I am looking for a function $g(x) \in [1, c], c\geq 1$ that satisfies
$$\alpha x(2-x) + (1-\alpha x)g(x) < g(x-\alpha x)$$ for all $0\leq x < 1$ and $0\leq \alpha <1$. Across such $g$, I want one that minimises $c$.
By tinkering with the inequality/trial and error, I proved that $g(x)=e^{x-1}+1-x$ satisfies the inequality. Here, $g\in [1, 1+e^{-1}]\approx [1, 1.3678]$.
Can this be improved?
For intuition on how I derived my $g$, I rearranged to get:
$$\frac{g(x)-g(x-\alpha x))}{\alpha x} = g(x)-2+x$$
Now, by the mean value theorem, we have $\frac{g(x)-g(x-\alpha x))}{\alpha x} \approx g'((1-\theta)x)$ for some $\theta \in (0, \alpha)$, so I "guessd" that the worst case is when $g'((1-\theta)x)\approx g'(x)$. Then you get $g'(x)=g(x)-2+x$, and after solving, you get $g(x)=ce^x + 1-x$. Reverse engineering this with the inequality gave me the answer.