I'm trying to calculate the subdifferential of the infinity norm, but I'm a bit stuck. The definition of the subdifferential $\partial f(x)$ of a function $f(x)$ that I am working with is $$\partial f(x) = \{g \colon f(y) \geq f(x) + \langle y - x, g \rangle \quad \forall y \in \text{dom} f\}.$$ In other words, the subdifferential of $f$ at $x$ contains all of the possible subgradients of $f$ at $x$.
I know that if $f(x) = \underset{i}{\max} \{f_{i}(x)\}$, then the subgradient of $f(x)$ is the convex hull of the union of the subgradients of $f_k(x)$ where $k \in \{k \colon f_{k}(x) = f(x)\}$. In other words $$\partial f(x) = \text{conv}\left(\underset{k\in\{k \colon f_{k}(x) = f(x)\}}{\bigcup}\partial f_{i}(x)\right).$$
I considered $f_{i}(x) = \left|x\right|=\left|e_{i}x\right|$, where $e_i$ is the $i$th standard basis vector in $\mathbb{R}^n$. I'm struggling to find the subdifferential of $f_{i}(x)$.
My intuition tells me that it should be an extension of the subdifferential for the absolute value function. In particular, I think it should be $\text{sgn}(x)e_i$ for all $x$ where $x_i \neq 0$ and $te_i$ where $t \in [-1, 1]$ when $x_i = 0$. I think this is a valid subgradient (using the definition of subgradient), but I don't know if/how to show this is the only subgradient of $f_i(x)$ using the definition of subdifferential/subgradient.