A problem with equality in a inequality for convex function

Question

Let $f:\rightarrow \mathbb R$ be a convex function on a convex subset $D$ of linear space $X$. Assume that for some pairwise disjoit $x_1,x_2,x_3\in D$ and some $t_1,t_2,t_3\in (0,1)$ such that $t_1+t_2+t_3=1$ the following equality holds: $$ f(t_1x_1+t_2x_2+t_3x_3)=t_1f(x_1)+t_2f(x_2)+t_3f(x_3). $$ Is it true that $$ f(s_1x_1+s_2x_2+s_3x_3)=s_1f(x_1)+s_2f(x_2)+s_3f(x_3) $$ for all $s_1,s_2,s_3\in (0,1)$ such that $s_1+s_2+s_3=1$? What does the last condition geometrically mean?

Answer 1

Let $A= \{ \lambda \in (0,1)^3 | \sum_k \lambda_k = 1 \}$ and define $\phi:A \to \mathbb{R}$ by $\phi(\lambda) = f(\sum_k \lambda_k x_k )-\sum_k \lambda_k f(x_k)$.

Note that $\phi$ is convex and since $f$ is convex on $D$, we have $\phi(\lambda) \le 0$ for $\lambda \in A$. By assumption, $\phi(t) = 0$, so $\phi$ has a maximum on $A$ at $t$.

It follows that $\phi(s) = 0$ for all $s \in A$.

To see this suppose $s \in A$. Then there is some $\lambda>1$ such that $p=s+\lambda(t-s) \in A$ (because $t \in \operatorname{ri} A$). We can write $t = (1-{1 \over \lambda})s+ {1 \over \lambda} p$ and we have $\phi(t) = 0 \le (1-{1 \over \lambda})\phi(s)+ {1 \over \lambda} \phi(p) \le 0$. Since $\phi(s),\phi(p)$ are non-positive and ${1 \over \lambda} \in (0,1)$ we see that $\phi(s) = 0$.

Note that the above analysis holds for $s \in \{ \lambda \in [0,1]^3 | \sum_k \lambda_k = 1 \}$.

The interpretation is that $f$ is 'locally affine' on the set $\operatorname{co} \{x_1,x_2,x_3\} \subset D$.