Convexity of the supremum of a family of convex functions Let $E\subset \mathbb R^n$ be a convex set and $f_k: E \to \overline {\mathbb R}$ be convex functions for $k \in \mathbb N$. Prove that $\sup_{k\in \mathbb N} f_k$ is also convex.
I know that if $f_i$ could only take values in $\mathbb R$ (instead than in $\overline {\mathbb R}$) I could write $\sup_k f_k(t x_1 + \alpha (1-t)x_2)\leq t \sup_k f_k(x_1) + (1-t)\sup_k f_k(x_2)$ and conclude that $\sup_{k\in \mathbb N} f_k$ is convex. However, if $f_i$ can take the values $+\infty$ or $-\infty$, I cannot write the same (because $\infty - \infty$ may show up). How is convexity proved in this case?
 A: From the wiki on convex functions:

The sum $-\infty +\infty$ is also undefined so a convex extended real-valued function is typically only allowed to take exactly one of $-\infty$ and $+\infty$ as a value.

So let's assume that for each of the $f_k$'s. We want to be able to write
$$ \begin{split}
& \sup_k f_k(t x_1 + (1-t)x_2) \\
\le & \sup_k \big(t f_k(x_1) + (1-t) f_k(x_2) \big) \\
\le &\ t \sup_k f_k(x_1) + (1-t)\sup_k f_k(x_2)\ .
\end{split} $$
If any $f_k$ doesn't take the value $-\infty$, then none of the sup's above can evaluate to $-\infty$, so everything is well-defined.
But we can run into problems. Let $E=[0,1]\subset\mathbb R$, and let
$$
f_k = \begin{cases} 
-\infty & x\in[0,1) \\ 
k & x=1 \end{cases} \ .
$$
Now $\sup f_k(0) = -\infty$ and $\sup f_k(1) = +\infty$, so the definition of convexity doesn't apply. So we either need an extension of the definition (of which I can imagine a couple of plausible variations), or to make some assumptions on the functions. Note, if there were only finitely many functions, we would be fine.
Also, any funny business can only happen at the boundary of $E$. The function value must be $>\!-\infty$ or $=\!-\infty$ on the whole interior of $E$ for each $f_k$, and therefore also for $\sup f_k$. So taking an open domain would solve the problem. Alternatively, I think it would be fair to call a function like $\sup f_k$ convex using any reasonable extended definition.
EDIT: With the epigraph-definition that gerw mentions, we can see from the above that it works out.
