I'm having trouble proving this theorem.
$\mathbf{Theorem.}$ Suppose $f:\mathbb R^m\to \mathbb R$ is a convex function and $A\in \mathbb R^{m\times n}, b \in \mathbb R^n$. If we let the mapping $h:\mathbb R^m\to \mathbb R$ be defined by $h(x) = f(Ax+b)$, then the subdifferential satisfies $$\partial h(x) = A^T \partial f(Ax+b).$$
$\mathbf{Proof.}$ If $g \in \partial f(Ax+b)$, then $$h(y) = f(Ay+b) \ge f(Ax + b) + g^T(Ay-Ax) = f(Ax+b) + (A^Tg)^T (y-x),$$ which shows that $A^Tg \in \partial h(x)$, and since $g$ was arbitrary, it follows that $$A^T\partial f(Ax+b) \subseteq \partial h(x).$$
However I am having trouble proving the other inclusion. If $g \in \partial h(x)$ that there exists $g_0 \in \partial f(Ax+b)$ such that $g = A^Tg_0$.
I only know that $$h(y) = f(Ay+b) \ge f(Ax+b) + g^T(Ay-Ax).$$