Let $A\subset\mathbb{R}^n$ be a convex set and $f$ be a function defined on $\mathbb{R}^n$.
Suppose $f$ is convex on $A$. Is $f$ still convex on the closure of $A$?
To add what conditions can we make $f$ convex on the closure of $A$?
For example, if $f$ is continuous?
No. Consider $f:[-1,1] \to \mathbb{R}$ defined by $$f(x) = \begin{cases} x^2 & \text{if } x \in (-1,1) \\ -1 & \text{if } x \in \{-1,1\}. \end{cases} $$
Edit. Suppose $f$ convex on $A$ and continuous on $\overline{A}$. Take two points $x,y \in \overline{A}$; since $\overline{A}$ is closed, there exist two sequences $\{x_n\}, \{y_n\} \subseteq A$ such that $x_n \to x $ and $y_n \to y$ for $n \to \infty$. Fix $\epsilon>0$; then by continuity $$f(tx+(1-t)y)\le f(tx_n +(1-t)y_n) + \epsilon $$if $n$ is big enough. Also $$f(tx_n +(1-t)y_n) + \epsilon \le t f(x_n) + (1-t)f(y_n) + \epsilon \le t f(x) + (1-t) f(y) + 3 \epsilon$$if $n$ is big enough. Since $\epsilon$ is arbitrary, you can conclude.