Let $g(x):=sup_{y\in R^n}f(x,y)$. Assume that for each fixed $y, f(x,y)$ is a convex function of x. Prove that $g(x)$ is a convex function.
I know that if ${\{f_i:i∈I\}}$ is a family of convex functions from a convex set $C $to $R$, then the same is true for $f=sup_{i∈I}f_i$. But how to prove it?
Note that, for any $x$ and any $i$, $f_i(x) \le f(x)$. Hence, if $x, y$ are elements of the domain and $\lambda \in [0, 1]$, then for any index $i$, $$f_i(\lambda x + (1 - \lambda)y) \le \lambda f_i(x) + (1 - \lambda)f_i(y) \le \lambda f(x) + (1 - \lambda)f(y).$$ Consider the set of values of $f_i(\lambda x + (1 - \lambda)y)$, as $i$ ranges over the index set. We have just shown that this set bounded above by $\lambda f(x) + (1 - \lambda)f(y)$. However, by definition of $f$, its least upper bound is $f(\lambda x + (1 - \lambda)y)$. Therefore, $$f(\lambda x + (1 - \lambda)y) \le \lambda f(x) + (1 - \lambda)f(y),$$ proving convexity.