Concavity formal definition in a negative space

Question

If we have a strictly concave $u$ with $\alpha∈(0,1)$ and $u(0)>0$, so that $$u(\alpha x)=u((1-\alpha)0+\alpha x)≥(1-\alpha)u(0)+\alpha u(x)>\alpha u(x)$$ What happens to the inequality if $\alpha=-1$?

Answer 1

Consider, for instance, $u(x)=1-x^2$ which is strictly concave and satisfies $u(0)>0$. Notice that $$u(-x)=1-(-x)^2=1-x^2=u(x).$$

It follows that,

\begin{align} x<-1 \;\;\text{or}\;\; x>1&\;\;\Rightarrow\;\; u(x)<0 \;\;\Rightarrow \;\;-u(x)>0>u(-x)\\ x\in(-1,1) & \;\;\Rightarrow\;\; u(x)>0 \;\;\Rightarrow \;\; -u(x)<0<u(-x)\\ x=-1 \;\;\text{or}\;\; x=1 &\;\;\Rightarrow \;\;u(x)=0 \;\;\Rightarrow \;\;-u(x)=0=u(-x). \end{align}

So, for $\alpha=-1$, the inequality can go in either direction depending on the value of $x$.