In W. Rudin's Principles of Mathematical Analysis, we read in Chapter 4 that real-valued functions defined on an open interval $(a,b)\subseteq\Bbb R$ are continuous (specifically, Exercise 23). I am wondering if this is true if the function is defined on a normed linear space?
My curiosity arises about it because I'm working on homework for a functional analysis course, and one of the exercises is referencing a function $\varphi:E\to\Bbb R$, where $E$ is a normed linear space. The problem here was to show that it is convex lower semi-continuous (l.s.c.), and I've shown that the function is convex without much difficultly. As I thought more and more about it, I convinced myself the function was also continuous (hence l.s.c.). But then calling a function convex l.s.c. seems redundant, so I'm not sure I've thought it through entirely correctly.
Best Answer
An example of a discontinuous finite-valued convex function on an (incomplete) normed linear space, posted by user127096.
Let $c_{00}$ be the subspace of $\ell^2$ consisting of the sequences with finitely many nonzero terms, with the norm inherited from $\ell^2$. The function $\varphi(x) = \sum_{n=1}^\infty n^2 x_n^2$ is convex on $c_{00}$ and takes values in $\mathbb R$ (no infinite values), yet is not continuous. Indeed, the sequence $n^{-1}e_n$ converges to $0$ in the norm of $X$, but $\varphi(n^{-1}e_n) = 1 \not\to 0 = \varphi(0)$. In particular, it is not lower semicontinuous. Also not locally bounded, which is related.