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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!

LinAlg
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Ariel Serranoni
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  • To apply Corollary 6.5.1, you already need $M$ to contain a point of ri(dom($g_1$)). – LinAlg Jun 05 '18 at 13:02
  • You cannot use Corollary 6.5.1. It does not apply to dom($g_1$)=${x \in \mathbb{R}^2 : x_1 \geq 0, x_2 = 0}$ and dom($g_2$)=${x \in \mathbb{R}^2 : x_1 = 0}$ (although the quoted argument from the book is still true). – LinAlg Jun 19 '18 at 23:44
  • I see.. I just wanted to know the reason why the quoted argument is true. – Ariel Serranoni Jun 20 '18 at 12:34

1 Answers1

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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 $x \in V$.

LinAlg
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  • I agree, but the difficult part should be to prove that $x\in U$ implies $x\in V$. – Ariel Serranoni Jun 20 '18 at 18:28
  • Not really. The intersection in $U$ is either one point or multiple points. The first case is straightforward, for the second case, the line segment between those points has to be in $V$. Intuitively speaking :) – LinAlg Jun 20 '18 at 22:19
  • And where did you use that $\text{dom}(g_1)$ is polyhedral? This condition is necessary for this claim, isn't it? However, I do not disagree with your argument. – Ariel Serranoni Jun 21 '18 at 00:34
  • It does not seem to be necessary for this particular claim. It's used later in the proof (top of p. 181). – LinAlg Jun 21 '18 at 01:44