Exercise 3.3

Proof. Suppose that $\phi$ is not convex. Then, for $a<x<b,a<y<b,0<\lambda <1$, $\phi (\phi (\mathit{\lambda x}+(1-\lambda )y)>\mathit{\lambda \phi }(x)+(1-\lambda )\phi (y))$ by the definition of convex. Let $z=\mathit{\lambda x}+(1-\lambda )y$. Assume, without a loss of generality, that $x<y$. Then construct $A=\{p\in [x,z):\phi (p)=\mathit{\lambda x}+(1-\lambda )y\}$. By construction, $A$ is bounded by $x$ below and $z$ above, and $\phi (x)\le \mathit{\lambda x}+(1-\lambda )y<\phi (z)$ implies that $A$ is not empty from the Intermediate Value Theorem since $\phi$ is continuous. The boundedness of $A$ means that we can let $\alpha =\sup <!--\; FUNCTION\; APPLICATION\; -->A$. Now construct $B=\{p\in (z,y]:\phi (p)=\mathit{\lambda x}+(1-\lambda )y\}$. By construction, $B$ is bounded by $z$ below and $y$ above, and $\phi (z)<\mathit{\lambda x}+(1-\lambda )y\le \phi (x)$ implies that $B$ is not empty from the intermediate value theorem since $\phi$ is continuous. The boundedness of $B$ means that we can let $\beta =\sup <!--\; FUNCTION\; APPLICATION\; -->B$. Next, by construction, we also know that $\alpha \le z\le \beta$. However, we know that both $\alpha =z$ and $\beta =z$ cannot be true because then we would have a simple discontinuity at $z$, which contradicts that $\phi$ is continuous. So, $\alpha <z<\beta$. By construction, we can then say that for all $c\in (\alpha ,\beta )$, $\phi (c)>\mathit{\lambda \phi }(\alpha )+(1-\lambda )\phi (\beta )$. Specifically for $\lambda =1∕2$, for all $c\in (\alpha ,\beta )$, $\phi (c)>(\phi (\alpha )+\phi (\beta ))∕2$. However, this is a contradiction since the values of $\phi$ over an interval cannot be strictly greater than the average value of $\phi$ over this interval since $\phi$ is continuous. Therefore, $\phi$ must be convex. □