Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 3, 767–770

A New Minimal Point Theorem in Product Spaces

A. Göpfert, Chr. Tammer and C. Zălinescu

Abstract. We derive a minimal point theorem for a subset A in a cone in product spaces under a weak assumption concerning the boundedness of the considered set A. Using this result we improve two vectorial variants of Ekeland’s variational principle. Finally, a new characterization of well-based cones is given.

Keywords: Minimal point theorems, Ekeland’s variational principle, well-based cones AMS subject classification: 49 J 40, 90 C 29, 46 A 40, 90 C 26

Assume that (X, d) is a complete metric space, Y is a separated locally convex space, Y^{∗} is its topological dual, K ⊂ Y is a convex cone, i.e. K + K ⊂ K and [0, ∞) · K ⊂ K,

K^{+} = {y^{∗} ∈ Y^{∗} : ⟨y, y^{∗}⟩ ≥ 0 for all y ∈ K}

is the dual cone of K and

K^{#} = {y^{∗} ∈ Y^{∗} : ⟨y, y^{∗}⟩ > 0 for all y ∈ K \ {0}}.

In this note we suppose that K is pointed, i.e. K ∩(−K) = {0}. The cone K determines an order relation on Y, denoted in the sequel by ≤_{K}; so, for y_{1}, y_{2} ∈ Y, y_{1} ≤_{K} y_{2} if y_{2} − y_{1} ∈ K. It is well known that “≤_{K}” is reflexive, transitive and antisymmetric. Let k^{0} ∈ K \ {0}; using the element k^{0} we introduce an order relation on X × Y, denoted by “⪯_{k^{0}}”, in the following manner:

(x_{1}, y_{1}) ⪯_{k^{0}} (x_{2}, y_{2})  iff  y_{1} + k^{0}d(x_{1}, x_{2}) ≤_{K} y_{2}.

Note that “⪯_{k^{0}}” is reflexive, transitive and antisymmetric. That is, our notations are those of [3].

The essential idea for the derivation of a minimal point theorem (cf. [2, 8]) in general product spaces X × Y, as well as of the vectorial Ekeland principle, consists in including the ordering cone K ⊂ Y in a “larger” cone B ⊂ Y: K \ {0} ⊂ int B. We will use B to define a suitable functional z_{B} : Y → ℝ. Moreover, we will replace the usual boundedness condition of the projection P_{Y} A of A onto Y by a weaker one.

A. Göpfert: Dept. Math. & Inform. Univ., D-06099 Halle Chr. Tammer: Dept. Math. & Inform. Univ., D-06099 Halle C. Zălinescu: Fac. Math. Univ. “Al. I. Cuza”, 6600 Iaşi, Romania ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

Theorem 1. Assume that there exists a proper convex cone B ⊂ Y such that K \ {0} ⊂ int B. Suppose that the set A ⊂ X × Y satisfies the condition

(H1) for every ⪯_{k^{0}}-decreasing sequence ((x_{n}, y_{n})) ⊂ A with x_{n} → x ∈ X there exists y ∈ Y such that (x, y) ∈ A and (x, y) ⪯_{k^{0}} (x_{n}, y_{n}) for every n ∈ ℕ

and that P_{Y} (A) ∩ (ỹ − int B) = ∅ for some ỹ ∈ Y. Then for every (x_{0}, y_{0}) ∈ A there exists (\overline{x}, \overline{y}) ∈ A, minimal with respect to ⪯_{k^{0}}, such that (\overline{x}, \overline{y}) ⪯_{k^{0}} (x_{0}, y_{0}).

Proof. Let

z_{B} : Y → ℝ, z_{B}(y) = inf{t ∈ ℝ : y ∈ tk^{0} − cl B}.

By [3: Lemma 7], z_{B} is a continuous sublinear function such that z_{B}(y+tk^{0}) = z_{B}(y)+t for all t ∈ ℝ and y ∈ Y, and for every λ ∈ ℝ

\begin{matrix}\{y ∈ Y : z_{B}(y) ≤ λ\} = λk^{0} − cl B \\ \{y ∈ Y : z_{B}(y) < λ\} = λk^{0} − int B. \end{matrix}

Moreover, if y_{2} − y_{1} ∈ K \ {0}, then z_{B}(y_{1}) < z_{B}(y_{2}). Observe that for (x, y) ∈ A we have that z_{B}(y − ỹ) ≥ 0. Otherwise for some (x, y) ∈ A we have z_{B}(y − ỹ) < 0. It follows that there exists λ > 0 such that y − ỹ ∈ −λk^{0} − cl B. Hence

y ∈ ỹ − (λk^{0} + cl B) ⊂ ỹ − (int B + cl B) ⊂ ỹ − int B

which is a contradiction. Since 0 ≤ z_{B}(y − ỹ) ≤ z_{B}(y) + z_{B}(−ỹ), it follows that z_{B} is bounded from below on P_{Y} (A). Let us construct a sequence ((x_{n}, y_{n}))_{n≥0} ⊂ A as follows: having (x_{n}, y_{n}) ∈ A we take (x_{n+1}, y_{n+1}) ∈ A, (x_{n+1}, y_{n+1}) ⪯_{k^{0}} (x_{n}, y_{n}), such that

z_{B}(y_{n+1}) ≤ inf \{z_{B}(y) : (x, y) ∈ A and (x, y) ⪯_{k^{0}} (x_{n}, y_{n})\} + \frac{1}{n + 1}.

Of course, the sequence ((x_{n}, y_{n})) is ⪯_{k^{0}}-decreasing. It follows that

y_{n+p} + k^{0}d(x_{n+p}, x_{n}) ≤_{K} y_{n} ∀ n, p ∈ ℕ^{∗}

so that

d(x_{n+p}, x_{n}) ≤ z_{B}(y_{n}) − z_{B}(y_{n+p}) ≤ \frac{1}{n} ∀ n, p ∈ ℕ^{∗}.

It follows that (x_{n}) is a Cauchy sequence in the complete metric space (X, d), and so (x_{n}) is convergent to some x̄ ∈ X. By condition (H1) there exists \overline{y} ∈ Y such that (\overline{x}, \overline{y}) ∈ A and (x̄, ȳ) ⪯_{k^{0}} (x_{n}, y_{n}) for every n ∈ ℕ.

Let us show that (\overline{x}, \overline{y}) is the desired element. Indeed, (x̄, ȳ) ⪯_{k^{0}} (x_{0}, y_{0}). Suppose that (x′, y′) ∈ A is such that (x′, y′) ⪯_{k^{0}} (x̄, ȳ) (⪯_{k^{0}} (x_{n}, y_{n}) for every n ∈ ℕ). Thus z_{B}(y′) + d(x′, \overline{x}) ≤ z_{B}(\overline{y}), whence

d(x′, x̄) ≤ z_{B}(ȳ) − z_{B}(y′) ≤ z_{B}(y_{n}) − z_{B}(y′) ≤ \frac{1}{n} ∀ n ≥ 1.

It follows that d(x′, x̄) = z_{B}(ȳ) − z_{B}(y′) = 0. Hence x′ = \overline{x}. As y′ ≤_{K} \overline{y}, if y′ ≠ \overline{y}, then \overline{y} − y′ ∈ K \ \{0\}, whence z_{B}(y′) < z_{B}(ȳ), which is a contradiction. Therefore (x′, y′) = (x̄, ȳ)

Comparing with [3: Theorem 4], note that the present condition on K is stronger (because in this case K^{#} ≠ ∅), while the condition on A is weaker (A may be not contained in a half-space). Note that when K and k^{0} are as in Theorem 1, Corollaries 2 and 3 from [3] may be improved. In the next result Y^{•} = Y ∪ {∞} with ∞ ∉ Y; we consider that y ≤_{K} ∞ for every y ∈ Y. We consider also a function f : X → Y^{•} and dom f = {x ∈ X : f(x) ≠ ∞}.

In the following corollary we derive a variational principle of Ekeland’s type for objective functions which take values in a general space Y (cf. [2, 3, 5 - 7]) under a weaker assumption with respect to the usual lower semicontinuity. For the case Y = ℝ, assumption (H4) in Corollary 2 is fulfilled for decreasingly semicontinuous real-valued functions as in the paper [4].

Corollary 2. Let f : X → Y^{•}. Assume that there exists a proper convex cone B ⊂ Y such that K \{0} ⊂ int B and f(X)∩(ỹ−B) = ∅ for some ỹ ∈ Y. Also, suppose that

(H3) {x′ ∈ X : f(x′) + k^{0}d(x′, x) ≤_{K} f(x)} is closed for every x ∈ X or (H4) for every sequence (x_{n}) ⊂ dom f with x_{n} → x and (f(x_{n})) ≤_{K}-decreasing,

f(x) ≤_{K} f(x_{n}) for every n ∈ ℕ, and K is closed in the direction k^{0}.

Then for every x_{0} ∈ dom f there exists \overline{x} ∈ X such that

f(\overline{x}) + k^{0}d(\overline{x}, x_{0}) ≤_{K} f(x_{0})

and

∀ x ∈ X : f(x) + k^{0}d(\overline{x}, x) ≤_{K} f(\overline{x}) ⟹ x = \overline{x}.

We say that K is closed in the direction k^{0} if K ∩ (y − ℝ_{+}k^{0}) is closed for every y ∈ K. The proof of Corollary 2 is similar to those of Corollaries 2 and 3 in [3].

As mentioned in [3], condition (H1) is verified if K is a well based convex cone, Y is a Banach space and A is closed. As usually (cf. [1]), a convex set S is said to be a base for a convex cone K ⊂ Y if

K = ℝ_{+}S = {λy : λ ≥ 0 and y ∈ S}  and  0 ∉ cl S.

The cone K is called well based if K has a bounded base S. Concerning well based convex cones in normed spaces we have the following characterization.

Proposition 3. Let Y be a normed vector space and K ⊂ Y a proper convex cone. Then K is well based if and only if there exist k^{0} ∈ K and z^{∗} ∈ K^{+} such that ⟨k^{0}, z^{∗}⟩ > 0 and

K ∩ S_{1} ⊂ k^{0} + {y ∈ Y : ⟨y, z^{∗}⟩ > 0}

where S_{1} = {y ∈ Y : ||y|| = 1} is the unit sphere in Y.

Proof. Suppose first that K is well based with bounded base S; therefore 0 ∉ cl S and K = [0, ∞) · S. Then there exists z^{∗} ∈ Y^{∗} such that 1 ≤ ⟨y, z^{∗}⟩ for all y ∈ S. Consider S̃ := {k ∈ K : ⟨k, z^{∗}⟩ = 1}. It follows that S̃ is a base of K; moreover, since S̃ ⊂ [0, 1] · S, S̃ is also bounded. Taking k^{1} ∈ K \ {0} we have K ∩ S_{1} ⊂ λk^{1} + B_{+} for some λ > 0, where B_{+} = {y ∈ Y : ⟨y, z^{∗}⟩ > 0}. Otherwise

∀ n ∈ ℕ^{∗} ∃ k_{n} ∈ K ∩ S_{1} : k_{n} ∉ \frac{1}{n} k^{1} + B_{+}.

Therefore ⟨k_{n}, z^{∗}⟩ ≤ \frac{1}{n}⟨k^{1}, z^{∗}⟩ for every n ≥ 1. But, because S̃ is a base, k_{n} = λ_{n}b_{n} with λ_{n} > 0 and b_{n} ∈ S̃; it follows that 1 = ||k_{n}|| = λ_{n}||b_{n}|| ≤ λ_{n}M with M > 0 (because S̃ is bounded). Therefore

M^{−1} ≤ λ_{n} = ⟨λ_{n}b_{n}, z^{∗}⟩ = ⟨k_{n}, z^{∗}⟩ ≤ n^{−1}⟨k^{1}, z^{∗}⟩ ∀ n ∈ ℕ^{∗}

whence M^{−1} ≤ 0, which is a contradiction. Thus there exists λ > 0 such that K ∩ S_{1} ⊂ λk^{1} + B_{+}. Taking k^{0} := λk^{1} the conclusion follows.

Suppose now that K ∩ S_{1} ⊂ k^{0} + B_{+} for some k^{0} ∈ K and z^{∗} ∈ K^{+} with ⟨k^{0}, z^{∗}⟩ = c > 0, where B_{+} is defined as above. Consider S = {k ∈ K : ⟨k, z^{∗}⟩ = 1}. Let k ∈ K \{0}. Then ∥k∥^{−1}k = k^{0} +y for some y ∈ B_{+}. It follows that ⟨k, z^{∗}⟩ > c∥k∥ > 0; therefore z^{∗} ∈ K^{#} and so k ∈ (0, ∞) · S. Since cl S ⊂ {y ∈ Y : ⟨k, z^{∗}⟩ = 1}, we have that S is a base of K. Let now y ∈ S (⊂ K). Then ∥y∥^{−1}y ∈ K ∩ S_{1}. There exists z ∈ B_{+} such that ∥y∥^{−1}y = k^{0} + z. We get

1 = ⟨y, z^{∗}⟩ = ∥y∥⟨k^{0} + z, z^{∗}⟩ ≥ c∥y∥

whence ∥y∥ ≤ c^{−1}. Therefore S is bounded, and so K is well-based

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Received 21.10.1998