TITLE: Prove the supremum of the set of affine functions is convex
QUESTION [12 upvotes]: Let $\langle f_i \rangle _{i \in I}$ be a family of affine functions on a convex and compact set $\Omega \subset \mathbb{R^d}$ such that $f_i = a_i.x +b_i$ for $x \in \Omega$. Prove that f, defined by $f(x) = sup_{i \in I}f_i(x)$ for $x \in \Omega$ is a convex function. Explain this geometrically.
I understand that since f is in $C^1(\Omega)$, f is convex if $sup(a_ix + b_i) \le sup(a_iy+b_i) + a_i(x-y)$, but i am having trouble proving this.

REPLY [2 votes]: I can't comment because I don't have enough reputation, hence posting as an answer. I don't think $f \in C^1(\Omega)$. Easiest example is take $\Omega = [0,1]$, $I = \{1,2\}$. $f_1(x) = \frac12$ and $f_2(x) = x$. Then, $f(x) = \frac12$ if $x \le \frac12$ and $x$ if $x > \frac12$. In particular, at $\frac12$ there is a kink.