Proof that $f(y) \ge f(x) + g^T(y-x)$, for all $y \in U \Rightarrow g \in \partial f(x)$ where U is a open neighborhood of x

Question

Can anyone help me this problem ?

For a convex function f, show that if $x \in U$ where $U$ is a open neighborhood in its domain, then
$f(y) \ge f(x) + g^T(y-x)$, for all $y \in U \Rightarrow g \in \partial f(x)$

$\partial f(x)$ mean the set of subgradient of f(x) at x.
This is the 2.c problem at homework 2 at this link. I got no idea to proof that because it is so abstract.

Answer 1

Let $z \notin U$
Cuz $U$ is an open set $\Rightarrow \exists \ 0 < t < 1 : y = tx + (1-t)z \in U$
$\Rightarrow f(y) \ge f(x) + g^T(y-x) \ \ \forall \ \ y \in U$
$\Rightarrow f(tx + (1-t)z) \ge f(x) + g^T[tx + (1-t)z - x]$
Cuz $f$ is convex $\Rightarrow f(tx+(1-t)z) \le tf(x) + (1-t)f(z)$
$\Rightarrow tf(x) + (1-t)f(z) \ge f(x) + g^T[tx + (1-t)z - x]$
$\Leftrightarrow (t-1)f(x) + (1-t)f(z) \ge g^T[(t-1)x +(1-t)z]$
$\Leftrightarrow (1-t)[f(z) - f(x)] \ge (1-t)g^T(z-x)$
$\Leftrightarrow f(z) \ge f(x) + g^T(z-x) \ \forall \ z \notin U$
$\Rightarrow g \in \partial f(x)$