I am trying to prove that a convex map $f: S \rightarrow \Bbb{R}$ where $S \subset \Bbb{R}^n$ is an open and convex set and $n \in \Bbb{N}$ arbitrary is continuous by generalizing this proof . This would be wonderful because the resulting proof would be clear and concise.
Therefore, I was wondering, if we suppose that $f$ is not continuous on some point $x_0 \in S$, can we use the convexity of $S$ to take another $x \in S$ and an increasing sequence $\{\lambda_n\} \subseteq (0,1)$ convergent to $1$ such that the sequence $x_n = \lambda_n x_0 + (1 - \lambda_n) x$ lies in $S$ and $f(x_n) \nrightarrow f(x_0)$ ? If it is the case, how we ensure its existence?
I understund that when a map is not continuous on a point, there is a sequence convergent to the point such that the image of the sequence does not converges to the image of the point, but I do not see clearly how to use this with the convexity to construct a sequence like the mentioned, so every possible answer would be appreciated.
