Journal of Convex Analysis Volume 2 (1995), No.1/2, 287–290

Existence of Zeros via Disconnectedness

Biagio Ricceri

Department of Mathematics, University of Catania, Viale A.Doria 6, 95125 Catania, Italy.

Received 16 May 1994 Revised manuscript received 25 November 1994

Dedicated to R. T. Rockafellar on his 60th Birthday

The purpose of this short paper is very modest if compared with the greatness of the mathematician to whom it is dedicated. Here, indeed, we wish only to signalize a new result on the existence of zeros for operators taking their values in the dual of a real topological vector space. We do hope, however, that the novelty together with the simplicity of its statement can draw the attention of other mathematicians, in order to discover a possible variety of intermediate conditions making easier its applicability to concrete situations.

The reader should think of our result as the extension to the general case of the following elementary observation: if f : IR → IR is a continuous function such that the set {(x, y) ∈ IR^{2} : f(x)y = 1} is disconnected, then f does vanish at some point. Indeed, if f never vanished, that set would coincide with the graph of the continuous function 1/f which is clearly connected. Our result is as follows:

Theorem 1.1. Let X be a connected topological space and let E be a real topological vector space, with topological dual E^{∗}. Then, any operator A : X → E^{∗} for which the set

{y ∈ E : x → ⟨A(x), y⟩ is continuous}

is dense in E and the set

{(x, y) ∈ X × E : ⟨A(x), y⟩ = 1}

is disconnected, does vanish at some point of X.

Proof. Denote by p_{X} the projection from X ×E onto X. Moreover, for any C ⊆ X ×E, x ∈ X, put

C_{x} = {y ∈ E : (x, y) ∈ C}.

Arguing by contradiction, assume that A(x) ≠ 0 for all x ∈ X. Denote by Γ the set

{(x, y) ∈ X × E : ⟨A(x), y⟩ = 1}.

ISSN 0944-6532 / $ 2.50 ◯^{c} Heldermann Verlag

288 B. Ricceri / Existence of zeros via disconnectedness

Since Γ is disconnected, there are two open sets Ω_{1}, Ω_{2} ⊆ X × E such that

Ω_{1} ∩ Γ ≠ ∅, Ω_{2} ∩ Γ ≠ ∅, Ω_{1} ∩ Ω_{2} ∩ Γ = ∅, Γ ⊆ Ω_{1} ∪ Ω_{2}.

We now prove that p_{X}(Ω_{1} ∩ Γ) is open in X. So, let (x_{0}, y_{0}) ∈ Ω_{1} ∩ Γ. Since E is locally connected ([4], p.35), there are a neighbourhood U_{0} of x_{0} in X and an open connected neighbourhood V_{0} of y_{0} in E such that U_{0} × V_{0} ⊆ Ω_{1}. Since ⟨A(x_{0}), ·⟩ is a non-null continuous linear functional, it has no local extrema. Consequently, since ⟨A(x_{0}), y_{0}⟩ = 1, the sets

{u ∈ V_{0} : ⟨A(x_{0}), u⟩ < 1},
{u ∈ V_{0} : ⟨A(x_{0}), u⟩ > 1}

are both non-empty and open. Then, thanks to our density assumption, there are u_{1}, u_{2} ∈ V_{0} such that the set

{x ∈ U_{0} : ⟨A(x), u_{1}⟩ < 1 < ⟨A(x), u_{2}⟩}

is a neighbourhood of x_{0}. Then, if x is in this set, due to the connectedness of V_{0}, there is some y ∈ V_{0} such that ⟨A(x), y⟩ = 1, and so, x actually lies in p_{X}(Ω_{1} ∩ Γ), as desired. Likewise, it is seen that p_{X}(Ω_{2} ∩ Γ) is open. Now, observe that, for any x ∈ X, the set {x} × Γ_{x} is non-empty and connected, and so it is contained either in Ω_{1} or in Ω_{2}. Summarizing, we then have that the sets p_{X}(Ω_{1} ∩Γ) and p_{X}(Ω_{2} ∩Γ) are non-empty, open, disjoint and cover X. Hence, X would be disconnected, a contradiction.

Once Theorem 1.1 has been obtained, we can state the following formally more complete result:

Theorem 1.2. Let X be a topological space, let E be a real topological vector space, and let A : X → E^{∗} be such that the set

{y ∈ E : x → ⟨A(x), y⟩ is continuous}

is dense in E. Then, the following assertions are equivalent:

(i) The set

{(x, y) ∈ X × E : ⟨A(x), y⟩ = 1}

is disconnected. (ii) The set X \ A^{−1}(0) is disconnected.

Proof. Let (i) hold. Since

{(x, y) ∈ X × E : ⟨A(x), y⟩ = 1} = {(x, y) ∈ (X \ A^{−1}(0)) × E : ⟨A(x), y⟩ = 1},

if X \ A^{−1}(0) were connected, we could apply Theorem 1.1 to A_{|(X\A^{−1}(0))}, and so A would vanish at some point of X \ A^{−1}(0), which is absurd. Conversely, if (ii) holds, then (i) follows at once observing that, with the notations of the proof of Theorem 1.1, one has X \ A^{−1}(0) = p_{X}(Γ).

Remark 1.3. It is worth pointing out that one can also prove Theorem 1.1 by applying some known results of set-valued analysis. Namely, assuming, as before, that A(x) ≠ 0 for all x ∈ X, we directly infer from Theorem 2.2 of [3] that the multifunction H : X → 2^{E} defined by H(x) = Γ_{x} (x ∈ X), is lower semicontinuous. Then, since X is connected and each H(x) is non-empty and connected, Theorem 3.2 of [1] ensures that the graph of H is connected too. But such graph is nothing else than Γ itself, and so we get a contradiction.

Remark 1.4. When X is a connected topological space, E is an infinite-dimensional real vector space (with algebraic dual E′), and A : X → E′ is a σ(E′, E)-continuous operator, one could try to apply Theorem 1.1 endowing E with the strongest vector topology ([2], p.53).

Remark 1.5. In Theorem 1.1, the role of the constant 1 can actually be assumed by any continuous real function on X. Precisely, we have the following

Proposition 1.6. Let X be a topological space, let E be a real topological vector space, and let A : X → E′. Assume that, for some continuous function α : X → IR, the set

Λ := {(x, y) ∈ X × E : ⟨A(x), y⟩ = α(x)}

is disconnected. Then, either A(x) = 0 for some x ∈ X, or the set

Γ := {(x, y) ∈ X × E : ⟨A(x), y⟩ = 1}

is disconnected.

Proof. Assume that A^{−1}(0) = ∅. So, p_{X}(Γ) = X. Consider the function f : X × E → X × E defined by putting f(x, y) = (x, α(x)y) for all (x, y) ∈ X × E. Of course, f is continuous. Arguing by contradiction, assume that Γ is connected. Then, f(Γ) is connected too. Now, observe that

Λ = ⋃_{x∈α^{−1}(0)}(f(Γ) ∪ ({x} × Λ_{x})).

Furthermore, note that, if x ∈ α^{−1}(0), then (x, 0) ∈ f(Γ) ∩ ({x} × Λ_{x}), and so f(Γ) ∪ ({x} × Λ_{x}) is connected. In turn, the sets f(Γ) ∪ ({x} × Λ_{x}) (x ∈ α^{−1}(0)) are clearly pairwise non-disjoint, and hence Λ is connected, a contradiction.

Remark 1.7. A further reasonable remark about possible uses of Theorem 1.1 arises from Theorem 1.2. In fact, in dealing with the existence of zeros for a given σ(E^{∗}, E)- continuous operator A : X → E^{∗}, if we were trying to apply Theorem 1.1 directly to A, we would have not only that A^{−1}(0) ≠ ∅, but also that X \ A^{−1}(0) is disconnected, which is a quite special property. In other words, such a direct application could imply rather severe assumptions on A. So, it would be reasonable to find another connected topological space T and a continuous function f : T → X in such a way to prove more easily the disconnectedness of the set {(t, y) ∈ T × E : ⟨A(f(t)), y⟩ = 1}. So doing, we would still have the requested information A^{−1}(0) ≠ ∅, the set X \A^{−1}(0) being, however, not necessarily disconnected.

Acknowledgment. The author wishes to thank one of the referees for a suggestion which led to inserting Remark 1.3.

References

[1] J.-B. Hiriart-Urruty : Images of connected sets by semicontinuous multifunctions, J. Math. Anal. Appl., 111, 1985, 407–422

[2] J.L. Kelley, I. Namioka : Linear topological spaces, Van Nostrand, 1963

[3] B. Ricceri : Sur la semi-continuité inférieure de certaines multifonctions, C.R.Acad.Sc. Paris, Série I, 294, 1982, 265–267.

[4] H.H. Schaefer : Topological vector spaces, Springer-Verlag, 1971.