# 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.

• Using relative interior is important: otherwise consider $f$ defined by $f(x)=-\infty$ for $x<0$, $=0$ for $x=0$, $=+\infty$ for $x>0$.
– daw
Commented Nov 16, 2021 at 16:08
• I see. If the points are from the relative interior of the domain, that will force all the $\lambda_i$, the weight for the convex combinations to be strictly less than 1. Commented Nov 16, 2021 at 19:40

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.
• I think the author meant "Since $f(x)$ is finite the epigraph is non-empty." Commented Aug 22, 2023 at 18:38