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Lions' representation theorem and applications

The Representation Theorem of Lions (RTL) is a version of the Lax--Milgram Theorem where completeness of one of the spaces is not complete. In this paper, RTL is deduced from an operator-theoretical version on normed space. The main point of the paper is a theory of derivations, based on RTL, for which well-posedness is proved. One application concerns non-autonomous evolution equations with a new initial-value and a periodic boundary condition for the time variable.


Introduction
A most elegant way to prove well-posedness of a non-autonomous evolutionary problem has been given by J. L. Lions [14, Chap. III, Théorème 1.1, p. 37] using a surprising version of the Lax-Milgram Theorem, where one of the spaces is not complete (see e.g. Theorem 2.1 below). We call it the Representation Theorem of Lions, RTL. The purpose of this article is to study a derivation problem with the help of Lions' result. A Hilbert space V is given, which we choose real for this introduction, as well as a derivation D : R → V * , where R is a dense subspace of V ; i.e. D satisfies Dr 1 , r 2 V * ,V + Dr 2 , r 1 V * ,V = 0 for all r 1 , r 2 ∈ R.
It turns out that this simple setting leads to an interesting structure and to interesting results on boundary conditions. In fact, D can be extended naturally to a Hilbert space W which embeds continuously into V . Given a coercive operator A ∈ L(V, V * ), we study the problem (1.1) Du + Au = f where f ∈ V * is given and u ∈ W is a solution to be found. We obtain well-posedness results if we impose boundary conditions which are expressed in terms of (strongly) admissible subspaces Z of W (see Section 3 and Section 4.2). Our main application concerns the evolution equations also considered by Lions (and which have seen a remarkable revival recently; see e.g. Ouhabaz and Spina [17], Haak and Ouhabaz [13], Ouhabaz [16], Auscher and Egert [9], Dier and Zacher [11], Achache and Ouhabaz [1,2] as well as [7,8]). For these evolution equations, we obtain a new result whose originality lies in the very general boundary conditions. It looks as follows. Let U, H be Hilbert spaces where U is continuously and densely embedded in H so that we have the Gelfand triple Let W = H 1 (0, T ; U ′ ) ∩ L 2 (0, T ; U). Then, as is well-known, W ⊂ C([0, T ], H). Let V = L 2 (0, T ; U) and let A ∈ L(V, V * ) be coercive. Typically, one chooses for A a non-autonomous differential operator. The following is proved in Section 5.
Theorem 1.1. Let Φ ∈ L(H) be a contraction, f ∈ L 2 (0, T ; U ′ ), y 0 ∈ H. Then there exists a unique u ∈ W such that Theorem 1.1 is known if the form a : V × V → R, given by a(u, v) = Au, v V * ,V is symmetric (see [20, III. Proposition 2.4]) but the estimates given here for the non-symmetric case are more demanding.
In Theorem 1.1, not only well-posedness but also the maximal regularity of the solution are remarkable: all three terms of the evolution equation u ′ , Au and f are in the space L 2 (0, T ; U ′ ) (see [6] for more information on such regularity properties).
The theory of derivations as developed in the present article has applications for completely different subjects. If we identify V and V ′ by the Riesz Theorem, then a densely defined operator S on V is symmetric if and only if iS is a derivation. It turns out that our results on boundary operators allow also a description of all selfajoint extensions of the symmetric operator S. In fact we refine a version of the theory of boundary triplets as known in the literature. The circle of these ideas is presented in [5].
We start this article by a review of the Representation Theorem of Lions (RTL). The new point is that we deduce it from a result on operators which are defined on normed space (incomplete, in general). This allows us to prove RTL also in a non-Hilbert context (in contrast with Lions [ In this section we describe (generalizations of) the representation theorems of Lions and give an application to dissipative operators.
2.1. Representation theorems of Lions. The representation theorems of Lions (abbreviated by RTL) may be expressed in terms of operators (which seems to be new), or in terms of bilinear forms or sesquilinear forms. They depend also heavily on the topological properties (completeness, reflexivity) of the involved normed vector spaces over K = R or C. The antidual of a normed space X is denoted by X ′ , whereas the dual space is denoted by X * . Theorem 2.1 (Representation Theorem of Lions, real version). Let X be a reflexive Banach space and Y be a normed space over R. Let a : X × Y → R be bilinear such that, for each y ∈ Y , x → a(x, y) is continuous on X. Let β > 0. The following assertions are equivalent: (i) sup x∈X, x X ≤1 a(x, y) ≥ β y Y ; (ii) for all L ∈ Y * , the dual of Y , there exists x L ∈ X such that Next we consider the complex version.
Theorem 2.2 (Representation Theorem of Lions, complex version). Let X be a reflexive Banach space and Y be a normed space over C. Let a : X ×Y → C be sesquilinear such that, for each y ∈ Y , x → a(x, y) is continuous on X. Let β > 0. The following assertions are equivalent: In Theorem 2.2, it is remarkable that Y does not need to be complete and a does not need to be continuous.
This theorem is a consequence of the following result involving operators instead of sesquilinear forms. Theorem 2.3. Let E and F be normed spaces over K, T : E → F linear and β > 0. The following assertions are equivalent: (ii) for all x * ∈ E * , there exists y * ∈ F * such that x * = y * • T and y * ≤ x * β .
(iii) for all x ′ ∈ E ′ , there exists y ′ ∈ F ′ such that Note that, in Theorem 2.3, T is not necessarily continuous. On the other side, if T ∈ L(E, F ), then x * = T * y * and (ii) (resp. (iii)) means that T * (resp. T ′ ) is surjective and Let x * ∈ E * . Then x * • T −1 ∈ Z * . From the Hahn-Banach theorem, there exists y * ∈ F * such that y * |Z = x * • T −1 and (ii)⇒(i): Let x ∈ E. Then, by the Hahn-Banach theorem, there exists x * ∈ E * , x * ≤ 1 with x * , x X * ,X = x . By (ii) there exists y * ∈ F * with y * ≤ 1 β and x * = y * • T . It follows that The equivalence with (iii) follows from similar arguments. |a(x, y)|.
Indeed, for each x ∈ X and y ∈ Y , there exists θ ∈ R such that a(x, y) = e iθ |a(x, y)|. Then Re a(e −iθ x, y) = |a(x, y)| with e −iθ x = x . Let E = Y and F = X ′ , the antidual of X. Define T : E → F by T w = a(., w). Then T is linear and since Re a(x, y) = Re a(x, y). By Theorem 2.3, Condition (ii) of Theorem 2.2 holds if and only if, for all L ∈ Y ′ , there exists Corollary 2.4. Let K = R or C. Let X be a reflexive Banach space and Y be a normed space such that Y embeds continuously in X. Let β > 0 and let a be a sesquilinear form on X × Y such that a) |a(y, y)| ≥ β y 2 Y for all y ∈ Y ; b) for all y ∈ Y , x → a(x, y) is continuous on X.
Let L ∈ Y ′ . Then there exists x L ∈ X such that a(x L , y) = L, y Y ′ ,Y for all y ∈ Y.
Proof. There exists c Y > 0 such that y X ≤ c Y y Y for all y ∈ Y . Thus for y ∈ Y \ {0}, 2.2. Dissipative operators. The following considerations are somehow a side order in this paper. Its aim is to show that Theorem 2.3 is useful also in a non-Hilbert space situation. We will give a new characterization of dissipativity (Theorem 2.6), a notion which will also be used in Section 3.
An operator B on a Banach space X is a linear mapping from dom(B) to X, where the domain dom(B) of B is a subspace of X. By x B := x X + Bx X we denote the graph norm of B. The operator B is called closed if its graph G(B) := {(x, Bx) : x ∈ dom(B)} is closed in X × X. This means that for x n ∈ dom(B) (2.1) x n → x and Bx n → y in X implies x ∈ dom(B) and Bx = y.
In other words, B is closed if and only if (dom(B), · B ) is complete.
An operator B is called closable if (0, y) ∈ G(B) ⊂ X × X implies y = 0. Then there exists a unique operator B on X, the closure of B such that G(B) = G(B).
We call B dissipative if If X is a Hilbert space, this is equivalent to Returning to the Banach space context, we say that B is m-dissipative if B is dissipative and (Id −tB) is surjective for one (equivalently all) t > 0. Then (Id − tB) −1 ∈ L(X) and (Id −tB) −1 ≤ 1 for all t > 0. A densely-defined dissipative operator is closable and its closure B is dissipative [3, Lemma 3.4.4].
Remark 2.5. The operator B generates a contractive C 0 -semigroup on X if and only if B is m-dissipative and has dense domain.
Theorem 2.3 allows us to give the following dual characterization of dissipativity, which seems to be new. Theorem 2.6. Let B be a densely defined operator on a Banach space X. The following assertions are equivalent.
(i) B is dissipative; (ii) for all t > 0 and for all y * ∈ X * there exists x * ∈ dom(B * ) such that This in turn is equivalent to y * ∈ dom(B * ) and tB * y * = y * − x * , i.e. y * − tB * y * = x * .
A densely defined operator B is m-dissipative if and only if B is closed, dissipative and Id −tB * is injective for some (equivalently all) t > 0. This is true since Id −tB has dense range if and only if (Id −tB * ) is injective. We illustrate Theorem 2.6 by the following example. if Ω is Wiener regular (see [4]). Note that C 0 (Ω) ′ = M(Ω) := {µ : µ is a signed measure on Ω} with µ the total variation of µ ∈ M(Ω).
Since B is dissipative, we see from Theorem 2.6 that for each t > 0 and each ν ∈ M(Ω), there exist µ ∈ dom(B * ) such that Moreover, Ω is Wiener regular if and only if the solution of (2.4) is unique for some (equivalently all) t > 0.

When Lions' Representation Theorem meets derivations
We start introducing the notion of derivation acting on a Hilbert space. Our aim is to study perturbations of a coercive operator by a derivation, which leads to define Weak and Strong Derivation Problems. In the remaining of the paper, V is a Hilbert space over K = R or C. By V ′ we denote the antidual of V , that is the space of all continuous antilinear forms on V . In the case K = R, V ′ = V * where V * is the dual space.

3.1.
Derivations. The following definition is central for the entire article.
Here, as previously, we denote by , V ′ ,V the duality between V ′ and V , whereas , V is the scalar product in V .
Remark 3.2. In general we do not want to identify V ′ and V . But if we do so, then D 0 is a derivation if and only if D 0 is antisymmetric, If K = C this is the same as saying that iD 0 is symmetric.
In the remain of this entire section, D 0 is a derivation according to Definition 3.1. We associate to D 0 a new Hilbert space W in the following way.
Since R is dense in V , given v ∈ W , there exists at most one f ∈ V ′ satisfying the above requirement. We then denote Dv := f .
The following is easy to see.
where V ′ carries the scalar product coming from the Riesz isomorphism between V and V ′ .
It follows from this definition that Moreover we set We call W the extended domain of D 0 , D the extension of D 0 , and b the associated boundary form.
If R is dense in W , then b is identically equal to 0. But in general b is different from 0 and will be used to define boundary conditions for a Derivation Problem associated with D.
Note that b(w, w) ∈ R for all w ∈ W .
3.2. Admissible subspaces. Another notion of great importance in the sequel is the notion of admissible space and strongly admissible space.
Since b : W × W → K is continuous, the closure of an admissible subspace Z is admissible, and, if Z is strongly admissible, then so is its closure Z.
3.3. The Weak Derivation Problem. Next we consider the weak derivation problem. Let Z ⊂ W be an admissible and closed subspace. We endow Z with the norm Then the weak derivation problem is the following: given L ∈ Z ′ find a vector v such that Theorem 3.6. Let Z ⊂ W be a closed admissible subspace and let L ∈ Z ′ . Then there exists a vector u such that (WDP) holds. If Z is strongly admissible, then the solution is unique.
We consider Z as a space of test functions. The solutions of (W DP ) depend on the choice of this space. We will make this dependance more transparent in Theorem 3.10, Proposition 4.3 and Theorem 4.17 below.
For the proof we need some preparation.
Lemma 3.7. Let X be a Banach space, B a closed operator on X and A ∈ L(X). Assume that there exists α > 0 such that Then there exists a constant β > 0 such that Proof. By assumption the space Z := dom(B) endowed with the norm x Z := x X + Bx X is complete. The mapping T := A−B : Z → X is continuous. In fact Thus T is injective and Y := T Z is closed in X. To see this let z n ∈ Z such that lim n→∞ T z n − y X = 0. It follows from (3.6) that Thus x := lim n→∞ z n exists in X. Since A ∈ L(X), it follows that lim n→∞ Az n = Ax. Thus lim n→∞ Bz n = Ax − y. Since B is closed it follows that x ∈ dom(B) = Z and Bx = Ax − y. Hence T x = y.
We have proved that Y is closed in X and hence complete. By the Theorem of the Continuous Inverse, T −1 ∈ L(Y, Z). Thus Choose β = 1 T −1 . On a Hilbert space V we obtain the following more special version.
Recall that an operator B on V is dissipative if and only if Re Bv, v V ≤ 0 for all v ∈ dom(B).
Proposition 3.8. Let B be a densely defined dissipative operator on a Hilbert space V and let A ∈ L(V ) be coercive, i.e. there exists α > 0 such that Then there exists β > 0 such that Proof. By [12, II. Proposition 3.14 (iv)] or [18, I. Theorem 4.5 (c)], B is closable and the closure B of B is dissipative again. Thus we can assume that B is closed. One has for It follows that the selfadjoint operator A + is positive and invertible. Let S be the (invertible and positive) square root of A + . Since Sv 2 Since Sv ≤ A + 1/2 v , we also get In addition, note that Using (3.9) we first get, for all v ∈ V , Note that In addition, since S −1 A + = S, Now we use the celebrated Peter-Paul inequality combined with the Cauchy-Schwarz to estimate, for all γ > 0, We get then 1+s where s > 0 be be fixed later. Then 1 − γ = s 1+s and 1 − 1/γ = −s. Then, by (3.9) and (3.10), it follows that Then we choose s such that s S −1 A − 2 = α 2 . Then we obtain In other words β does not depend on B.

Proof of Theorem 3.6. By the Riesz Representation Theorem there exists an operator
Similarly there exists B : Z → V such that The operator B is closed since Z is complete for the norm Since Z is admissible B is dissipative. The operator A * is coercive. By Lemma 3.7 there exists β > 0 such that By RTL (Corollary 2.4) there exists v ∈ V such that For the uniqueness, we now assume that Z is strongly admissible. Let u ∈ V be the difference of two solutions. Then Taking z ∈ R we see that u ∈ W and Du + Au = 0 in V ′ .
In particular, Substracting (3.12) from (3.13) we deduce that Thus u ∈ Z b . Since Z is strongly admissible, it follows that 3.4. The Strong Derivation Problem. Next we consider the strong derivation problem. For that we have to assume that the antilinear form L on Z has a special form. Lemma 3.9. Let Z be a subspace of W endowed with the norm · Z = · W such that R ⊂ Z, and let L ∈ Z ′ . The following assertions are equivalent: (i) there exists a constant c ≥ 0 such that | L, r V | ≤ c r V for all r ∈ R; (ii) there exist f ∈ V ′ and g ∈ Z ′ such that g, r V = 0 for all r ∈ R and L = f + g.
and g(r) = 0 for all r ∈ R.
Let Z ⊂ W be an admissible subspace and let L ∈ Z ′ be of the form L = f + g where f ∈ V ′ and g ∈ Z ′ such that g(r) = 0 for all r ∈ R. The Strong Derivation Problem consists in finding a vector u such that (SDP) u ∈ W, Du + Au = f and b(u, z) = − g, z Z ′ ,Z , z ∈ Z.
Thus, the functional g expresses a kind of boundary condition. In the examples (Section 5), this will become even more transparent. Theorem 3.10. Let Z ⊂ W be an admissible subspace. Let L = f + g with f ∈ V ′ , g ∈ Z ′ such that g, r Z ′ ,Z = 0 for all r ∈ R. For u ∈ V the following are equivalent: (i) u is a solution of (WDP); (ii) u ∈ W and u is a solution of (SDP). Consequently, there exists a solution of (SDP). Moreover it is unique if Z is strongly admissible.
Remark 3.11. Letting g = 0, Theorem 3.10 proves that the operator D + A : Z b → V ′ is invertible, i.e. it is injective with a closed and dense range. This means that there exists β ′ > 0 such that, for all u ∈ Z b , and that, for any w ∈ V , Indeed (3.17) is equivalent to D + A is injective with closed range, whereas (3.18) is equivalent to D + A has dense range using the Hahn-Banach theorem combined with the reflexivity of our spaces. Note that the existence of β ′ follows directly from Lemma 3.7. Indeed, for all u ∈ Z b , b(u, u) ≥ 0, and the second property follows along the same lines as the proof of (i) ⇒ (ii) above (first prove that w ∈ W , then that w ∈ Z bb = Z, and finally that Re A ⋆ w, w V ′ ,V ≤ 0).

Admissible subspaces and boundary operators
If we consider D as a time derivative, then (SDP) may be interpreted as an evolution equation with a boundary condition with respect to time.
This can be made more concrete when we make further assumptions on the associated boundary form b and, of course, in the examples we give below. 4.1. Maximal admissible subspaces. We start considering maximal admissible subspaces. This maximality is a handy criterion for proving strong admissibility. Proof. Let M := {Z 1 ⊂ W : Z 0 ⊂ Z 1 , Z 1 admissible} ordered by inclusion. If (Z i ) i∈I is a chain in M, then Z := ∪ i∈I Z i is a subspace of W containing Z 0 . If w ∈ Z, then there exists i ∈ I such that w ∈ Z i . Consequently b(w, w) ≤ 0. Thus Z is admissible and hence an upper bound for the chain (Z i ) i∈I . Now by Zorn's Lemma M has a maximal element.
Let Z 1 be a maximal admissible subspace and let u ∈ W such that b(u, w) = 0 for all w ∈ Z 1 . Suppose that b(u, u) < 0. Then u ∈ Z 1 . Thus Z 2 := Z 1 ⊕ Ku is a subspace of W and for w = This contradicts the maximality of Z 1 . We have proved that b(u, u) ≥ 0 for all u ∈ Z b 1 ; i.e. Z 1 is strongly admissible. Let Z ⊂ W be a strongly admissible subspace and L ∈ Z ′ , where Z carries the norm of W . Then, by Theorem 3.6, (WDP) has a unique solution u ∈ V . Let Z ⊂ W be a maximal admissible subspace such that Z ⊂ Z. By the Hahn-Banach Theorem there exists L ∈ Z ′ such that L |Z = L. Since Z is also strongly admissible, there exists a unique u ∈ V solution for (WDP) with respect to Z. Proof. One has −Dz, u + Au, z = L, z Z ′ ,Z for all z ∈ Z and −Dz, u + A u, z = L, z Z ′ , Z for all z ∈ Z.
In particular Now it follows from Theorem 3.6 applied to L = 0 that u − u = 0.
In view of the preceding abservations, our aim is now to study the maximal admissible subspaces in the next subsection.

Boundary operators.
In this section we continue to study the abstact framework from Section 3. But we give more structure to the associated boundary form b.
Let V, R, D, b be given as in Section 3 and let W be defined by (3.1) and b : W × W → K by (3.3). Recall that b is a continuous, symmetric sesquilinear form. Now we make the following assumption.   Our aim is to describe the maximal admissible subspaces in terms of B 0 and B 1 . We need several auxiliary results.
Proof. a) Since by our assumption W = ker B 0 + ker B 1 , it follows that  Proof. a) R ⊂ ker B 1 by Lemma 4.6. b) Let w ∈ ker B 1 . Then b(w, w) = − B 0 w 2 H ≤ 0. c) Let Z be admissible and such that ker B 1 ⊂ Z. Let w ∈ Z. By our assumptions there exists a decomposition w = w 0 + w 1 ∈ ker B 0 + ker B 1 .
The following observation will be useful in the next proof .
Remark 4.9. The operator −D 0 is also a derivation and its extended domain is W . The boundary form associated with −D 0 is −b and the extended operator −D. In particular We now show that B 0 and B 1 have closed images. This is surprising and due to the special construction of b. For the proof we will use the well-posedness of the Derivation problem with respect to Z = ker B 1 . The result will be important in the sequel.  Proof. Since Z is admissible, Thus, letting Φ 0 B 0 w := B 1 w for all w ∈ Z yields a well-defined mapping Φ 0 : B 0 Z → B 1 Z. Then Φ 0 is linear and contractive. Hence Φ 0 has a unique continuous extension Φ 1 : Proof. a) Since B 1 r = B 0 r = 0 for all r ∈ R by Lemma 4.6, one has Let Z be admissible and such that Z Φ ⊂ Z. We show that Z Φ = Z. In fact, by Lemma 4.11, there exists a contraction Φ 1 : In fact, let w ∈ W . By Lemma 4.8 there exists u ∈ W such that The fact that each maximal subspace of W is of the form Z Φ for some linear contraction Φ : Ran B 0 → Ran B 1 follows from Lemma 4.11.
Next we compute the initial value condition.
Proposition 4.13 also shows that Z Φ is strongly admissible (without using Proposition 4.2. In fact, by (4.5) Proof. By Theorem 4.12, any maximal admissible subspace of W is of the form Z Φ for some linear contraction Φ : Ran B 0 → Ran B 1 . By Proposition 4.13, we have Z b Φ = Z Φ * and therefore Z bb 4.3. Boundary conditions. Next we compare maximal and strongly admissible subspaces. We already know that maximal admissible subspaces are strongly admissible. Recall from Theorem 4.12 that the maximal admissible subspaces of W are exactly the spaces Z Φ where Φ : Ran B 0 → Ran B 1 is a contraction.
Proposition 4.15. Let Z ⊂ W be strongly admissible. Then there exists exactly one maximal admissible subspace Z Φ such that Z ⊂ Z Φ . Moreover, Note also that −D is a derivation with extended space W and associated with −b. Thus, by Theorem 4.12, Z Φ * is a maximal admissible subspace for −b. Since Z is strongly admissible, the space Z b is admissible for −D. Since Z φ * ⊂ Z b , it follows that Z Φ * = Z b . We have shown the second assertion.
Newt we show uniqueness. Let Ψ : Ran B 0 → Ran B 1 be another contraction such that Z ⊂ Z Ψ . Then by what we have just proved. We show that Ψ * = Φ * (which implies that Ψ = Φ). Let w 1 ∈ Ran B 1 , w 0 = Φ * w 1 ∈ Ran B 0 . By Lemma 4.8, there exists w ∈ W such that w 1 = B 1 w and w 0 = B 0 w. Thus An obvious corollary is the following.
Thus, in view of Proposition 4.3, the solutions of (WDP) with respect to Z and to Z Φ are the same. Now in view of Theorem 4.12 we obtain from Theorem 3.6 a theorem on existence and uniqueness of the solution to a Derivation Problem.
Let Φ : Ran B 0 → Ran B 1 be linear such that Φ ≤ 1. Consider the strongly admissible space Proof. Let Z = Z Φ and let us define g ∈ Z ′ by g, z Z ′ ,Z = y 0 , B 0 z H . Then g is such that g, r Z ′ ,Z = 0 for all r ∈ R. By Theorem 3.10, there exists u such that (SDP) holds, that is u ∈ W, Du + Au = f and b(u, z) = − g, z Z ′ ,Z , z ∈ Z.
Theorem 3.10 also shows the uniqueness of u since Z Φ is strongly admissible.

Non-autonomous evolution equations
In this section we establish existence and uniqueness for an evolutionary problem with very general boundary conditions concerning the time variable which can prescribe as well an initial value problem or a periodicity condition.
Let T > 0, U a Hilbert space over K = R or C which is continuous and densely embedded into another Hilbert space H, i.e. U  Thus it is possible to define B 1 , B 0 : W → H by B 0 w = w(0) and B 1 w = w(T ). Indeed, we have that for all u ∈ W, u(t) = t T u(t)+ T −t T u(t) with (t → t T u(t)) ∈ ker B 0 and (t → T −t T u(t)) ∈ ker B 1 . So we have W = ker B 0 + ker B 1 where B 0 and B 1 are selfadjoint. Now let A ∈ L(V, V ′ ) be coercive.
Then A ∈ L(V, V ′ ) is coercive.
In the following it is important to know the trace space of W .  For Φ = 0, Theorem 5.4 establishes well-posedness for an initialvalue problem. It is due to J. L. Lions [14], see also [20,  If Φ = Id and y 0 = 0 we obtain a periodic problem. If A is selfadjoint then Theorem 5.4 is given in [20, II. Proposition 2.4] where selfadjointness is used in the proof on page 112. The general case given here is new and depends on our perturbation result Proposition 3.8.