About the subdifferential of a convex function

Suppose $f$ is a convex function over $\mathbb R$, and there exists two points $b_1 < b_2$, $a_1 \leq 0\leq a_2$ such that

$$a_1 \in \partial f(b_1), $$ $$ a_2 \in \partial f(b_2).$$ Then $\arg\min f(x) \in [b_1,b_2].$

Im actually asking this to understand the proof in here. The proof is immediate if we assume $f$ differentiable, but i dont know really how to work with subdifferentials.

Solution 1:

You know that $$f(x) \ge a_1(x-b_1)+f(b_1)$$ $$f(x) \ge a_2(x-b_2)+f(b_2)$$ for all $x$.

In particular, the first inequality implies $f(x) \ge f(b_1)$ if $x < b_1$, and the second inequality implies $f(x) \ge f(b_2)$ if $x > b_2$. So the minimizer must be in $[b_1, b_2]$.