I have been self-studying Proximal Algorithms by Neal Parikh and Stephen Boyd. It provides a definition of closed proper convex functions without any examples. The definition is given below. Convex and proper seem clear to me. However, I am not sure how to check for closeness. Does continuous imply closeness? I wonder if $l_1$ norm is a closed proper convex function? What about $\| \Sigma x\|_1$ for some positive semidefinite $\Sigma$? Thank you so much!
Yes, continuity implies closedness. If $f : \Bbb{R}^n \to \Bbb{R}$ is continuous, then the map $$g : \Bbb{R}^n \times \Bbb{R} \to \Bbb{R} : (x, y) \mapsto y - f(x)$$ is continuous, and the inverse image of the closed set $[0, \infty)$ is the epigraph of $f$, which therefore must be closed.
It's also true that, in finite-dimensions, if $f$ is closed and $x$ lies in the (algebraic) interior of the domain of $f$, then $f$ must be continuous at $x$. So, closedness differs from continuity primarily at the boundary of the domain. Note that continuity kind of "breaks down" at these points in the traditional sense, since there are points with infinite values in the vicinity.
In particular, this means that for convex functions with full domain, being closed and continuous are equivalent! This includes all the examples you specified.
As for an example of a closed, proper convex map without full domain, you could take $$f : \Bbb{R} \to \Bbb{R} \cup \{\infty\} : x \mapsto \begin{cases} x^2 & \text{if } x \ge 0 \\ 0 & \text{otherwise.} \end{cases}$$ But, if you increase the value at $0$, e.g. $$g : \Bbb{R} \to \Bbb{R} \cup \{\infty\} : x \mapsto \begin{cases} x^2 & \text{if } x > 0 \\ 4 & \text{if } x = 0 \\ 0 & \text{otherwise,} \end{cases}$$ then you get a convex function that is not closed. If you draw the epigraph, you'll notice that the line segment from $(0, 0)$ to $(0, 4)$ is in the boundary of the epigraph, but not in the epigraph itself (except the point $(0, 4)$).