Let $f:\mathbb R\to \mathbb R$ a convex function. I would like to prove that $$f(x)=\sup_{(a,b)\in E}(ax+b),$$ where $$E=\{(a,b)\in\mathbb Q^2\mid \forall x, f(x)\geq ax+b\}.$$
Let $f'_r$ and $f'_l$ the derivative at left and at right. Let $a\in\mathbb R$. Since $$f(x)\geq \min\{f'_r(a),f'_l(a)\}(x-a)+f(a),$$
for all $x\in\mathbb R$, we get that $E\neq \emptyset$ and upper bounded. So the supremum is well defined. By definition, $f(x)\geq \sup_{(a,b)\in E}(ax+b)$. But how can I prove the converse inequality ? I mean, how can I get $$f(x)\leq \sup_{(a,b)\in\mathbb Q^2}(ax+b) \ \ ?$$ Any idea ?