By Proposition 3.40(3,4),

$$A^{\ast}(f)=(H+{u_{0}})^{\ast}(f)=H^{\ast}(\tau_{-u_{0}}f),\quad A_{\ast}g=(H+{u_{0}})_{\ast}g=\tau_{u_{0}}(H_{\ast}g).$$

Then, except for the last assertion, the result follows by combining this with the case when $A$ is a linear map, treated in [Roc70, Theorem 16.3].

To prove the last assertion of the proposition, we first note that the concave function

$$(f^{\vee}-u_{0})|_{(H^{\vee})^{-1}(y)}$$

attains its maximum at a point $x$ if and only if its sup-differential at $x$ contains $0$. We fix a point $x_{0}$ in $(H^{\vee})^{-1}(y)$ and we consider the affine inclusion

$$\iota\colon\operatorname{Ker}(H^{\vee})\hookrightarrow M_{\mathbb{R}},\quad z\mapsto z+x_{0}.$$

We denote by $\iota^{\vee}\colon N_{\mathbb{R}}\to N_{\mathbb{R}}/\operatorname{im}(H)$ the dual of the linear part of $\iota$. Set $F=\iota^{*}(f^{\vee}-u_{0})$, then for $z\in\operatorname{Ker}(H^{\vee})$, by Proposition 3.45, we have

$$\partial F(z)=\iota^{\vee}(\partial f^{\vee}(z+x_{0})-u_{0})$$

and so $0\in\partial F(z)$ if and only if $\partial f^{\vee}(z+x_{0})\cap\operatorname{im}(A)\not=\emptyset$. Hence $x=z+x_{0}$ realizes the maximum if and only if $x\in\partial f(Av)$ for some $v\in Q_{\mathbb{R}}$ such that $y\in\partial(A^{*}f)(v)$, as stated. ∎