show that $1-x$ is the only extreme point in the set of all twice differentiable convex function

Question

Suppose $A = \{f(x): f''(x)\geq0, f(0)=1, f(1)=0\}$, it is easy to show that $A$ is a convex set and $1-x$ is an extreme point of $A$, could you please show that $1-x$ is the only extreme point of this convex set?

Answer 1

Consider the discontinuous function $\phi(x)$ on $[0,1]$ with $\phi(0)=1$ and with $\phi(x)=0$ for $x \in ]0, 1]$.

For every $\epsilon$, one can find in $A$ an approximation $\phi_{\epsilon}(x)$ of $\phi(x)$ "of quality $\epsilon$"; i.e., $\phi_{\epsilon}(x) < \epsilon$ for $x > \epsilon$. (Sketch this.)

Now, suppose some $g(x)$ in $A$ is an extreme point in $A$ other than $1 - x$. It follows that $g(x) \leq (1-x)$, and the inequality is strict for at least some $x$. Need to show that $A$ contains functions that are everywhere $< g(x)$.

For every $p \in ]1, 0[$ and every small $\epsilon > 0$ form the function $$ g_{p, \epsilon}(x) = p \, g(x) + (1 - p) \phi_{\epsilon}(x). $$ This function is still in $A$, but it can be shown (using the compactness of $[0, 1]$) that $p$ and $\epsilon$ can be chosen small enough to have $$ g_{p, \epsilon}(x) < g(x) \mbox{ for $x \in [0,1]$}. $$