Supremum of convex lipschitz functions. Let $f_i:K\to R, i\in I$ be a family of convex, equi-Lipschitz functions on some compact subset $K$ of $\Bbb R^n$. Is it true that $\sup f_i$ is also Lipschitz continuous(assuming that the sup exists)? Thank you

That is true. The $\sup$ even exists everywhere, if it exists at at least one point $x_0 \in K$.
Let us first derive for $x,y \in K$ arbitrary:
$$f_i(x) = f_i(y) + f_i(x) - f_i(y) \leq f_i(y) + L \cdot |x-y|,$$
where $L$ is the joint(!) Lipschitz constant for the $f_i$.
Taking the supremum yields
$$\sup_{i \in I} f_i(x) \leq \sup_{j\in I} f_j(y) + L \cdot |x-y|$$
for all $x,y \in K$. In particular, the left hand side is finite if the right hand side is. Putting $y = x_0$, where $\sup_{j \in J} f_j(x_0) < \infty$, we get the existence of the $\sup$ everywhere.
Furthermore, the above implies
$$\sup_i f_i(x) - \sup_j f_j(y) \leq L \cdot |x-y|.$$
By swapping $x,y$, we also get the "inverse" estimate and hence Lipschitz-continuity of the $\sup$-function.
