Problem Statement
let $f: \mathbb{E}\mapsto \bar{\mathbb{R}}$ be a convex function, show that if there exists a point $x\in \text{ri}(\text{dom}(f))$ with $f(x)$ taking a finite value, then $f$ must be proper.
This is an exercise problem that is highly relevant to Rockafellar's Textbook on Variational Analysis. There was a similar exercise related to improper function, but professor tweaked it.
My Take
So far this is my take, I don't think it's correct, or rigorous.
$\text{epi}(f)$ is convex, then
$$
f\left(
\sum_{i = 1}^n \lambda_i x_i
\right) \le \sum_{i = 1}^n \lambda_i f(x_i) \quad \lambda \in \Delta_n, x_i \in \text{dom}(f)
$$
Convexity of Epigraph implies an inequality, relating to the convex combinations of points in the effective domain of the function $f(x)$. Now consider fixing value of $a \in \text{dom}(f)$ such that $f(a)$ is finite, then (Not very rigours here):
$$
a = \sum_{i = 1}^n \lambda_i x_i
$$
A can be represented as a convex combination of points in the affective domain of the function. Then:
$$
-\infty < f(a) = f\left(
\sum_{i = 1}^n \lambda_i x_i
\right) \le \sum_{i = 1}^n \lambda_i f(x_i)
$$
Therefore, for any $x_i$, $f(x_i)$ is not $-\infty$, the function is proper.
However, I don't understand why the original statement stated the existence of finite value inside the Relative Interior of the Affective Domain, instead of just the effective domain?
I am not sure how many points I should choose to construct the convex combinations to be equal to $a$, nor I am sure whether is possible.
Appreciate it.
Best Answer
Two conditions must be true for a convex function to be proper: (i) the epigraph must be non empty and (ii) never takes the value $-\infty$.
Since $f(x)$ is finite the epigraph is non empty.
If there is some value $y$ such that $f(y) = -\infty$ then it must be the case that $f(t) = -\infty$ for $t \in \operatorname{ri}(\operatorname{dom} f)$ (See Rockafellar, "Convex Analysis", Theorem 7.2. Since $f(x)$ is finite, we must have $f(t) > -\infty$ for all $t$.
To see that latter, suppose $f(y) = -\infty$. Since $x \in \operatorname{ri}(\operatorname{dom} f)$ there is some $z \in \operatorname{ri}(\operatorname{dom} f)$ and $\lambda \in (0,1)$ such that $x = \lambda y+(1-\lambda)z$ and hence $f(x) \le \lambda f(y) +(1-\lambda)f(x) = -\infty$ which is a contradiction.