# Convex Functions are Continuous

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.

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See here. –  John Moeller Feb 20 at 3:56
What function is it? Convex functions are often allowed to take on extended real number values, in which case they may be lower semicontinuous but not continuous. For example, the indicator function of a closed convex set is l.s.c, but not continuous at the boundary. –  littleO Feb 20 at 4:04
@littleO: We are given a function $F:\Bbb R\to(-\infty,\infty]$ with $F$ convex l.s.c. and $F(0)=0$. Our task is to prove $\varphi(x)=F(\Vert x\Vert)$ is convex l.s.c. –  Clayton Feb 20 at 4:14
@JohnMoeller: I see that the top-voted answer claims that convex functions need not be continuous on general topological vector spaces; are normed vector spaces not restrictive enough? Thanks for the reference; I searched the site but was unable to find anything. –  Clayton Feb 20 at 15:00
Do you have a Banach space? There were comments about that. –  John Moeller Feb 20 at 17:43

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.