If a function $g: \mathbb{R} \rightarrow \mathbb{R} $ is continuous, concave, and strictly increasing (not necessary need $g$ is differentiable), then for every $a, b \in \mathbb{R}$, and $c \in \mathbb{R}_{+}$, how to prove that $\left| g (a + c) - g (b + c) \right| \leq \left| g (a ) - g(b) \right| $ is valid?
Actually, I proved this statement by using the differentiation of $g$. However, in this statement, we don't know whether $g$ is differentiable, so is there anyone could prove this statement without using the differentiation of $g$?
Does right-differentiation of $g$ reach the purpose?
Anyway, I attached my orignal proof. Hopefully it could be helpful to you.
Proof: Without loss of generality, we assume $a \leq b$. Since $g$ is increasing, $ g(a) \leq g(b)$. To prove above inequality will be equivalent to show that $g(b+c) - g(b) \leq g(a+c) - g(a)$.
To this end, we set a function $f(x) := g(x + c) - g(x)$, for all $ c$ in $\mathbb{R}_{+}$.
Taking the derivative of $f$ w.r.t $x$, we get $f'(x) = g '(x + c) - g '(x)$. Because $g$ is concave, its first-order derivative is non-increasing, namely, $ g '(x + c) \leq g '(x) $, and thus, $f'(x) \leq 0$ ($f$ is decreasing).
So, $a \leq b$ implies $ f(a) \geq f(b)$, that is, $ g(a+c) - g(a) \geq g(b+c) - g(b) $. The stated property follows.