Rockafellar's Theorem 20.1

Question

I am trying to understand the proof of Theorem 20.1 from Rockafellar's classic book "Convex Analysis".

My issue is the argument:

..., and hence $$\text{dom(}g_1)\cap\text{ri(dom}(g_2))\not=\varnothing.$$ This implies that, for the affine hull M of $\text{dom}(g_2)$, $$\text{ri(dom(}g_1)\cap M)\cap\text{ri(dom(}g_2))\not=\varnothing.$$

My thought is, by Corollary 6.5.1, $$\text{ri(dom(}g1)\cap M)=\text{ri(dom(}g_1))\cap M.$$ Thus, since $\text{dom}(g_2)\subseteq M$, $$\text{ri(dom(}g_1)\cap M)\cap\text{ri(dom(}g_2))\\= \text{ri(dom(}g_1))\cap M\cap\text{ri(dom(}g_2))\\=\text{ri(dom(}g_1))\cap\text{ri(dom(}g_2)).$$

And then we have nothing to say (?)

Any guidance will be really helpful!

Answer 1

I think the proof goes along these lines. Define $$ \begin{align} L &= \text{aff dom(}g_1\text{)} \\ M &= \text{aff dom(}g_2\text{)} \\ S &= \text{dom(}g_1)\cap\text{ri(dom}(g_2)) \\ T &= \text{ri(dom(}g_1))\cap\text{ri(dom(}g_2))\\ U &= [\text{dom(}g_1)\backslash\text{ri(dom(}g_1))]\cap\text{ri(dom(}g_2))\\ V &= \text{ri(dom(}g_1)\cap M)\cap\text{ri(dom(}g_2)). \end{align} $$ Note that $x\in T$ implies $x\in V$ by corollary 6.5.1. We therefore only have to consider the case $S\not=\emptyset$ but $T=\emptyset$. Then $x \in U$ implies $x \in V$.