proof of criterion for convexity II

If $f$ was not convex, then there was a point $\xi \in (a,b)$ suchthat $f(\xi )>h(x)=\frac{f(v)-f(u)}{v-u}(x-u)+f(u)$ for some 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” theline 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 constructanother 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,

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

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

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 $\lambda$.