# proof of criterion for convexity II

If $f$ was not convex, then there was a point $\xi\in(a,b)$ such that $f(\xi)>h(x)=\frac{f(v)-f(u)}{v-u}(x-u)+f(u)$ for some $u in $(a,b)$. Since $f$ is continuous, there would be a neighborhood $(\xi-\delta,\xi+\delta),\delta>0$, of $\xi$ 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 $s=\xi-\delta,t=\xi+\delta$.

Using the two points $A=(s,f(s)),B=(t,f(t))$, we construct another line segment $\overline{AB}$ whose equation is given by $g(x)=\frac{f(s)-f(t)}{2\delta}(x-s)+f(s)$; we have $f(x)>g(x)$ for $x\in(s,t)$. In particular,

 $\displaystyle f(\xi)=f\left(\frac{s+t}{2}\right)>g(\xi)=\frac{f(s)+f(t)}{2}.$ (1)

(One easily verifies $g(\xi)=(f(s)+f(t))/2$.) This contradicts hypothesis.

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

Title proof of criterion for convexity II ProofOfCriterionForConvexityII 2013-03-22 18:25:25 2013-03-22 18:25:25 yesitis (13730) yesitis (13730) 5 yesitis (13730) Proof msc 52A41 msc 26A51 msc 26B25