Let f, g be real-valued functions defined on X × Y with the following properties: (2)(0) f (x, y

2 Minimax theorems under η-connectedness

Let X be a nonempty set in a topological space, Y a nonempty η-connected set in a vector space for some η, and f : X × Y −→ R be a real-valued function. Then f is said to be lower semicontinuous on the η of Y if for each x ∈ X and y¹, y₂ ∈ Y , the function

t −→ f(x, η(y¹, y₂, t))

is a lower semicontinuous function of t on [0, 1]. If f is a lower semicontinuous function on Y , then f is clearly lower semicontinuous on the η of Y .

In this section, our main result involves only the η-connectedness instead of convexity.

Before proceeding this, we require the following technical lemma of Farka’s type.

Lemma 2.1. Let X be a nonempty compact set of a topological space and let Y be a nonempty η-connected set. Let f, g be real-valued functions defined on X × Y with the following properties:

(0) f (x, y) ≤ g(x, y) for all (x, y) ∈ X × Y ;

(i) f (·, y) and g(·, y) are upper semicontinuous on X for each y ∈ Y ; (ii) For any y₀, y₁ ∈ Y ,

g(x, η(y₀, y₁, t)) ≤ max{g(x, y⁰), f (x, y₁)}

for all x ∈ X, for all t ∈ [0, 1];

(iii) f is lower semicontinuous on the η of Y ;

(iv) For any y1, · · · , yⁿ ∈ Y , any λ ∈ R, the set ∩ⁿi=1{x ∈ X; g(x, yⁱ) ≥ λ} is either connected or empty.

Then for any λ ∈ R, we have the following alternative:

Either there exists x0 ∈ X such that g(x⁰, y) ≥ λ for all y ∈ Y or there exists y₀ ∈ Y such that f(x, y⁰) < λ for all x ∈ X.

Proof. For each λ ∈ R, y ∈ Y , let

Ug(y) = {x ∈ X; g(x, y) ≥ λ}, and

U_f(y) = {x ∈ X; f(x, y) ≥ λ}.

Fixed λ ∈ R, if for some y⁰ ∈ Y, U^f(y₀) = ∅, then f(x, y⁰) < λ for all x ∈ X. Thus, we may assume that

U_f(y) 6= ∅. (1)

for all y ∈ Y . We are going to show that

\

y∈Y

U_g(y) 6= ∅.

Since X is compact, we need only to show that the family {U^g(y)}y∈Y has the finite inter- section property, that is,

\

y∈F

U_g(y) 6= ∅ for all finite subsets F ⊂ Y.

If |F |=1. By (1), U^f(y) 6= ∅ for all y ∈ Y . From condition (0) , U^g(y) ⊃ U^f(y), so U_g(y) 6= ∅ for all y ∈ Y .

Let F = {y⁰, y1} ⊂ Y, y⁰ 6= y¹. We want to show that Ug(y0) ∩ U^g(y1) 6= ∅ By condition(0) and

Ug(y0) ∩ U^f(y1) = \

>0

(Ug(y0, λ − ) ∩ U^f(y1, λ − )), it is sufficent to show that for any  > 0,

U_g(y₀, λ − ) ∩ U^f(y₁, λ − ) 6= ∅ (2)

Fixed  > 0, for each t ∈ [0, 1], consider the following sets Sg(t) ≡ U^g(η(y0, y1, t), λ − )

= {x ∈ X; g(x, η(y⁰, y₁, t)) ≥ λ − }, and

S_f(t) ≡ U^f(η(y₀, y₁, t), λ − )

= {x ∈ X; f(x, η(y⁰, y₁, t)) ≥ λ − }.

Notice that, for all t ∈ [0, 1], S^g(t) is a nonempty compact connected subset of X by (1), conditions (i) and (iv), and S_f(t) is nonempty closed subset of X by (1) and condition (i).

Let

A₀ ≡ {t ∈ [0, 1]; S^f(t) ⊂ S^g(0)}, and

A₁ ≡ {t ∈ [0, 1]; S^f(t) ⊂ S^f(1)},

If S_g(0) ∩ S^f(1) were empty set, we claim (α), (β) and (γ) as follows.

(α) A₀∩ A¹ = ∅.

This is clear by the construction of these two sets.

(β) A₀∪ A¹ = [0, 1].

Indeed, by (ii), we have, for all x ∈ X, for all t ∈ [0, 1],

g(x, η(y₀, y₁, t)) ≤ max{g(x, y⁰), f (x, y₁)}.

For each t ∈ [0, 1], let x ∈ S^f(t), so x ∈ S^g(t) by condition (0) . Then g(x, η(y₀, y₁, t)) ≥ λ−. It follows that g(x, y⁰) ≥ λ− or f(x, y¹) ≥ λ− by condition (ii). This implies that x ∈ S^g(0) or x ∈ S^f(1), and hence S_g(t) ⊂ S^g(0) ∪ S^f(1). Since S_g(t) is connected, S_g(0) and S_f(1) are nonempty closed sets and S_g(0)∩S^f(1) = ∅, we have either S^g(t) ⊂ S^g(0) or S_g(t) ⊂ S^f(1) and hence S_f(t) ⊂ S^g(0) or S_f(t) ⊂ S^f(1). It follows that t ∈ A⁰ or t ∈ A¹. This shows (β).

(γ) Both A₀ and A₁ are closed subsets of [0, 1].

Let {tⁿ} be a sequence in A⁰ with t_n−→ t⁰ ∈ [0, 1]. Let x ∈ S^f(t₀).

Then

f (x, η(y₀, y₁, t)) ≥ λ − . (3)

From (iii), since f is lower semicontinuous on the η of Y , the function F defined by F (t) = f (x, η(y₀, y₁, t)) is lower semicontinuous on [0, 1]; thus lim inf_nF (t_n) ≥ F (t⁰). It follows that there exits m ∈ N such that F (t^m) ≥ F (t⁰). That is, f (x, η(y₀, y₁, t_m)) ≥ f (x, η(y₀, y₁, t₀)). By (3), this implies that x ∈ S^f(t_m). Since t_m ∈ A⁰, we have x ∈ S^g(0).

Therefore, S_f(t₀) ⊂ S^g(0). It follows that t₀ ∈ A⁰, so A₀ is closed. Similarly, A₁ is closed.

From (α), (β), (γ), we get a contradiction about the connectedness of [0, 1], since A₀ and A₁ are nonempty sets (0 ∈ A⁰ and 1 ∈ A¹). Therefore, S_g(0) ∩ S^f(1) 6= ∅. Hence, (2) holds.

Now, we proceed the proof by induction on |F |, and suppose that ∩y∈FU_g(y) 6= ∅ for all F ⊂ Y, 2 ≤ |F | ≤ k.

Choose {y⁰, · · · , y^k} ⊂ Y, yⁱ 6= y^j if i 6= j. From the above argument, in order to show that ∩^ki=0U_g(y_i) 6= ∅, it is sufficent to show, for each  > 0,that

Ug(y0, λ − )\

Uf(y1, λ − )\

(∩^ki=2Ug(yi, λ − )) 6= ∅. (4) Let  > 0 and H = ∩^ki=2U_g(y_i, λ − ). Notice that, by induction hypothesis, since

T

>0[H ∩ U^g(y, λ − )] = (∩^ki=2U_g(y_i))T

U_g(y) 6= ∅ for all y ∈ Y, H ∩ U^g(y, λ − ) 6= ∅ for all y ∈ Y . In particular,

H ∩ S^g(t) 6= ∅ for all t ∈ [0, 1].

Also, by (iv), H ∩ S^g(t) is connected for all t ∈ [0, 1]. For t ∈ [0, 1], we let B₀ ≡ {t ∈ [0, 1]; H ∩ S^f(t) ⊂ H ∩ S^g(0)},

and

B₁ ≡ {t ∈ [0, 1]; H ∩ S^f(t) ⊂ H ∩ S^f(1)}.

If (H ∩ S^g(0)) ∩ (H ∩ S^f(1)) were empty set, then by the same argument as above, we would have

(α⁰) B₀∩ B¹ = ∅.

(β⁰) B0∪ B¹ = [0, 1].

(γ⁰) Both B₀ and B₁ are nonempty closed subsets of [0, 1].

This again contradicts the connectedness of [0, 1]. Therefore, (4) holds and hence ∩^ki=0U_g(y_i) 6=

∅.

Remark 2.2. The condition (iv) of Lemma 2.1 is a weaker form introducted by Geraghty and Lin [6] than a connected condition used by Terkelsen [7].

Using the above technique Lemma, we now establish our main result as follow.

Theorem 2.3. Let X be a nonempty compact set of a topological space and let Y be a nonempty η-connected set. Let f, g be real-valued functions defined on X × Y with the following properties:

(0) f (x, y) ≤ g(x, y) for all (x, y) ∈ X × Y ;

(i) f (·, y) and g(·, y) are upper semicontinuous on X for each y ∈ Y ; (ii) For any y₀, y₁ ∈ Y ,

g(x, η(y0, y1, t)) ≤ max{g(x, y⁰), f (x, y1)}

for all x ∈ X, for all t ∈ [0, 1];

(iii) f is lower semicontinuous on the η of Y ;

(iv) For any y₁, · · · , yⁿ ∈ Y , any λ ∈ R, the set ∩ⁿi=1{x ∈ X; g(x, yⁱ) ≥ λ} is either connected or empty.

Then

inf

y∈Y sup

x∈Xf (x, y) ≤ sup

x∈X

inf

y∈Yg(x, y) Proof. Let λ ∈ R with

sup

x∈X

inf

y∈Yg(x, y) < λ.

Thus, there exists y₀ ∈ Y such that g(x, y⁰) < λ for all x ∈ X. By Lemma 2.1, there exists y⁰ ∈ Y such that f(x, y⁰) < λ for all x ∈ X. This implies that infy∈Y sup_x∈Xf (x, y) < λ.

It follows that

y∈Yinf sup

x∈Xf (x, y) ≤ sup

x∈X

y∈Yinf g(x, y)

The η-connected property uniformly on Y can be replaced by pointwise η-connectedness defined below. Let Y be a nonempty set. Define

G(y₀, y₁) ≡



η : Y × Y × [0, 1] −→ Y ; η is continuous and η(y₀, y₁, 0) = y₀ η(y0, y1, 1) = y1



We say Y is pointwise η-connected if for any y₀, y₁ ∈ Y , there exists η in G(y⁰, y₁) such that

g(x, η(y₀, y₁, t)) ≤ max{g(x, y⁰), f (x, y₁)}

for all x ∈ X, for all t ∈ [0, 1].

By a similar process of proof in Lemma 2.1, we can get the following theorem imme- diately.

Theorem 2.4. Let X be a nonempty compact set of a topological space, let Y be a nonempty pointwise η-connected set. Let f, g be real-valued functions defined on X × Y with the following properties:

(0) f (x, y) ≤ g(x, y) for all (x, y) ∈ X × Y ;

(i) f (·, y) and g(·, y) are upper semicontinuous on X for each y ∈ Y ; (ii) For any y₀, y₁ ∈ Y , there exists η in G(y⁰, y₁) such that

g(x, η(y₀, y₁, t)) ≤ max{g(x, y⁰), f (x, y₁)}

for all x ∈ X, for all t ∈ [0, 1], and

f is lower semicontinuous on the η of Y

(iii) For any y₁, · · · , yⁿ ∈ Y , any λ ∈ R, the set ∩ⁿi=1{x ∈ X; g(x, yⁱ) ≥ λ} is either connected or empty. Then for any λ ∈ R, we have the following alternative:

Either there exists x₀ ∈ X such that g(x⁰, y) ≥ λ for all y ∈ Y or there exists y₀ ∈ Y such that f(x, y⁰) < λ for all x ∈ X.

Theorem 2.5. Let X be a nonempty compact set of a topological space, let Y be a nonempty pointwise η-connected set. Let f, g be real-valued functions defined on X × Y with the following properties:

(0) f (x, y) ≤ g(x, y) for all (x, y) ∈ X × Y ;

(i) f (·, y) and g(·, y) are upper semicontinuous on X for each y ∈ Y ; (ii) For any y₀, y₁ ∈ Y , there exists η in G(y⁰, y₁) such that

g(x, η(y₀, y₁, t)) ≤ max{g(x, y⁰), f (x, y₁)}

for all x ∈ X, for all t ∈ [0, 1], and

f is lower semicontinuous on the η of Y

(iii) For any y₁, · · · , yⁿ ∈ Y , any λ ∈ R, the set ∩ⁿi=1{x ∈ X; g(x, yⁱ) ≥ λ} is either connected or empty.

Then

inf

y∈Y sup

x∈Xf (x, y) ≤ sup

x∈X

inf

y∈Yg(x, y)