Reading Evan's, in the example $(51)$ at page 136, there's written $-|x|$ is a semiconcave function, according to the definition
\begin{align} g(x+z) -2g(x)+g(x-z) \leq C |z|^2 \end{align}
for every $x,z \in \mathbb{R}^n$ and $C>0$. I know this is equivalent to as for the concavity of the function $z \mapsto g(z)-\frac{C}{2}|z|^2$
But in my example the problem is at $x=0$, which is not a differentiable point for my function. Up to now, I know that $|x|$ in not semiconcave in every open set containing $x=0$, since by taking $x=0$ in the definition I can't find a $C>0$ that satisfies the bound.
I'd like to be sure that my reasoning is correct for the $-|x|$ case.
Again, applying the definition, I have to show that exists some $C>0$ such that
\begin{align} |x+z|-2|x|+|x-z| > C|z|^2 \end{align}
but I don't know how.
Of course, I know that $-|x|$ is concave ($|x|$ is convex) and hence it's semiconcave, but I'd like to use the definition.
If $f$ is convex, we have $f(x) = f({1 \over 2} ((x+z)+(x-z)) \le {1 \over 2} (f(x+z)+f(x-z))$.
Letting $f(x) = |x|$ we obtain $|x| \le {1 \over 2} (|x+z|+|x-z|)$, and so $2g(x) \ge g(x+z)+g(x-z)$ from which you obtain the desired answer with $C=0$.