Is the minimum of a parametric convex function convex again? 
Let $I$ and $J$ be compact intervals. 
  Let $f:I\times J\to\mathbb R$ be differentiable and strictly convex. 
  Is the function $g:I\to\mathbb R$ defined by 
  $$ g(x) = \min_{y\in J} f(x,y) $$
  convex?

Remarks:


*

*I know that minimum of convex functions is in general not convex. However, I can't find a counter example in which $f$ is convex.

*The regularity ensures that the minimizer $y^*(x)$ of $f(x, \cdot)$ is unique. 

*Assume $y^*$ as function is convex, $y^*$ maps $I$ into an interval $J^*$, and $f(x, \cdot)$ is increasing on $J^*$ for every $x\in I$. Then, $g$ is convex.
Thanks for any input :)
 A: It is convex!
Your first statement that the minimum of convex functions is in general not convex is true, but here you have a lot more structure! In a sense you are projecting onto $x$. In fact, $g$ is also called the inf-projection of $f$. Let $\lambda \in (0,1)$ and $y_1, y_2 \in J$ arbitrary:
$$
\begin{aligned}
g(\lambda x_1 + (1-\lambda) x_2) &= \min_{y} f(\lambda x_1 + (1-\lambda)x_2, y) \\
&\leq  f(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2)\\
&\leq \lambda f(x_1, y_1) + (1-\lambda) f(x_2,y_2)\\
\end{aligned}
 $$
Now first minimize with respect to $y_1$, then with respect to $y_2$ to finally get:
$$g(\lambda x_1 + (1-\lambda) x_2) \leq \lambda g(x_1) + (1-\lambda) g(x_2)$$
Also notice that you do not need the regularity conditions you imposed onto $f$.
A: Theorem: Let $X,Y$ be real linear spaces and $f\colon X\times Y\to [-\infty,+\infty]$ be convex. Then 
$$
\phi(x)=\inf_{y\in Y}f(x,y)
$$
is convex.
Proof: Let $E$ be the image of $\text{epi}(f)$ under the projection $(x,y,\alpha)\to (x,\alpha)$. Then by definition of infimum
$$
\text{epi}(\phi)=\{(x,\alpha)\in X\times\mathbb{R}\colon \ (x,\beta)\in E,\ \forall\beta>\alpha\}.\tag1
$$
The epigraph of $f$ is convex, then $E$ is convex (as a linear image of the convex set), and $(1)$ yields then that $\text{epi}(\phi)$ is convex as an intersection of convex sets
$E_\epsilon=E-(0,\epsilon)$, i.e. 
$$
\text{epi}(\phi)=\bigcap_{\epsilon>0}E_\epsilon.
$$

P.S. Since a convex function $f$ on any set $S\subset X\times Y$ can be extended to the whole space by (re)defining $f=+\infty$ outside $S$, the generality is not lost by assuming that $f$ is defined on $X\times Y$.
