Generalizations of the Lax-Milgram theorem

We prove a linear and a nonlinear generalization of the Lax-Milgram theorem. In particular we give sufficient conditions for a real-valued function defined on the product of a reflexive Banach space and a normed space to represent all bounded linear functionals of the latter. We also give two applications to singular differential equations.


Introduction
The following generalization of the Lax-Milgram Theorem was proved recently by An, Du, Duc and Tuoc in [1]. Theorem 1.1. Let X be a reflexive Banach space over R, {X n } n∈N be an increasing sequence of closed subspaces of X and V = n∈N X n . Suppose that is a real-valued function on X × V for which the following hold: (a) A n = A | Xn×Xn is a bounded bilinear form, for all n ∈ N.
for all v ∈ V . Then, for each bounded linear functional v * on V , there exists x ∈ X such that In this paper our aim is to prove a linear and a nonlinear extension of Theorem 1.1. In the linear case we use a variant of a theorem due to Hayden [5,6] and thus manage to substitute the coercivity condition in (c) of the previous theorem with a more general inf-sup condition. In the nonlinear case we appropriately modify the notion of type M operator and use a surjectivity result for monotone, hemicontinuous, coercive operators. We also present two examples to illustrate the applicability of our results.
All Banach spaces considered are over R. Given a Banach space X, X * will denote its dual and ·, · their duality product. Moreover if M is a subset of X, then M ⊥ will denote its annihilator in X * and if N is a subset of X * , then ⊥ N will denote its preannihilator in X.

The linear case
To prove our main result for the linear case we need the following lemma which is a variant of [5,Theorem 12 for all x ∈ X.
Then, for every y * ∈ Y * , there exists a unique x ∈ X with A(x, y) = y * , y , Obviously T is a bounded, linear map. Since, by (b), T x ≥ c x , for all x ∈ X, T is one-to-one. To complete the proof we need to show that T is onto.
Since A is non-degenerate with respect to the second variable, we have that Hence ( ⊥ T (X)) ⊥ = Y * and so, by [7, Proposition 2.6.6], Thus to show that T maps X onto Y * we need to prove that T (X) is w * -closed in Y * . To see that let {T x λ } λ∈Λ be a net in T (X) and y * be an element of Y * such that Without loss of generality we may assume, using the special case of the Krein-Smulian Theorem on w * -closed linear subspaces (see [ Remark 2.2. An alternative proof of the previous lemma can be obtained using the Closed Range Theorem.
We are now in a position to prove our main result for the linear case.
is a function for which the following hold: So {x λ } λ∈Λ is a bounded net in X. Since X is reflexive, there exist a subnet {x λµ } µ∈M of {x λ } λ∈Λ and x in X such that {x λµ } µ∈M converges weakly to x.
We are going to prove that The following example illustrates the possible applicability of Theorem 2.3. Example 2.4. Let a ∈ C 1 (0, 1) be a decreasing function with lim t→0 a(t) = ∞ and a(t) ≥ 0, for all t ∈ (0, 1). We will establish the existence of a solution for the following Cauchy problem: which is equivalent to the original Sobolev norm, and Y = L 2 (0, 1). Note that X is a reflexive Banach space, being a closed subspace of H 1 (0, 1). Let {α n } n∈N be a decreasing sequence in (0, 1) with lim n→∞ α n = 0. Define X n = {u ∈ H 1 (α n , 1) | u(α n ) = 0} and Y n = L 2 (α n , 1) (we can consider X n and Y n as closed subspaces of X and Y respectively, by extending their elements by zero outside (α n , 1)).
A is well-defined and A(·, v) is a bounded linear functional on X for any v ∈ V .
Let A n = A| Xn×Yn . A n is a bounded, bilinear form since where M n is the bound of a on [α n , 1]. It should be noted that A is not bounded on the whole of X × V . To show that A n is non-degenerate let v ∈ Y n and assume that A n (u, v) = 0, for all u ∈ X n , i.e. It is easy to see that the above implies that We next show that Define T n : X n −→ Y * n by T n u, v = A n (u, v). T n is a well-defined bounded linear operator and T n u = u ′ + a(t)u. Hence Xn , since u(α n ) = 0, a is decreasing and a(t) ≥ 0, for all t ∈ (0, 1).
All the hypotheses of Theorem 2.3 are hence satisfied and so if F ∈ V * is defined by F Thus u satisfies (2.1).

The nonlinear case
We start by recalling some well-known definitions: Definition 3.1. Let T : X −→ X * be an operator. We say that T is: (ii) hemicontinuous if, for all x, y ∈ X, T (x + ty) We also need the following generalization of the notion of type M operator (for the classical definition see [3] or [8]). Definition 3.2. Let X be a Banach space, V be a linear subspace of X and Our result is the following: Theorem 3.3. Let X be a reflexive Banach space, Λ be a directed set, {X λ } λ∈Λ be an upwards directed family of closed subspaces of X and V = λ∈Λ X λ . Suppose that A : X × V −→ R is a function for which the following hold: (a) A is of type M with respect to V . , y), for all x, y ∈ X λ , is monotone and hemicontinuous, for all λ ∈ Λ. Then for each v * ∈ V * there exists x ∈ X such that Proof. As in the proof of Theorem 2.3, for each λ ∈ Λ, let v * λ = v * | X λ . By the Browder-Minty Theorem (see [8,Theorem 26.A]), a monotone, coercive and hemicontinuous operator, from a real reflexive Banach space into its dual, is onto. Thus, by (b) and (d), for each λ ∈ Λ, the operator T λ is onto and so there exists λ , x λ and hence, by (b), we get that the net {x λ } λ∈Λ is bounded. Continuing as in the proof of Theorem 2.3 and applying the fact that A is of type M with respect to V we get the required result.
Remark 3.4. It should be noted that, since a crucial point in the above proof is the existence and boundedness of the net {x λ } λ∈Λ , variants of the previous theorem could be obtained using in (b) and (d) alternative conditions corresponding to other surjectivity results.
We now apply Theorem 3.3 to a singular Dirichlet problem.

Example 3.5.
Let Ω be a bounded domain in R N . We consider the Dirichlet problem where a ∈ L ∞ loc (Ω) and there exists c 1 > 0 such that a(x) ≥ c 1 a.e. on Ω and f : Ω × R −→ R is a monotone increasing (with respect to its second variable for each fixed x ∈ Ω), Carathéodory function, for which there exist h ∈ L 2 (Ω) and c 2 > 0 such that (3.2) |f (x, u)| ≤ h(x) + c 2 |u|, for all x ∈ Ω and u ∈ R.
We will show that if the above hypotheses on a and f hold, then problem (3.1) has a weak solution, i.e. that there exists a function u ∈ H 1 0 (Ω) with To this end let X = H 1 0 (Ω), {Ω n } n∈N be an increasing sequence of open subsets of Ω such that Ω n ⊆ Ω n+1 and ∞ n=1 Ω n = Ω and X n = H 1 0 (Ω n ), for each n ∈ N. Observe that we can consider each X n as a closed subspace of X by extending its elements by zero outside Ω n and let

Finally let
A : X × V −→ R be the function defined by By a(x) ≥ c 1 a.e. on Ω, the monotonicity of f and the growth condition (3.2), we have . Since by the Poincaré inequality ∇u L 2 (Ω) is equivalent to the norm of X it follows that A is coercive.
Let A n = A| Xn×Xn . Then, since a ∈ L ∞ loc (Ω) it follows that a ∈ L ∞ (Ω n ), for all n ∈ N. Combining this with (3.2), we have that where c(u, n) is a positive constant depending on n and u. So the operator T n : X n −→ X * n , with T n u, v Xn = A n (u, v) is well-defined, for all n ∈ N. Let T 1,n , T 2,n : X n −→ X * n be the operators defined by Then T 1,n is a monotone bounded linear operator. Using the monotonicity of f , it is easy to see that T 2,n is monotone. Finally, recalling that the Nemytskii operator corresponding to f is continuous (see for example [8,Proposition 26.7]) and that the embedding of X n into L 2 (Ω n ) is compact we have that T 2,n is hemicontinuous. Thus T n = T 1,n + T 2,n is monotone and hemicontinuous, for all n ∈ N.
To finish the proof let u n w −→ u in X. Then since, for all v ∈ V , u −→ Ω a(x)∇u∇vdx is a bounded linear functional and, by the continuity of the Nemytskii operator and the compactness of the embedding of X into L 2 (Ω), for all v ∈ V , we get that Thus A is of type M with respect to V . Applying now Theorem 3.3 we get that there exists u ∈ X such that A(u, v) = 0, for all v ∈ V . Observing that C ∞ 0 (Ω) is contained in V we get that u is the required weak solution of (3.1).