Example of a striclty convex real valued function that's discontinuous

Question

So I've been learning about convex functions, and I know it is the case that if $f$ is diffrentiable and $f'$ is increasing then $f$ is convex.

However is it possible for a strictly convex function to be discontinuous or even just non diffrentiable on an interval?

In the case of convex functions, I can think of the example of the function made from choosing various points on a convex function graph and "joining them up" with straight lines, which would be non-diffrentiable at any of the "links" (though it would still be continuous)

And for strictly convex functions I can think of the case of the lower unit semicircle, which is not differentiable at the end points of the interval.

However I would like to know of a less trivial counterexample for strictly convex functions (since the unit semicircle only loses differentiability at the boundary)

Also I would like to know of a strictly convex (or even just convex) function that is discontinuous (again not including the trivial case of at the endpoint of the interval)

Answer 1

What about $f$ defined on $[-1,1]$ as

$$f(x)=\begin{cases} 1 & x = -1\\ x^2 - 1 & -1 \lt x \lt 1\\ 1 & x = 1 \end{cases}$$

Note: a convex map is always continuous on an open interval.