Let $C\subset\mathbb{R}^n$ be a convex set and let $\phi_1(x),\ldots,\phi_m(x)$ be convex functions over C. Let U be the following subset of $\mathbb{R}^m$:
$U=\{y\in\mathbb{R}^m:\phi_1(x)\leq y_1,\ldots,\phi_m(x)\leq y_m,\mbox{for some x in C}\}$
Prove $U$ is convex.
My thoughts were that $U$ is an intersection of $Lev(\phi_i,y_i)$ $(Lev(f,a)=\{x:\in\mathbb{R}^n:f(x)\leq a\}$ and we know that the level set of convex function is convex and intersection of convex sets is convex.
Let $y_1=(y_1^1, \dots, y_m^1), y_2=(y_1^2, \dots, y_m^2)$ be elements of $U$ and $\lambda \in [0,1]$. By definition, it exists $x^1,x^2 \in C$ such that:
$$\begin{cases} \phi_1(x^1)\leq y_1^1,\ldots,\phi_m(x^1)\leq y_m^1\\ \phi_1(x^2)\leq y_1^2,\ldots,\phi_m(x^2)\leq y_m^2 \end{cases}$$
As all the $\phi_i$ are supposed to be convex, for $1 \le i \le m$:
$$\phi_i((1-\lambda)x^1 + \lambda x^2) \le (1-\lambda)\phi_i(x^1) + \lambda \phi_i(x^2) \le (1-\lambda)y_i^1 + \lambda y_i^2.$$
As $C$ is supposed to be convex, $(1-\lambda)x^1 + \lambda x^2$ belongs to $C$. Which implies that $(1-\lambda)y^1 + \lambda y^2$ belong to $U$.
So by the very definition of convexity, $U$ is convex.