I studying convex analysis and in my book I have the following statement and proof:
Lets assume that $f:S\rightarrow \mathbb{R}, \;S\subset \mathbb{R}^n$ is a convex function. Then $f$ is continuous in the interior $int \;S$ of its domain.
Proof: Lets assume, that $\textbf{x}^{*} \in int \;S$. We have to show that, given $\epsilon > 0$, there exists $\delta>0$, such that
$$||\textbf{x}-\textbf{x}^*||\leq\delta\;\;\rightarrow\;\;|f(\textbf{x})-f(\textbf{x}^*)| \leq \epsilon.$$
Because $\textbf{x}^*\in int\;S$, then there exists $\delta'>0$, such that
$$||\textbf{x}-\textbf{x}^*||\leq\delta' \;\;\rightarrow \;\; \textbf{x}\in S.$$
Let us next define $\theta$ in the following way:
$$\theta = \max_{1\leq i\leq n}\{ \max[f(\textbf{x}^*+\delta'\textbf{e}_i)-f(\textbf{x}^*), f(\textbf{x}^*-\delta'\textbf{e}_i)-f(\textbf{x}^*)]\}.$$
Now because $f$ is convex we have $0 \leq\theta < \infty.$
...
I didn't write the rest of the proof, because I'm only interested on the last line. Why did this $0\leq\theta < \infty$ follow from $f$ being convex?
Thank you for any help =)
This $\theta$ is the maximum of $2n$ numbers hence finite. By convexity, for every $i$, $$(f(x^*+\delta'e_i)-f(x^*))+(f(x^*-\delta'e_i)-f(x^*))\geqslant0,$$ hence $f(x^*+\delta'e_i)-f(x^*)\geqslant0$ or $f(x^*-\delta'e_i)-f(x^*)\geqslant0$, in particular, $$\max(f(x^*+\delta'e_i)-f(x^*),f(x^*-\delta'e_i)-f(x^*))\geqslant0.$$ Thus $\theta\geqslant0$.