Convexity and unique minimum points Josef Berger and Gregor Svindland February 17, 2018 Abstract We show constructively that every quasi-convex, uniformly continuous function f : C R with at most one minimum point has a minimum point, where C is a convex compact subset of a finite dimensional normed space. Applications include a result on strictly quasi-convex functions, a supporting hyperplane theorem, and a short proof of the constructive fundamental theorem of approximation theory. 1 Introduction Let (X, d) be a compact metric space. The infimum of a uniformly continuous function f : X R is given by inf f = inf {f(x) x X}. An element x of X is a minimum point of f if f(x) = inf f. The function f has at most one minimum point if d(x, y) > 0 inf f < f(x) inf f < f(y) for all x, y X. In Bishop s constructive mathematics [6, 7, 8], the framework of this paper, the following statements are equivalent: (I) Brouwer s fan theorem for detachable bars. (II) Every positive-valued, uniformly continuous function on a compact metric space has positive infimum. 1
(III) Every uniformly continuous function on a compact metric space which has at most one minimum point has a minimum point. The equivalence of (I) and (II) was proved in [2, 10] and the equivalence of (I) and (III) was proved in [1, 2, 11]. In [3], we proved (II) for quasi-convex functions whose domain C is a convex compact subset of R m. Quasi-convex means that f(λx + (1 λ)y) max (f(x), f(y)) for all λ [0, 1] and x, y C. That result was crucial for the constructive treatment of the fundamental theorem of asset pricing in [4] and corresponds to a constructively valid version of the fan theorem, see [5]. In this paper which is a sequel of [3] we show (III) for quasi-convex functions whose domain C is a convex compact subset of R m (Theorem 1) and generalise this result to finite-dimensional normed spaces (Theorem 2). As applications we obtain a supporting hyperplane theorem (Proposition 1) and a result on strictly quasi-convex functions (Proposition 2). Moreover, we obtain a new short proof of the fundamental theorem of approximation theory (Proposition 3). 2 Unique minimum points In this section, we prove the following theorem. Theorem 1. Let C be a convex compact subset of R m. Then every quasiconvex, uniformly continuous function f : C R with at most one minimum point has a minimum point. To this end, let C be a convex compact subset of R m. Fix a quasi-convex, uniformly continuous function f : C R which has at most one minimum point. Without loss of generality, we may assume that inf f = 0. For a subset S of C let f S denote the restriction of f to S. Lemma 1. Fix convex compact subsets A, B of C such that d(a, b) > 0 for all a A and b B. Then inf(f A ) > 0 or inf(f B ) > 0. Proof. The set A B is a convex compact subset of R 2m, and the function F : A B R, (a, b) max (f(a), f(b)) is positive-valued, quasi-convex, and uniformly continuous. By [3, Theorem 1] we can conclude that ε := inf F > 0. Since inf(f A ) > 0 inf(f A ) < ε 2
and inf(f B ) > 0 inf(f B ) < ε, this implies that inf(f A ) > 0 or inf(f B ) > 0. For x R m and j {1,..., m} the j-th component of x is denoted by x j. Lemma 2. For each ε > 0 there exists C C such that (i) C convex and compact (ii) inf(f C ) = 0 (iii) x, y C j (x j y j < ε). Proof. Consider the first coordinate. Since the projection mapping x x 1 is uniformly continuous and uniformly continuous images of totally bounded sets are totally bounded [8, Proposition 2.2.6], the set D = {x 1 x C} is totally bounded. Set ι = inf D and η = sup D. If η ι < ε, set C = C. We may therefore assume that ι < η. Define s = ι + 1(η ι) and t = ι + 2 (η ι). 3 3 By [3, Proof of Lemma 4], the sets and A = {x C x 1 s} B = {x C x 1 t} are convex and compact. For a A and b B we have d(a, b) > 0. By Lemma 1, we can conclude that In the first case, set and in the second case set inf(f A ) > 0 or inf(f B ) > 0. (1) C = {x C x 1 s} C = {x C x 1 t}. The the set C fulfils the properties (i) and (ii). Iterating this procedure, also over the coordinates, we eventually obtain a set C which fulfills (iii) as well. The diameter of a compact set S is defined by diam S = sup {d(x, y) x, y S}. 3
of Theorem 1. By Lemma 2, we can construct a sequence (C n ) of compact subsets of C such that (a) n (C n+1 C n ) (b) lim n diam C n = 0 (c) n (inf(f C n ) = 0). For each n, fix x n C n with f(x n ) < 1/n. The sequence (x n ) is a Cauchy sequence and its limit is a minimum point of f. 3 Applications 3.1 Finite-dimensional normed spaces A normed space V is finite-dimensional if either V = {0} or else there exist b 1,..., b m V such that the linear mapping κ : R m V, λ m λ i b i i=1 is bijective. (Injective in the sense that λ > 0 implies κ(λ) > 0.) In this case, both κ and its inverse κ 1 are uniformly continuous. See [6, 7, 8] for more information on finite-dimensional normed spaces. In view of the definition of a finite-dimensional normed space, we obtain a straightforward generalisation of Theorem 1. Theorem 2. Let C be a convex compact subset of a finite-dimensional normed space. Then every quasi-convex, uniformly continuous function f : C R with at most one minimum point has a minimum point. 3.2 Supporting hyperplanes A subset C of a normed space X is strictly convex if λa + (1 λ)b C for all a, b C with d(a, b) > 0 and all λ ]0, 1[. The set C, the interior of C, is defined as usual: x C ε > 0 y X (d(y, x) < ε y C) The following lemma is easy to prove. 4
Lemma 3. Fix a subset C of X. (a) If C is convex and open, then it is strictly convex. (b) If C is strictly convex and closed, then it is convex. Proposition 1. Let C be a compact, strictly convex subset of a finite dimensional normed space V. Let g : V R be a linear function, and v an element of V with g(v) > 0. Then the restriction of g to C has a minimum point z. Proof. Let f denote the restriction of g to C. Note that linear functions are quasi-convex. Fix a, b with d(a, b) > 0. Set c = (a + b)/2. Since C is strictly convex, there exists δ > 0 such that c δv C. We obtain f(c δv) < f(c) max (f(a), f(b)). Thus f has at most one minimum point. By Theorem 2, f has a minimum point. In the situation of Proposition 1, the set {x V g(x) = g(z)} is called a supporting hyperplane of C. 3.3 Strictly quasi-convex functions Let C be a convex subset of a normed space. A function f : C R is strictly quasi-convex if f(λx + (1 λ)y) < max (f(x), f(y)) for all λ ]0, 1[ and x, y C such that x y > 0. Lemma 4. Every strictly quasi-convex function is quasi-convex. Proof. Fix a strictly quasi-convex function f : C R. As a first step, we show that x, y C λ ]0, 1[ (f(λx + (1 λ)y) max (f(x), f(y))). (2) To this end, fix x, y C and λ ]0, 1[ and assume that f(λx + (1 λ)y) > max (f(x), f(y)). (3) 5
If x y > 0, we have f(λx + (1 λ)y) < max (f(x), f(y)) by the strict quasi-convexity of f. This contradicts (3), thus we can conclude that x = y. This implies that (3) is false. Therefore, we have proved (2). As a second step, we show that f is quasi-convex. Fix x, y C and λ [0, 1]. We either have λ < 1 or else λ > 0. Without loss of generality, we may assume that λ < 1. Assume (3). If λ > 0, then (2) implies that f(λx + (1 λ)y) max (f(x), f(y)), which contradicts (3). Thus λ = 0. This implies that (3) is false, therefore we have shown f(λx + (1 λ)y) max (f(x), f(y)). Remark 1. In a constructive proof, case distinctions are only allowed when we can decide which one of the cases holds. However, this restriction no longer applies when we prove a negated statement. This follows from the scheme ( A A). (4) Having (4) in mind, we observe that the proof of Lemma 4 kan be rephrased as follows. Assume that f is strictly quasi-convex. Fix x, y C and λ [0, 1]. We have to show that f(λx + (1 λ)y) max (f(x), f(y)). (5) This is the negation of (3). If x y = 0 or λ {0, 1}, the statement (5) holds anyway. In the case λ ]0, 1[ and x y > 0, the statement (5) follows from the quasi-convexity of f. Since strictly quasi-convex functions have at most one minimum point, Theorem 2 and Lemma 4 imply the following result. Proposition 2. Let C be a convex compact subset of a a finite-dimensional normed space. Then every strictly quasi-convex, uniformly continuous function f : C R has a minimum point. 6
3.4 Approximation theory Let Y be a subset of a normed linear space X. Let g : Y R be a function. An element y of Y is a minimum point of g if x Y (g(y) g(x)). The function g has at most one minimum point if x, y Y (d(x, y) > 0 z Y (g(z) < max (g(x), g(y)))). These definitions are compatible with those given in Section 1. For a X let fa Y be the function : Y y d(y, a). f Y a The set Y is quasiproximinal if for every a X the implication f Y a has at most one minimum point f Y a has a minimum point is valid. As a consequence of Theorem 2, we obtain the the constructive fundamental theorem of approximation theory from [9]. Proposition 3. Every finite-dimensional subspace V of a normed space X is quasiproximinal. Proof. Since {0} is quasiproximinal, we may assume that the dimension of V is positive. Fix a X. Set f = fa V and suppose that f has at most one minimum point. Since V is a positive-dimensional linear space, there exists b V such that ε := f(b) = d(b, a) > 0. The triangle inequality yields Set v V (d(v, b) > 2 ε f(v) > f(b)). (6) V 0 = {v V d(v, b) 3 d(a, b)}. The set V 0 is compact, see [8, Corollary 4.1.7], and convex. The function g = f V 0 is uniformly continuous and quasi-convex. By (6), g has at most one minimum point. Theorem 2 implies that g has a minimum point v 0. Applying (6) once again, we can conclude that v 0 is a minimum point of f. 7
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