What is the closure of the level set of convex function $g(x):R^n\rightarrow R,$ $\{x:g(x)<a\}$.

Question

suppose that $A:=\{x:g(x)<a\}\neq \emptyset$ and $g(x)$ is a convex function. What is the closure of $A$? My conjecture is $\bar{A}=\{x:g(x)\leq a\}$ but I do not know how to prove it..

Answer 1

It is known that a convex function $g:\mathbb{R}^n\to \mathbb{R}$ is continuous. In particular, $\bar{A}\subset \{x:g(x)\leq a\}.$ On the other hand, if $x\in\mathbb{R}^n$ is such that $g(x)=a,$ then take $y\in\mathbb{R}^n$ such that $g(y)<a$ and let $x_t=tx+(1-t)y,$ $0<t<1.$ By convexity, $g(x_t)<a.$ By continuity, $g(x_t)\to g(x)$ as $t\to 1.$

Answer 2

Hint: Use that convex functions on open sets are continuous.