Abstract

Given the abstract evolution equation ,   with scalar type spectral operator in a complex Banach space, found are conditions necessary and sufficient for all weak solutions of the equation, which a priori need not be strongly differentiable, to be strongly infinite differentiable on . The important case of the equation with a normal operator in a complex Hilbert space is obtained immediately as a particular case. Also, proved is the following inherent smoothness improvement effect explaining why the case of the strong finite differentiability of the weak solutions is superfluous: if every weak solution of the equation is strongly differentiable at , then all of them are strongly infinite differentiable on .

“Curiosity is the lust of the mind.”
Thomas Hobbes

1. Introduction

We find conditions on a scalar type spectral operator in a complex Banach space necessary and sufficient for all weak solutions of the evolution equationwhich a priori need not be strongly differentiable, to be strongly infinite differentiable on . The important case of the equation with a normal operator in a complex Hilbert space is obtained immediately as a particular case. We also prove the following inherent smoothness improvement effect explaining why the case of the strong finite differentiability of the weak solutions is superfluous: if every weak solution of the equation is strongly differentiable at 0, then all of them are strongly infinite differentiable on .

The found results develop those of paper [1], where similar consideration is given to the strong differentiability of the weak solutions of the equationon and .

Definition 1 (weak solution). Let be a densely defined closed linear operator in a Banach space and be an interval of the real axis . A strongly continuous vector function is called a weak solution of the evolution equationif, for any , where is the domain of an operator, is the operator adjoint to , and is the pairing between the space and its dual (cf. [2]).

Remarks 2. (i)Due to the closedness of , a weak solution of (3) can be equivalently defined to be a strongly continuous vector function such that, for all ,  where is an arbitrary fixed point of the interval , and is also called a mild solution (cf. [3, Ch. II, Definition 6.3], see also [4, Preliminaries]).(ii)Such a notion of weak solution, which need not be differentiable in the strong sense, generalizes that of classical one, strongly differentiable on and satisfying the equation in the traditional plug-in sense, the classical solutions being precisely the weak ones strongly differentiable on .(iii)As is easily seen is a weak solution of (1) iff is a weak solution of (2) and is a weak solution of the equation(iv)When a closed densely defined linear operator in a complex Banach space generates a strongly continuous group of bounded linear operators (see, e.g., [3, 5]), i.e., the associated abstract Cauchy problem (ACP) is well-posed (cf. [3, Ch. II, Definition 6.8]), the weak solutions of (1) are the orbits with (cf. [3, Ch. II, Proposition 6.4], see also [2, Theorem]), whereas the classical ones are those with (see, e.g., [3, Ch. II, Proposition 6.3]).(v)In our discourse, the associated ACP may be ill-posed, i.e., the scalar type spectral operator need not generate a strongly continuous group of bounded linear operators (cf. [6]).

2. Preliminaries

Here, for the reader’s convenience, we outline certain essential preliminaries.

Henceforth, unless specified otherwise, is supposed to be a scalar type spectral operator in a complex Banach space with strongly -additive spectral measure (the resolution of the identity) assigning to each Borel set of the complex plane a projection operator on and having the operator’s spectrum as its support [7, 8].

Observe that, in a complex finite-dimensional space, the scalar type spectral operators are all linear operators on the space, for which there is an eigenbasis (see, e.g., [7, 8]) and, in a complex Hilbert space, the scalar type spectral operators are precisely all those that are similar to the normal ones [9].

Associated with a scalar type spectral operator in a complex Banach space is the Borel operational calculus analogous to that for a normal operator in a complex Hilbert space [7, 8, 10, 11], which assigns to any Borel measurable function a scalar type spectral operator (see [7, 8]).

In particular,( is the set of nonnegative integers, , is the identity operator on ), andThe properties of the spectral measure and operational calculus, exhaustively delineated in [7, 8], underlie the entire subsequent discourse. Here, we underline a few facts of particular importance.

Due to its strong countable additivity, the spectral measure is bounded [8, 12], i.e., there is such an that, for any Borel set ,Observe that the notation is used here to designate the norm in the space of all bounded linear operators on . We adhere to this rather conventional economy of symbols in what follows also adopting the same notation for the norm in the dual space .

For any and , the total variation measure   of the complex-valued Borel measure is a finite positive Borel measure with(see, e.g., [13, 14]).

Also (Ibid.), for a Borel measurable function , , , and a Borel set ,In particular, for , andObserve that the constant in (15)–(17) is from (14).

Further, for a Borel measurable function , a Borel set , a sequence of pairwise disjoint Borel sets in , and , ,Indeed, since, for any Borel sets , [7, 8], for the total variation measure, Whence, due to the nonnegativity of (see, e.g., [15]),The following statement, allowing characterizing the domains of Borel measurable functions of a scalar type spectral operator in terms of positive Borel measures, is fundamental for our discourse.

Proposition 3 ([16, Proposition 3.1]). Let be a scalar type spectral operator in a complex Banach space with spectral measure and be a Borel measurable function. Then iff (i)for each , ;(ii) ,  , where is the total variation measure of .

The succeeding key theorem provides a description of the weak solutions of (2) with a scalar type spectral operator in a complex Banach space.

Theorem 4 ([16, Theorem 4.2] with ). Let be a scalar type spectral operator in a complex Banach space . A vector function is a weak solution of (2) iff there is an such that the operator exponentials understood in the sense of the Borel operational calculus (see (13)).

Remark 5. Theorem 4 generalizes [17, Theorem 3.1], its counterpart for a normal operator in a complex Hilbert space.

We also need the following characterizations of a particular weak solution’s of (2) with a scalar type spectral operator in a complex Banach space being strongly differentiable on a subinterval of .

Proposition 6 ([1, Proposition 3.1] with ). Let and be a subinterval of . A weak solution of (2) is times strongly differentiable on iff in which case

Subsequently, the frequent terms “spectral measure” and “operational calculus” are abbreviated to s.m. and o.c., respectively.

3. General Weak Solution

Theorem 7 (general weak solution). Let be a scalar type spectral operator in a complex Banach space . A vector function is a weak solution of (1) iff there is an such thatthe operator exponentials understood in the sense of the Borel operational calculus (see (13)).

Proof. As is noted in the Introduction, is a weak solution of (1) iffis a weak solution of (2) andis a weak solution of (8).
Applying Theorem 4, to and , we infer that this is equivalent to the fact

Remarks 8. (i)More generally, Theorem 4 and its proof can be easily modified to describe in the same manner all weak solution of (3) for an arbitrary interval of the real axis .(ii)Theorem 7 implies, in particular,(a)that the subspace of all possible initial values of the weak solutions of (1) is the largest permissible for the exponential form given by (25), which highlights the naturalness of the notion of weak solution;(b)that associated ACP (9), whenever solvable, is solvable uniquely.(iii)Observe that the initial-value subspace of (1), containing the dense in subspace , where  which coincides with the class of entire vectors of of exponential type [18], is dense in as well.(iv)When a scalar type spectral operator in a complex Banach space generates a strongly continuous group of bounded linear operators, [6], and hence, Theorem 7 is consistent with the well-known description of the weak solutions for this setup (see (10)).(v)Clearly, the initial-value subspace of (1) is narrower than the initial-value subspace of (2) and the initial-value subspace of (8); in fact it is the intersection of the latter two.

4. Differentiability of a Particular Weak Solution

Here, we characterize a particular weak solution’s of (1) with a scalar type spectral operator in a complex Banach space being strongly differentiable on a subinterval of

Proposition 9 (differentiability of a particular weak solution). Let and be a subinterval of . A weak solution of (1) is times strongly differentiable on iffin which case,

Proof. The statement immediately follows from the prior theorem and Proposition 6 applied to for an arbitrary weak solution of (1).

Remark 10. Observe that, as well as for Proposition 6, for , the subinterval can degenerate into a singleton.

Inductively, we immediately obtain the following analog of [1, Corollary 3.2].

Corollary 11 (infinite differentiability of a particular weak solution). Let be a scalar type spectral operator in a complex Banach space and be a subinterval of . A weak solution of (1) is strongly infinite differentiable on () iff, for each , in which case

5. Infinite Differentiability of Weak Solutions

In this section, we characterize the strong infinite differentiability on of all weak solutions of (1) with a scalar type spectral operator in a complex Banach space.

Theorem 12 (infinite differentiability of weak solutions). Let be a scalar type spectral operator in a complex Banach space with spectral measure . Every weak solution of (1) is strongly infinite differentiable on iff there exist and such that the set , where is bounded (see Figure 1).

Figure 1 

Proof. “If” part: suppose that there exist and such that the set is bounded and let be an arbitrary weak solution of (1).
By Theorem 7, Our purpose is to show that , which, by Corollary 11, is attained by showing that, for each ,Let us proceed by proving that, for any and via Proposition 3.
For any , , and an arbitrary ,Indeed, and due to the boundedness of the sets the continuity of the integrated function on , and the finiteness of the measure .
Further, for any , , and an arbitrary ,Finally, for any , and an arbitrary ,Also, for any , , and an arbitrary ,Indeed, since, due to the boundedness of the setsand the continuity of the integrated function on , the setsandare empty for all sufficiently large , we immediately infer that, for any and ,andFurther, for any , , and an arbitrary ,Finally, for any , , and an arbitrary ,By Proposition 3 and the properties of the o.c. (see [8, Theorem XVIII.2.11 (f)]), (40) and (46) jointly imply that, for any and ,which further implies that, for each , Whence, by Corollary 11, we infer that which completes the proof of the “if” part.
“Only if” part: let us prove this part by contrapositive assuming that, for any and , the set is unbounded. In particular, this means that, for any , unbounded is the set Hence, we can choose a sequence of points in the complex plane as follows: The latter implies, in particular, that the points , , are distinct (, ).
Since, for each , the set is open in ; along with the point , it contains an open diskcentered at of some radius , i.e., for each ,Furthermore, we can regard the radii of the disks to be small enough so thatWhence, by the properties of the s.m., where 0 stands for the zero operator on .
Observe also, that the subspaces , , are nontrivial sincewith being an open set in .
By choosing a unit vector for each , we obtain a sequence in such thatwhere is the Kronecker delta.
As is easily seen, (65) implies that the vectors , , are linearly independent.
Furthermore, there is an such thatIndeed, the opposite implies the existence of a subsequence such that Then, by selecting a vectorsuch thatwe arrive atwhich is a contradiction proving (66).
As follows from the Hahn-Banach Theorem, for any , there is an such thatLet us consider separately the two possibilities concerning the sequence of the real parts : its being bounded or unbounded.
First, suppose that the sequence is bounded, i.e., there is such an thatand consider the element which is well defined since ( is the space of absolutely summable sequences) and , (see (65)).
In view of (65), by the properties of the s.m.,For any and an arbitrary ,Also, for any and an arbitrary ,Similarly to (75), for any and an arbitrary ,Similarly to (76), for any and an arbitrary ,By Proposition 3, (75), (76), (77), and (78) jointly imply that and hence, by Theorem 7, is a weak solution of (1).
Letthe functional being well defined since and , (see (71)).
In view of (71) and (66), we haveHence,By Proposition 3, (83) implies that which, by Proposition 9 (, ) further implies that the weak solution , , of (1) is not strongly differentiable at 0.
Now, suppose that the sequence is unbounded.
Therefore, there is a subsequence such that Let us consider separately each of the two cases.
First, suppose that Then, without loss of generality, we can regard thatConsider the elements well defined since, by (87),and , (see (65)).
By (65),andFor any and an arbitrary ,Indeed, for all sufficiently large so thatin view of (87),For any and an arbitrary ,Indeed, for all , in view of , and hence, in view of (87),Similarly to (92), for any and an arbitrary ,Similar to (95), for any and an arbitrary ,By Proposition 3, (92), (95), (98), and (99) jointly imply that and hence, by Theorem 7, is a weak solution of (1).
Since, for any , , by (62), (87), and, by (61), we infer that, for any , , Using this estimate, for the functional defined by (81), we haveBy Proposition 3, (83) implies that which, by Proposition 9 (, ), further implies that the weak solution , , of (1) is not strongly differentiable at 0.
The remaining case of is symmetric to the case of and is considered in absolutely the same manner, which furnishes a weak solution of (1) such that and hence, by Proposition 9 (, ), not strongly differentiable at 0.
With every possibility concerning considered, we infer that assuming the opposite to the “if” part’s premise allows to find a weak solution of (1) on that is not strongly differentiable at 0, much less strongly infinite differentiable on .
Thus, the proof by contrapositive of the “only if” part is complete and so is the proof of the entire statement.