# Continuity of Concave and Convex Functions

A real function p on a subset of a Euclidean space is convex if, for each pair of points a, b and every X e [0, 1], p(Xa + (1 — X)b) > Xp(a) + (1 — X)p(b). For example, p(x) = x2 is a convex function on the real line, and so is p(x) = exp(x). Similarly, p is concave if, for each pair of points a, b and every к e [0, 1], p(ka + (1 — k)b) < kp(a) + (1 — k)p(b).

I will prove the continuity of convex and concave functions by contradiction. Suppose that p is convex but not continuous in a point a. Then

p(a+) = lim p(b) = p(a) (II.6)

b^a

or

p(a—) = lim p(b) = p(a), (II.7)

bfa

or both. In the case of(II.6) we have

p(a+) = lim p(a + 0.5(b — a)) = lim p(0.5a + 0.5b)

b^a b^a

< 0.5p(a) + 0.5 lim p(b) = 0.5p(a) + 0.5p(a+);

b^a

hence, p(a+) < p(a), and therefore by (II.6), p(a+) < p(a). Similarly, if (II.7) is true, then p(a-) < p(a). Now let 8 > 0. By the convexity of p, it follows that

p(a) = p(0.5(a — 8) + 0.5(a + 8)) < 0.5p(a — 8) + 0.5p(a + 8),

and consequently, letting 8 I 0 and using the fact that p(a+) < p(a), or p(a—) < p(a), or both, we have p(a) < 0.5p(a—) + 0.5p(a+) < p(a). Because this result is impossible, it follows that (II.6) and (II.7) are impossible; hence, p is continuous.

If p is concave, then — p is convex and thus continuous; hence, concave functions are continuous.

An (open) covering of a subset © of a Euclidean space Kk is a collection of (open) subsets U(a), a e A, of Kk, where A is a possibly uncountable index set such that © c Uae AU (a). A set is described as compact if every open covering has a finite subcovering; that is, if U(a), a e A is an open covering of © and © is compact, then there exists a finite subset B of A such that © c UaeB U (a).

The notion of compactness extends to more general spaces than only Euclidean spaces. However,

Theorem II.2: Closed and bounded subsets of Euclidean spaces are compact.

Proof: I will prove the result for sets © in К only. First note that boundedness is a necessary condition for compactness because a compact set can always be covered by a finite number of bounded open sets.

Next let © be a closed and bounded subset of the real line. By boundedness, there exist points a and b such that © is contained in [a, b]. Because every open covering of © can be extended to an open covering of [a, b], we may without loss of generality assume that © = [a, b]. For notational convenience, let © = [0, 1]. There always exists an open covering of [0, 1] because, for arbitrary є > 0, [0, 1] c Uo<x<i(x – є, x + є). Let U(а), а є A, be an open covering of [0, 1]. Without loss of generality we may assume that each of the open sets U(a) takes the form (a(a), b(a)). Moreover, if for two different indices a and в, a(a) = a(e), then either (a(a), b(a)) c (a(P), b(fi)), so that (a(a), b(a)) is superfluous, or (a(a), b(a)) э (a(fi), b(P)), so that (a(fi), b(P)) is superfluous. Thus, without loss of generality we may assume that the a(a)’s are all distinct and can be arranged in increasing order. Consequently, we may assume that the index set A is the set of the a(a)’s themselves, that is, U (a) = (a, b(a)), a є A, where A is a subset of К such that [0, 1] c UaeA(a, b(a)). Furthermore, if a1 < a2, then b(aj) < b(a2), for otherwise (a2, b(a2)) is superfluous. Now let 0 є (ai, b(a1)), and define for n = 2, 3, 4,…,an = (an-1 + b(an—1))/2. Then [0, 1] c UJ=1(an, b(an)). This implies that 1 є U^=1(an, b(an)); hence, there exists an n such that 1 є (an, b(an)). Consequently, [0, 1] c Un=1(aj, b(aj)). Thus, [0,1] is compact. This argument extends to arbitrary closed and bounded subsets of a Euclidean space. Q. E.D.

A limit point of a sequence xn of real numbers is a point x* such that for every є > 0 there exists an index n for which | xn — x* | < є. Consequently, a limit point is a limit along a subsequence. Sequences xn confined to an interval [a, b] always have at least one limit point, and these limit points are contained in [a, b] because limsupn^TO xn and liminfn^TO xn are limit points contained in [a, b] and any other limit point must lie between liminfn^TO xn and limsupn^TO xn. This property carries over to general compact sets:

Theorem II.3: Every infinite sequence вn of points in a compact set © has at least one limit point, and all the limit points are contained in ©.

Proof: Let © be a compact subset of a Euclidean space and let ©k, к = 1, 2,… be a decreasing sequence of compact subsets of © each containing infinitely many en’s to be constructed as follows. Let ©0 = © and k > 0. There exist a finite number of points в* j, j = 1,…,mk such that, with Uk (в *) = {в : ||в — в*|| < 2—к}, ©к is contained in и"= 1 Uk(в* j). Then at least one of these open sets contains infinitely many points вп, say Uk(в*^). Next, let

©k+1 = {в : ||в — в*лЦ <2—k}n©k,

which is compact and contains infinitely many points вп. Repeating this construction, we can easily verify that n£=0 ©k is a singleton and that this singleton is a limit point contained in ©. Finally, if a limit point в * is located outside ©, then, for some large к, Uk(в *) П © = 0, which contradicts the requirement that Uk(в*) contain infinitely many вп’s. Q. E.D.

Theorem II.4: Let в n be a sequence ofpoints in a compact set ©. If all the limit points of в n are the same, then limn^me n exists and is a point in ©.

Proof: Let в* e © be the common limit point. If the limit does not exist, then there exists a 8 > 0 and an infinite subsequence вщ such that | ent – в* | > 8 for all к. But вщ also has limit point в*, and thus there exists a further subsequence вщ(m) that converges to в*. Therefore, the theorem follows by contradiction. Q. E.D.

TheoremII.5: For a continuous function g on a compact set ©, supe e© g(e) = maxвe© g(e) and inf в e© g(e) = minвe© g(в). Consequently, argmaxв e© g(в) e © and argminee© g(в) e ©.

Proof: It follows from the definition of sup,, e© g(в) that for each к > 1 there exists a point вк e © such that g(вк) > sup, e© g(в) – 2-t; hence, limt^TO ge) = supee© g(e). Because © is compact, the sequence вк has a limit point в* e © (see Theorem II.3); hence, by the continuity of g, g(e*) = supee© g(e)• Consequently, sup, e© g(e) = max, e© g(e) = g(e*). Q. E.D.

Theorem II.6: Let g be a continuous function on a compact set ©, and let в0 = argminee© g(e) be unique. Then there exists a 8 > 0 such that for all 8 e (0,8), infee©.ye-в0 у>8 g(e) > g(eo). Similarly, if во = argmaxe e© g(e) is unique, then there exists a 8 > 0 such that for all 8 e (0, 8), supee©. ув-в0 У> g(e) < g(,0).

Proof: It follows from Theorem II.5 that в0 = argminee© g(e) e ©. Let ©8 = {в e ©: ув – в0|| > 8} for 8 > 0. If ©8 is nonempty, then it is compact. To see this, let {©a, a e A} be an open covering of ©8 : ©8 c UaeA ©a, and let ©* = {в : ув – в0 У < 8}. Then © c ©* U (UaeA ©a), and thus by the compactness of © there exists a finite subcovering © c Un=1 © j. Without loss of generality we may assume that ©* = ©0 and thus that ©8 c Un=0 © j; hence, ©8 is compact. Then by Theorem II.5, в8 = argminee©s g(e) e ©8 c ©• Because в0 is unique we have g(e0) < g(e8)• The argmax case follows by a similar argument. Q. E.D.

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