Calculate the subdifferential of a function $f: \mathbb{R}^3 \to \mathbb{R}$ defined by $$f(x) = \vert x_1 - x_2\vert + \vert x_2 - x_3\vert$$
My attempt
Applying the sum rule we have $$\partial f(x) = \partial g(x) + \partial h(x),$$ where $g(x)=\vert x_1-x_2 \vert,\ h(x)=\vert x_2-x_3\vert$.
Firstly, we calculate $\partial g(x)$. $$\partial g(x) =\left\{v \in E: g(y)-g(x)\ge \langle v,y-x\rangle\right\} = \left\{v \in E:\vert y_1-y_2\vert-\vert x_2-x_1\vert \ge \sum_{k=1}^{3}\langle v_k,y_k-x_k\rangle\right\}.$$ Since $y \in E$ arbitrary then we can choose $y_3=x_3$. Thus we have $$\partial g(x) = \left\{v \in E : \vert y_1-y_2\vert-\vert x_1-x_2\vert \ge \sum_{k=1}^{2}\langle v_k,y_k-x_k\rangle\right\}.$$ Case 1: $x_1 \ge x_2$. We can choose $y_1 > y_2$ and $y_2 = x_2$. Thus we have $$\partial g(x) = \left\{v \in E: y_1-x_1 \ge v_1(y_1-x_1)\right\} = \left[-1,1\right].$$ Case 2: $x_2<x_1$. Like case 1, we have $\partial g(x) = \left[-1,1\right]$.
In summary, we have $\partial g(x)=\left[-1,1\right]$.
Secondly, similarly,we can calculate $\partial h(x) = \left[-1,1\right]$.
I wonder that my solution is right?