I am trying to prove the following result: If $u\in \mathrm{C}([0,1])$ such that $u'\geq 0$ in the viscosity sense in $(0,1)$, i.e., for any $x\in (0,1)$, if $p\in D^-u(x)$ then $p\geq 0$, then $u$ is non-decreasing.
Here $D^-u(x)$ is the set of super-differentials of $u$ at $x$, defined as
$$D^-u(x) = \left\lbrace p\in \mathbb{R}: \liminf_{y\rightarrow x} \frac{u(y)-u(x) - p(y-x)}{|y-x|}\geq 0 \right\rbrace.$$
It can be understood also in term of touching from below, i.e., $p\in D^-u(x)$ iff there is a smooth function $\varphi$ such that $u-\varphi$ has a local minimum at $x$, with $p = \varphi'(x)$.
With that in mind, assuming that $x_1<x_2<x_3$ and $u(x_1)>u(x_2)>u(x_3)$, I am trying to construct a function $\varphi$ such that $u-\varphi$ has a minimum at $x_0\in(0,1)$ with $\varphi'(x_0) < 0$ to get a contradiction. I have been trying with interpolation, like $$\varphi(x) = u(x_2)\frac{(x-x_1)(x-x_3)}{(x_1-x_3)(x_2-x_3)} - \frac{(x-x_2)^2}{\varepsilon}$$ but not successful.
Oh, I figured it out that we can do interpolation for this one. Let $\varepsilon > 0$, consider \begin{equation} \varphi(x) = u(x_1)\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)} + (u(x_2)+\varepsilon)\frac{(x-x_1)(x-x_3)}{(x_2-x_2)(x_2-x_3)} + u(x_3)\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)} \end{equation} where $u(x_1) > u(x_2)+\varepsilon > u(x_3)$. As $\varphi$ is a second-order polynomial, and $\varphi(x_1)>\varphi(x_2)<\varphi(x_3)$, we deduce that $\varphi'(x)<0$ for all $x\in (x_1,x_3)$. Now $(u-\varphi)(x_1) = (u-\varphi)(x_3) = 0$ while $(u-\varphi)(x_2) = -\varepsilon < 0$, we see that $u-\varphi$ must have a minimum at $x_\varepsilon\in (x_1,x_3)$, thus \begin{equation} 0>\varphi'(x_\varepsilon) \geq 0 \qquad\text{since}\; \varphi'(x_\varepsilon)\in D^-u(x_\varepsilon), \end{equation} which is a contradiction.