(t, ξ′)dμ_{ξ′} − u(t, ξ) ∫_{Ω} k(ξ, ξ′)dμ_{ξ′}, (1.1)

where dμ_{ξ} is an appropriate measure in the state space, and k(ξ′, ξ) is the transition rate (probability density of transitions per unit time) of the change between the states ξ′ and ξ (that is the probability of the change from ξ′ to ξ to occur in the time interval dt is approximately k(ξ′, ξ)dξdt.

A heuristic justification of this equation goes as follows. The expected number of particles around ξ at time t + dt is given by u(t + dt, ξ)dμ_{ξ} is equal to u(t, ξ)dμ_{ξ}, plus the number of particles that within dt changed from the state ξ′ into the state ξ: dt(k(ξ′, ξ)dξ)u(t, ξ′)dμ_{ξ′}, integrated over all possible states ξ′, and minus the number of particles at the state ξ which changed into the state ξ′: dt(k(ξ, ξ′)dξ′)u(t, ξ)dμ_{ξ}, integrated over all the possible states ξ′.

An intrinsic property of the above process is that all the particles must be accounted for or, in other words, the total number of particles doesn’t change:

∫_{Ω} u(t, ξ)dμ_{ξ} = ∫_{Ω} u(0, ξ)dμ_{ξ},  (1.2)

for any time t. Therefore from the physical point of view the natural spaces for studying such problems are L_{1} spaces (or l_{1} if ξ takes only discrete values). Equation (1.2) can be checked formally by integrating (1.1) and taking into account that

∫_{Ω} ∫_{Ω} [k(ξ′, ξ)u(t, ξ′) − k(ξ, ξ′)u(t, ξ)]dμ_{ξ′}dμξ = ∫_{Ω} ∫_{Ω} k(ξ′, ξ)u(t, ξ′)dμξ′dμ_{ξ} − ∫_{Ω} ∫_{Ω} k(ξ, ξ′)u(t, ξ)dμξ′dμ_{ξ} = 0

which follows by changing ξ′ into ξ in one of the integrals.

This process (or in general, a combination of such processes) may take place simultaneously with an evolution in the physical space so that in general we will be concerned with evolution equations of the form

∂_{t}u = A_{0}u + A_{1}u + Bu,  (1.3)

where A_{0} typically is the free streaming operator: A_{0}u = −v · ∇_{x}u, or the diffusion operator, whereas A_{1}u = −mu with a nonnegative, measurable, and almost everywhere finite function m, and B is an integral operator.

The first equation of this type, namely the Kolmogoroff equation, was first studied in the semigroup theory framework by Kato in his seminal paper [6]. His results were extended to a more abstract framework and extensively applied to a range of problems arising mainly in the kinetic theory by Voigt, Arlotti, Desch, Mokhtar-Karroubi, and recently by the author (see e.g. [13, 1, 2, 9, 3, 4]).

An interesting feature of these problems is that the semigroup (G(t))_{t≥0} solving the Cauchy problem related to (1.3) is constructed by a limiting procedure which doesn’t allow to control the domain of the generator; in general case one can only prove that the generator T of (G(t))_{t≥0} is an extension of A_{0} + A_{1} + B. In most applications it is not enough as typically we are interested in the cases when the solution semigroup (G(t))_{t≥0} is a transition (or stochastic) semigroup, that is, when ∥G(t)x∥ = ∥x∥ for all t ≥ 0 and x ≥ 0 (see also Remark 2.2). In a number of applications (see [6]) this could happen if and only if T = \overline{A_{0} + A_{1} + B}; in general the last equality is a sufficient conditions for (G(t))_{t≥0} to be stochastic. Thus it is important to be able to determine when T = \overline{A_{0} + A_{1} + B}. We shall present some sufficient conditions, which are a modification of those introduced in [2], and apply them to the fragmentation model [14, 8] which will be introduced in Section 4. We shall see that for same values of the parameter the generator T of the semigroup is the closure of the operator appearing on the right-hand side of the equation (and that this result is sharp, that is T ≠ A_{0} + A_{1} + B). However, in the last section of the paper we shall demonstrate that for a range of parameter values the generator T is a proper extension of \overline{A_{0} + A_{1} + B} and that the semigroup generated by T is not stochastic.

Acknowledgment. The author is grateful to Dr W. Lamb for communicating the fragmentation model to him, and for a number of stimulating discussions.

2 Main Theorem

In this section we shall formulate and provide proofs of several results which are scattered in the literature ([13, 1, 2]), and which are very often considered only in a specific context.

Let (Ω, μ) be a measure space. By X we denote the Banach space L_{1}(Ω, μ) endowed with the standard norm ∥ · ∥. For any subspace Z ⊂ X by Z_{+} we denote the cone of nonnegative (a.e.) elements of Z. Let (G(t))_{t≥0} be a strongly continuous semigroup on X. We say that (G(t))_{t≥0} is a substochastic semigroup if for each t ≥ 0, G(t) ≥ 0 and ∥G(t)∥ ≤ 1. It is called a stochastic semigroup if additionally ∥G(t)f∥ = ∥f∥ for X_{+}.

In accordance with the discussion in the introduction, we shall consider three linear operators in X: A_{0}, A_{1}, and B which have the following properties:

1. A_{0} and A = A_{0} + A_{1} (defined on D(A) = D(A_{1}) ∩ D(A_{0})) generate substochastic semigroups denoted, respectively, by (G_{A_{0}}(t))_{t≥0} and (G_{A}(t))_{t≥0},

2. D(B) ⊃ D(A) and Bf ≥ 0 for any f ∈ D(B)_{+},

3. for any f ∈ D(A_{1})_{+}

∫_{Ω} (A_{1}f + Bf)dμ = 0.  (2.1)

Let us observe that the above list yields the following properties of the operators involved.

Lemma 2.1 (i) For any f ∈ D(A)_{+}

∫_{Ω} (Af + Bf)dμ ≤ 0.  (2.2)

(ii) The operator B(λI − A)^{−1} is a bounded positive operator in X.

(iii) If A_{1}f = −mf for some a.e. positive measurable m, then for any f ∈ D(A_{1})

∥Bf∥ ≤ ∥A_{1}f∥  (2.3)

Proof. ad. (i) Since A_{0} generates a contraction semigroup, it is dissipative, hence by (2.1)

∫_{Ω} (Af + Bf)dμ = ∫_{Ω} A_{0}fdμ + ∫_{Ω} (A_{1} + B)fdμ ≤ 0.

ad. (ii) Since A generates a contraction semigroup, the resolvent is well-defined for λ > 0. From the properties of B we see that B(λI − A)^{−1} is defined on X and positive, hence it is bounded ([12], II, Theorem 5.3).

ad. (iii) From (2.1) we obtain for f ∈ D(A_{1})_{+}

∥Bf∥ = ∫_{Ω} Bfdμ = ∫_{Ω} mfμ = ∥A_{1}f∥.

Let us take arbitrary f ∈ D(A_{1}); then also |f| ∈ D(A_{1}) and |Bf| ≤ B|f| and

∥Bf∥ ≤ ∥B|f|∥ = ∫_{Ω} B|f|dμ = ∫_{Ω} m|f|dμ = ∫_{Ω} |mf|dμ = ∥A_{1}f∥.

The next lemma strengthens Lemma 1.2 of [13] and is also more general then Lemma 1 of [1].

Lemma 2.2 For any f ∈ D(A) the function t → BG_{A}(t)f is continuous and

∫_{0}^{t} ∥BG_{A}(s)f∥ds ≤ ∥f∥ − ∥G_{A}(s)f∥.  (2.4)

Proof. Let f ∈ D(A). We can write

BG_{A}(t)f = B(I − A)^{−1}G_{A}(t)(I − A)f

hence the continuity follows from Lemma 2.1, item (ii).

Next we prove (2.4). Using the facts that for f ∈ D(A)_{+} we have 0 ≤ G_{A}(s)f ∈ D(A) ⊂ D(B) and that B is positive on its domain, by (2.1) we obtain for such f

∫_{0}^{t} ∥BG_{A}(s)f∥ds = ∫_{0}^{t} ∫_{Ω} BG_{A}(s)fdμds ≤ − ∫_{0}^{t} ∫_{Ω} AG_{A}(s)fdμds
= − ∫_{Ω} ∫_{0}^{t} AG_{A}(s)fdsdμ = ∫_{Ω} fdμ − ∫_{Ω} G_{A}(t)fdμ = ∥f∥ − ∥G_{A}(t)f∥.

Let us take now arbitrary f ∈ D(A). Since in general |f| ∉ D(A), we consider the regularization R_{n}|f| = n ∫_{0}^{1/n}G_{A}(s)|f|ds. Since |f| ∈ X, 0 ≤ R_{n}|f| ∈ D(A) ([10], Theorem 2.4 ), and |R_{n}f| ≤ R_{n}|f|. Thus, as above

∫_{0}^{t} ∥BG_{A}(s)R_{n}f∥ds ≤ ∫_{0}^{t} ∥BG_{A}(s)R_{n}|f|∥ds ≤ ∥R_{n}|f|∥ − ∥G_{A}(t)R_{n}|f|∥.

Using again Theorem 2.4 of [10], we obtain that R_{n}|f| → |f| in X, whereas R_{n}f → f in D(A). Passing thus to the limit, we obtain

∫_{0}^{t} ∥BG_{A}(s)f∥ds ≤ ∫_{0}^{t} ∥BG_{A}(s)R_{n}|f|∥ds ≤ ∥|f|∥ − ∥G_{A}(t)|f|∥,

where to be able to pass to the limit in the first integral we used the Lebesgue dominated convergence theorem and the estimate

∥BG_{A}(t)R_{n}f∥ = ∥B(I − A)^{−1}G_{A}(t)(I − A)n ∫_{0}^{1/n}G_{A}(s)fds∥ ≤ M (∥f∥ + ∥n(G_{A}(1/n)f − f)∥)
≤ M (∥f∥ + ∥T (δ)Af∥) ≤ M(∥f∥ + ∥Af∥),

for some 0 < δ < 1/n. In the above we used again Theorem 2.4 of [10].

Next, since G_{A}(t)|f| ≥ |G_{A}(t)f|, we obtain −∥G_{A}(t)|f|∥ ≤ −∥G_{A}(t)f∥, and the lemma is proved.

The following generalizes Theorem 4 of [1] to a more abstract setting though the key ingredients of the proof remain the same.

Theorem 2.1 Under the above assumptions there exists the smallest substochastic semigroup (G(t))_{t≥0} generated by an extension T of A + B. This semigroup satisfies the integral equation

G(t)f = G_{A}(t)f + ∫_{0}^{t} G_{A}(t − s)BG_{A}(s)fds  (2.5)

for any f ∈ D(A) and t ≥ 0.

Proof. The construction of the semigroup is through the Phillips-Dyson expansion. Let us define the following operators by the recursive formula:

S_{0}(t)f = G_{A}(t)f,
S_{n}(t)f = ∫_{0}^{t} S_{n−1}(t − s)BG_{A}(s)fds, n > 0,  (2.6)

for f ∈ D(A) and t ≥ 0. First we shall show that S_{n} can be extended to a strongly continuous families of bounded operators in X satisfying

∥S_{n}(t)f∥ ≤ ∥f∥ − ∑_{k=0}^{n−1} ∥S_{k}(t)f∥.  (2.7)

We emphasize that this inequality holds for any fixed t and any fixed f. The statement is clearly true for n = 0. Assume thus that it is valid for some n − 1. We see that S_{n}(t)f is well-defined for f ∈ D(A), as t → BG_{A}(s)f is continuous by Lemma 2.2, S_{n−1}(t − s) is strongly continuous and the composition is continuous by e.g. [5], p. 482. To be able to extend S_{n}(t) to a bounded operator on X we prove the estimate (2.7). Let f ∈ D(A), then by (2.4)

∥S_{n}(t)f∥ ≤ ∫_{0}^{t} ∥S_{n−1}(t − s)BG_{A}(s)f∥ds
≤ ∫_{0}^{t} (∥BG_{A}(s)f∥ − ∑_{k=0}^{n−2} ∥S_{k}(t − s)BG_{A}(s)f∥) ds
≤ ∥f∥ − ∥G_{A}(t)f∥ − ∑_{k=0}^{n−2} ∫_{0}^{t} ∥S_{k}(t − s)BG_{A}(s)fds∥
≤ ∥f∥ − ∥G_{A}(t)f∥ − ∑_{k=0}^{n−2} ∥ ∫_{0}^{t} S_{k}(t − s)BG_{A}(s)fds∥
= ∥f∥ − ∑_{k=0}^{n−1} ∥S_{k}(t)f∥

where we used the inverted estimate: − ∫_{0}^{t} ∥ · ∥ds ≤ −∥ ∫_{0}^{t} ·ds∥. Hence, we see that the operators S_{n} can be extended to bounded (contractive) operators on X (though in general they will be no longer given by (2.6)).

Next we shall prove that they are strongly continuous. Let us first consider f ∈ D(A) and t′ > t′′

S_{n}(t′)f − S_{n}(t′′)f = ∫_{0}^{t′} S_{n−1}(t′ − s)BG_{A}(s)fds − ∫_{0}^{t′′}S_{n−1}(t′′ − s)BG_{A}(s)fds
= ∫_{0}^{t′} S_{n−1}(v)BG_{A}(t′ − v)fdv − ∫_{0}^{t′′}S_{n−1}(v)BG_{A}(t′′ − v)fdv
= ∫_{t′′}^{t′} S_{n−1}(v)BG_{A}(t′ − v)fds + ∫_{0}^{t′′}S_{n−1}(v)B(G_{A}(t′ − v)f − G_{A}(t′′ − v)f)ds
= ∫_{t′′}^{t′} S_{n−1}(v)BG_{A}(t′ − v)fds + ∫_{0}^{t′′}S_{n−1}(t′′ − s)BG_{A}(s)(G_{A}(t′ − t′′) − I)fds
= ∫_{t′′}^{t′} S_{n−1}(v)BG_{A}(t′ − v)fds + S_{n}(t′′)(G_{A}(t′ − t′′) − I)f

Therefore, by (2.7) we obtain

∥S_{n}(t′)f − S_{n}(t′′)f∥ ≤ ∫_{t′′}^{t′} ∥S_{n−1}(v)BG_{A}(t′ − v)f∥dv + ∥G_{A}(t′ − t′′)f − f∥
≤ |t′ − t′′|∥f∥_{D(A)} + ∥G_{A}(t′ − t′′)f − f∥

which gives strong continuity for f ∈ D(A). The extension to X follows from the density and the estimate ∥S_{n}(t)∥ ≤ 1.

Next we observe that the estimate (2.7) yields the convergence in X of the series

G(t)f = ∑_{n=0}^{∞} S_{n}(t)f  (2.8)

for any f ∈ X and t ≥ 0 and Banach-Steinhaus theorem shows that (G(t))_{t≥0} is a family of bounded operators (contractions). It is also seen that this series is uniformly convergent (Weierstrass theorem), thus t → G(t) is strongly continuous. Using the calculations of the Miyadera-Voigt perturbation theorem ([5], pp. 182–183) we see that (G(t))_{t≥0} is a semigroup and the theorem is proved.

Remark 2.1 In [13] the author proved, generalizing the idea of Kato [6], that the semigroup (G(t))_{t≥0} can be obtained as a strong limit of the semigroups (G_{r}(t))_{t≥0} generated by A + rB, when 0 < r → 1^{−}. Moreover, if r ≤ r′, then G_{r}(t) ≤ G_{r′}(t) for each t. This fact will be used later in Proposition 5.1.

Remark 2.2 The drawback of Theorem 2.1 is that it doesn’t provide any characterization of the domain of the generator T. The ideal situation would be, of course, if D(T) = D(A) = D(A + B). However, as we mentioned in the introduction, the case when T = \overline{A + B} is also physically acceptable since in such a case the semigroup is still stochastic. In fact, if A_{0} is conservative, then by assumption (2.1) for u ∈ D(A+B) = D(A) we have

∫_{Ω} (A + B)udμ = 0.

If T = \overline{A + B}, then for u ∈ D(T) there exists a sequence (u_{n})_{n∈ℕ} of elements of D(A) such that u_{n} → u and (A + B)u_{n} → T u in X as n → ∞, thus

∫_{Ω} T udμ = lim_{n→∞} ∫_{Ω} (A + B)u_{n}dμ = 0.

This in turn shows that if 0 ≤u^{∘} ∈ D(T), then 0 ≤ u(t) = G(t)u^{∘} ∈ D(T) for any t ≥ 0 and

\frac{d}{dt}∥u(t)∥ = ∫_{Ω} \frac{du(t)}{dt}dμ = ∫_{Ω} T G(t)u^{∘} dμ = 0.

On the other hand, if T is a bigger extension of A + B, then the above property may not hold and there may be a loss of particles in the evolution. We shall see that such a situation is possible in Proposition 5.2.

The next proposition is related to the above discussion. We note that for a particular case of Kolmogoroff’s equation similar results, though through a different method, have been proved in [6].

Proposition 2.1 Under the assumption of this section we have (i) the generator T is characterized by

(I − T)^{−1}f = ∑_{n=0}^{∞}(I − A)^{−1}[B(I − A)^{−1}]^{n}f,  (2.9)

for every f ∈ X.

(ii) If ∑_{n=0}^{∞}[B(I − A)^{−1}]^{n}f converges for any f ∈ X, then T = A + B.

(iii) If lim n → ∞∥[B(I − A)^{−1}]^{n}f∥ = 0 for any f ∈ X, then T = \overline{A + B}.

Proof. ad. (i) Taking the Laplace transform of (2.8) with λ = 1 we obtain

(I − T)^{−1}f = ∑_{n=0}^{∞} L(S_{n}(t)f)(1).  (2.10)

Now, for f ∈ D(A)

L(S_{n}(t)f)(1) = L(S_{n−1}(t)f)(1)L(BG_{A}(t)f)(1),

where we used Corollary C.17 of [5] (applicable by Lemma 2.2). The next step requires some care as we don’t know whether B is a closed operator. However, repeating the approach of Lemma 2.2 we obtain

L(BG_{A}(t)f)(1) = L(B(I − A)^{−1}G_{A}(t)(I − A)f)(1) = B(I − A)^{−1}f,

by the boundedness of B(I − A)^{−1}. Therefore, for any f ∈ D(A)

L(S_{0}(t)f)(1) = (I − A)^{−1}f
L(S_{n}(t)f)(1) = L(S_{n−1}(t))B(I − A)^{−1}f = (I − A)^{−1}[B(I − A)^{−1}]^{n}f,

and since L(S_{n}(t)f)(1) is a bounded operator, we can extend the above to the whole of X. Combining this with (2.10) we obtain (2.9).

ad. (ii) If the series converges in X, then we clearly have

(I − T)^{−1}f = ∑_{n=0}^{∞}(I − A)^{−1}[B(I − A)^{−1}]^{n}f = (I − A)^{−1} ∑_{n=0}^{∞}[B(I − A)^{−1}]^{n}f,

which shows that D(A) ⊃ D(T). Since A ⊂ T, we have the equality.

ad. (iii) Assume that for any f ∈ X, ∥(B(I − A)^{−1})^{n}f∥ → 0 as n → 0. Denote

z_{N} = (I − A)^{−1} ∑_{n=0}^{N}(B(I − A)^{−1})^{n}f ∈ D(A).

Clearly, z_{N} → z = (I − T)^{−1}f ∈ D(T) in X. Now,

(I − (A + B))z_{N} = ∑_{n=0}^{N}(B(I − A)^{−1})^{n}f − B(I − A)^{−1} ∑_{n=0}^{N}(B(I − A)^{−1})^{n}f
= ∑_{n=0}^{N}(B(I − A)^{−1})^{n}f − ∑_{n=1}^{N+1}(B(I − A)^{−1})^{n}f
= f − (B(I − A)^{−1})^{N}f,

thus, if the assumption is satisfied, then (I − (A + B))z_{N} converges in X and consequently z ∈ D(\overline{A + B}). Since z is an arbitrary member of D(T), we obtain D(T) ⊂ D(\overline{A + B}). On the other hand, T, as a generator, is a closed extension of A + B, thus \overline{A + B} ⊂ T.

From this theorem we see that the extension T of A + B generates a substochastic semigroup. In particular, this is a semigroup of contractions and, consequently, T is a dissipative operator. However, any restriction of a dissipative operator is also dissipative, thus A + B is dissipative. This allows a simple corollary.

Corollary 2.1 Under the assumptions of the Theorem 2.1

1. T = \overline{A + B} if and only if the range of λ − (A + B), Rg(λ − (A + B)), is dense in X for some (all) λ > 0.

2. If additionally A + B is a closed operator, then T = A + B.

Proof. ad. 1) Let us assume that Rg(λ − (A + B)) is dense in X. By [5], Theorem 3.15, (Lumer–Phillips Theorem), this statement is equivalent to saying that \overline{A + B} is a generator (that A + B is densely defined follows from the fact that A itself is a generator). Since T a closed extension of A + B (as a generator), we must have \overline{A + B} ⊂ T. Since both are generators, they must coincide. The converse is obvious.

Item 2 follows immediately.

3 Sufficient condition for T = \overline{A + B}

In this section we shall sketch an approach of [2] (with some modification due to the author, first given for a specific situation in [4]) which enables a better characterization of the generator T.

As the first step we extend all the discussed operators in the following way.

Let E denote the set of all the extended real-valued measurable functions defined on Ω; clearly X ⊂ E. We define the subset F ⊂ E by the following condition: f ∈ F if and only if for every nonnegative and nondecreasing sequence of functions (f_{n})_{n∈ℕ} satisfying sup_{n} f_{n} = |f| we have sup_{n}(I − A)^{−1}f_{n} ∈ X.

Before proceeding any further we adopt the assumption that f ∈ D(B) if and only if f^{+} = max{f, 0}, f^{−} = max{−f, 0} both belong to D(B). Through B we construct another subset of E, say G, defined as the set of all functions f ∈ X such that for any nonnegative, nondecreasing sequence (f_{n})_{n∈ℕ} of elements of D(B) such that sup_{n} f_{n} = |f|, we have sup_{n} Bf_{n} < +∞ almost everywhere. It is easy to check that D(A_{0}) ⊆ D(A) ⊆ G ⊆ X ⊆ F ⊆ E.

It can be proved that we can correctly define the mapping L : F_{+} → X_{+} by

Lf = sup_{n}(I − A)^{−1}f_{n}

where 0 ≤ f_{n} ≤ f_{n+1} for any n, and sup_{n} f_{n} = f.

To proceed, in [2] the assumptions (1)–(3) of Section 2 were supplemented by two more:

(4) The mapping L is injective.

(5) If f ∈ G_{+} and (f′_{n})_{n∈ℕ}, (f′′_{n})_{n∈ℕ} are two nondecreasing sequences of elements of D(A)_{+} satisfying f = sup_{n} f′_{n} = sup_{n} f′′_{n}, then sup_{n} Bf′_{n} = sup_{n} Bf′′_{n}. The common value will be denoted by Bf.

We extend the mappings L and B onto F and G, respectively, by linearity.

The above construction of L is not always easy to apply. We shall interpret it from the point of view of Sobolev towers (see e.g. [5], Section II.5). This interpretation will also show that the assumption (4) above is spurious.

Thus, according to [5], we define the space X^{−1} as the completion of X with respect to the norm

∥f∥_{−1} = ∥(I − A)^{−1}f∥_{X}.

Then the semigroup (G_{A}(t))_{t≥0} extends continuously to a semigroup (G_{A,−1}(t))_{t≥0} in X^{−1}, which is generated by the closure of A in X^{−1}. This closure, denoted by A_{−1} is defined on the domain D(A_{−1}) = X ⊂ X_{−1}. The resolvent extends then by density to a bounded one-to-one operator R(λ, A_{−1}) : X^{−1} → X^{−1} with the range exactly equal to X. We have the following lemma.

Lemma 3.1 The operator L is a restriction of R(1, A_{−1}). As a consequence, the assumption (4) is satisfied.

Proof. Let g ∈ X_{+} satisfies g = Lf. This means that g = sup_{n}(I − A)^{−1}f_{n} for a non-decreasing sequence of nonnegative functions f_{n} ∈ X_{+} such that sup_{n} f_{n} = f. Since (I − A)^{−1} is a positive operator, the sequence ((I − A)^{−1}f_{n})_{n=0}^{∞} is also non-decreasing and g ≥ (I − A)^{−1}f_{n} for any n ∈ ℕ. Because g is integrable, we obtain

lim_{n→∞} ∫_{Ω} (I − A)^{−1}f_{n}dμ = ∫_{Ω} gdμ

and

lim_{n→∞} ∫_{Ω} |g − (I − A)^{−1}f_{n}| = ∫_{Ω} gdμ − lim_{n→∞} ∫_{Ω} (I − A)^{−1}f_{n}dμ = 0

This shows that ((I − A)^{−1}f_{n})_{n∈ℕ} converges in X and therefore g = Lf = R(1, A_{−1})f. The extension for arbitrary f is done by linearity.

Corollary 3.1 If Af = A_{1}f = −mf, where m is a nonnegative, measurable, and almost everywhere finite function, then

F = X^{−1} = {f ∈ E; (1 + m)^{−1}f ∈ X},  (3.1)

and Lf = (1 + m)^{−1}f.

Proof. Since (I − A)^{−1}f = (1 + m)^{−1}f, then by the definition of F, f ∈ F provided sup_{n}(1 + m)^{−1}f_{n} ∈ X for any any nondecreasing sequence of nonnegative functions f_{n} such that sup_{n} f_{n} = |f|.

Denote g = (I − T)h ∈ X. Using (2.9) and replacing the operators involved there by their extensions we see that h can be expressed by

h = ∑_{k=0}^{∞} L(BL)^{k}g.  (3.2)

Let us denote, for any g ∈ X

f_{n} = ∑_{k=0}^{n}(BL)^{k}g

and

h_{n} = Lf_{n}

We note that for a nonnegative g we can consider limits of both sequences in the sense of monotonic convergence almost everywhere, as L and B are positive operators. We denote by f and h the respective limits provided they exist. In this case we obtain Lf = h. The result of [2] which we shall apply here reads as follows.

Theorem 3.1 Let all the above assumptions be satisfied. If for any g ∈ X_{+} the limit of the following real convergent sequence satisfy

lim_{n→∞} ∫_{Ω} A_{0}h_{n}dμ ≤ ∫_{Ω} (h − f + Bh)dμ,  (3.3)

then T = \overline{A + B}.

This theorem is not immediately useful as the function h is not explicitly known. However, the following corollary is useful in many instances.

Corollary 3.2 If Af = A_{1}f = −mf for some measurable, nonnegative, and almost everywhere finite function m, and for any H ∈ X_{+} such that −mH + BH ∈ X we have

∫_{Ω} (−mH + BH)dμ ≥ 0,  (3.4)

then T = \overline{A + B}.

Proof. Inequality (3.4) coincides with (3.3) re-written in the context of this corollary, hence if we prove then that any h defined by (3.2) satisfies the assumptions imposed on H, then (3.4) will yield (3.3). Clearly, h ∈ X_{+}. Next, since f_{n} ∈ X, we have h_{n} ∈ D(A) ⊂ D(B) and Bh_{n} = f_{n+1} − g. Thus sup_{n} Bh_{n} = f − g. On the other hand, we have f = sup_{n} f_{n} and h = sup_{n} h_{n} = sup_{n}(I − A)^{−1}f_{n} = Lf ∈ X, thus f ∈ F = X^{−1} and from the assumption on m it follows that f is bounded almost everywhere. This implies that h ∈ G and Bh = f − g. Therefore −mh + Bh = h − f + f − g = h − g ∈ X. Hence, if (3.4) is satisfied for any H stipulated in the corollary, then (3.3) is satisfied for any h defined by (3.2) and the proof is complete.

Remark 3.1 The theory sketched in the previous two section has been successfully applied to a number of problems ranging from the linear Boltzmann equation with external field in the context of both standard [1, 2] and extended [3] kinetic theory, Kolmogoroff’s and cell growth equations [2] and kinetic-diffusion equations arising in the theory of hydrodynamic limits in the extended kinetic theory [4]. In the next section we shall apply it to the fragmentation equation.

4 Fragmentation model

An equation of a simple fragmentation model (polymer degradation) [14] can be written as:

∂_{t}u = K_{α}u = A_{α}u + B_{α}u,  (4.1)

where, for x > 0,

A_{α}u = −x^{α}u(x),

and

(B_{α}u)(x) = 2 ∫_{x}^{∞}y^{α−1}u(y)dy.

Here u is the number density of particles with mass x. The parameter α is an arbitrary real number. The problem splits into three distinct cases depending on whether α = 0, α < 0, α > 0. When α = 0, then the operators K_{α} and B_{α} are bounded and the problem is easy; one can obtain even the closed form solution to (4.1), [7]. Therefore in the analysis below we shall always assume that α ≠ 0.

The basic space in our considerations will be the space

X_{1} = L_{1}([0, ∞[; xdx),

with the natural norm

∥f∥ = ∫_{0}^{∞}x|f(x)|dx.

The natural domain for K_{α} is then

D(K_{α}) = D(A_{α}) = D_{α} = {f ∈ X_{1}; x^{α}f ∈ X_{1}}.

The reason for the introduction of the space X_{1} is that for 0 ≤ u ∈ D_{α}

∫_{0}^{∞}(B_{α}u)(x)xdx = 2 ∫_{0}^{∞}x (∫_{x}^{∞}u(y)y^{α−1}dy) dx = 2 ∫_{0}^{∞}y^{α−1}u(y) (∫_{0}^{y} xdx) dy = ∫_{0}^{∞}(K_{α}u)(x)xdx.

This equation in the present context expresses the mass conservation law.

We see immediately that we can apply Theorem 2.1 to get the following result.

Theorem 4.1 Let α ∈ ℝ be arbitrary. There exists the smallest substochastic semigroup (G_{α}(t))_{t≥0} in X_{1} generated by an extension T_{α} of K_{α}.

Below we shall prove that the generator of (G_{α}(t))_{t≥0} coincides with the closure of K_{α} if α > 0 and identify the generator (which turns out to be a proper extension of \overline{K_{α}}) for α < 0.

Since the operator A_{α} is the multiplication by −x^{α} we can apply Corollary 3.2. Thus the space F coincides with the first associated space of the Sobolev tower, that is, explicitly

F = X^{−1} = L_{1}(ℝ_{+}, x(1 + x^{α})^{−1}dx).

The set G is the set of functions f ∈ X_{1} for which the function

F (x) = (B_{α}f)(x) = 2 ∫_{x}^{∞}y^{α−1}f(y)dy

is finite almost everywhere. Clearly, the only point where F (x) could be infinite is x = 0. Now Corollary 3.2 reads that if for any 0 ≤ H ∈ X_{1} such that −x^{α}H + B_{α}H ∈ X_{1} we have

∫_{0}^{∞}(−x^{α}H + B_{α}H)xdx ≥ 0,

then T_{α} = \overline{K_{α}}. Technical details relevant to the problem at hand are given below.

Proposition 4.1 If α > 0, then T_{α} = \overline{K_{α}}.

Proof. Since H ∈ X_{1}, −x^{α}H ∈ L_{1}([0, n], xdx) for any n and, since by assumption −x^{α}H + B_{α}H ∈ X_{1}, we have also B_{α}H ∈ L_{1}([0, n], xdx). Therefore

∫_{0}^{∞}(−x^{α}H(x) + (B_{α}H)(x))xdx = lim_{n→∞} ∫_{0}^{n}(−x^{α}H(x) + (B_{α}H)(x))xdx
= lim_{n→∞} (∫_{0}^{n}(−x^{α}H(x)xdx + ∫_{0}^{n}(B_{α}H)(x))xdx) = lim_{n→∞} (∫_{0}^{n}(−x^{α}H(x)xdx
+ 2 ∫_{n}^{∞}(∫_{0}^{n} xdx) y^{α−1}H(y)dy + 2 ∫_{0}^{n} (∫_{0}^{y} xdx) y^{α−1}H(y)dy)
= lim_{n→∞} n^{2} ∫_{n}^{∞}y^{α−1}H(y)dy.

Since the limit is finite by the definition of H, it must be nonnegative and the statement is proved.

If we look now at the case α < 0, then a similar argument shows that

∫_{0}^{∞}(−x^{α}H(x) + (B_{α}H)(x))xdx = lim_{ϵ→0} ∫_{ϵ}^{∞}(−x^{α}H(x) + (B_{α}H)(x))xdx

= lim_{ϵ→0} (∫_{ϵ}^{∞}(−x^{α}H(x)xdx + ∫_{ϵ}^{∞}(B_{α}H)(x)xdx)
= lim_{ϵ→∞} (∫_{ϵ}^{∞}(−x^{α}H(x)xdx + 2 ∫_{ϵ}^{∞}(∫_{ϵ}^{y} xdx) y^{α−1}H(y)dy)
= − lim_{ϵ→0} ϵ^{2} ∫_{ϵ}^{∞}y^{α−1}H(y)dy.

Here, unfortunately, the limit can be either zero or negative, which is inconclusive. Although it is possible to find a function H satisfying the assumptions of Corollary 3.2, and for which the above limit is strictly negative (H should behave as 1 close to zero), we don’t know whether such a function belong to D(T_{α}).

Remark 4.1 This short section shows some advantages as well as limitations of the method presented earlier in this paper. Apart from the last inconclusive result, we don’t know whether Proposition 4.1 gives the best answer – in principle it is possible that T_{α} = K_{α}. Fortunately, for the model discussed in this section it is possible to find explicitly the resolvent of the generator and study it directly to find answers to all the relevant questions. As we shall see in the next section, the results obtained here are the best possible: for α > 0 indeed T_{α} = \overline{K_{α}} ≠ K_{α} and for α < 0 the generator T_{α} is a proper extension of \overline{K_{α}}.

5 Direct study of the fragmentation model

For reasons which will become clear later (see also Remark 2.1) we shall consider a more general operator

K_{α,r}u = −A_{α}u + rB_{α}u.

for 0 < r ≤ 1. We adopt the convention that K_{α,1} = K_{α}. To find the resolvent of K_{α,r}, we start by solving the equation

λu + x^{α}u − 2r ∫_{x}^{∞}y^{α−1}u(y)dy = f.

Proceeding at this moment purely formally, we differentiate the above equation with respect to x to get

(λ + x^{α})u′ − (α + 2r)x^{α−1}u = f′,

and, after standard manipulations, we obtain

u = \frac{f}{λ + x^{α}} + \frac{1}{(λ + x^{α})^{1+2r/α}} (C − 2r ∫_{0}^{x} f(s)(λ + s^{α})^{−1+2r/α}s^{α−1}ds) .  (5.1)

The multiplier (λ + x^{α})^{−1} is bounded for any α. Precisely, we have for α > 0

(λ + x^{α})^{−1} = \{ \begin{matrix}O(x^{−α}) & as & x → +∞, \\ O(1) & as & x → 0^{+}, \end{matrix}  (5.2)

and for α < 0

(λ + x^{α})^{−1} = \{ \begin{matrix}O(x^{−α}) & as & x → 0^{+}, \\ O(1) & as & x → +∞. \end{matrix}  (5.3)

Let us also write down the asymptotics of the kernel of the integral operator. Since we know that xf is integrable, we shall rather investigate (λ + x^{α})^{−1+2r/α}x^{α−2}. We have for α > 0

(λ + x^{α})^{−1+2r/α}x^{α−2} = \{ \begin{matrix}O(x^{2(r−1)}) & as & x → +∞, \\ O(x^{α−2}) & as & x → 0^{+}, \end{matrix}  (5.4)

and for α < 0

(λ + x^{α})^{−1+2/α}x^{α−2} = \{ \begin{matrix}O(x^{α−2}) & as & x → +∞, \\ O(x^{2(r−1)}) & as & x → 0^{+}. \end{matrix}  (5.5)

Hence, the first term in (5.1) doesn’t create any problems. Let us consider first the case with α > 0. By (5.2), the multiplier (λ + x^{α})^{−1−2r/α} is O(x^{−α−2r}) as x → ∞, and bounded as x → 0^{+}. Since u should be integrable with the weight x^{1+α}, we see that it is necessary to require that

C − 2r ∫_{0}^{x} f(s)(λ + s^{α})^{−1+2r/α}s^{α−1}ds → 0 as x → ∞,

hence C = 2 ∫_{0}^{∞}f(s)(λ + s^{α})^{−1+2r/α}s^{α−1}ds and the solution takes the form

u = \frac{f}{λ + x^{α}} + \frac{2r}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}f(s)(λ + s^{α})^{−1+2r/α}s^{α−1}ds.  (5.6)

Let us suppose now that α < 0. The multiplier (λ + x^{α})^{−1−2r/α} is O(x^{−α−2r}) as x → 0^{+}, and bounded away from zero as x → +∞. We have to have u ∈ X_{1}, thus as above the only way to ensure this is to make the bracket in (5.1) vanish at infinity, which produces the same solution given by (5.6).

Thus we introduce the operators

R_{α,r}(λ)f = R_{1}(λ)f + R_{2}(λ)f = \frac{f}{λ + x^{α}} + \frac{2r}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}f(s)(λ + s^{α})^{−1+2r/α}s^{α−1}ds,  (5.7)

and, as above, we adopt the convention that R_{α,1} = R_{α}. In all the considerations below we assume that λ > 0.

Lemma 5.1 The operators R_{α,r}(λ) extend to bounded operators on X_{1}.

Proof. We need to focus only on R_{2}:

∥R_{2}(λ)f∥ ≤ ∫_{0}^{∞} \frac{x}{(λ + x^{α})^{1+2r/α}} |∫_{x}^{∞}f(s)(λ + s^{α})^{−1+2r/α}s^{α−1}ds| dx
≤ ∫_{0}^{∞}(∫_{0}^{s} \frac{x}{(λ + x^{α})^{1+2r/α}} dx) |sf(s)|(λ + s^{α})^{−1+2r/α}s^{α−2}ds.

We shall analyse the behaviour of the inner integral for cases α > 0 and α < 0 separately. Let first α > 0. Then the inner integrand behaves as x close to zero and as x^{−α−2r+1} as x → ∞. Thus for small s the inner integral is of order of s^{2}, and since by (5.4) the kernel (λ + s^{α})^{−1+2r/α}s^{α−2} behaves there as s^{α−2}, we obtain the integrability at zero. For large s the inner integral is of order of s^{−α−2r+2} and since the kernel behaves as s^{2r−2}, we obtain that the integral is finite.

Let us look now at α < 0. The inner integrand is of order of x^{−α−2r+1} close to zero, thus the integral is of order of s^{−α−2r+2}. Since by (5.5) the kernel is of order of s^{2r−2} at zero, the product is of order of s^{−α} and bounded, as α < 0; thus we obtain the integrability at zero. Next, for large x the inner integrand behaves as x, thus the integral increases as s^{2}, but the kernel is of order of s^{α−2}, hence again the product is bounded and the integral is finite.

Lemma 5.2 For each 0 < r ≤ 1 and all α, the operator R_{α,r}(λ) is a left-inverse of λ − K_{α,r}.

Proof. We have to prove that for any u ∈ D_{α} we have R_{α,r}(λ)(λ + A_{α} − rB_{a})u = u. Clearly,

R_{1}(λ)(λ + A_{α})u = u.

Next,

rR_{1}(λ)(B_{α})u = \frac{2r}{λ + x^{α}} ∫_{x}^{∞}s^{α−1}u(s)ds,

moreover

R_{2}(λ)(λ + A_{α})u = \frac{2r}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}s^{α−1}(λ + s^{α})^{2r/α}u(s)ds,

and

−rR_{2}(λ)B_{α}u = \frac{−4r^{2}}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}s^{α−1}(λ + s^{α})^{−1+2r/α} (∫_{s}^{∞}y^{α−1}u(y)dy) ds
= \frac{−4r^{2}}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}y^{α−1}u(y) (∫_{x}^{y} s^{α−1}(λ + s^{α})^{−1+2r/α}ds) dy
= \frac{−2r}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}y^{α−1}u(y) ((λ + y^{α})^{2r/α} − (λ + x^{α})^{2r/α}) dy
= \frac{−2r}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}y^{α−1}(λ + y^{α})^{2r/α}u(y)dy + \frac{2r}{λ + x^{α}} ∫_{x}^{∞}y^{α−1}u(y)dy
= −R_{2}(λ)(λ + A_{α})u + R_{1}(λ)(rB_{α})u

Thus

R_{α,r}(λ)(λ + A_{α} − rB_{α})u = R_{1}(λ)(λ + A_{α})u − R_{1}(λ)(rB_{α})u + R_{2}(λ)(λ + A_{α})u − R_{2}(λ)(rB_{α})u
= u − rR_{1}(λ)B_{α}u + R_{2}(λ)(λ + A_{α})u − (−rR_{1}(λ)B_{α}u + R_{2}(λ)(λ + A_{α})u
= u

and the lemma is proved.

Lemma 5.3 If 0 < r < 1 and all α, then

R_{α,r}(λ)X_{1} = D_{α}.  (5.8)

If r = 1 and α ≠ 0, then

R_{α}(λ)X_{1} ≠ D_{α}  (5.9)

Proof. It is clear that for any f ∈ X_{1}, α and 0 < r ≤ 1 we have R_{1}(λ)f ∈ X_{1} so that we have to consider only R_{2}(λ). Let us take f ∈ X_{1} and r < 1. We obtain

∥x^{α}R_{2}(λ)f∥_{X_{1}} ≤ ∫_{0}^{∞} \frac{x^{α+1}}{(λ + x^{α})^{1+2r/α}} ∫_{x}^{∞}(s|f(s)|)(λ + s^{α})^{−1+2r/α}s^{α−2}dsdx
= ∫_{0}^{∞}(∫_{0}^{s} \frac{x^{1+α}}{(λ + x^{α})^{1+2r/α}} dx) (s|f(s)|)(λ + s^{α})^{−1+2r/α}s^{α−2}ds.  (5.10)

Let us first consider the case with α > 0. By (5.2) the inner integrand is of order of x^{1+α} for small x and behaves as x^{1−2r} for large x, thus the integral is of order of s^{2+α} for small s and of order of s^{2−2r} (2−2r > 0) for large s. Since by (5.4) we know that (λ + s^{α})^{−1+2r/α}s^{α−2} is O(s^{α−2}) for small s and O(s^{2−2r}) for large s, we see that the integral converges; consequently R_{2}(λ)f ∈ D_{α}.

On the other hand, if α < 0, then by (5.3) the inner integral is O(s^{2−2r}) (as 1 − 2r ≠ −1) for small s, and O(s^{2+α}) for large s so that the multiplier of s|f(s)| is bounded on ℝ_{+}, and the integral converges. Hence (5.8) is proved.

Let us consider now the case r = 1 and α > 0. Suppose 0 ≤ f ∈ X_{1}; then we obtain as above

∥x^{α}R_{2}(λ)f∥_{X_{1}} = ∫_{0}^{∞} \frac{x^{α+1}}{(λ + x^{α})^{1+2/α}} ∫_{x}^{∞}(sf(s))(λ + s^{α})^{−1+2/α}s^{α−2}dsdx
= ∫_{0}^{∞}(∫_{0}^{s} \frac{x^{1+α}}{(λ + x^{α})^{1+2/α}} dx) (sf(s))(λ + s^{α})^{−1+2/α}s^{α−2}ds.  (5.11)

It is easy to see that for large s the inner integral behaves as ln s. By (5.4) we know that (λ+s^{α})^{−1+2/α}s^{α−2} is bounded away from zero for large s, there exist functions in X_{1} for which the above integral is divergent (for example one can take sf(s) = 1/s ln^{2} s). Since all the functions are nonnegative, the original integral is also divergent by Tonelli’s theorem. It is also clear that if f has the support in [0, n], n < ∞, then the above integral is convergent. Therefore there exists a dense subspace X̂_{1} such that R_{α}(λ)X̂_{1} ⊂ D_{α}.

Let as consider now the case with r = 1 and α < 0. Then

∥x^{α}R_{2}(λ)f∥_{X_{1}} = ∫_{0}^{∞} \frac{x^{α+1}}{(λ + x^{α})^{1+2/α}} ∫_{x}^{∞}(sf(s))(λ + s^{α})^{−1+2/α}s^{α−2}dsdx,  (5.12)

and we see that the term \frac{x^{α+1}}{(λ+x^{α})^{1+2/α}} behaves as x^{−1} at zero (and as x^{α+1} at infinity) and therefore the above integral is divergent unless at least ∫_{0}^{∞}(sf(s))(λ + s^{α})^{−1+2/α}s^{α−2}ds = 0. (In the Appendix we will see another proof of this fact based on the Fredholm alternative argument.)

Lemma 5.4 For arbitrary, 0 < r ≤ 1, α ∈ ℝ, if for some f ∈ X_{1} we have R_{α,r}(λ)f ∈ D_{α}, then (λ − K_{α,r})R_{α,r}(λ)f = f. Therefore

(a) R_{α,r}(λ) is the resolvent of (K_{α,r}, D_{α}) for 0 < r < 1 and arbitrary α,

(b) if α > 0, then the range of (λ − K_{α}, D_{α}) is dense in X_{1},

(c) if α < 0, then the range of (λ − K_{α}, D_{α}) is a dense subspace of a proper closed subspace of X_{1}.

Proof. Since we know that for any f ∈ X_{1}, R_{1}(λ)f ∈ D_{α}, we see that we must have also R_{2}(λ)f ∈ D_{α}. Thus clearly

(λ + A_{α})R_{1}(λ)f = f

and

rB_{α}R_{1}(λ)f = 2r ∫_{x}^{∞} \frac{y^{α−1}}{λ + y^{α}} f(y)dy.

Moreover

(λ + A_{α})R_{2}(λ)f = \frac{2r}{(λ + x^{α})^{2r/α}} ∫_{x}^{∞}y^{α−1}(λ + y^{α})^{−1+2r/α}f(y)dy

and finally

rB_{α}R_{2}(λ)f = 4r^{2} ∫_{x}^{∞} \frac{y^{α+1}}{(λ + x^{α})^{1+2r/α}} ∫_{y}^{∞}s^{α−1}(λ + s^{α})^{−1+2r/α}f(s)dsdy
= 4r^{2} ∫_{x}^{∞}(∫_{x}^{s} \frac{y^{α+1}}{(λ + x^{α})^{1+2/α}} dy) s^{α−1}(λ + s^{α})^{−1+2r/α}f(s)ds
= −2r ∫_{x}^{∞}((λ + s)^{−2r/α} − (λ + x)^{−2r/α}) s^{α−1}(λ + s^{α})^{−1+2r/α}f(s)ds
= −rB_{α}R_{1}(λ)f + (λ + A_{α})R_{2}(λ)f

which proves the first statement.

From Lemma 5.3 we know that for 0 < r < 1 we have R_{α,r}(λ)X_{1} ⊂ D_{α}, thus R_{α,r}(λ) is right and left inverse to λ − A_{α,r} and thus it is the resolvent of A_{α,r}.

Let r = 1 and α > 0. In Lemma 5.3 we showed that there exists a dense subspace X̂_{1} for which we have R_{α}(λ)X̂_{1} ⊂ D(K_{α}). The first part of this lemma gives (λ − K_{α})R_{α}(λ)X̂_{1} = X̂_{1} ⊂ Rg(λ − K_{α}) (the range of (λ − K_{α}, D_{α})), and consequently the range of λ − K_{α} is dense in X_{1} (and not equal to X_{1}).

Finally, let us consider the case with α < 0 and r = 1. Suppose f ∈ Rg(λ − K_{α}), then f = (λ − K_{α})g for some g ∈ D_{α}. Lemma 5.2 implies then that R_{α}(λ)f ∈ D_{α}, but by Lemma 5.3 we must have ∫_{0}^{∞}(sf(s))(λ + s^{α})^{−1+2/α}s^{α−2}ds = 0. In fact, since the function (sf(s))(λ + s^{α})^{−1+2/α}s^{α−2} is integrable on

[0, ∞[, the function F (x) = ∫_{x}^{∞}(sf(s))(λ + s^{α})^{−1+2/α}s^{α−2}ds is continuous and therefore it has a limit at zero, F (0). If this limit is nonzero, then by continuity, F (x) is of constant sign at some interval [0, δ], and since x^{α+1}(λ + x^{α})^{−1−2/α} is positive and of order of x^{−1} close to zero we obtain that the integral

∫_{0}^{∞} \frac{x^{α+1}}{(λ + x^{α})^{1+2/α}} F (x)dx  (5.13)

diverges. Thus F (0) = 0 or, in other words, for f to belong to Rg(λ − K_{α}), it is necessary that ∫_{0}^{∞}(sf(s))(λ + s^{α})^{−1+2/α}s^{α−2}ds = 0. Hence, f must belong to the subspace of codimension 1 which is annihilated by the functional generated by the function (λ + s^{α})^{−1+2/α}s^{α−1}; this subspace will be denoted by X̃_{1}. (Note that for α < 0 this function belongs to X_{1}^{∗} = {f ∈ L_{∞}([0, ∞[); x^{−1}f ∈ L_{∞}([0, ∞[)}. However, X̃_{1} ≠ Rg(λ−K_{α}). This can be established by noting that if F (x) behaves as 1/ ln x at zero, then the integral in (5.13) diverges as ∫_{0}^{0.5} \frac{dx}{x ln x} = −∞. On the other hand we can prove that X̃_{1} = \overline{Rg(λ − K_{α})}. Firstly, we observe that if f ∈ X̃_{1} is continuous at x = 0, then f ∈ Rg(λ − K_{α}). In fact, then F′ is continuous at zero and the claim follows from

∫_{0}^{δ} \frac{1}{x}F (x)dx = ∫_{0}^{δ} \frac{F (x) − F (0)}{x}dx < +∞

for sufficiently small δ > 0. Having this in mind, for a given f ∈ X̃_{1} we construct the sequence

f_{n}(x) = \{ \begin{matrix}f(x) & for & x > n^{−1}, \\ −2n(λ + x^{α})^{1−2/α}x^{−α+1} ∫_{n^{−1}}^{∞} f(s)(λ + s^{α})^{−1+2/α}s^{α−1}ds & for & (2n)^{−1} ≤ x ≤ n^{−1}, \\ 0 & for & 0 ≤ x < (2n)^{−1} \end{matrix}

We see that

∫_{0}^{∞}f_{n}(x)(λ + x^{α})^{−1+2/α}x^{α−1}dx = ∫_{n^{−1}}^{∞} f(x)(λ + x^{α})^{−1+2/α}x^{α−1}

−2n ∫_{(2n)^{−1}}^{n^{−1}} ( ∫_{n^{−1}}^{∞} f(s)(λ + s^{α})^{−1+2/α}s^{α−1}ds) dx = 0.

Moreover, since f_{n}(x) = 0 for 0 ≤ x ≤ (2n)^{−1}, f_{n} ∈ Rg(λ − A_{α}). We shall prove that f_{n} → f in X_{1} as n → ∞. We have

∥f − f_{n}∥_{X_{1}} ≤ ∫_{0}^{n^{−1}}x|f(x)|dx + 2n ( ∫_{(2n)^{−1}}^{n^{−1}} x(λ + x^{α})^{1−2/α}x^{−α+1}dx) ( ∫_{n^{−1}}^{∞} f(s)(λ + s^{α})^{−1+2/α}s^{α−1}ds) .

The first term tends to zero since f ∈ X_{1}. In the second term we see that the integrand x(λ+x^{α})^{1−2/α}x^{−α+1} is bounded close to zero, thus the product tends to zero due to the behaviour of ∫_{n^{−1}}^{∞} f(s)(λ+s^{α})^{−1+2/α}s^{α−1}ds. Hence, \overline{Rg(λ − A_{α})} = X̃_{1}. This ends the proof of the lemma.

Corollary 5.1 Let α ≠ 0. The operator K_{α} is closeable but not closed. If α > 0, then the extension T_{α} of K_{α} which generates the semigroup of Theorem 4.1 is given by

T_{α} = \overline{K_{α}}.

If α < 0, then T_{α} is a proper extension of \overline{K_{α}}.

Proof. It is clear that K_{α} is densely defined. By the comment before Corollary 2.1 we know that K_{α} is dissipative for any α, thus from [5], Proposition II.3.14 (iii-iv) and the lemma above we infer that K_{α} is closeable, but not closed. Hence for neither α ≠ 0, K_{α} is the generator of a semigroup.

Assume now that α > 0. Since we have proved that Rg(λ − K_{α}) is a proper dense subspace of X_{1}, using Corollary 2.1 we obtain that T_{α} = \overline{A_{α} + B_{α}} is the generator.

Consider next the case α < 0. Using again Proposition II.3.14 (iv) of op. cit we see that \overline{Rg(λ − K_{α})} = Rg(λ − \overline{K_{α}}). By the lemma above \overline{Rg(λ − K_{α})} ≠ X_{1}, hence in this case \overline{K_{α}} cannot be m-dissipative and therefore the generator T_{α} must be a proper extension of \overline{K_{α}}.

In the last step we shall identify T_{α} in the case when α < 0.

Proposition 5.1 Let α < 0. Then

T_{α} = λ − R_{α}(λ).  (5.14)

Proof. It is enough to prove that R_{α}(λ) is the resolvent of the generator of the semigroup (G_{α}(t))_{t≥0}. Let us observe that by Remark 2.1 (or [13]), the semigroup (G_{α}(t))_{t≥0} can be obtained as

G_{α}(t)f = lim_{r→1^{−}} G_{α,r}(t)f, f ∈ X_{1},

where (G_{α,r}(t))_{t≥0} are semigroups generated by K_{α,r}. Moreover, for r ≤ r′ we have G_{α,r}(t) ≤ G_{α,r′}(t) for any t ≥ 0. This implies immediately that for f ∈ X_{1,+}

R_{α,r′}(λ)f = ∫_{0}^{∞}e^{−λt}G_{α,r′}(t)fdt ≥ ∫_{0}^{∞}e^{−λt}G_{α,r}(t)fdt = R_{α,r}(λ)f,

so that by (5.7) the resolvents R_{α,r}(λ)f monotonically converge for all f to R_{α}(λ)f. Using again the monotonicity we obtain

∥R_{α,r′}(λ)f − R_{α}(λ)f∥_{X_{1}} = ∫_{0}^{∞}((R_{α}(λ)f)(x) − (R_{α,r′}(λ)f)(x)) xdx,

and the right-hand side tends to zero by the Lebesgue monotone convergence theorem. Thus, for any f ∈ X_{1,+}, R_{α,r′}(λ)f → R_{α}(λ)f in X_{1} norm. On the other hand, it is clear that the resolvents R_{α,r}(λ)f converge to the resolvent R(λ) of (G_{α}(t))_{t≥0} in X_{1}. Thus, R(λ)f = R_{α}(λ)f for all positive f and by linearity this extends to the whole of X_{1}.

In the last step we shall prove that the semigroup (G_{α}(t))_{t≥0}, α < 0 is not a stochastic semigroup. This result could also be proved modifying argument of [6] but in our opinion the direct proof offered here has some instructive values.

Proposition 5.2 Let α < 0. There exists f ∈ X_{1,+} such that

∥G_{α}(t)f∥ ≠ ∥f∥  (5.15)

for some t > 0. Therefore, (G_{α}(t))_{t≥0} is not a stochastic semigroup.

Proof. If (5.15) were valid for all t ≥ 0, then using the additivity of the L_{1} norm we would have for any f ∈ X_{1,+}

∥R_{α}(1)f∥ = ∫_{0}^{∞}e^{−t}∥G_{α}(t)f∥dt = ∥f∥.  (5.16)

On the other hand, we have

∥R_{α}(1)f∥ = ∫_{0}^{∞} \frac{xf(x)dx}{1 + x^{α}} + 2 ∫_{0}^{∞}(\frac{x}{(1 + x^{α})^{1+2/α}} ∫_{x}^{∞}f(s)(1 + s^{α})^{−1+2/α}s^{α−1}ds) dx.  (5.17)

Evaluating the second integral we obtain

2 ∫_{0}^{∞}(\frac{x}{(1 + x^{α})^{1+2/α}} ∫_{x}^{∞}f(s)(1 + s^{α})^{−1+2/α}s^{α−1}ds) dx

= 2 ∫_{0}^{∞}(f(s)(1 + s^{α})^{−1+2/α}s^{α−1} ∫_{0}^{s} \frac{x}{(1 + x^{α})^{1+2/α}} dx) ds

= ∫_{0}^{∞}f(s)(1 + s^{α})^{−1+2/α}s^{α−1} (\frac{s^{2}}{(1 + s^{α})^{2/α}} − 1) ds

= ∫_{0}^{∞}f(s)(1 + s^{α})^{−1}s^{α+1}ds − ∫_{0}^{∞}f(s)(1 + s^{α})^{−1+2/α}s^{α−1}ds

Inserting the above into (5.17) and simplifying we obtain

∥R_{α}(1)f∥ = ∫_{0}^{∞}xf(x)dx − ∫_{0}^{∞}f(x)(1 + x^{α})^{−1+2/α}x^{α−1}dx

thus if f is not an element of the closure of the range of K_{α}, then ∥G_{α}(t)f∥ ≠ ∥f∥ for some t > 0.

A Appendix: the Fredholm alternative for λ − K_{α}

In author’s opinion it is instructive to look at the solvability of the equation

λu(x) + x^{α}u(x) − 2 ∫_{x}^{∞}y^{α−1}u(y)dy = f(x),  (A.1)

using the Fredholm alternative type of argument. Let us recall that the adjoint space X_{1}^{∗} can be identified with M_{−1} = {g ∈ L_{∞}([0, ∞[); x^{−1}g ∈ L_{∞}([0, ∞[)} with the natural pairing

< f, g >= ∫_{0}^{∞}f(x)g(x)dx.

Assume that there exists a solution to (A.1) in D_{α}, so that each term at the left-hand side is in X_{1}. Multiplying (A.1) by g ∈ M_{−1} and integrating both sides we get

λ ∫_{0}^{∞}u(x)g(x)dx + ∫_{0}^{∞}x^{α}u(x)g(x)dx − 2 ∫_{0}^{∞}g(x) (∫_{x}^{∞}y^{α−1}u(y)dy) dx = ∫_{0}^{∞}f(x)g(x)dx,  (A.2)

It follows that if u ∈ D_{α}, then we can change the order of integration in the iterated integral getting

∫_{0}^{∞}(∫_{x}^{∞}y^{α−1}u(y)dy) dx = ∫_{0}^{∞}y^{α−1}u(y) (∫_{0}^{y} g(x)dx) dy,

which yields

∫_{0}^{∞}u(x) (λg(x) + x^{α}g(x) − 2x^{α−1} ∫_{0}^{x} g(y)dy) dx = ∫_{0}^{∞}f(x)g(x)dx.  (A.3)

Thus, if there exists a nontrivial solution g ∈ M_{−1} to

λg(x) + x^{α}g(x) − 2x^{α−1} ∫_{0}^{x} g(y)dy = 0,  (A.4)

then the necessary condition for the existence of a solution u ∈ D_{α} is that ∫_{0}^{∞}f(x)g(x)dx = 0. Note that if g solves (A.4), then it follows that g is continuous on ]0, ∞[ and consequently it is continuously differentiable there, so that all the solutions to (A.4) are among the solutions to the differential equation

(λx^{−α+1} + x)g′(x) + (λ(1 − α)x^{−α} − 1)g(x) = 0.  (A.5)

It can be found that the general solution to this equation is given by

g(x) = Cx^{α−1}(λ + x^{α})^{−1+2/α},  (A.6)

where C is an arbitrary constant, which agrees perfectly with (5.12). However, this solution is valid for all α ≠ 0 and since we haven’t encountered any problem for α > 0, the matter calls for a further analysis.

The explanation is very basic: g, as defined in (A.6) is the solution to (A.5) for all α, but to (A.4) only for α < 0. In fact, we have

∫_{0}^{x} g(x)dx = \{ \begin{matrix}−\frac{λ}{2} + \frac{1}{2}(λ + x^{α})^{2/α} & for & α > 0, \\ \frac{1}{2}(λ + x^{α})^{2/α} & for & α < 0, \end{matrix}

and the claim follows by direct substitution.

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