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 $\theta\geqslant0$.

  • $\begingroup$ Thank you for your reply! =) So it's not possible that any of these $2n$ numbers could equal $\infty$? =) $\endgroup$ – jjepsuomi Oct 23 '14 at 7:04
  • 1
    $\begingroup$ It should not be possible (otherwise the result is wrong), which is why one should either assume that $$||\textbf{x}-\textbf{x}^*||\leq\delta' \;\;\rightarrow \;\; \textbf{x}\in \text{int}S,$$ or that, for some $\delta''$, $$||\textbf{x}-\textbf{x}^*||\leq\delta'' \;\;\rightarrow \;\; \textbf{x}\in S$$ and then consider $\delta'\lt\delta''$. $\endgroup$ – Did Oct 23 '14 at 7:11

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