# Closedness and Lower Semicontinuity

1. Jan 11, 2006

Given a real-valued function $$f:X \mapsto \mathbb{R}$$ where $$X$$ is a subset of $$\mathbb{R}^n$$. Consider a closed extended real-valued function $$\tilde{f}:\mathbb{R}^n \mapsto (-\infty, \infty]$$ defined as

$$\tilde{f}(x) = \left \{ \begin{array}{ll} f(x) & \quad , \textup{ if } x \in X \\ \infty & \quad , \textup{ otherwise} \end{array} \right..$$

I want to show that if $$X$$ is not closed, then for a sequence $$\{x_k\} \subset X$$ converging to a point $$\bar{x}$$ on the boundary of $$X$$ that does not belong to $$X$$ necessarily results in $$\liminf_{k \to \infty} f(x_k) = \infty$$. I use the following definition for closed function (definition used by my text book): A function is said to be closed if its epigraph is closed.

The following is my proof, but I am a bit unsure of its correctness. Hope someone can give me some feedback.

Suppose otherwise that $$\liminf_{k \to \infty} f(x_k) = M$$ for some $$M < \infty$$. Since $$\tilde{f}$$ is closed, it is lower semicontinuous. Hence,

$$\tilde{f}(\bar{x}) \leq \liminf_{k \to \infty} \tilde{f}(x_k) = \liminf_{k \to \infty} f(x_k) = M < \infty.$$

But $$\tilde{f}(\bar{x}) = \infty$$, thus a contradiction.

Last edited: Jan 11, 2006