Convex function is proper when it has at least one finite value in the relative interior of its effective domain

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.

Solution 1:

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 empty.

If there is some value $y$ such that $f(x) = -\infty$ 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.