Let $f: \mathbb R^n \rightarrow \mathbb R$ be a convex function. By a subdifference of $f$ in $x\in \mathbb R^n$ we mean an $h\in \mathbb R^n$ such that
$f(x) \geq f(p)+<x-p,h>$ for all $x\in \mathbb R^n$
($<,>$ is the standard scalar product in $\mathbb R^n$).
Let $D^*f(x)$ be the set of all subdifferentials of $f$ at $x$. By properties of convex functions this set is nonempty.
Let $f: \mathbb R^n \rightarrow \mathbb R$ be a convex function and $x,y \in \mathbb R^n$, $x\neq y$, $t\in (0,1)$.
I wish to show that the following conditions are equivalent:
1). $f(tx+(1-t)y)=tf(x)+(1-t)f(y)$,
2). there is a $h\in \mathbb R^n$ such that
$h\in D^*f(x) \cap D^*f(y)$.
We call $h$ a subgradient of $f$ at $x$ iff $f(y) -f(x) \ge \langle h, y-x \rangle$ for all $y$. The usual notation for the set of subgradients at $x$ is $\partial f(x)$.
The implication 2) $\Rightarrow$ 1) is straightforward.
Suppose $h \in \partial f(x) \cap \partial f(y)$. Then we have $f(x) -f(y) \ge \langle h, x-y \rangle$ and $f(y) -f(x) \ge \langle h, y-x \rangle$ which gives $f(y)-f(x) = \langle h, y-x \rangle$. If $t \in (0,1)$, we have $f(y+t(x-y)) -f(y) \le t (f(x)-f(y))$, and since $f(y+t(x-y)) -f(y) \ge t \langle h, x-y \rangle = t(f(x)-f(y))$, we obtain the desired conclusion.
The implication 1) $\Rightarrow$ 2) is geometrically straightforward, but analytically tedious (for me, at least).
I need a few facts about the epigraph of a convex function: Since $f$ is convex and finite everywhere, $f$ is continuous, the epigraph $\operatorname{epi} f$ has a non-empty interior, and we have $\overline{(\operatorname{epi} f)^\circ} = \operatorname{epi} f$. Furthermore, $(\operatorname{epi} f)^\circ = \{ (x,\alpha) | f(x) < \alpha \}$.
Since $f$ is convex, and $f(y+t(x-y)) = f(y) + t (f(x)-f(y))$ for $t \in (0,1)$, we have $f(y+t(x-y)) \ge f(y) + t (f(x)-f(y))$ for all $t \in \mathbb{R}$.
If we let $L \subset \mathbb{R}^n \times \mathbb{R}$ be the line $L = \{(y + t (x-y), f(y)+t (f(x)-f(y))) \}_{t \in \mathbb{R}}$, we see that $L$ does not intersect $(\operatorname{epi} f)^\circ$. Hence there exists some hyperplane that separates $L$ and $(\operatorname{epi} f)^\circ$.
That is, there exists some $\bar{h} = (h,h_0)$ and $\beta$ such that $\langle \bar{h}, e \rangle > \beta \ge \langle \bar{h}, l \rangle$ for all $e \in (\operatorname{epi} f)^\circ, l \in L$.
It is easy to see that we must have $h_0 >0$, hence without loss of generality, we can take $h_0 = 1$.
Since $L$ is a line, the quantity $\langle \bar{h}, l \rangle$ must be a constant (otherwise it would be unbounded above and below), so we may take $\beta = \langle \bar{h}, l \rangle$. Since $(x,f(x)), (y,f(y)) \in L$, we have $\beta = f(x) + \langle h, x \rangle = f(y) + \langle h, y \rangle $.
Since $\bar{h}$ separate the two sets, we have $\alpha + \langle h, z \rangle > \beta$ for all $\alpha > f(z)$, letting $\alpha \to f(x)$ shows $f(z) + \langle h, z \rangle \ge \beta$ for all $z$.
Then $f(z) \ge f(x) + \langle h, x-z \rangle$ for all $z$ and similarly for $y$. Hence $h \in\partial f(x) \cap \partial f(y)$.
(It always bothers me that the geometrically 'obvious' is so much work for me to prove analytically.)