Define a function $F:[0, \infty) \to \mathbb R$ by $$F(s) := \begin{cases} 2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\ 1-2s, & \text{ if }\, s \le \frac{1}{4} \end{cases} $$
Is $F$ convex? This seems plausible from looking at its graph.
It is not hard to see that the restrictions $F|_{[0,\frac{1}{4}]},F|_{[\frac{1}{4},\infty)}$ are convex. Is there an elegant way to show that the combined definition also produces a convex function on the entire domain $[0, \infty)$?
*My motivation for studying this function comes from this optimization problem.
We have $F\in\mathcal C^1(\mathbb R)$ and $$F'(s)=\begin{cases} -2, & s\le\frac{1}{4}, \\ 2\left(1-\frac1{\sqrt{s}}\right), & s\geq\frac14 \end{cases}$$
which is non-decreasing so that $F$ is convex.