A THEOREM ON VARIATIONAL INEQUALITIES FOR AFFINE MAPPINGS

JEAN SAINT RAYMOND

Abstract. We give a proof of a conjecture by B. Ricceri on variational inequalties which extends a partial result of N.D. Yen. We need an additional hypothesis of compactness for a convex set, and refute the conjecture if this additional hypothesis is removed.

1. Introduction

In his paper [3], B. Ricceri shows the following theorem, in a slightly more general form :

Theorem 1 (B. Ricceri). Let E be a topological linear Hausdorff space, X be a closed convex subset of E with nonempty interior, A : X → E^{∗} be a continuous mapping to the dual of E equipped with the weak* topology, K be a compact subset of X and K_{1} ⊂ K be a closed finite-dimensional set.

Assume that for every x ∈ X \K : sup_{y∈K_{1}}⟨A(x), x−y⟩ > 0. Then there exists some x^{∗} ∈ K such that sup_{y∈X}⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0.

In particular this statement has the following corollary cited in [4] and [5]:

Theorem 2. Let H be an infinite-dimensional Hilbert space, X be a closed convex subset of H with nonempty weak interior, A : H → H be a weakly continuous mapping, K be a weakly compact subset of X and K_{1} ⊂ K be a closed finitedimensional set.

Assume that for every x ∈ X \K : sup_{y∈K_{1}}⟨A(x), x−y⟩ > 0. Then there exists some x^{∗} ∈ K such that sup_{y∈X}⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0.

As showed in [1] the hypothesis of nonempty interior on X cannot be removed: Frasca and Villani construct an example of a continuous affine function A on a Hilbert space satisfying sup_{y∈B}⟨Ax, x − y⟩ > 0 for all x in the unit ball B of H.

In [4] (Problem 1) and again in [5] (Problem 1.1) B. Ricceri asked whether Theorem 1 still holds without the hypothesis that K_{1} be finite-dimensional. And he conjectured that the following statement, would fail even with A affine.

Conjecture 3. Let H be an infinite-dimensional Hilbert space, X be a closed convex subset of H with nonempty weak interior, A : H → H be a weakly continuous mapping and K be a weakly compact subset.

2010 Mathematics Subject Classification. 47J20, 49J45, 46E10.

Key words and phrases. Variational inequalities, convex compact sets.

Assume that for every x ∈ X \ K : sup_{y∈K}⟨A(x), x − y⟩ > 0. Then there exists some x^{∗} ∈ K such that sup_{y∈X}⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0.

In fact, as proved by Yen in [6], this is not the case if A is affine and X is the whole H. Indeed:

Theorem 4 (N.D. Yen). Let H be an infinite-dimensional Hilbert space, A : H → H be an affine mapping and K be a weakly compact subset of H.

Assume that for every x ∈ H \ K : sup_{y∈K}⟨A(x), x − y⟩ > 0. Then there exists some x^{∗} ∈ K such that A(x^{∗}) = 0 (equivalently ⟨A(x^{∗}), x^{∗} −y⟩ = 0 for all y ∈ H).

The proof of Yen is based on a theorem of [2] founded on the monotonicity of A, which shows that the hypothesis of continuity for A is not needed. In fact in the first part of his proof, Yen shows that A is monotone on H and the following is known (see [7], Proposition 26.4)

Theorem 5. Let E be a Banach space, E^{∗} be its dual and φ : E → E^{∗} be a monotone linear operator. Then φ is continuous.

So the continuity of the operator A follows from the hypotheses. In fact we can weaken slightly the hypotheses of Yen’s theorem.

Theorem 6. Let E be an infinite-dimensional Banach space, A be an affine mapping from E to E^{∗} and K be a weakly compact subset of E. Assume that for every x ∈ E \K : A(x) ≠ 0 and sup_{y∈K}⟨A(x), x−y⟩ ≥ 0. Then there exists some x^{∗} ∈ K such that A(x^{∗}) = 0 (equivalently ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ E).

In fact after this paper was written down B. Ricceri informed me that he could show by a simpler proof using Kneser’s minimax theorem the following version of previous thorem :

Theorem 7 (B. Ricceri). Let E be a real Hausdorff locally convex topological vector space, K ⊂ E a compact set and A : E → E^{∗} an affine operator such that sup_{y∈K}⟨A(x), x−y⟩ ≥ 0 for all x ∈ E \K. Assume also that \overline{conv}(K) is compact. Then there exists x^{∗} ∈ \overline{conv}(K) such that A(x^{∗}) = 0.

We will also prove the following extension of Yen’s result:

Theorem 8. Let E be Banach space space, X be a closed convex subset of E with nonempty weak interior, A : E → E^{∗} be a continuous affine mapping and K be a weakly compact subset of X.

Assume that for every x ∈ X \ K : sup_{y∈K}⟨A(x), x − y⟩ > 0. Then there exists some x^{∗} ∈ K such that ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X.

Theorem 8 will in fact appear as a corollary of the following theorem which answers Conjecture 3 with the additional hypothesis that \overline{conv}(K) is compact in E, and is the main positive result of this paper, aside a negative answer to Problem 1 of [4].

Theorem 9. Let E be a locally convex Hausdorff vector space, X be a closed convex subset of E with non-empty interior for σ(E, E^{∗}), K ⊂ X be a compact subset and A : E → E^{∗} a continuous affine mapping for the weak* topology. Assume moreover that there exists some compact convex subset K̃ of E containing K and that sup_{y∈K}⟨A(x), x − y⟩ > 0 for all x ∈ X \ K.

Then there exists a point x^{∗} ∈ K such that ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X.

Of course for deducing Theorem 8 from Theorem 9 we equip E with the weak topology and the completeness of E ensures that the weak compact set K has a relatively weakly compact convex hull in E.

In Section 2 we will prove our main result. Number of lemmas in this section are known but their proofs are quite simple and we will give them for sake of self-containedness.

In Section 3 we construct an example where the conclusion of Theorem 9 fails but the hypotheses of it all hold except for the existence of the compact convex set K̃ containing K. So it gives a negative answer to Problem 1 of [4].

In Section 4 we extend the result of Section 2 to the case where E is a Banach space and the strict inequality “ sup_{y∈K}⟨Ax, x − y⟩ > 0” is replaced as in the statement of Theorem 6 by the large inequality “sup_{y∈K}⟨Ax, x − y⟩ ≥ 0” for all x ∈ X \ K.

In section 5 an auxiliary result is proved for the case where A has finite rank, and finally in Section 6 we prove that in Theorem 8 the continuity of the operator A does not follow from the hypothesis “ sup_{y∈K}⟨A(x), x − y⟩ > 0 for every x in X \K ” but the conclusion “ there exists some x^{∗} ∈ K such that ⟨Ax^{∗}, x^{∗} −y⟩ ≤ 0 for all y ∈ X” still holds as in Theorem 8 without the hypothesis of continuity of A. Nevertheless in this frame we must keep the hypothesis of strict inequality “sup_{y∈K}⟨Ax, x − y⟩ > 0” as shown by a counterexample.

2. The main result

It is clear that the statement of Theorem 9 is invariant under translations in E. So without loss of generality we can and do assume that X is a weak neighborhood of 0. Furthermore we can replace if necessary the compact convex set K̃ by X ∩ K̃ which is compact and convex too and contains K. So we will assume that K ⊂ K̃ ⊂ X.

Lemma 2.1. If X is a closed convex neighborhood of 0 for the weak topology σ(E, E^{∗}), there exists a closed cofinite-dimensional subspace W of E, a complement Z of W in E and a closed convex neighborhood Y of 0 in Z such that E = W ⊕ Z and X = W +Y.

We will denote by P and Q the projections on W and Z respectively; then P and Q are continuous mappings from E to itself.

Proof. By definition of the weak topology σ(E, E^{∗}) there are finitely many f_{j} in E^{∗} (j = 1, 2, … d) such that X ⊃ {x ∈ E : sup_{j} |⟨f_{j}, x⟩| ≤ 1}. Define W = ⋂_{j} ker(f_{j}) which is a closed subspace of E of finite codimension k ≤ d and choose any finite-dimensional subspace Z of dimension k such that W ⊕ Z = E. If x ∈ X, w ∈ W and n is any integer then 2^{n}w ∈ W and

x_{n} = (1 − 2^{−n})x + w = (1 − 2^{−n})x + 2^{−n}(2^{n}w) ∈ conv({x} ∪ W) ⊂ X

And since X is closed then x+w = lim_{n} x_{n} ∈ X. It follows that for every w ∈ W: x ∈ X ⟺ x + w ∈ X. Set Y = X ∩ Z which is a closed convex subset of Z and a neighborhood of 0 in Z. For each x ∈ E there is a unique w ∈ W such that x − w ∈ Z; so x ∈ X ⟺ x − w ∈ Y, and X = Y + W. □

In the sequel of this section we will always silently assume that X is a closed weak neighborhood of 0 in E and that W, Z, P and Q are as in previous lemma.

We assume now that K ⊂ X is compact, that A : E → E^{∗} is affine and that sup_{y∈K}⟨A(x), x − y⟩ > 0 for all x ∈ X \ K. There exists a linear mapping φ : E → E^{∗} and some a = −A(0) ∈ E^{∗} such that A(x) = φ(x) − a for all x ∈ E.

Lemma 2.2. The mapping φ_{|W} is monotone on W, i.e. ⟨φ(w), w⟩ ≥ 0 for all w ∈ W.

Proof. Let w ∈ W \ {0}. Since K is compact there is some positive real R such that t.w ∉ K for t > R. For t > R there exists by hypothesis some y_{t} ∈ K such that ⟨A(t.w), t.w − y_{t}⟩ > 0. It follows that

0 < ⟨A(t.w), t.w − y_{t}⟩ = ⟨φ(t.w) − a, t.w − y_{t}⟩
= t^{2}⟨φ(w), w⟩ − t⟨a, w⟩ − t⟨φ(w), y_{t}⟩ + ⟨a, y_{t}⟩
≤ t^{2}⟨φ(w), w⟩ + t|⟨a, w⟩| + t|⟨φ(w), y_{t}⟩| + |⟨a, y_{t}⟩|

Since a and φ(w) belong to E^{∗}, their absolute values are bounded on the compact set K by m_{a} et m_{w} respectively, and one gets

⟨φ(w), w⟩ ≥ sup_{t>R}(−\frac{|⟨a, w⟩|}{t} − \frac{m_{w}}{t} − \frac{m_{a}}{t^{2}} ) = 0

what is the expected result since ⟨φ(w), w⟩ = 0 for w = 0. Notice that this lemma needs not φ to be continuous. If E is a Banach space and X = E (hence W = E) this implies the continuity of A by Theorem 5. □

From now on in this section we assume that A is continuous, hence also the linear mapping φ : x ↦ A(x) − A(0).

Definition 2.3. An affine mapping f : E → E^{∗} will be called almost monotone if there exists some continuous linear mapping g : E → E^{∗} with finite rank such that f − g is monotone.

Lemma 2.4. Let f : E → E^{∗} be a continuous affine mapping and W be a closed cofinite-dimensional subspace of E. If f_{|W} is monotone then f is almost monotone.

Proof. Denote by φ the continuous linear mapping x ↦ f(x) − f(0). Let Z be a complement of W in E, P and Q : E → E be the projections onto W and Z respectively (so P + Q is the identity mapping of E).

For any linear subspace L of E, L^{⊥} will denote the weak*-closed linear subspace {f ∈ E^{∗} : ∀z ∈ L ⟨f, z⟩ = 0} of E^{∗}. The continuous linear mappings P^{∗} and Q^{∗} from E^{∗} to itself are respectively the projections on Z^{⊥} and W^{⊥}. We have dim(W^{⊥}) = dim(Z) = k and P^{∗} + Q^{∗} = (P + Q)^{∗} = Id_{E}^{∗} = Id_{E^{∗}}.

Hence

φ = φ(P + Q) = φP + φQ = (P^{∗} + Q^{∗})φP + φQ = P^{∗}φP + Q^{∗}φP + φQ .

Define the linear mapping ψ = Q^{∗}φP + φQ which has finite rang since so do Q and Q^{∗} and is continuous since so is φ.

Then with φ̃ = φ−ψ = P^{∗}φP, ⟨φ̃(x), x⟩ = ⟨P^{∗}φ(P x), x⟩ = ⟨φ(P x), P x⟩ which is non-negative since A_{|W} is monotone, hence also φ_{|W} ; and this completes the proof, since f̃ = f − ψ is then affine and monotone. □

Corollary 2.5. A is almost monotone.

Proof. This follows immediately from Lemma 2.4. □

Lemma 2.6. Let f : E → E^{∗} be an almost monotone affine mapping and B be a bounded subset of E. Then the function x ↦ ⟨f(x), x⟩ is weakly lower semicontinuous on B.

Proof. Let ψ : E → E^{∗} be a continuous linear mapping with finite rank such that f − ψ is monotone and 𝔘 be a filter on B which converges weakly to x̂. We have to prove that we have ⟨f(x̂), x̂⟩ ≤ lim inf_{x,𝔘}⟨f(x), x⟩. Let a = f(0) and φ : x ↦ f(x) − a. Since we have lim_{x,𝔘}⟨a, x⟩ = ⟨a, x̂⟩ we have to show that

⟨φ(x̂), x̂⟩ ≤ lim inf_{x,𝔘}⟨φ(x), x⟩ .

And since the continuous function ψ has finite rank there are finitely many (f_{i})_{i≤n} and (g_{i})_{i≤n} in E^{∗} such that ψ(x) = ∑_{i≤n}⟨f_{i}, x⟩g_{i}. So we have

lim_{x,𝔘}⟨ψ(x), x⟩ = ∑_{i≤n} lim_{x,𝔘}⟨f_{i}, x⟩⟨g_{i}, x⟩ = ∑_{i≤n} lim_{x,𝔘}⟨f_{i}, x⟩ lim_{x,𝔘}⟨g_{i}, x⟩

= ∑_{i≤n}⟨f_{i}, x̂⟩⟨g_{i}, x̂⟩ = ⟨ψ(x̂), x̂⟩

So it remains only to prove that ⟨φ̃(x̂), x̂⟩ ≤ lim inf_{x,𝔘}⟨φ̃(x), x⟩. Since φ̃ is monotone we have ⟨φ̃(x) − φ̃(x̂), x − x̂⟩ ≥ 0, hence

⟨φ̃(x), x⟩ ≥ ⟨φ̃(x), x̂⟩ + ⟨φ̃(x̂), x⟩ − ⟨φ̃(x̂), x̂⟩

and since the right-hand side converges to

⟨φ̃(x̂), x̂⟩ + ⟨φ̃(x̂), x̂⟩ − ⟨φ̃(x̂), x̂⟩ = ⟨φ̃(x̂), x̂⟩ ,

we get lim inf_{x,𝔘}⟨φ̃(x), x⟩ ≥ ⟨φ̃(x̂), x̂⟩. □

Lemma 2.7. Let B be a compact convex subset of E , V be a finite-dimensional linear subspace of E meeting B and f : V → E^{∗} a continuous mapping. Then there is a point x ∈ B ∩ V such that ⟨f(x), x − y⟩ ≤ 0 for all y ∈ B ∩ V.

Proof. Assume the contrary : then the set U_{y} = {x : ⟨f(x), x − y⟩ > 0} would be open in B ∩ V for all y ∈ B ∩ V, and the family (U_{y})_{y∈B∩V} would cover the compact set B ∩ V. So there should be a finite continuous partition of the unity (χ_{y})_{y∈J} such that for all y ∈ J : χ_{y} > 0 ⟹ x ∈ U_{y}. Define then the continuous function g : B ∩ V → B ∩ V by

g(x) = ∑_{y∈J} χ_{y}(x).y .

It follows that for all x ∈ B∩V : ⟨f(x), x − g(x)⟩ = ∑_{y∈J} ψ_{y}(x)⟨f(x), x − y⟩ > 0. Thus the function g could not have any fixed point, and this would contradict Brouwer’s theorem since B ∩ V is finite-dimensional, non-empty, compact and convex. □

Lemma 2.8. Let B be a non-empty compact convex subset of E and f : E → E^{∗} be an almost monotone affine mapping. Then there is a point x̂ ∈ B such that ⟨f(x̂), x̂ − y⟩ ≤ 0 for all y ∈ B. Moreover if K ⊂ B and sup_{y∈K}⟨f(x), x − y⟩ > 0 for all x ∈ B \ K, this point x̂ has to be in K.

Proof. Let V be the set of all finite-dimensional linear subspaces of E meeting B and 𝔉 a filter on V such that for all V ∈ V the set {V′ ∈ V : V ⊂ V′} belongs to 𝔉. For each V ∈ V there is by Lemma 2.7 some x_{V} ∈ B ∩ V such that ⟨f(x_{V}), x_{V} ⟩ ≤ ⟨f(x_{V}), y⟩ for all y ∈ B ∩ V. By compactness we can find a cluster point x̂ of x_{V} along 𝔉.

For every y ∈ B the set {V ∈ V : y ∈ V } is in 𝔉. By Lemma 2.6 we have

⟨f(x̂), x̂⟩ ≤ lim sup_{V,𝔉}⟨f(x), x⟩ ≤ lim sup_{V,𝔉}⟨f(x), y⟩ ≤ ⟨f(x̂), y⟩

what is the expected inequality.

If moreover K ⊂ B it is impossible that such a x̂ lies in B \ K : if not by hypothesis there should be an y ∈ K ⊂ B such that ⟨f(x̂), x̂ − y⟩ > 0, in contradiction with the choice of x̂. □

Let F be any finite-dimensional linear subspace of W and F_{1} be a generating symmetric convex compact subset of F, and choose a compact convex subset Y_{0} of Y. Consider the compact convex subset B of X defined by

B = conv(K̃ ∪ Y_{0}) + F_{1}  (∗)

which is clearly contained in X + F ⊂ X + W = X, and contains K.

Lemma 2.9. There exists at least a point x̂ ∈ K such that ⟨A(x̂), x̂ − y⟩ ≤ 0 for all y ∈ Y_{0} + F.

Proof. Following Corollary 2.5 the mapping A is almost monotone. Applying Lemma 2.8 to the compact convex set B defined by (∗) we get some x̂ in K such that ⟨A(x̂), x̂−y⟩ ≤ 0 for all y ∈ B, a fortiori for all y ∈ Y_{0}. Moreover, for z ∈ F_{1}, the points y_{1} = x̂ + z and y_{2} = x̂ − z both belong to B. Thus for j = 1, 2:

⟨A(x̂), x̂ − y_{j}⟩ = (−1)^{j}.⟨A(x̂), z⟩ ≤ 0

from what we deduce ⟨A(x̂), z⟩ = 0 for all z ∈ F_{1} hence for all z ∈ F. So for y_{0} ∈ Y_{0} and z ∈ F :

⟨A(x̂), x̂ − (y_{0} + z)⟩ = ⟨A(x̂), x̂ − y_{0}⟩ − ⟨A(x̂), z⟩ = ⟨A(x̂), x̂ − y_{0}⟩ ≤ 0

as claimed. □

We now can prove our main result (Theorem 9 in the introduction):

Theorem 2.10. Let E be a locally convex Hausdorff vector space, X be a closed convex subset of E with non-empty interior for σ(E, E^{∗}), K ⊂ X be a compact subset and A : E → E^{∗} a continuous affine mapping for the weak* topology. Assume moreover that there exists some compact convex subset K̃ of E containing K and that sup_{y∈K}⟨Ax, x − y⟩ > 0 for all x ∈ X \ K. Then there exists a point x^{∗} ∈ K such that ⟨Ax^{∗}, x^{∗} − y⟩ ≤ 0 for all y ∈ X.

Proof. For all finite-dimensional linear subspace F of W and every compact convex subset Y_{0} of Y, previous lemma yields a x̂_{Y_{0},F} ∈ K satisfying the inequality ⟨Ax^{∗}, x^{∗} − y⟩ ≤ 0 for all y ∈ Y_{0} + F. Choose a filter 𝔉 on the set 𝒫 of such pairs (Y_{0}, F), such that the set {C, G) ∈ 𝒫 : Y_{0} ⊂ C and F ⊂ G} belongs to 𝔉 for all (Y_{0}, F) then a cluster value x^{∗} ∈ K of x̂_{Y_{0},F} along 𝔉 since K is compact. And refining 𝔉 into 𝔘 if necessary we can assume that x^{∗} = lim_{(Y_{0},F),𝔘} x̂_{Y_{0},F}.

For y ∈ Y + W the set {(Y_{0}, F) ∈ 𝒫 : (Qy, P y) ∈ Y_{0} × F } belongs to 𝔘 and it follows from Lemma 2.6 that

⟨A(x^{∗}), x^{∗} − y⟩ = ⟨A(x^{∗}), x^{∗}⟩ − ⟨A(x^{∗}), y⟩

≤ lim sup_{(Y_{0},F),𝔘}⟨A(x̂_{Y_{0},F}), x̂_{Y_{0},F} ⟩ − lim_{(Y_{0},F),𝔘}⟨A(x̂_{Y_{0},F}), y⟩

= lim sup_{(Y_{0},F),𝔘}⟨A(x̂_{Y_{0},F}), x̂_{Y_{0},F} − y⟩ ≤ 0

and since X = Y + W the proof is complete. □

Proof of Theorem 6. Recall that a Banach space E is said to be weakly compactly generated (w.c.g. for short) if it contains a weakly compact subset B which is total in E (i.e. the closed linear subspace generated by B is the whole E). It is easy to see that any reflexive or separable Banach space is w.c.g.

Assume first that E is w.c.g. Then there exists a weakly compact symmetric subset B of E which is total in E. For ε > 0 we set K_{ε} = K + ε.B. Then K_{ε} is weakly compact.

It follows that p(ξ) = sup_{u∈B}⟨ξ, u⟩ > 0 for all ξ ∈ E^{∗} \ {0}. For all ε > 0 and all x ∈ E \ K_{ε} :

sup_{y∈K_{ε}}⟨A(x), x − y⟩ = sup_{y∈K}⟨A(x), x − y⟩ + ε. sup_{u∈B}⟨A(x), u⟩

= sup_{y∈K}⟨A(x), x − y⟩ + ε.p(A(x)) ≥ ε.p(A(x)) > 0

since A(x) ≠ 0 for x ∉ K_{ε}.

It follows then from Theorem 8 applied to X = E that there exists some x_{ε}^{∗} ∈ K_{ε} such that ⟨A(x_{ε}^{∗}), x_{ε}^{∗} − y⟩ ≤ 0 for all y ∈ E, it is A(x_{ε}^{∗}) = 0. And since we assumed that A(x) ≠ 0 for all x ∈ E \ K we conclude that x_{ε}^{∗} ∈ K. Again since A is monotone on the Banach space E by Lemma 2.2 (notice that W = E) A is continuous. Then if x^{∗} is any weak cluster point of x_{ε}^{∗}, x^{∗} lies in K and A(x^{∗}), cluster point of A(x_{ε}^{∗}), is 0.

In the general case denote by E_{0} the closed linear subspace of E generated by K, by B the unit ball of E and by ℱ the set of finite-dimensional subspaces of E ordered by inclusion. It is clear that if F ∈ ℱ then K + (B ∩ F) is weakly compact and generates the Banach space E_{0} + F. Thus E_{0} + F is w.c.g. and it follows from what precedes applied to E_{0} + F that there is some x_{F}^{∗} ∈ K such that ⟨A(x_{F}^{∗}), y⟩ = 0 for all y ∈ E_{0} + F. Again if x^{∗} is a cluster point of (x_{F}^{∗})_{F ∈ℱ} the continuity of the monotone mapping A implies that ⟨A(x^{∗}), y⟩ = 0 for all y ∈ E, it is A(x^{∗}) = 0. □

3. The negative result

We now prove by constructing a counterexample that Theorem 9 fails when removing the hypothesis that K is contained in some compact convex subset K̃ of E. Of course such a space E cannot be complete because of Theorem 4.

Theorem 3.1. There exists a prehilbertian space E, a continuous affine mapping A from E to E^{∗} and a norm-compact subset K of E such that both inequalities

sup_{y∈K}⟨A(x), x − y⟩ > 0 for all x ∈ E \ K  (1)
sup_{y∈E}⟨A(x), x − y⟩ > 0 for all x ∈ K .  (2)

hold.

Let 𝕋 be the unit circle of the complex plane which is a compact group and equip with the normalized Haar measure λ, so that ∫_{𝕋} ds = 1 and ∫_{𝕋} z^{m} dλ(z) = 0 for m ∈ ℤ \ {0}.

The space E will be a linear subspace of the Sobolev space H^{−1}(𝕋) of distri-

butions f on 𝕋 whose Fourier series f̂ satisfies ∑_{n∈ℤ} \frac{|f̂(n)|^{2}}{1 + n^{2}} < ∞ equipped with

the norm ∥f∥_{−1} = (∑_{n∈ℤ} \frac{|f̂(n)|^{2}}{1 + n^{2}} )^{1/2}. This space is a Hilbert space whose dual

can be identified with the space H^{1}(𝕋) of distributions g on 𝕋 such that g and its derivative g′ are both in L^{2}(𝕋). We will respectively denote by ∥.∥_{−1} and ∥.∥_{1} the norms on H^{−1}(𝕋) and H^{1}(𝕋).

Lemma 3.2. For each s ∈ 𝕋 the Dirac measure δ_{s} at s is in H^{−1}(𝕋). Moreover the mapping s ↦ δ_{s} is continuous (and even Hölder) from 𝕋 to H^{−1}(𝕋).

Proof. Let e_{n} be the function z ↦ z^{−n}. We have δ̂_{s}(n) = ⟨e_{n}, δ_{s}⟩ = s^{−n}, and

∑_{n∈ℤ} \frac{|δ̂_{s}(n)|^{2}}{1 + n^{2}} = ∑_{n∈ℤ} \frac{1}{1 + n^{2}} < +∞ .

We have for any m ∈ ℤ, s ≠ t ∈ 𝕋,

|s^{m} − t^{m}| = |s^{|m|} − t^{|m|}| = |(s − t)( ∑_{p+q=|m|−1} s^{p}t^{q})| ≤ |m|.|s − t|

hence

∥δ_{s} − δ_{t}∥_{−1}^{2} = ∑_{n∈ℤ} \frac{|δ̂_{s}(n) − δ̂_{t}(n)|^{2}}{1 + n^{2}} = ∑_{n∈ℤ} \frac{|s^{−n} − t^{−n}|^{2}}{1 + n^{2}} = ∑_{n≠0} \frac{|s^{−n} − t^{−n}|^{2}}{1 + n^{2}}

≤ ∑_{0<|n|≤p} \frac{n^{2}|s − t|^{2}}{1 + n^{2}} + ∑_{|n|>p} \frac{4}{1 + n^{2}}

≤ 2p.|s − t|^{2} + 4 ∑_{n=p+1}^{∞}(\frac{1}{n − 1} − \frac{1}{n}) ≤ 2p.|s − t|^{2} + \frac{4}{p}

Take p such that 1 ≤ p ≤ θ := \frac{2}{|s − t|} < p + 1. We get

∥δ_{s} − δ_{t}∥_{−1}^{2} ≤ |s − t|(2p · \frac{2}{θ} + \frac{4}{p} · \frac{θ}{2}) ≤ |s − t|(4 + 2 · \frac{p + 1}{p}) ≤ 8|s − t|

hence ∥δ_{s} − δ_{t}∥ ≤ \sqrt{8|s − t|}. □

It follows from previous lemma that K = {δ_{s} : s ∈ 𝕋} is a norm-compact subset of H^{−1}(𝕋).

Let (α_{m})_{m≥0} be a series such that sup_{m}(m^{2} + 1)α_{m} < +∞ and α_{m} > 0 for all m (e.g. α_{m} = 2^{−m}) then consider the continuous function

h : z ↦ α_{0} + 2 ∑_{m>0} α_{m}.ℜ(z^{m}) = ∑_{m∈ℤ} α_{|m|}.z^{m}

from 𝕋 to ℝ. We want to define φ from H^{−1}(𝕋) to H^{1}(𝕋) by φ(f) = f ∗ h.

Lemma 3.3. The mapping φ is linear and continuous from H^{−1}(𝕋) to H^{1}(𝕋).

Proof. The mapping φ is clearly linear. By Fourier transform it is enough to show

that ∥f ∗ h∥_{1}^{2} ≤ θ^{2}∥f∥_{−1}^{2} for some constant θ. We have ∥f∥_{−1}^{2} = ∑_{m∈ℤ} \frac{|f̂(m)|^{2}}{m^{2} + 1} ,

{ \atop f ∗ h}(m) = f̂(m).ĥ(m), ĥ(m) = α_{|m|} and ∥h ∗ f∥_{1}^{2} = ∑_{m∈ℤ}(m^{2} + 1)|{ \atop f ∗ h}(m)|^{2}

hence

∥h ∗ f∥_{1}^{2} = ∑_{m∈ℤ}(m^{2} + 1)|{ \atop f ∗ h}(m)|^{2} ≤ ∑_{m∈Z} α_{|m|}^{2}(m^{2} + 1)^{2} \frac{|f̂(m)|^{2}}{m^{2} + 1} ≤ θ^{2}∥f∥_{−1}^{2}

with θ = sup_{m≥0}(m^{2} + 1)α_{m}. □

Let us define E ⊂ H^{−1}(𝕋) as the linear space spanned by the (δ_{s})_{s∈𝕋} and the map A by

A(f) = φ(f) − α_{0}.1l

where 1l denotes the constant function on 𝕋 with value 1, which belongs to H^{1}(𝕋). Thus K is a norm-compact subset of E and A is an affine continuous mapping.

Lemma 3.4. There exists no δ_{s} ∈ K such that ⟨A(δ_{s}), δ_{s} − f⟩ ≤ 0 for all f ∈ E; so A satisfies (2).

Proof. If for some s ∈ 𝕋 this inequality were true for all f ∈ E we would have for all integer n ∈ ℤ:

⟨A(δ_{s}), δ_{s}) ≤ ⟨A(δ_{s}), nf⟩

hence necessarily ⟨A(δ_{s}), f) = 0 for all f ∈ E and in particular for f = δ_{s} − δ_{t} for all t ∈ 𝕋. Since ⟨A(δ_{s}), δ_{s} − δ_{t}⟩ = ⟨h ∗ δ_{s}, δ_{s} − δ_{t}⟩ = h(0) − h(s − t) we should have

0 = ∫_{𝕋}(h(0) − h(s − t)) dt = h(0) − ∫_{𝕋} h(t) dt = h(0) − α_{0} = 2 ∑_{m>0} α_{m} > 0

and this contradiction achieves the proof. □

Lemma 3.5. Let f = ∑_{j} λ_{j}δ_{s_{j}} ∈ E. Then ⟨φ(f), f) = ∑_{m=0}^{∞} α_{|m|}|f̂(m)|^{2} where f̂(m) = ⟨e_{m}, f⟩ is the m^{th} Fourier coefficient of f.

Proof. Indeed:

⟨φ(f), f⟩ = ∑_{j,k} λ_{j}λ_{k}h(s_{j}.s_{k}^{−1}) = ∑_{m∈ℤ} α_{|m|} ∑_{j,k} λ_{j}λ_{k}s_{j}^{−m}s_{k}^{m}

= ∑_{m∈ℤ} α_{|m|}(∑_{j} λ_{j}s_{j}^{−m})(∑_{k} λ_{k}s_{k}^{m})

= ∑_{m∈ℤ} α_{|m|}(∑_{j} λ_{j}s_{j}^{−m})(\overline{∑_{k} λ_{k}s_{k}^{−m}}) = ∑_{m∈ℤ} α_{|m|}|∑_{j} λ_{j}s_{j}^{−m}|^{2}

= ∑_{m∈ℤ} α_{|m|}|⟨e_{m}, f⟩|^{2} = ∑_{m∈ℤ} α_{|m|}|f̂(m)|^{2}

what is the expected result. Notice that f̂(0) = ⟨1l, f⟩ = ∑_{j} λ_{j}. □

Lemma 3.6. Let f = ∑_{j} λ_{j}δ_{s_{j}} ∈ E \ K. Then there exists at least one s ∈ 𝕋 such that ⟨A(f), f − δ_{s}⟩ > 0; so A satisfies (1).

Proof. Since the function s ↦ ⟨A(f), f − δ_{s}⟩ is continuous on 𝕋, it is enough to

show that ∫_{𝕋}⟨A(f), f − δ_{s}⟩ ds > 0 for f ∈ E \ K. Indeed, following Lemma 3.5

⟨A(f), f⟩ = ⟨φ(f), f⟩ − α_{0}⟨1l, f⟩ = ∑_{m∈ℤ} α_{|m|}|f̂(m)|^{2} − α_{0}f̂(0)

and ⟨A(f), δ_{s}⟩ = ⟨φ(f), δ_{s}⟩ − α_{0}⟨1l, δ_{s}⟩ = ⟨φ(f), δ_{s}⟩ − α_{0}. Thus

∫_{𝕋}⟨A(f), δ_{s}⟩ ds = ∑_{j} λ_{j} ∫ h(s.s_{j}^{−1}) ds − α_{0}

= (∑_{m∈ℤ} ∑_{j} α_{|m|}λ_{j} ∫_{𝕋} z^{−m}s_{j}^{m} dz) − α_{0}

= α_{0}(∑_{j} λ_{j}) − α_{0} = α_{0}f̂(0) − α_{0}

It follows that

∫_{𝕋}⟨A(f), f − δ_{s}⟩ ds = ∑_{m∈ℤ} α_{m}|f̂(m)|^{2} − α_{0}f̂(0) − (α_{0}f̂(0) − α_{0})

= ∑_{m∈ℤ,m≠0} α_{m}|f̂(m)|^{2} + α_{0}f̂(0)^{2} − α_{0}f̂(0) − (α_{0}f̂(0) − α_{0})

= 2 ∑_{m=1}^{∞} α_{m}|f̂(m)|^{2} + α_{0}|f̂(0) − 1|^{2} ≥ 0

Moreover if we had ∫_{𝕋}⟨A(f), f − δ_{s}⟩ ds ≤ 0 we would necessarily have f̂(m) = 0

for all m ≠ 0 and f̂(0) = 1. But the only element in H^{−1}(𝕋) having these Fourier coefficients is the normalized Haar measure which does not belong to E. It follows that

sup_{s∈𝕋}⟨A(f), f − δ_{s}⟩ ds ≥ ∫_{𝕋}⟨A(f), f − δ_{s}⟩ ds > 0

for all f ∈ E \ K. □

4. The non-strict case

The aim of this section is to prove the following :

Theorem 4.1. Let E be a Banach space, X be a closed convex subset of E with non-empty interior, K be a weakly compact convex subset of X and A : E → E^{∗} be an almost monotone affine function.

Assume that for all x ∈ X, sup_{y∈K}⟨A(x), x − y⟩ ≥ 0. Then there exists some x^{∗} ∈ K such that ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X.

Moreover if W is a closed cofinite-dimensional linear subspace of E such that X + W = X and K_{0} = {y ∈ K : A(y)_{|W} = 0} then K ⊂ K_{0} + W.

Notice that in this statement the strict inequality sup_{y∈K}⟨A(x^{∗}), x^{∗} − y⟩ > 0 which was assumed in Theorem 9 has been replaced by the large inequality sup_{y∈K}⟨A(x^{∗}), x^{∗} − y⟩ ≥ 0. In order to prove this last statement we need some preliminary results. Next lemma is probably well-known.

Lemma 4.2. Let E be a Banach space, K′ be a weakly compact subset of E, f be a non-zero continuous linear functional on E and μ = sup_{x∈K}⟨f, x⟩. Then the intersection of the hyperplane H = f^{−1}(μ) and the closed convex hull K_{1} of K′ is the closed convex hull of K′ ∩ H.

Proof. Since K′ ∩ H is contained in H and in \overline{conv}(K′) it is clear that the weakly compact convex set K′′ = \overline{conv}(K′ ∩H) is contained in H ∩\overline{conv}(K′). Conversely if there were some point a ∈ H \ K′′ there would be some g ∈ E^{∗} such that g(a) > sup_{z∈K′}⟨g, z⟩. Consider the weakly compact sets

L_{n} = {x ∈ K′ : 2^{n}(f(x) − μ) + g(x) − g(a) ≥ 0}

Since f(x) ≤ μ for x ∈ K′, we have for x ∈ L_{n+1} :

0 ≤ μ − f(x) ≤ 2^{−n−1}(g(x) − g(a))

hence g(x) ≥ g(a) and 2^{−n−1}(g(x) − g(a)) ≤ 2^{−n}(g(x) − g(a)). It follows that L_{n+1} ⊂ L_{n}. If x_{0} ∈ ⋂_{n} L_{n} we have necessarily f(x_{0}) ≤ μ hence f(x_{0}) = μ and x_{0} ∈ H ∩ K′ and also g(x_{0}) ≥ g(a) > sup_{z∈K∩H}⟨g, z⟩. This contradiction implies that ⋂_{n} L_{n} = ∅ hence by compactness that L_{m} = ∅ for some integer m. Thus the linear functional g + 2^{m}f separates a from K′ and a ∉ H ∩ \overline{conv}(K′), so K′′ = H ∩ \overline{conv}(K′). □

Lemma 4.3. Let E be a w.c.g. Banach space, K ⊂ E be a weakly convex compact set, a ∈ E \ K, B a weakly compact symmetric and total subset of E and ε > 0 such that (a + εB) ∩ K = ∅. Then the set

K′ = {(1 − t)x + ta + εt(1 − t)u : t ∈ [0, 1], x ∈ K, u ∈ B}

is a weakly compact set contained in C_{ε} = conv(K ∪ (a + εB)) and containing K ∪ {a}. Moreover if p denotes the norm f ↦ sup_{u∈B}⟨f, u⟩ on E^{∗} and γ(f) denotes

sup_{0≤t≤1}((1 − t) sup_{z∈K}⟨f, z⟩ + tf(a) + εt(1 − t)p(f))

for f ∈ E^{∗} then the set K_{1} = {x ∈ E : ∀f ∈ E^{∗} f(x) ≤ γ(f)} is the closed convex hull of K′.

Proof. For t ∈ [0, 1] and u ∈ B the point a + ε(1 − t)u belongs to a + εB. Thus for x ∈ K : (1 − t)x + ta + εt(1 − t)u = (1 − t)x + t(a + ε(1 − t)u) ∈ C_{ε}. And using t = 0 or t = 1 it is clear that K ⊂ K′ and a ∈ K′. Since [0, 1], K and B are weakly compact, so are C_{ε} and K′.

For f ∈ E^{∗} it is clear that

sup_{z∈K′}⟨f, z⟩ = sup_{0≤t≤1}((1 − t) sup_{z∈K}⟨f, z⟩ + tf(a) + εt(1 − t)p(f)) = γ(f)

It follows that K′ ⊂ K_{1} and that K_{1} is closed and convex. We now prove that K_{1} is the closed convex hull of K′: if b ∈ K_{1} \ \overline{conv}(K′) there exists by Hahn-Banach an f ∈ E^{∗} such that f(b) > sup_{z∈ \overline{conv}(K′)}⟨f, z⟩ = sup_{z∈K′}⟨f, z⟩ = γ(f). And f(b) > γ(f) contradicts b ∈ K_{1}. In particular K_{1} ⊂ C_{ε} hence K_{1} is weakly compact. □

Lemma 4.4. If f ∈ E^{∗} is not 0, it attains its maximum on K_{1} at some point x, and if sup_{z∈K}⟨f, z⟩ = γ(f) then x ∈ K.

Proof. Since K′ is compact and f continuous there is some point x′ ∈ K′ ⊂ K_{1} such that m := ⟨f, x′⟩ = sup_{z∈K′}⟨f, z⟩ = γ(f). And since {z ∈ E : ⟨f, z⟩ ≤ m} is closed convex and contains K′, it contains K_{1}. So sup_{z∈K_{1}}⟨f, z⟩ = m = f(x′).

Assume f ≠ 0 attains its maximum on K_{1} at x and denote μ = sup_{z∈K}⟨f, z⟩. We have

f(x) = γ(f) = sup_{0≤t≤1}((1 − t)μ + tf(a) + εt(1 − t)p(f))

Since f ≠ 0 we have p(f) > 0 and the function

ρ : t ↦ (1 − t)μ + tf(a) + εt(1 − t)p(f)

is strictly concave on [0, 1]: so it attains its maximum m on [0, 1] at a unique point s. If γ(f) = μ = ρ(0) we necessarily have s = 0. Since x′ ∈ K′ there are θ ∈ [0, 1], y ∈ K and u ∈ B such that x′ = (1 − θ)y + θa + εθ(1 − θ)u. Therefore

γ(f) = f(x′)
= (1 − θ)f(y) + θf(a) + εθ(1 − θ)f(u)
≤ (1 − θ)μ + θf(a) + εθ(1 − θ)p(f) = ρ(θ)
≤ γ(f) = μ

hence ρ(θ) = μ. This implies that θ = 0 and x′ = y ∈ K. So K′ ∩ f^{−1}(μ) ⊂ K. And since K_{1} = \overline{conv}(K′) it follows from Lemma 4.2 that

x ∈ K_{1} ∩ f^{−1}(μ) ⊂ \overline{conv}(K′ ∩ f^{−1}(μ)) ⊂ \overline{conv}(K) = K ,

what is the desired conclusion. □

Theorem 4.5. Let E be a w.c.g. Banach space, X be a closed convex subset of E, K ⊂ X be a weakly compact convex set and a ∉ K an interior point of X. Then there exists a convex weakly compact set K_{1} such that K ⊂ K_{1} ⊂ X and that for all f ∈ E^{∗} the set M(f) := {x ∈ K_{1} : f(x) = sup_{z∈K_{1}} f(z)} satisfies M(f) ⊂ K or M(f) ∩ K = ∅.

Proof. This follows immediately from Lemmas 4.3 and 4.4. □

Theorem 4.6. Let E be a w.c.g. Banach space, X be a closed convex subset of E with non-empty interior, K be a weakly compact convex subset of X and A : E → E^{∗} be an almost monotone affine function.

Assume that for all x ∈ X, sup_{y∈K}⟨A(x), x − y⟩ ≥ 0. Then there exists some x^{∗} ∈ K such that ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X.

Proof. Assume first that A(x) ≠ 0 for all x ∈ X \ K.

Claim 4.7. Let J = {a_{0}, a_{1}, …, a_{m−1}} be a finite subset of the interior X_{0} of X. Then there exists some x̂_{J} ∈ K such that ⟨A(x̂_{J}, x̂_{J} − y⟩ ≤ 0 for all y ∈ K ∪ J.

Proof. Using Theorem 4.5 we construct inductively a finite non-decreasing sequence (K_{n})_{n≤m} of convex weakly compact subsets of X such that K_{0} = K, a_{n} ∈ K_{n+1} and for all f ≠ 0 in E^{∗} and 0 ≤ n ≤ m

M_{n}(f) ⊂ K or M_{n}(f) ∩ K = ∅  (†)

where M_{n}(f) denotes the set {x ∈ K_{n} : ⟨f, x⟩ = γ_{n}(f) := sup_{z∈K_{n}}⟨f, z⟩}. Indeed (†) holds for K_{0}. Assume 0 ≤ n < m, K_{n} is constructed and (†) holds for K_{n}.

If a_{n} ∈ K_{n} set K_{n+1} = K_{n}. And if a_{n} ∉ K_{n} choose ε_{n} > 0 small enough for a_{n} + ε_{n}B ⊂ X \ K_{n}. Then the convex weakly compact set K_{n+1} given by Lemma 4.3 from K_{n} and a_{n} + ε_{n}B is contained in conv(K_{n} ∪ (a_{n} + ε_{n}B)) hence in X. Moreover for f ≠ 0 in E^{∗}

- either M_{n+1}(f) ∩ K_{n} = ∅ and a fortiori M_{n+1}(f) ∩ K = ∅,

- or M_{n+1}(f) ⊂ K_{n}, so γ_{n+1}(f) = γ_{n}(f) and

M_{n}(f) ⊂ M_{n+1}(f) ⊂ {x ∈ K_{n} : ⟨f, x⟩ = γ_{n}(f))} = M_{n}(f)

And since (†) holds for K_{n},

M_{n+1}(f) = M_{n}(f) ⊂ K or M_{n+1}(f) ∩ K = M_{n}(f) ∩ K = ∅

hence (†) holds for K_{n+1}. Since A is almost monotone it follows from Lemma 2.8 that one can find some x̂_{J} = x̂_{m} ∈ K_{m} such that ⟨A(x̂_{J}), x̂_{J} − y⟩ ≤ 0 for all y ∈ K_{m} and in particular ⟨A(x̂_{J}), x̂_{J} − y⟩ ≤ 0 for all y ∈ K ∪ J. □

By hypothesis the linear functional f = −A(x̂_{J}) satisfies

∃y_{0} ∈ K ⟨f, y_{0}⟩ ≥ ⟨f, x̂_{J}⟩ and ∀y ∈ K_{m} ⟨f, x̂_{J}⟩ ≥ ⟨f, y⟩

it is x̂_{J} ∈ M_{m}(f) and y_{0} ∈ K ∩ M_{m}(f). So by (†) we get x̂_{J} ∈ K or f = 0, hence x̂_{J} ∈ K since A(x) ≠ 0 for x ∈ X \ K. And if x^{∗} is a cluster value in K of the x̂_{J}’s for finite J ⊂ X \ K we get by Lemma 2.6 : ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all finite J ⊂ X_{0} \ K and all y ∈ K ∪ J hence for all y ∈ X since X_{0} is dense in X.

If there exists some x̂ ∈ K such that A(x̂) = 0, this x̂ satisfies ⟨A(x̂), x̂−y⟩ = 0 for all y ∈ X. Finally, if the closed convex set N = {x ∈ X : A(x) = 0} is non-empty and disjoint from K, one can find by Hahn-Banach some h ∈ E^{∗} and some α ∈ ℝ such that sup_{x∈K}⟨h, x⟩ < α < inf_{y∈N}⟨h, x⟩.

Then replace X by X_{α} = {y ∈ X : ⟨h, y⟩ ≤ α} ⊃ K: we get A(x) ≠ 0 for all x ∈ X_{α} and sup_{y∈K}⟨A(x), x − y⟩ ≥ 0 for all x ∈ X_{α}; and it follows from what precedes that one can find some x^{∗} ∈ K such that ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X_{α}. Then for y ∈ X \ X_{α} put y_{t} = x^{∗} + t(y − x^{∗}) for 0 < t ≤ 1.

Since ⟨h, y_{t}⟩ = ⟨h, x^{∗}⟩ + t⟨h, y − x^{∗}⟩ we get ⟨h, y_{t}⟩ = α hence y_{t} ∈ X_{α} for

t = \frac{α − ⟨h, x^{∗}⟩}{(⟨h, y⟩ − α) + (α − ⟨h, x^{∗}⟩)} ∈]0, 1[ then x^{∗} − y_{t} = t(x^{∗} − y) and

⟨A(x^{∗}), x^{∗} − y⟩ = \frac{1}{t}⟨A(x^{∗}), x^{∗} − y_{t}⟩ ≤ 0

thus ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X. □

Proof of Theorem 4.1. If E is w.c.g. this is exactly the statement of Theorem 4.6. If not, as in the proof of Theorem 6, denote by E_{0} the closed linear space generated by K, by ℱ the set of finite-dimensional linear subspaces of E and by B_{E} the unit ball of E. For all F ∈ ℱ, E_{0} + F is a w.c.g. Banach space since K + F ∩ B_{E} is weakly compact and total in E_{0} + F, and X_{F} = X ∩ (E_{0} + F) is a closed convex neighborhood of K in E_{0} + F. Moreover A_{|E_{0}+F} is almost monotone and sup_{y∈K}⟨A(x), x − y⟩ ≥ 0 for all x ∈ X ∩ (E_{0} + F). It follows from Theorem 4.6 that one can find some x̂_{F} ∈ K such that ⟨A(x̂_{F}), x̂_{F} − y⟩ ≤ 0 for all y ∈ X_{F}. Then by Lemma 2.6 any weak cluster value x^{∗} in K of the x̂_{F} ’s will satisfy ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X.

Let Z be a complement of W in E, Q : E → Z be the linear projection with kernel W and W^{o} the polar set of W, i.e. the set {f ∈ E^{∗} : ∀w ∈ W ⟨f, w⟩ = 0}. Since A is affine and continuous, K_{0} = K ∩ A^{−1}(W^{o}) is convex and compact, and so is Q(K_{0}) ⊂ Z.

Replace X by X′ = K + W. For any linear functional ℓ on Z and any integer n consider α = sup_{x∈K}⟨ℓ, Q(x)⟩ = sup_{x∈X′}⟨ℓ, Q(x)⟩ and the affine mapping A′ : E → E^{∗} defined by

A′(x) = A(x) + 2^{n}(⟨Q^{∗}(ℓ), x⟩ − α).Q^{∗}(ℓ) .

We claim that for all x ∈ X′ \ K we have sup_{y∈K}⟨A′(x), x − y⟩ ≥ 0. Indeed if y_{0} ∈ K satisfies ⟨A(x), x − y_{0}⟩ = sup_{y∈K}⟨A(x), x − y⟩ ≥ 0 we have

⟨A′(x), x − y_{0}⟩ = ⟨A(x), x − y_{0}⟩ + 2^{n}(⟨ℓ, Q(x) − α⟩).⟨ℓ, Q(x) − Q(y_{0})⟩
≥ 2^{n}(⟨ℓ, Q(x)⟩ − α).⟨ℓ, Q(x) − Q(y_{0})⟩ ≥ 0

since ⟨ℓ, Q(x)⟩ ≤ α and ⟨ℓ, Q(x)⟩ ≤ ⟨ℓ, Q(y_{0})⟩. So A′ satisfies the conditions of Theorem 4.6 and there exists x′ ∈ K such that ⟨A′(x′), x′ − y⟩ ≤ 0 for all y ∈ X′ and in particular for all y ∈ x′ + W. Hence A′(x′)_{|W} = 0, and since Q_{|W} = 0 we have A(x′)_{|W} = A′(x′)_{|W} = 0, thus x′ ∈ K_{0}. Moreover,

0 ≥ ⟨A′(x′), x′ − y_{0}⟩ = ⟨A(x′), x′ − y_{0}⟩ + 2^{n}(⟨ℓ, Q(x′)⟩ − α).⟨ℓ, Q(x′) − Q(y_{0})⟩

≥ −∥A(x′)∥.∥x′ − y_{0}∥ + 2^{n}(α − ⟨ℓ, Q(x′)⟩)^{2}

Since the set {A(x) : x ∈ K} is weak*-compact in E^{∗}, it is bounded and M^{2} = sup_{x∈K} ∥A(x)∥. diam(K) < +∞. So α − ⟨ℓ, Q(x′)⟩ ≤ 2^{−n/2}.M and

sup_{y∈K_{0}}⟨ℓ, Q(y)⟩ ≥ ⟨ℓ, Q(x′)⟩ ≥ α − 2^{−n/2}.M = sup_{x∈K}⟨ℓ, Q(x)⟩ − 2^{−n/2}.M

It follows that for all ℓ ∈ Z^{∗} we get sup_{z∈Q(K_{0})}⟨ℓ, z⟩ = sup_{z∈Q(K)}⟨ℓ, z⟩. So Q(K_{0}) = Q(K) by Hahn-Banach’s theorem whence K ⊂ K + W = K_{0} + W. □

5. The finite-rank case

We now examine the case where the (non-necessarily continuous) affine operator A on E = W × Z takes values in W^{o} = {f ∈ E^{∗} : ∀w ∈ W ⟨f, w⟩ = 0}, with W Banach space and Z finite-dimensional space.

Again we assume that K is a weakly compact subset of E, Y is a closed convex subset of Z, K ⊂ X = W × Y and that for all x ∈ X \ K the inequality

sup_{y∈K}⟨A(x), x − y⟩ > 0  (∗)

holds. And we want to prove that there exists some x^{∗} ∈ K such that

sup_{y∈X}⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 .  (∗∗)

As previously we denote by P and Q the projections on W and Z. We also denote by M the finite-dimensional compact convex set Q(K) ⊂ Z that we look as a subset of E by identifying z ∈ Z with (z, 0) ∈ E. Let φ : W → Z^{∗} be the linear mapping w ↦ A(w)−A(0) ; in particular we have A(x+w) = A(x)+φ(w) for all x ∈ E and all w ∈ W.

Lemma 5.1. If W is infinite-dimensional there exist points x_{∗} in K and x in X \ K such that Q(x) = Q(x_{∗}) and ⟨A(x), x_{∗}⟩ = sup_{y∈K}⟨A(x), y⟩.

Proof. Since Z is finite-dimensional the affine mapping h : Z → Z^{∗} defined by h(z) = −A(z), is continuous. So applying Lemma 2.7 to M and h we can find u^{∗} ∈ M such that sup_{v∈M}⟨h(u^{∗}), u^{∗} −v⟩ ≤ 0, so in fact sup_{v∈M}⟨h(u^{∗}), u^{∗} −v⟩ = 0 since sup_{v∈M}⟨h(u^{∗}), u^{∗} − v⟩ ≥ ⟨h(u^{∗}), u^{∗} − u^{∗}⟩ = 0. This means that

sup_{v∈M}⟨A(u^{∗}), v⟩ = ⟨A(u^{∗}), u^{∗}⟩ + sup_{v∈M}⟨A(u^{∗}), v − u^{∗}⟩

= ⟨A(u^{∗}), u^{∗}⟩ + sup_{v∈M}⟨h(u^{∗}), u^{∗} − v⟩ = ⟨A(u^{∗}), u^{∗}⟩

We claim that there are x ∈ X \ K and x_{∗} ∈ K such that Q(x) = Q(x_{∗}) = u^{∗}. Indeed u^{∗} ∈ M = Q(K) means exactly that there exists x_{∗} ∈ K such that u^{∗} = Q(x_{∗}). Moreover since Z is finite-dimensional and W is not, the linear mapping φ cannot be one-to-one, and there exists w_{0} ∈ ker(φ) \ {0}. Since K is bounded one can find some real number λ large enough for u^{∗} + λw_{0} ∉ K. Then x = u^{∗} + λw_{0} ∈ X \ K.

For y ∈ X and v = Q(y) ∈ M we have w_{1} = x − u^{∗} ∈ W, w_{2} = y − v ∈ W and get since A(x) ∈ W^{o} :

⟨A(u^{∗}), u^{∗}⟩ = ⟨A(x − λw_{0}), x − λw_{0}⟩ = ⟨A(x) − λφ(w_{0}), x − λw_{0}⟩
= ⟨A(x), x − λw_{0}⟩ = ⟨A(x), x⟩ − λ⟨A(x), w_{0}⟩ = ⟨A(x), x⟩

and similarly

⟨A(x), y⟩ = ⟨A(u^{∗} + λw_{0}), v + w_{2}⟩ = ⟨A(u^{∗}) + λφ(w_{0}), v + w_{2}⟩
= ⟨A(u^{∗}), v + w_{2}⟩ = ⟨A(u^{∗}), v⟩ + ⟨A(u^{∗}), w_{2}⟩ = ⟨A(u^{∗}), v⟩ .

It follows that

sup_{y∈K}⟨A(x), y⟩ = sup_{v∈M}⟨A(u^{∗}), v⟩ = ⟨A(u^{∗}), u^{∗}⟩ = ⟨A(x), x⟩

and this completes the proof. □

Theorem 5.2. Under the above hypotheses there exists some x^{∗} ∈ K such that ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X.

Proof. It follows from Lemma 5.1 that if W were not finite-dimensional A could not satisfy (∗). Thus W is finite-dimensional and E = W × Z too. Thus A is continuous and the result follows from Theorem 9. □

6. Getting Rid of Continuity

The aim of this section is to show that in Theorem 8 the hypothesis of continuity on A does not follow automatically from the hypothesis

sup_{y∈K}⟨A(x), x − y⟩ > 0 for all x ∈ X \ K  (∗)

but that the conclusion still holds without this assumption.

As observed in the introduction Yen’s theorem (Theorem 4) needs no continuity hypothesis on the affine mapping A : H → H because the hypothesis

sup_{y∈K}⟨A(x), x − y⟩ ≥ 0 for all x ∈ H \ K

implies the monotonicity of A which in turn implies the continuity of A and

∃x^{∗} ∈ K sup_{y∈X}⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0  (∗∗)

follows from Theorem 9. Nevertheless, in general, when X ≠ E the hypothesis (∗) does not imply the continuity of A as shown by the trivial following example.

Example 6.1. There exists a Hilbert space H, a weakly compact subset K of H, a closed convex subset X of H with non-empty weak interior containing K and an affine discontinuous mapping A : H → H^{∗} ≃ H satisfying (∗).

Let W be the Hilbert space ℓ^{2}, (e_{n})_{n} its canonic orthonormal basis, H = W ×ℝ, X = W × [−1, 1], K = {0} × [−1, 1]. Equip H with the norm : (w, t) ↦ (∥w∥^{2} + t^{2})^{1/2} for which it is a Hilbert space. Choose using the Axiom of Choice some linear functional f : W → ℝ satisfying ⟨f, e_{n}⟩ = n. Then f is discontinuous. Define the linear operator A : H → H by:

A(w, t) = (w, f(w) + t) .

If A were continuous so would be f since f(w) = ⟨A(w, 0), (0, 1)⟩. Thus A is discontinuous but for x = (w, t) ∈ X \ K and y = (0, t) ∈ K we have w ≠ 0 hence

⟨Ax, x − y⟩ = ⟨w, w − 0⟩ + (f(w) + t).(t − t) = ∥w∥^{2} > 0

and this shows that (∗) holds. Of course x^{∗} = (0, 0) ∈ K satisfies for all element y = (w, t) of X:

⟨Ax^{∗}, x^{∗} − y⟩ = ⟨0, −w⟩ − t.(f(0) + 0) = 0

hence (∗∗) holds too. □

The following example shows that if we replace in (∗) the strict inequality by a large inequality as in section 4 the conclusion does not longer hold without assuming the continuity of A.

Example 6.2. There exists a Hilbert space ℋ , a weakly compact subset K of ℋ , a closed convex subset X with non-empty weak interior of ℋ containing K and an affine discontinuous mapping A : ℋ → ℋ ^{∗} ≃ ℋ satisfying sup_{y∈K}⟨A(x), x− y⟩ ≥ 0 for all x ∈ ℋ \ K but sup_{y∈X}⟨A(x), x − y⟩ > 0 for all x ∈ K.

Proof. We embed as in section 3 the Radon measures on the unit circle 𝕋 ⊂ ℂ in the Hilbert space ℋ = H^{−1}(𝕋) and take for K the set of probability Radon measures on 𝕋. As in section 3 we have for μ and ν in K :

∥μ − ν∥_{−1}^{2} = ∑_{p∈ℤ} \frac{|μ̂(p) − ν̂(p)|^{2}}{p^{2} + 1}

and |μ̂(p) − ν̂(p)| ≤ 2 for all integer p. It follows that the embedding of K into H^{−1}(𝕋) is continuous when K is equipped with the weak*-topology σ(K, 𝒞 (T)) and H^{−1}(𝕋) with the norm topology, hence that K is norm-compact in H^{−1}(𝕋).

The functions ℜ : x + iy ↦ x and ℑ : x + iy ↦ y from 𝕋 to ℝ are both in H^{1}(𝕋); so the linear mapping Q : T ↦ ⟨T, ℜ + iℑ⟩ is continuous on H^{−1}(𝕋) to ℂ ≃ ℝ^{2} and it is easy to check that Q(K) is the closed unit disk 𝔻 since for μ ∈ K: |Q(μ)| ≤ ∫_{𝕋} |z| dμ(z) = 1 and Q(δ_{u}) = u for u ∈ 𝕋. Moreover if μ ∈ K

and Q(μ) = u ∈ 𝕋 then

0 = ℜ(1 − ūQ(μ)) = ∫_{𝕋} ℜ(1 − ū.z) dμ(z)

while the continuous function on 𝕋 : z ↦ ℜ(1 − ū.z) is non-negative and vanishes only at u : it follows that μ = δ_{u} is the only solution in K of the equation Q(μ) = u. Let X = Q^{−1}(𝔻).

We now consider the affine function χ_{0} on K defined by χ_{0}(μ) = μ({1}) and extend it (using the Axiom of Choice) to a (necessarily discontinuous) real linear function χ on H^{−1}(𝕋).

Then for S and T ∈ H^{−1}(𝕋) we define A(T) = ℜ + χ(T).ℑ ∈ H^{1}(𝕋), so :

⟨A(T), S⟩ = ⟨S, ℜ⟩ + χ(T).⟨S, ℑ⟩ = ⟨S, ℜ + χ(T).ℑ⟩ ∈ ℝ

This yields an affine function A from ℋ = H^{−1}(𝕋) to its dual ℋ ^{∗} = H^{1}(𝕋). Notice that whenever Q(S) = 0 then ⟨S, ℜ⟩ = ⟨S, ℑ⟩ = 0 hence ⟨A(T), S⟩ = 0 for all T ∈ ℋ . So with W = ker(Q) ⊂ ℋ , we get A(T) ∈ W^{o}.

Claim 6.3. We have

sup_{S∈K}⟨A(T), T − S⟩ ≥ 0 for all T ∈ X ,

sup_{S∈X}⟨A(T), T − S⟩ > 0 for all T ∈ K .

Proof. For T ∈ X = ℚ^{−1}(𝔻) we can choose some probability measure μ on 𝕋

such that ∫_{𝕋} z dμ(z) = Q(T) ∈ 𝔻. Then μ ∈ K satisfies Q(T − μ) = 0 hence

⟨A(T), T − μ⟩ = 0, thus

sup_{S∈K}⟨A(T), T − S⟩ ≥ ⟨A(T), T − μ⟩ = 0 .

Finally since ⟨A(T), δ_{1}⟩ = ⟨δ_{1}, ℜ + χ(T).ℑ⟩ = 1 it is clear that A(T) ≠ 0 for all T ∈ ℋ and that if T^{∗} ∈ K satisfies ⟨A(T^{∗}), T^{∗} − S⟩ ≤ 0 for all S ∈ X we have for S = T^{∗} + αℜ + βℑ

Q(S) = Q(T^{∗}) + α + iβ ∈ 𝔻 ⟹ ⟨A(T^{∗}), αℜ + βℑ⟩ ≤ 0 .

Thus if w = u + iv = Q(T^{∗}) ∈ 𝔻

|w + α + iβ| ≤ 1 ⟹ ⟨A(T^{∗}), αℜ + βℑ⟩ = α + βχ(T^{∗}) ≤ 0

This means that the linear function (α, β) ↦ α + βχ(T^{∗}) attains its maximum 0 on the disk D(−w, 1) at 0.

For λ ∈ ℝ and ω = 1 + iλ the function u = α + iβ ↦ α + λβ = ℜ(ω̄.u) attains

its maximum on D(−w, 1) at −w + \frac{ω}{|ω|}, and this is 0 iff w = \frac{1 + iλ}{|1 + iλ|}. It follows

that |w| = |Q(T^{∗})| = 1 for T^{∗} ∈ K hence that T^{∗} = δ_{w} and χ(δ_{w}) = 1 if w = 1,

χ(δ_{w}) = 0 if w ≠ 1. So \frac{1 + iχ(δ_{w})}{|1 + iχ(δ_{w})|} ≠ w for all w ∈ 𝕋. □

We conclude that sup_{S∈X}⟨A(T), T −S⟩ > 0 for all T ∈ K, and that Theorem 4.1 does not hold without any hypothesis of continuity. □

Nevertheless we can show the following.

Theorem 6.4. Let E be a Banach space, X a closed convex subset of E with non-empty weak interior and A : E → E^{∗} a (non necessarily continuous) affine operator. Assume that A satisfies (∗) with respect to K and X. Then there exists some point x^{∗} ∈ K such that ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 for all y ∈ X.

For the proof of this theorem we need the following lemmas.

Lemma 6.5. With the notations of section 2, Condition (∗∗) for x^{∗} ∈ K is equivalent to

i) P^{∗}A(x^{∗}) = 0 ii) ∀z ∈ Y ⟨A(x^{∗}), Q(x^{∗}) − z⟩ ≤ 0

Proof. If (∗∗) holds for x^{∗} ∈ K then for all w ∈ W and y = x^{∗} − w ∈ X we have ⟨A(x^{∗}), x^{∗} − y⟩ = ⟨A(x^{∗}), w⟩ ≤ 0. And applying to −w we get ⟨A(x^{∗}), −w⟩ ≤ 0 hence ⟨A(x^{∗}), w⟩ = 0. So for all u ∈ E, w = P (u) ∈ W and

⟨P^{∗}A(x^{∗}), u⟩ = ⟨A(x^{∗}), P (u)⟩ = ⟨A(x^{∗}), w⟩ = 0

thus P^{∗}A(x^{∗}) = 0. Moreover, with w = P (x^{∗}) ∈ W, z ∈ Y and y = z+P (x^{∗}) ∈ X

⟨A(x^{∗}), Q(x^{∗}) − z⟩ = ⟨A(x^{∗}), x^{∗} − (P (x^{∗}) + z)⟩ = ⟨A(x^{∗}), x^{∗} − y⟩ ≤ 0 .

Conversely, if i) and ii) hold then for any y ∈ X there is some z ∈ Y such that w = z − y ∈ W. So P (w) = w and

⟨A(x^{∗}), x^{∗} − y⟩ = ⟨A(x^{∗}), Q(x^{∗}) + P (x^{∗}) + w − z⟩
= ⟨A(x^{∗}), Q(x^{∗}) − z⟩ + ⟨P^{∗}A(x^{∗}), x^{∗} + w⟩
= ⟨A(x^{∗}), Q(x^{∗}) − z⟩ ≤ 0  □

Lemma 6.6. Let F be a finite-dimensional normed space, V be a normed space, B be its unit ball, and f : V → F be a linear mapping. Let G ⊂ V × F be the graph of f and F_{0} = \{y ∈ F : (0, y) ∈ \overline{G}\}. Then F_{0} = ⋂_{n∈ℕ} f(2^{−n}B) and f is continuous if and only if F_{0} = {0}.

Morover if g is a linear mapping from F to a normed space E and g_{|F_{0}} = 0 then g∘f is continuous.

Proof. Denote F′_{0} = ⋂_{n∈ℕ} f(2^{−n}B). If f is continuous, there exists M such that ∥f(x)∥ ≤ M.∥x∥. Then f(2^{−n}B) ⊂ B_{F} (0, 2^{−n}.M) and

F′_{0} ⊂ {y ∈ F : ∥y∥ ≤ inf_{n} M.2^{−n}} = {0} .

If y ∈ F′_{0} there exists a sequence (v_{n}) in V such that f(v_{n}) = y and ∥v_{n}∥ ≤ 2^{−n}. Then (v_{n}, y) ∈ G and (v_{n}, y) → (0, y) : so y ∈ F_{0}.

The linear space f(V) is finite-dimensional hence closed in F. Thus if y ∈ F_{0}, there is a sequence (v_{n}, y_{n}) ∈ G which converges to (0, y) : so y ∈ \overline{f(V)} = f(V).

We can find a linear section ρ : f(V) → V which is continuous since f(V) is finite-dimensional. Put v′_{n} = v_{n} + ρ(y − y_{n}) ∈ V. Then

f(v′_{n}) = y_{n} + f ∘ρ(y − y_{n}) = y_{n} + (y − y_{n}) = y

and v′_{n} → 0 since ρ(y − y_{n}) → ρ(y − y) = 0. Thus y ∈ F′_{0}. So F_{0} = F′_{0}.

We now prove by induction on the rank k = rk(f) that f is continuous if F_{0} = {0}. For k = 0 nothing is to prove. Assume the statement holds for rk(f) < k and F = f(V) has dimension k. Choose any x_{0} ∈ V \ ker(f) and set y_{0} = f(x_{0}) ≠ 0, then fix n such that y_{0} ∉ W = f(2^{−n}B). Since W is convex symmetric and generates F, the interior W_{0} of W in F contains 0 and by Hahn-Banach’s theorem we can find some linear (necessarily continuous) functional ξ on F such that ⟨ξ, y⟩ < 1 = ⟨ξ, y_{0}⟩ for all y ∈ W_{0}. Since W_{0} is dense in W we get ⟨ξ, y⟩ ≤ 1 for y ∈ f(2^{−n}B), hence ∥ξ∘f∥ ≤ 2^{n} and ξ∘f(x_{0}) = 1 ≠ 0.

Thus ξ∘f is continuous on V, the kernel V_{0} = ker(ξ∘f) is closed in V and rk(f_{|V_{0}}) < k. Since ⋂_{n} f(2^{−n}B ∩ V_{0}) ⊂ ⋂_{n} f(2^{−n}B) = F_{0} = {0} it follows from the induction hypothesis that f_{|V_{0}} is continuous. Finally we get for x ∈ V :

f(x) = f_{|V_{0}}(x − ξ∘f(x) · x_{0}) + ξ∘f(x) · y_{0}

hence f = f_{|V_{0}}∘(I − π) + ξ∘f ∘π · y_{0}, where π is the continuous linear mapping x ↦ ξ∘f(x) · x_{0} from V to itself. And this shows the continuity of f.

Assume now that g : F → E is linear and g = 0 on F_{0}. Up to replacing E by g(F) we can also assume that E is finite-dimensional and g is onto. Denote by K the kernel of g and by G′ ⊂ V × E the graph of g∘f. We have

G′ = \{(v, g∘f(v)) : v ∈ V \} = (I × g)(G) ⊂ (I × g)(\overline{G})
⊂ (I × g)(G + ({0} × K)) ⊂ (I × g)(G)

since \overline{G} = G + (\{0\} × F_{0}) ⊂ G + (\{0\} × K), hence G′ = (I × g)(G + (\{0\} × K)). Since F and E are finite-dimensional, it is easy to check that I × g is an open mapping from V × F onto V × E. Moreover

G′ = (I × g)(G + (\{0\} × K)) = (V × E) \ (I × g)((V × F) \ (\overline{G} + (\{0\} × K)))

Since \overline{G} is closed and \{0\} × K finite-dimensional then \overline{G} + (\{0\} × K) is closed too, and this shows that G′ is closed in V × E.

It follows that \overline{G′} ∩ (\{0\} × E) = G′ ∩ (\{0\} × E) = \{0\}, and the above proof shows that g∘f is continuous. □

We still define φ, W, Z, Y, P, Q as in section 2, and B_{W} as the unit ball of W. Since E is a Banach space, so is its closed linear subspace W, and Lemma 2.2 proves that φ_{|W} is monotone, hence continuous by Theorem 5. Since Z is finitedimensional we can equip it with an euclidean norm ∥.∥_{2} and renorm E ≃ W × Z in an equivalent way by defining |||w + z||| = ∥w∥ + ∥z∥_{2}. Then E^{∗} is identified to W^{∗} × Z, P and Q : E → E are the projections (w, z) ↦ (w, 0) and (w, z) ↦ z, P^{∗} and Q^{∗}: E^{∗} → E^{∗} are the mappings (ξ, z) ↦ (ξ, 0) and (ξ, z) ↦ (0, z). In the whole sequel of this section we assume silently that (∗) holds.

The linear space F = Q^{∗}φP (E) = Q^{∗}φ(W) ⊂ Z^{∗} ≃ Z is a finite-dimensional subspace of E^{∗}. Then F_{1} = ⋂_{n} 2^{−n} · Q^{∗}φ(B_{W}) is a (necessarily closed) linear subspace of F; we consider the orthogonal projection π from Z onto F_{1} and π′ = Id_{Z} − π is the projection from Z onto F_{0} = F_{1}^{⊥}. Recall that K̃ is the closed convex hull of K, which is weakly compact and contained in X.

Theorem 6.7. Assume that K is convex and weakly compact in E and that sup_{y∈K}⟨A(x), x − y⟩ ≥ 0 for all x ∈ X \ K.

Then the set K_{0} = {x ∈ K : A(x)_{|W} = 0} is convex and weakly compact, and Q(K_{0}) = Q(K).

Proof. Since X′ = K+W ⊂ X we have sup_{y∈K}⟨A(x), x−y⟩ ≥ 0 for all x ∈ X′\K. So we can and do replace X by X′ and get Q(X) = Q(K).

For any z ∈ F_{1} consider the sets E(z) = (πQ)^{−1}(z) , X(z) = X ∩ E(z) and K(z) = K ∩ E(z) which is a convex weakly compact subset of X(z). It is clear that x + W ⊂ X(z) as soon as x ∈ X(z). So X(z) has non-empty weak interior in the affine space it spans. Moreover it follows from Lemma 6.6 that A_{|E(z)} is

continuous: indeed ⋂_{n}(A(x + 2^{−n}B_{E}) − A(x)) ∩ ker(πQ) = {0}.

Claim 6.8. Assume K(z) ≠ ∅. Then for every x ∈ X(z) \ K(z) the inequality sup_{y∈K(z)}⟨A(x), x − y⟩ ≥ 0 holds.

Proof. Assume to a contradiction that

sup_{y∈K(z)}⟨A(x_{∗}), x_{∗} − y⟩ < 0

for some x_{∗} ∈ X(z). Then the weakly compact convex set K is disjoint from the closed affine subspace H = {y ∈ E(z) : ⟨A(x_{∗}), x_{∗} − y⟩ = 0} of E since K ∩ H ⊂ K(z). It follows from Hahn-Banach’s theorem that there exists a continuous linear functional ℓ ∈ E^{∗} which separates K from H: sup_{x∈H}⟨ℓ, x⟩ < inf_{x∈K}⟨ℓ, x⟩.

In particular the linear functional ℓ is bounded from above on the linear space H −x^{∗} = ker(A(x_{∗}))∩ker(πQ) hence vanishes on ker(A(x_{∗}))∩ker(πQ). It follows that ℓ = αA(x_{∗}) + ℓ_{1}, where α ∈ ℝ and ℓ_{1} ∈ F_{1}. Since Q(x_{∗}) ∈ Q(X) = Q(K), there exists some x′ ∈ K such that Q(x^{∗}) = Q(x′). Thus we have x_{∗} ∈ H and x′ ∈ K(z) thus

0 > ⟨ℓ, x^{∗}⟩ − ⟨ℓ, x′⟩ = α⟨A(x^{∗}), x^{∗} − x′⟩ + ⟨ℓ_{1}, x^{∗} − x′⟩ = α⟨A(x^{∗}), x^{∗} − x′⟩

since πQ(x_{∗} − x′) = 0. Then A(x_{∗}, x_{∗} − x′) < 0 so α > 0. And replacing ℓ by ℓ′ = α^{−1}ℓ we get

sup_{x∈H}⟨ℓ′, x⟩ = sup_{x∈H}(⟨A(x_{∗}), x⟩ + α^{−1}⟨ℓ_{1}, x⟩) = ⟨A(x_{∗}), x_{∗}⟩ + α^{−1}⟨ℓ_{1}, x_{∗}⟩

< inf_{y∈K}⟨ℓ′, y⟩ = inf_{y∈K}(⟨A(x_{∗}, y⟩ + α^{−1}⟨ℓ_{1}, y⟩)

hence −δ = sup_{y∈K}⟨A(x_{∗}) + α^{−1}ℓ_{1}, x_{∗} − y⟩ < 0.

Since α^{−1}ℓ_{1} ∈ F_{1} we can find a sequence (w_{n}) ∈ W such that ∥w_{n}∥ < 2^{−n} and πQ(w_{n}) → α^{−1}ℓ_{1}. Then x_{n} = x_{∗} + w_{n} → x_{∗} and x_{n} ∈ X \ K for n large enough.

Moreover

A(x_{n}) = A(x_{∗}) + φ(w_{n}) = A(x_{∗}) + P^{∗}φ(w_{n}) + π′Q^{∗}φ(w_{n}) + πQ^{∗}φ(w_{n})
→ A(x_{∗}) + α^{−1}ℓ_{1} = ℓ

since φ′ = (P^{∗}φ + π′Q^{∗}φ)_{|W} = (φ − πQ^{∗}φ)_{|W} is continuous. It follows that for y ∈ K

⟨A(x_{n}), x_{n} − y⟩ ≤ ⟨ℓ, x_{n} − y⟩ + ∥A(x_{n}) − ℓ∥.∥x_{n} − y∥
≤ ⟨ℓ, x_{∗} − y⟩ + ∥ℓ∥.∥w_{n}∥ + ∥A(x_{n}) − ℓ∥.(d(x_{n}, K) + diam(K))
≤ −δ + 2^{−n}∥ℓ∥ + ∥A(x_{n}) − ℓ∥.(d(x_{∗}, K) + diam(K) + ∥w_{n}∥)

hence sup_{y∈K}⟨A(x_{n}), x_{n} − y⟩ ≤ −δ/2 for n large, in contradiction with the hypothesis of the Theorem. And this completes the proof of the claim. □

Then it follows from the continuity of A_{|E(z)} and from Theorem 4.1 that for all y ∈ Q(K) such that π(y) = z we have y ∈ Q(K_{0}). And this achieves the proof of Theorem 6.7. □

Define now

E_{0} = {x ∈ E : A(x) ∈ W^{o}} = {x ∈ E : A(x)_{|W} = 0} = {x ∈ E : P^{∗}A(x) = 0}

which is a closed affine manifold of E since P^{∗}A is affine and continuous. Then we put W_{0} = W ∩ E_{0}, X_{0} = X ∩ E_{0} and K_{0} = K ∩ E_{0} as before. Up to a translation we can assume 0 ∈ K hence 0 ∈ Q(K) = Q(K_{0}) by Theorem 6.7, and even 0 ∈ K_{0} ⊂ E_{0}. Then E_{0} appears as a Banach space, and we define the affine mapping A_{0} : E_{0} → E_{0}^{∗} by ⟨A_{0}(x), y⟩ = ⟨A(x), y⟩ for x and y in E_{0}.

Lemma 6.9. Assume that K is convex and weakly compact in E and that sup_{y∈K}⟨A(x), x − y⟩ ≥ 0 for all x ∈ X \ K. Then Q(X_{0}) = Y = Q(X).

Proof. Let a be any point in X and replace K by K′ = conv(K ∪ {a}) ⊂ X: the condition sup_{y∈K′}⟨A(x), x − y⟩ ≥ 0 is clearly satisfied for all x ∈ X \ K′ and we conclude from Theorem 6.7 applied to K′ that a ∈ K′ ⊂ K′_{0} + W hence that there exists some a′ ∈ X satisfying Q(a′) = Q(a) and A(a′)_{|W} = 0. Thus Y = Q(X_{0}). □

Lemma 6.10. The mapping A_{0} satisfies (∗) with respect to K_{0} and X_{0}.

Proof. Let x_{0} ∈ X_{0} \ K_{0}. Since A satisfies (∗) and

x_{0} ∈ X_{0} \ K_{0} ⊂ X ∩ (E_{0} \ K_{0}) = X ∩ (E_{0} \ E_{0} ∩ K) ⊂ X \ K

there exists some y ∈ K such that ⟨A(x_{0}), x_{0} − y⟩ > 0 and since Q(K) = Q(K_{0}) one can find some y_{0} ∈ K_{0} such that Q(y) = Q(y_{0}) hence w = y − y_{0} ∈ W. Since x_{0} ∈ E_{0} we have A(x_{0}) ∈ W^{o} hence ⟨A(x_{0}), w⟩ = 0 and

⟨A_{0}(x_{0}), x_{0} − y_{0}⟩ = ⟨A(x_{0}), x_{0} − y_{0}⟩ = ⟨A(x_{0}), x_{0} − y⟩ + ⟨A(x_{0}), w⟩
= ⟨A(x_{0}), x_{0} − y⟩ > 0  □

Proof of Theorem 6.4. It is clear that A_{0} takes values in W_{0}^{o} : indeed for any x_{0} ∈ E_{0} and any w_{0} ∈ W_{0} = W ∩ E_{0} we have A(x_{0}) ∈ W^{o}, hence ⟨A(x_{0}), w_{0}⟩ = 0 and ⟨A_{0}(x_{0}), w_{0}⟩ = 0. Then we deduce from Theorem 5.2 that for some x_{0}^{∗} ∈ K_{0} we have ⟨A_{0}(x_{0}^{∗}), x_{0}^{∗} − y_{0}⟩ ≤ 0 for all y_{0} ∈ X_{0}. Since K_{0} = K ∩ E_{0}, then x_{0}^{∗} ∈ K and A_{0}(x_{0}^{∗}) = A(x_{0}^{∗}), hence ⟨A(x_{0}^{∗}), x_{0}^{∗} − y⟩ ≤ 0 for all y ∈ X_{0}.

If y ∈ X we have Q(y) ∈ Y = Q(X_{0}) by Lemma 6.9, and we can find y_{0} ∈ X_{0} such that w = y − y_{0} ∈ W. Again ⟨A(x_{0}^{∗}), w⟩ = 0 since x_{0}^{∗} ∈ K_{0} ⊂ E_{0}, and

⟨A(x_{0}^{∗}), x_{0}^{∗} − y⟩ = ⟨A(x_{0}^{∗}), x_{0}^{∗} − y_{0}⟩ − ⟨A(x_{0}^{∗}), w⟩ = ⟨A(x_{0}^{∗}), x_{0}^{∗} − y_{0}⟩
= ⟨A_{0}(x_{0}^{∗}), x_{0}^{∗} − y_{0}⟩ ≤ 0

what shows that x_{0}^{∗} satisfies the requested properties for (∗∗) and completes the proof of Theorem 6.4. □

References

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Jean Saint Raymond, Sorbonne Université, Université Paris Diderot, CNRS, Institut de Mathématiques de Jussieu-Paris Rive Gauche, IMJ-PRG, F-75005, Paris, France

  E-mail address: jean.saint-raymond@imj-prg.fr