Let $f_1(x), f_2(x), \ldots, f_n(x)$ be a set of convex functions. We define $f(x)$ as
$$ f(x) = \underset{i}{\text{max}} \left\{ f_i(x) \right\}. $$
How do I show that $f(x)$ is also convex, and the domain of $f$ is intersection of the domains of $f_i$?
Best Answer
Well, the domain of $f$ must be the intersection of the domains of the constituent functions, otherwise the $\max$ cannot be defined.
More generally, suppose $f_\alpha :C \to \mathbb{R}$ is a family of convex functions, with $\alpha \in A$, some index set, and let $f(x) = \sup_{\alpha \in A} f_{\alpha}(x)$.
Then, for any $\alpha \in A$, $\lambda \in [0,1]$, \begin{eqnarray} f_\alpha(\lambda x + (1-\lambda) y) &\le& \lambda f_\alpha(x) + (1-\lambda) f_\alpha(y) \\ & \le & \sup_{\alpha' \in A} (\lambda f_{\alpha'}(x) + (1-\lambda) f_{\alpha'}(y)) \\ & \le & \sup_{\alpha' \in A} \lambda f_{\alpha'}(x) + (1-\lambda) \sup_{\alpha' \in A}f_{\alpha'}(y) \\ &=& \lambda f(x) + (1-\lambda) f(y) \end{eqnarray}
How taking the supremum of the left hand side gives the desired result: $$f(\lambda x + (1-\lambda) y) \le \lambda f(x) + (1-\lambda) f(y)$$
Alternatively, you could note that $\operatorname{epi} f = \cap_{\alpha \in A} \operatorname{epi} f_\alpha$, and that the intersection of convex sets is convex.