From concavity to convexity

Question

Suppose I have a function $g$ on $[-1,1]$ which is increasing, concave and such that $g(-1)=0$. I set $h(t)=\frac{g(t)}{1+t}$. Is it true that $h$ is decreasing and convex?

Answer 1

Take $g(x)=\cos\left(\frac{\pi x}{4}-\frac{\pi}{4}\right)$

Hence, with $h(x)=\frac{\cos\left(\frac{\pi x}{4}-\frac{\pi}{4}\right)}{x+1}$ there are problems.