proof of criterion for convexity II

If f was not convex, then there was a point ξ(a,b) such that f(ξ)>h(x)=f(v)-f(u)v-u(x-u)+f(u) for some u<v in (a,b). Since f is continuousMathworldPlanetmathPlanetmath, there would be a neighborhoodMathworldPlanetmath (ξ-δ,ξ+δ),δ>0, of ξ such that f(x)>h(x) for all x in this neighborhood. (I.e., f(x) was “above” the line segment joining f(u) and f(v).) Let s=ξ-δ,t=ξ+δ.

Using the two points A=(s,f(s)),B=(t,f(t)), we construct another line segment AB¯ whose equation is given by g(x)=f(s)-f(t)2δ(x-s)+f(s); we have f(x)>g(x) for x(s,t). In particular,

f(ξ)=f(s+t2)>g(ξ)=f(s)+f(t)2. (1)

(One easily verifies g(ξ)=(f(s)+f(t))/2.) This contradicts hypothesisMathworldPlanetmath.

Note that we have tacitly used the fact that h(x)=λf(v)+(1-λ)f(u) for some λ and g(x)=λf(s)+(1-λ)f(t) for some λ.

Title proof of criterion for convexity II
Canonical name ProofOfCriterionForConvexityII
Date of creation 2013-03-22 18:25:25
Last modified on 2013-03-22 18:25:25
Owner yesitis (13730)
Last modified by yesitis (13730)
Numerical id 5
Author yesitis (13730)
Entry type Proof
Classification msc 52A41
Classification msc 26A51
Classification msc 26B25