If $a^Tz \ge a^Tx$ for all $z \in dom f$ then $x \not \in dom f$.

Question

$a^Tz \ge a^Tx$ for all $z \in \text{dom} f$ then $x \not \in dom f$. Where $f: R^n \to R$ is a convex function.

The question is why $a^Tz \ge a^Tx$ should imply that $x \not \in \text{intdom} f$?

Answer 1

You need additional assumption that $a \ne 0$.

Guide:

If $x$ is in the interior, then $x - \delta a$ is in the domain for sufficiently small positive $\delta$.

Try to use this to argue for contradiction.