Fix $x_{0}\in{\operatorname{dom}}(\partial f^{\vee})$ such that $C=C_{x_{0}}$ and $u_{0}\in\operatorname{ri}(C)$. Let $x\in\partial f(u_{0})$. Then

(3.29)

$$\langle x,v-u_{0}\rangle\geq f(v)-f(u_{0})\quad\text{for all }v\in N_{\mathbb{R}}.$$

Let $u\in C$. By (3.25), we have $f(u)-f(u_{0})=\langle x_{0},u-u_{0}\rangle$ and so the above inequality implies $\langle x,u-u_{0}\rangle\geq\langle x_{0},u-u_{0}\rangle.$ The fact $u_{0}\in\operatorname{ri}(C)$ implies $u_{0}+\lambda(u_{0}-u)\in C$ for some small $\lambda>0$. Applying the same argument to this element we obtain the reverse inequality $\langle x,u-u_{0}\rangle\leq\langle x_{0},u-u_{0}\rangle$ and so

(3.30)

$$\langle x-x_{0},u-u_{0}\rangle=0.$$

In particular, $f(u)-f(u_{0})=\langle x_{0},u-u_{0}\rangle=\langle x,u-u_{0}\rangle$ and from (3.29) we obtain

$$\langle x,v-u\rangle=\langle x,v-u_{0}\rangle+f(u_{0})-f(u)\geq f(v)-f(u)\quad\text{for all }v\in N_{\mathbb{R}}.$$

Hence $x\in\bigcap_{u\in C}\partial f(u)$ and so $\partial f(u_{0})\subset\bigcap_{u\in C}\partial f(u)$, which implies the stated equality. ∎