A general theorem on generation of moments-preserving cosine families by Laplace operators in C[0,1]

We use Kelvin’s method of images (Bobrowski in J. Evol. Equ. 10(3):663–675, 2010; Semigroup Forum 81(3):435–445, 2010) to show that given two non-negative integers $i\not= j$ there exists a unique cosine family generated by a restriction of the Laplace operator in C[0,1], that preserves the moments of order i and j about 0, if and only if precisely one of these integers is zero.


Introduction
Following the seminal work of J.R. Cannon [6], a semigroup-theoretical study of diffusion and wave equations associated with one-dimensional Laplace operators equipped with integral conditions has recently been commenced in [9], where an abstract framework for studying such problems in Hilbert spaces has been proposed. Paper [5] presents a different approach, applicable apparently in a broader context: it shows that the recently developed Lord Kelvin's method of images [3,4] provides natural tools for constructing moments-preserving cosine families. In particular, the main theorem of [5] states that there is a unique cosine family generated by a Laplace operator in C[0, 1] that preserves the moments of order zero and 1 (about 0).
In this context, a natural question arises of whether and to what extent the main result of [5], may be generalized to the case of two arbitrary moments of order, say i and j , where i < j are two non-negative integers. Our main theorem (Theorem 2.1) provides the following answer to this question: a necessary and sufficient condition for existence of a cosine family generated by a Laplace operator in C[0, 1] that preserves the moments of order i and j (about 0) is that i = 0. In particular, there are no Laplace-operator-generated cosine families that preserve two moments of order larger than 0. Moreover, the cosine family that preserves the moments of order 0 and j ≥ 1 is uniquely determined, and can be constructed (almost) explicitly by means of the abstract Kelvin formula (3.1). Extensions to non-integer moments are also discussed.

Preservation of moments about 0
Let C[0, 1] be the Banach space of continuous functions on the unit interval, and let C 2 [0, 1] be its subspace of twice continuously differentiable functions. (In what follows we think of real-valued functions, but this is merely to fix attention; the same analysis can be performed in the space of complex functions, as well.) By L c we denote the class of restrictions of the Laplace operator Lf = f , D(L) = C 2 [0, 1] to various domains that generate (strongly continuous) cosine families in C[0, 1]. In other words, a member of L c is a closed linear operator A that generates a strongly continuous cosine family in C[0, 1] and on its domain coincides with L. The cosine family generated by A will be denoted C A = (C A (t)) t∈R . We recall (see, e.g., [1, Proof of Theorem 3.14.17] or [8,Theorem 8.7]) that A ∈ L c generates also a strongly continuous semigroup S A = (S A (t)) t≥0 ; the latter semigroup is given by the abstract Weierstrass formula Given A ∈ L c we say that the related cosine family C A preserves the ith moment iff for all t ∈ R and f ∈ C This theorem will be proved in two steps. First, following [5] we will relate preservation of moments with boundary conditions to show that a generator A ∈ L s of a semigroup that preserves two moments of order i, j ≥ 1 could not be densely defined (thus establishing (b)). Then, in the next section, we will construct the moments preserving cosine family of point (a).
Before continuing, we note that in case (a), a generation theorem for momentspreserving semigroups has been obtained in [9,Theorem 3.4].

Proposition 2.2 Let
A be a member of L c and let i ∈ N. The following statements are equivalent.
we have: Proof If we assume that C A preserves F i , then since F i is bounded,

by (2.1) and because
∞ 0 e −τ 2 /4t dτ = √ πt. This proves that (a) implies (b). In order to prove that (b) implies (c) let f ∈ D(A) and consider The equivalence of (c) and (d) is evident since holds for all i ≥ 1 and all f ∈ C 1 [0, 1]. Finally, assume condition (c) holds. We show that the cosine family C A pre- Then, similarly as in the proof of (b) ⇒ (c), Proof Since (c) and (d) in Proposition 2.2 are equivalent, where H i,j is a bounded linear functional on C[0, 1] given by and for i ≥ 2, Hence, the corollary follows since Ker H i,j is closed and not equal to This corollary clearly implies (b) in Theorem 2.1. For, if the semigroup generated by A ∈ L s preserves moments F i and F j then, by Proposition 2 3 Proof of the case i = 0, j ≥ 1 Let C(R) be the Fréchet space of continuous functions on R with topology of almost uniform convergence, and let (C(t)) t∈R be the basic cosine family in C(R) given by the D'Alembert formula, Also, let F j be the linear functional on C(R) defined by The fact that two distinct objects, the functional in C(R) defined here, and the functional on C[0, 1] defined in (2.2), are denoted by the same letter, should not lead to misunderstanding. Clearly, F i is continuous both on C[0, 1] and C(R) for all i ∈ N.
In the theory of semigroups of linear operators and the related theory of cosine families, Lord Kelvin's method of images can be thought of as a way of constructing families of operators generated by an operator with a boundary condition by means of families generated by the same operator in a larger space, where no boundary conditions are imposed (cf. [3,4]). In our particular context, the method boils down to constructing a cosine family where 'mp' stands for 'moments-preserving' and, more importantly, f ∈ C(R) is a certain extension of f , chosen in such a way that C mp preserves both F 0 and F j . To be more specific: Given f ∈ C[0, 1], and j ≥ 1, we are looking for an f : Existence of such an extension is secured by Proposition 3.2, later on. Before proceeding, we need to introduce some notations. For a function f defined on [0, 1] let f e be its symmetric reflection about 1 2 , that is 1] .
For the proof of Proposition 3.2, we need the following lemma.
Proof Let us define the Bielecki-type norm [2,7] 1] with λ > 0. We note that this norm is equivalent to the usual supremum norm in this space. Consider the operator T : Hence, λ can be chosen so large that In order to prove the second part of the lemma recall that the unique f ∈ C[0, 1] constructed above satisfies f = lim n→∞ f n in C[0, 1], where f n = T n g = g + n k=1 φ k * g, and φ 1 := φ, φ k+1 := φ k * φ, k ≥ 1. Observe that for φ ∈ C 1 [0, 1] we have φ k ∈ C 1 [0, 1],k ≥ 1, by induction. Hence, if we assume that g is twice continuously differentiable, then so is f n , n ≥ 1. Moreover, since φ k+1 = φ(0)φ k + φ * φ k , k ≥ 1, we have Therefore, (f n ) n≥1 converges in C[0, 1], for Since L is closed, f = lim n→∞ f n is twice continuously differentiable, which completes the proof. 1], an extension f that fulfills conditions (A)-(C), listed above, exists and is uniquely determined.
Proof It suffices to find for all n ∈ N, functions g n , h n ∈ C[0, 1] related to f as follows: Since we want f to be well-defined and continuous, these functions must satisfy compatibility conditions: for t ≥ 0. Differentiating with respect to t and then writing t = n + x, x ∈ [0, 1], we see that this is equivalent to which is satisfied if and only if for x ∈ [0, 1] and n ∈ N. Finally, since h n+1 = g n + h n − g n+1 by (3.4), condition (C) holds if and only if g n+1 − j (k j −1 + k e j −1 ) * g n+1 = j (g e n * k j −1 ) e − j k j −1 * (g n + h n ) − j (h e n * k e j −1 ) e + h n , n∈ N. We note that the extension operator Differentiating (3.5) with n = 0 we obtain Since f ∈ D j , by the equivalence of conditions (c) and (d) in Proposition 2.2, we see that Turning to the second equality we consider the cases j = 2 and j > 2 separately. If j = 2, then (3.5) gives Similarly, if j > 2, then (1). Finally, by (3.4) with n = 0, it follows that which completes the proof. Fix f ∈ C[0, 1] and s ∈ R. Clearly, C(s)Ef extends RC(s)Ef and, by the cosine equation for C and the definition of Ef , we have By uniqueness of integral extensions, this shows that C(s)Ef is the integral extension of RC(s)Ef : Using this and the cosine equation for C, we check that i.e., that C mp is a cosine family. This family is strongly continuous, i.e., we have lim t→0 RC(t)Ef = f for all f ∈ C[0, 1], since Ef , as restricted to any compact interval, is a uniformly continuous function, and on [0, 1] it coincides with f .
Turning to the characterization of the generator: Lemma 3.4 and the Taylor formula imply that for f ∈ D j , lim the limit is uniform in x ∈ [0, 1] since f is uniformly continuous in any compact subinterval of (−1, 2). By (3.9) this proves that f belongs to D(A) and we have Af = f .
Finally, we observe that Proposition 2.2 implies D(A) ⊂ D j , which shows that D(A) = D j , and completes the proof.
When combined with Proposition 2.2, Theorem 3.5 proves not only existence of the cosine family that preserves moments of order 0 and j ≥ 1, but also its uniqueness (in the class of cosine families generated by members of L c ). For, by Proposition 2.2, the domain of the generator of a cosine family preserving these moments is contained in D j . Since no member of L c is a proper extension of another member, this generator must coincide with the generator described in Theorem 3.5. In particular, we have completed the proof of Theorem 2.1.
We conclude this section with a remark on symmetries in the moments-preserving cosine families. We say that a function f ∈ C[0, 1] is symmetric about 1 2 if f = f e and similarly, we say that f is asymmetric about 1 2 if f = −f e . By C even [0, 1] and C odd [0, 1] we denote the spaces of symmetric and asymmetric functions, respectively.
In [5,Proposition 3.2] it is proved that in the case i = 0, j = 1, the momentspreserving cosine family C mp leaves the spaces C even [0, 1] and C odd [0, 1] invariant. This allows decomposition of C mp into 'smaller' pieces which are easier to handle. (For example, one of the pieces is the cosine family related to the Neumann boundary conditions.) As we shall see now, such a decomposition is not possible in general. More specifically, for j ≥ 2 the space C odd [0, 1] is invariant for C mp (the reason for that is formula (3.4) showing that integral extensions of asymmetric functions are asymmetric) but the space C even [0, 1] does not possess this property.
Indeed, suppose that, contrary to our claim, C mp C even [0, 1] ⊂ C even [0, 1]. If A is the generator of C mp , then the generator of C mp restricted to C even [0, 1] is A p , the part of A in C even [0, 1]. Hence, for f ∈ D(A p ) the first and the third conditions in (2.3) hold and, f being even, f (1) = f (0) = 0. Therefore D(A p ) ⊂ Ker H j , where H j is the bounded linear functional on C even [0, 1] given by Since D(A p ) is dense in C even [0, 1] and Ker H j is closed, Ker H j = C even [0, 1]. This contradicts the fact that for f ∈ C even [0, 1] given by completing the proof of the claim. In this context it is natural to ask whether there is a subspace of C[0, 1] that is complementary to C odd [0, 1] and invariant for C mp . However, at present an answer to this question eludes us. In fact, as we have noted above, the requirement of preservation of the first moment forces the cosine family to leave the space C odd [0, 1] invariant, yet in general it is unclear how to relate a moment to be preserved with an invariant subspace for the cosine family.

Extensions
The article focuses on non-negative integer moments. However, the main theorem (Theorem 2.1) may be extended to the case of non-negative real moments, as follows.
Because k i is integrable over Next, for real i, j ≥ 1, Corollary 2.3 remains valid, but for i or j in (−1, 1), it does not (see below). Finally, for real positive j , unless j = 1 or j = 2, the argument used in Lemma 3.4 requires existence of F j −3 f , and does not work for j ∈ (0, 1) ∪ (1, 2). To recapitulate, we have the following theorem.  Except for the claim that D i,j of Corollary 2.3 is dense for i or j in (−1, 1), all the statements presented above are proved precisely as in the case where i and j are integers. Hence, we restrict ourselves to showing the former.
Given f ∈ C 2 [0, 1], let c > 0 be chosen so that c + j − 1 < 0. For k > 1 we define  and similarly F j f k = 0, that is f k ∈ D i,j . Furthermore f −f k C[0,1] = 1 k c h k C[0,1] < 1 k c CK, where C depends merely on f , and K depends on i and j . Since k > 1 is arbitrary, this shows that D i,j is dense in C 2 [0, 1], and the claim follows in this case.
For the proof of the other case, where i ∈ (−1, 1) and j ≥ 1, we claim first that for f ∈ C[0, 1] and k > 1, there is f j ∈ C 2 [0, 1] such that F j f j = 0 and f − f j C[0,1] < 1 k . Since C 2 [0, 1] is dense in C[0, 1], it suffices to show this for f ∈ C 2 [0, 1]. Given k > 1 let g a ∈ C 2 [0, 1], a ≥ 1, be defined by g a (x) = 1 k x a . Then g a C[0,1] < 1 k and F j g a = 1 k a(a−1) a−1+j . Hence, a → F j g a ∈ R + is a continuous function satisfying F j g 1 = 0 and lim a→∞ F j g a = +∞. Thus, by the intermediate value theorem there exists a 0 ≥ 1 such that F j g a 0 = |F j f |. Finally, let f j = f − sgn(F j f ) g a 0 . Then f j ∈ C 2 [0, 1], f − f j C[0,1] ≤ 1 k and F j f j = 0, as desired.
To complete the proof, we fix k > 1. Defining f k via (4.1) with c > 0, c +i −1 < 0, C := |F i f j |, α = k c+i−1 F i f j , and β = 0, and with f replaced by f j , we obtain f k ∈ C 2 [0, 1] and F i f k = 0 as before, and F j f k = F j f j − k 1−i−c k 0 h k (y)y j dy = F j f j = 0, since β = 0. Thus f k ∈ D i,j , and f − f k C[0,1] < 1 k CK, where K is defined as before. This completes the proof of the claim.
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