DIFFERENTIABILITY OF SOLUTIONS OF THE ABSTRACT CAUCHY PROBLEM

In this note we establish a result of differentiability for the mild solution of the inhomogeneous abstract Cauchy problem when the underlying space is reflexive. 2000 Mathematics Subject Classification : 47D09, 34G10.


Introduction.
In this work we are concerned with regularity properties of solutions of the first order abstract Cauchy problem (in short, ACP).We refer the reader to [3,10] for the theory of strongly continuous semigroup operators and the associated ACP.
Let X be a Banach space endowed with a norm • .Henceforth T (t) is a strongly continuous semigroup of operators on X with infinitesimal generator A.
The existence of solutions of the first order abstract Cauchy problem x (t) = Ax(t) + h(t), 0 ≤ t ≤ a, (1.1) x(0) = x 0 , (1.2) it has been treated in several works.We only mention here the texts [3,10] and the references cited therein.Similarly, the existence of solutions of the semilinear abstract Cauchy problem it has been discussed in [1,9].
Assuming that h : [0, a] → X is integrable the function given by is said mild solution of (1.1)-(1.2).In the case in which h is continuous , the function The existence of classical solutions of (1.1)-(1.2) as well as some weaker forms of differentiability of solutions have been studied in a number of works.We refer the reader to [2,8,10,12,13,14] and the references therein indicated.
The purpose of this note is to establish a new condition in order to the mild solution x(•) turn to be a classical solution .
Next the notation C([0, a]; X) stands for the space of continuous functions from [0, a] into X, whilst BV ([0, a]; X) represents the space of functions with bounded variation from [0, a] into X.For a function h ∈ BV ([0, a]; X) we denote by V (h) the variation of h on [0, a] and by v(t, h) the variation of h on [0, t], for 0 ≤ t ≤ a.Additional termi-nology and notations are those generally used in functional analysis.In particular, X * denotes the dual space of X.

Results.
In this section h denotes a continuous function of bounded variation on a fixed interval [0, a], a > 0. We define the translation of h by We introduce the following condition for a function h ∈ C([0, a]; X)∩ BV ([0, a]; X).
Example 2. Let h : [0, 1] → IR be the function defined in [4], Exercise 4.19.Let E be a perfect nowhere dense set with measure 0 included in

and let
∞ k=1 c k be a convergent series of positive number with sum equal to 1.For each x ∈ [0, 1] let It is clear that h(0) = 0, h(1) = 1.Moreover, h is continuous and nondecreasing with h = 0, a.e.Thus h is a singular function .Now we establish that h does not satisfy (H 0 ).In fact, from (2.1) it follows easily that for each t > 0 and 0 which implies that µ(t, h) does not converge to 0 as t → 0 + .Example 3. Let h : [0, 1] → IR be the singular function defined in [6], Example 18.8.As above, h(0) = 0, h(1) = 1, h is continuous and strictly increasing and h = 0, a.e.We will show that this function satisfies the assumption (H 0 ).Initially, for completeness we include here the construction carried out in [6].
Let (t n ) n be a sequence in (0, 1).Set F 1 (0) = 0, F 1 (1) = 1, and define F 1 to be linear on [0, 1  2 ] and [ and complete the definition of F n+1 as a continuous linear func- [6] that (F n ) n is a nondecreasing sequence.Thus this sequence converges to a function h which satisfies the properties already mentioned.Applying now the Dini's theorem ( [11]) we obtain that the convergence of (F n ) n is uniform.Since (T t F n − F n ) n converges uniformly to T t h − h, as n → ∞, and this convergence is also uniform on t ≥ 0, it is follows that µ(t, F n ) → µ(t, h), n → ∞, and this convergence is uniform on t.In view of F n is absolutely continuous , from Example 1 we infer that µ(t, F n ) → 0, as t → 0 + , which implies that µ(t, h) → 0, as t → 0 + .
To study the regularity of solutions of the abstract Cauchy problem (1.1)-(1.2) we begin by establishing some preliminary lemmas.
In the sequel we denote by M a positive constant such that T (t) ≤ M, 0 ≤ t ≤ a.Moreover, for a fixed h, we use the notation Assume that X is a reflexive space .Let T (•) be a strongly continuous semigroup of operators on X and let h : [0, a] → X be a continuous function of bounded variation which satisfies the assumption (H 0 ).Then the Riemann-Stieltjes integral exists in the weak topology and define a continuous function w : [0, a] → X.

Proof.
Let Λ : X * → I C be defined by The Riemann-Stieltjes integral in the above expression exists because T (•) * x * is a continuous function ( [10]) and h has bounded variation ( [7]).Moreover, Λ is linear and Consequently, Λ ∈ X * * and in view of that X is reflexive we infer the existence of w(t) ∈ X such that Λ(x * ) =< x * , w(t) >, for all x * ∈ X * .
On the other hand, for t < 1 and τ small enough, from the relations we deduce that Since µ(τ, h) → 0, τ → 0, because the condition (H 0 ) holds and v(τ, h) → 0, τ → 0, by the Proposition I.2.9 in [7]) the previous estimation shows that w(•) is right continuous at t. Similarly, one can prove that w is left continuous at t > 0.
Next we denote by χ E the characteristic function of a set E. where Then the function u given by (2.3) is piecewise smooth, u(t) ∈ D(A), Au(•) is continuous on [0, a] and u (t) = Au(t) + h(t), t / ∈ P, where P denotes the set formed by the extreme points of intervals

Proof.
Applying the linearity of u in terms of h, it is sufficient to prove the assertion for a function h = xχ I where I is an interval of type [t 1 , t 2 ].In fact, in this case, u(t) is given by From the properties of semigroups we infer that u(t) ∈ D(A) and that This shows that Au( Proof. We consider a sequence (h n ) n of step functions , where each h n is given by In this expression we have denoted where the points t i have been chosen as It is clear that the sequence (h n ) n converge uniformly to h.Let u n be the function given by (2.3), with h n instead of h.Then, u n → u, n → ∞, uniformly on [0, a].Moreover, by Lemma 2.2 we have that u n (t) ∈ D(A) and, if we fix 0 ≤ t ≤ a and n ∈ IN , then t ∈ I k , for some k = 1, • • • , n. From our definitions we can write This shows that (Au n (t)) n is a bounded sequence.Consequently, there is a subsequence which converges to z(t) ∈ X in the weak topology.Moreover, from (2.4)   A similar result holds for the second order abstract Cauchy problem ( [5]).
This shows that u(•) is a function of class C 1 that satisfies (1.1)-(1.2).
•) is continuous .Moreover, it is immediate to verify that u (t) = Au(t) + h(t), t = t 1 , t 2 .Now we can prove the main result of this note.
it follows that z(t) = w(t) − h(t) + T (t)h(0).An standard argument shows that the full sequence (Au n (t)) n converges to w(t).As A is a closed operator this implies that u(t) ∈ D(A) and z(t) = Au(t).An application of Lemma 2.1 yields that Au(•) is a continuous function .On the other hand, from Lemma 2.2 we haveu n (t) = Au n (t) + h n (t), n ∈ IN, t = i/n, i = 1, • • • , n − 1,so that for each x * ∈ X * we obtain < x * , u n (t) >= , Au n (s) + h n (s) > ds and taking limit as n → ∞, it follows that < x * , u(t) >= *