A convex function which intersects a chord at three points is affine Let  $F:[a,b] \to \mathbb R$ be a convex function.
Let $c \in [a,b]$, and let $x < c < y$. Let $\lambda \in (0,1)$ satisfy $c = \lambda \, x + (1-\lambda)\, y$.
Suppose that  $F(c)=\lambda \, F(x) + (1-\lambda) \, F(y)$.
How to prove that $F|_{[x,y]}$ is affine?
This "gap-domination" property implies that
$F(c)=\tilde \lambda \, F(\tilde x) + (1-\tilde \lambda) \, F(\tilde y)$ holds for every $\tilde x < c < \tilde y$ and $\tilde \lambda \in [0,1]$ satisfying $c =\tilde \lambda \, \tilde x + (1-\tilde\lambda)\, \tilde y$, but I don't see how that implies the claim, since we need to consider combinations whose mean is different from $c$.
 A: Given a function $f$ and $u,v$ in the domain of $f,$ let $s(u,v)$ be the slope of the chord $[(u,f(u)),(v,f(v))].$
A standard result for convex functions: Slopes of consecutive chords increase. I.e., if $f$ is convex and $u<v<w,$ then $s(u,v)\le s(v,w).$
In your problem, consider the line $L$ through $(x,f(x))$ and $(y,f(y)).$ We are told that $f(c)=L(c)$ for some $c, x<c<y.$ Let $d\in (x,c).$ Certainly $f(d)\le L(d)$ by convexity. Could $f(d)<L(d)?$ No, because then $s(d,c)> s(c,y),$ violating the "increasing slopes" property. A similar argument shows that if $c<d'<y,$ then $f(d')=L(d').$ Thus $f=L$ on $[x,y]$ as desired.
A: First, let's notice that the hypothesis on $c$ can be rewritten this way :
$$F(c) = \frac{F(y)-F(x)}{y-x}(c-x) + F(x)$$
Let's take $z \in (x,c)$. Because $F$ is convex on $[x,y]$, one has
$$F(z) \leq \frac{F(y)-F(x)}{y-x}(z-x) + F(x) \quad \quad (1)$$
But because $F$ is convex on $[z,y]$, and because $c \in (z,y)$, one must also have
$$F(c) \leq \frac{F(z)-F(y)}{z-y}(c-y) + F(y) \quad \quad (2)$$
Introducing $(2)$ in $(1)$, you  deduce that
$$F(c) \leq \frac{\frac{F(y)-F(x)}{y-x}(z-x) + F(x) -F(y)}{z-y}(c-y) + F(y) $$
$$ =\frac{(F(x)-F(y))\left(\frac{z-x}{y-x} -1 \right)}{z-y}(c-y) + F(y) $$
$$= \frac{(F(x)-F(y))\left(z-y  \right)}{(z-y)(y-x)}(c-y) + F(y) $$
$$=\frac{(F(x)-F(y))}{(y-x)}(c-y) + F(y) = F(c)$$
So all the inequalities are in fact equalities. In particular, the equation $(1)$ is
$$F(z) = \frac{F(y)-F(x)}{y-x}(z-x) + F(x) \quad \quad (1)$$
So $F$ is affine on $[x,c]$.
The exact same can be done to prove that $F$ is affine on $[c,y]$.
A: If $I$ is an interval and the convex function $F: I \to \Bbb R$ intersects a linear function $L: I \to \Bbb R$ at two points $x < y$ then $F \le L$ on $[x, y]$ and $F \ge L$ on $I \setminus [x, y]$.
Now if $F$ intersects $L$ at three points $a < c< b$ then we can apply the above to $[a, c]$ and $[c, b]$, so that

*

*$F \le L$ on $[a, c]$ and $F \ge L$ on $[c, b]$,

*$F \le L$ on $[c, b]$ and $F \ge L$ on $[a, c]$.

Combining these inequalities gives $F=L$ on $[a, b]$.
