Suppose that $g: Y \rightarrow ]-\infty,+\infty]$ and let $A: X \rightarrow Y$ be a linear operator (i.e., an $m \times n$ matrix if X and Y are $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively). Let $\bar{x}\in X$ be such that $A\bar{x}\in \operatorname{dom} g$. Show that $A^{*}\partial g(A\bar{x})\subseteq \partial(g \circ A)(\bar{x})$, where $A^{*}: Y \rightarrow X$ stands for the adjoint operator (i.e., transpose matrix) of A.
This is what I have so far, but I feel like my conclusion does not follow. Looking for suggestions to improve this proof. Thank you.
\begin{align} v\in A^{*}\partial g(A\bar{x})&\iff g(A\bar{x})+\langle vA^{*},x-A\bar{x}\rangle\leq g(A\bar{x})\;\;\;(\forall A,x\in \operatorname{dom}(g))\\ &\iff g(A\bar{x})+ vA^{*}x- vA^{*}A\bar{x}\leq g(A\bar{x})\\ &\iff (g \circ A)(\bar{x})+\langle vA^{*},x-\bar{x}\rangle\leq(g \circ A)(x)\\ &\iff v\in\partial(g \circ A)(\bar{x}). \end{align}
Let $X$ and $Y$ be euclidean spaces (for simplicity of the exposition), $g: X \rightarrow (-\infty, +\infty]$ be convex and $A : Y \rightarrow X $ be affine, say $Ax \equiv Lx + \text{cte}$, where $L : Y \rightarrow X$ is linear. Finally, let $x \in Y$. We'll show that $\partial (g \circ A)(x) \supseteq L^* \partial g(A(x))$.
Indeed, let $v \in \partial L^* \partial g(A(x))$. Then there exists $u \in \partial g(A(x))$ such that $ v = L^*u$. Now for any $ z \in Y$, we have \begin{equation} \begin{split} (g \circ A)(x) + \langle v, z - x\rangle &= g(A(x)) + \langle L^*u, z - x\rangle\\ &= g(A(x) + \langle u, Lz - Lx\rangle\\ &= g(A(x)) + \langle u, Az - Ax\rangle\\ &\le g(A(z)),\text{ since }u \in \partial g(A(x))\\ &= (g \circ A)(z). \end{split} \end{equation} Thus $v \in \partial (g \circ A)(x)$, and we're done.