Let $P_{f}(u,x)=f(u)+f^{\vee}(x)-\left<u,x\right>$ be the function introduced in (3.20). For each $x\in\operatorname{stab}(f)$ write $P_{f,x}(u)=P(u,x)$. Let $C_{x}$ be as in Definition 3.23. By Lemma 3.24,

$$C_{x}=\{u\in{\operatorname{dom}}(f)\mid P_{f,x}(u)=0\}.$$

Write $P^{\prime}(v)=\operatorname{rec}(f)(v)-\left<u,x\right>$. Then $P^{\prime}=\operatorname{rec}(P_{f,x})$.

We claim that, for each $x\in\operatorname{stab}(f)$,

$$\operatorname{rec}(C_{x})=\{v\in{\operatorname{dom}}(\operatorname{rec}(f))\mid P^{\prime}(v)=0\}.$$

Let $v\in\operatorname{rec}(C_{x})$. Clearly $v\in{\operatorname{dom}}(\operatorname{rec}(f))$ and, since $x\in\operatorname{stab}(f)$, the set $C_{x}$ is non-empty. Let $u_{0}\in C_{x}$. Then, for each $\lambda>0$, $u_{0}+\lambda v\in C_{x}$. Therefore,

$$P^{\prime}(v)=\lim_{\lambda\to\infty}\frac{P_{f,x}(u_{0}+\lambda v)-P_{f,x}(u_{0})}{\lambda}=0.$$

Conversely, let $v\in{\operatorname{dom}}(\operatorname{rec}(f))$ satisfying $P^{\prime}(v)=0$ and $u\in C_{x}$. On the one hand, by the properties of the function $P_{f}$, we have $P_{f,x}(u+v)\leq 0$. On the other hand, since $P^{\prime}=\operatorname{rec}(P_{f,x})$,

$$P_{f,x}(u+v)-P_{f,x}(u)\geq P^{\prime}(v)=0.$$

Thus $P_{f,x}(u+v)\geq P_{f,x}(u)=0$ and finally $P_{f,x}(u+v)=0$. This implies that, if $u\in C_{x}$ then $u+v\in C_{x}$, showing $v\in\operatorname{rec}(C_{x})$. Hence the claim is proved.

By definition $\Pi(f)=\{C_{x}\}_{x\in\operatorname{stab}(f)}$. Hence $\operatorname{rec}(\Pi(f))=\{\operatorname{rec}(C_{x})\}_{x\in\operatorname{stab}(f)}$. For each $x\in\operatorname{stab}(\operatorname{rec}(f))$, write

$$C^{\prime}_{x}=\{v\in{\operatorname{dom}}(\operatorname{rec}(f))\mid P^{\prime}(v)=0\}.$$

Then $\Pi(\operatorname{rec}(f))=\{C^{\prime}_{x}\}_{x\in\operatorname{stab}(\operatorname{rec}(f))}$. The result follows from the previous claim and the fact that $\operatorname{stab}(f)=\operatorname{stab}(\operatorname{rec}(f))$ by (3.62). ∎