Suppose $\phi:\mathbb{R}^3\to\mathbb{R}$ is a smooth, strictly convex function such that $\lim_{|y|\to\infty}\phi(x,y)=\infty$ for all $x\in\mathbb{R}$. And define $F:\mathbb{R}\to\mathbb{R}$ $$F(x)=\min_{y\in\mathbb{R}}\phi(x,y)$$
I am wondering whether $F$ a convex function? I tried but failed to show $$F(tx_1+(1-t)x_2)\le tF(x_1)+(1-t)F(x_2)$$ If $F$ is not a convex function, is there any counterexamples?
By the assumptions, the minimum with respect to $y$ exists and is unique, so $F:\mathbb R\to \mathbb R$ is well-defined.
Let $x_1,x_2$, $\lambda\in (0,1)$ be given. Then $$ \begin{split} F( \lambda x_1 + (1-\lambda)x_2) & = \inf_{y\in \mathbb R} \phi(\lambda x_1 + (1-\lambda)x_2, y) \\ & = \inf_{y_1,y_2\in \mathbb R} \phi(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2) \\ & \le \inf_{y_1,y_2\in \mathbb R}( \lambda \phi(x_1,y_1) + (1-\lambda)\phi(x_2,y_2))\\ & = \lambda\inf_{y_1\in \mathbb R} \phi(x_1,y_1) + (1-\lambda)\inf_{y_2\in \mathbb R}\phi(x_2,y_2)\\ & = \lambda F(x_1) + (1-\lambda) F(x_2). \end{split} $$