Existence of zeros for operators taking their values in the dual of a Banach space

We prove, under certain conditions, the existence of zeros for a weakly continuous operator on a paracompact topological space into the dual of a Banach space.


BIAGIO RICCERI
Throughout the sequel, E denotes a reflexive real Banach space and E * its topological dual. We also assume that E is locally uniformly convex. This means that for each x ∈ E, with x = 1, and each ǫ > 0 there exists δ > 0 such that, for every y ∈ E satisfying y = 1 and x − y ≥ ǫ, one has x + y ≤ 2(1 − δ). Recall that any reflexive Banach space admits an equivalent norm with which it is locally uniformly convex ( [1], p. 289). For r > 0, we set B r = {x ∈ E : x ≤ r}.
Moreover, we fix a topology τ on E, weaker than the strong topology and stronger than the weak topology, such that (E, τ ) is a Hausdorff locally convex topological vector space with the property that the τ -closed convex hull of any τ -compact subset of E is still τ -compact and the relativization of τ to B 1 is metrizable by a complete metric. In practice, the most usual choice of τ is either the strong topology or the weak topology provided E is also separable.
The aim of this short paper is to establish the following result and present some of its consequences: THEOREM 1. -Let X be a paracompact topological space and A : X → E * a weakly continuous operator. Assume that there exist a number r > 0, a continuous function for all x ∈ X, a closed set C ⊂ X, and a τ -continuous function g : C → B r satisfying for all x ∈ C, in such a way that, for every τ -continuous function ψ : X → B r satisfying ψ |C = g, there exists x 0 ∈ X such that Then, there exists x * ∈ X such that A(x * ) = 0.
For the reader's convenience, we recall that a multifunction F : S → 2 V , between topological spaces, is said to be lower semicontinuous at s 0 ∈ S if, for every open set Ω ⊆ V meeting F (s 0 ), there is a neighbourhood U of s 0 such that F (s) ∩ Ω = ∅ for all s ∈ U . F is said to be lower semicontinuous if it is so at each point of S.
The following well-known results will be our main tools.
Then, for each closed set C ⊂ X and each τ -continuous function g : C → B 1 satisfying g(x) ∈ F (x) for all x ∈ C, there exists a τ -continuous function ψ : X → B 1 such that ψ |C = g and ψ(x) ∈ F (x) for all ∈ X. THEOREM B ( [4]). -Let X, Y be two topological spaces, with Y connected and locally connected, and let f : X × Y → R be a function satisfying the following conditions: (a) for each x ∈ X, the function f (x, ·) is continuous, changes sign in Y and is identically zero in no nonempty open subset of Y ; Then, the multifunction x → {y ∈ Y : f (x, y) = 0 and y is not a local extremum for f (x, ·)} is lower semicontinuous, and its values are nonempty and closed.
Proof of Theorem 1. Arguing by contradiction, assume that A(x) = 0 for all x ∈ X.
For each x ∈ X, y ∈ B 1 , put Since A is weakly continuous, the function x → A(x) E * , as supremum of a family of continuous functions, is lower semicontinuous. From this, it follows that the set X 0 is open.
For each x ∈ X 0 , the function f (x, ·) is continuous and has no local, nonabsolute, extrema, being affine. Moreover, it changes sign in is continuous for all y ∈ B 1 , we then realize that the restriction of f to X 0 × B 1 satisfies the hypotheses of Theorem B, B 1 being considered with the relativization of the strong topology. Hence, the multifunction F |X 0 is lower semicontinuous. Consequently, since X 0 is open, the multifunction F is lower semicontinuous at each point of and ǫ > 0. Clearly, since y 0 is an absolute extremum of A(x 0 ) in B 1 , one has y 0 = 1. Choose δ > 0 so that, for each y ∈ E satisfying y = 1 and y − y 0 ≥ ǫ, one has y + y 0 ≤ 2(1 − δ). By semicontinuity, the function x → ( A(x) E * ) −1 is bounded in some neighbourhood of x 0 , and so, since the functions α and A(·)(y 0 ) are continuous, it follows that So, there is a neighbourhood U of x 0 such that ¿From this choice, it follows, of course, that the segment joining y 0 and z meets the hyperplane (A(x)) −1 ( α(x) r ). In other words, there is λ ∈ [0, 1] such that So, if we put y = λz + (1 − λ)y 0 , we have y ∈ F (x) and We claim that y − y 0 < ǫ. This follows at once from (3) if λ < ǫ 2 . Thus, assume λ ≥ ǫ 2 . In this case, to prove our claim, it is enough to show that since (4) implies z − y 0 < ǫ. To this end, note that, by (2), one has and so, from (1), it follows that Suppose A(x)(z) = A(x) E * . Then, from (5), we get On the other hand, we also have So, (4) follows from (6) and (7). Now, suppose A(x)(z) = − A(x) E * . Then, from (5), we get On the other hand, we have So, in the present case, (4) is a consequence of (8) and (9). In such a manner, we have proved that F is lower semicontinuous at x 0 . Hence, it remains proved that F is lower semicontinuous in X with respect to the strong topology, and so, a fortiori, with respect to τ . Since F is also with nonempty τ -closed convex values, and g r is a τ -continuous selection of it over the closed set C, by Theorem A, F admits a τ -continuous selection ω in X such that ω |C = g r . At this point, if we put ψ = rω, it follows that ψ is a τ -continuous function, from X into B r , such that ψ |C = g and A(x)(ψ(x)) = α(x) for all x ∈ X, against the hypotheses. This concludes the proof. △ We now indicate two reasonable ways of application of Theorem 1. The first one is based on the Tychonoff fixed point theorem.
THEOREM 2. -Assume that E is a separable Hilbert space, with inner product ·, · . Let r > 0 and let A : B r → E be a continuous operator from the weak to the strong topology. Assume that there exist a weakly continuous function α : B r → R satisfying |α(x)| ≤ r A(x) E * for all x ∈ B r , and a weakly continuous function g : Then, there exists x * ∈ B r such that A(x * ) = 0. PROOF. Identifying E with E * , we apply Theorem 1 taking X = B r , with the relativization of the weak topology of E, and taking as τ the weak topology of E. Due to the kind of continuity we are assuming for A, the function x → A(x), x turns out to be weakly continuous (see the proof of Theorem 4), and so the set C is weakly closed. Now, let ψ : B r → B r be any weakly continuous function such that ψ |C = g. By the Tychonoff fixed point theorem, there is x 0 ∈ B r such that ψ(x 0 ) = x 0 . Since g ha no fixed points in C, it follows that x 0 / ∈ C, and so Hence, all the assumptions of Theorem 1 are satisfied, and the conclusion follows from it. △ It is worth noticing the following consequence of Theorem 2.
-Let E and A be as in Theorem 2. Assume that for each x ∈ B r , with A(x) > r, one has Then, the operator A has either a zero or a fixed point. PROOF. Define the function α : B r → R by Clearly, the function α is weakly continuous and satisfies |α(x)| ≤ r A(x) for all x ∈ B r . Put Note that if x ∈ C then A(x) ≤ r. Indeed, otherwise, we would have A(x), x = r A(x) , and so, necessarily, x = rA(x) A(x) , against (10). Hence, we have A(x), A(x) = α(x) for all x ∈ C. At this point, the conclusion follows at once from Theorem 2, taking g = A |C . △ REMARK 1. -It would be interesting to know whether Theorem 3 can be improved assuming that A is a continuous operator with relatively compact range.
The second application of Theorem 1 is based on connectedness arguments. For other results of this type we refer to [5] (see also [2]). THEOREM 4. -Let X be a connected paracompact topological space and A : X → E * a weakly continuous and locally bounded operator. Assume that there exist r > 0, a closed set C ⊂ X, a continuous function g : C → B r and an upper semicontinuous function β : for all x ∈ X \ C and y ∈ B r \ g(C). Then, there exists x * ∈ C such that A(x * ) = 0. PROOF. First, note that the function x → A(x)(g(x)) is continuous in C. To see this, let x 1 ∈ C and let {x γ } γ∈D be any net in C converging to x 1 . By assumption, there are M > 0 and a neighbourhood U of x 1 such that A(x) E * ≤ M for all x ∈ U . Let γ 0 ∈ D be such that x γ ∈ U for all γ ≥ γ 0 . Thus, for each γ ≥ γ 0 , one has |A(x γ )(g(x γ )) − A(x 1 )(g(x 1 ))| ≤ M g(x γ ) − g(x 1 ) + |A(x γ )(g(x 1 )) − A(x 1 )(g(x 1 ))| from which, of course, it follows that lim γ A(x γ )(g(x γ )) = A(x 1 )(g(x 1 )). Next, observe that the multifunction x → [β(x), r A(x) E * ] is lower semicontinuous and that the function x → A(x)(g(x)) is a continuous selection of it in C. Hence, by Michael's theorem, there is a continuous function α : X → R such that α(x) = A(x)(g(x)) for all x ∈ C and β(x) ≤ α(x) ≤ r A(x) E * for all x ∈ X. Now, let ψ : X → B r be any continuous function such that ψ |C = g. Since X is connected, ψ(X) is connected too. But then, since g(C) is disconnected and g(C) ⊂ ψ(X), there exists y 0 ∈ ψ(X) \ g(C). Let x 0 ∈ X \ C be such that ψ(x 0 ) = y 0 . So, by hypothesis, we have A(x 0 )(ψ(x 0 )) = A(x 0 )(y 0 ) < β(x 0 ) ≤ α(x 0 ) .
Hence, taking as τ the strong topology of E, all the assumptions of Theorem 1 are satisfied, and the conclusion follows from it. △ REMARK 2. -Observe that when X is first-countable, the local boundedness of A follows automatically from its weak continuity. This follows from the fact that, in a Banach space, any weakly convergent sequence is bounded.
It is worth noticing the corollary of Theorem 4 which comes out taking X = B r , β = 0 and g=identity: THEOREM 5. -Let E be a Hilbert space, with inner product ·, · . Let r > 0 and let A : B r → E be a continuous operator from the strong to the weak topology. Assume that the set C = {x ∈ B r : A(x), x ≥ 0} is disconnected and that, for each x, y ∈ B r \ C, one has A(x), y < 0.
Then, there exists x * ∈ C such that A(x * ) = 0.