Let $f_1,f_2$ be convex function and let $f(x)=\max\{f_1(x), f_2(x)\}$. It is clear to me that if $f_1(x) = f_2(x)$, then $[\nabla f_1(x), \nabla f_2(x)] \subseteq \partial f(x)$, but why do we also have that $[\nabla f_1(x), \nabla f_2(x)] = \partial f(x)$?
Thank you!
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Elaborating on @max_zorn 's answer, it may be helpful to consider a specific case to see why the reverse inclusion holds true. As he said, consider $f(x) = |x|=\max\{-x,x\}, x\in\mathbb{R}$, at $x=0$. This means we eventually wish to show that $[\nabla f_1(x), \nabla f_2(x)] = [-1,1] = \partial f(0)$, where that equality is equality of sets. Recall if $f: \mathbb{R}^n\to\overline{\mathbb{R}}$ is a convex function, and $\bar{x}$ is a point in the domain of $f$ such that $f(\overline{x})<\infty$, then an element $v\in\mathbb{R}^n$ is called a subgradient of $f$ at $\bar{x}$ if:
$$ \langle v, x-\bar{x}\rangle \leq f(x)-f(\bar{x})\quad \forall x\in\mathbb{R}^n$$ The collection of all subgradients of $f$ at $\bar{x}$ is called the subdifferential of $f$ at $\bar{x}$, and is denoted $\partial f(\bar{x})$. In our case, fix any $v\in \partial f(0)$, then by definition: $$\langle v, x-0\rangle \leq f(x)-f(0)\quad \forall x\in\mathbb{R}^n $$ $$\langle v, x\rangle \leq |x|\quad \forall x\in\mathbb{R}^n $$ In particular this is true for $x=v$, so we have $$ \langle v,v \rangle \leq |x|$$ $$ |x|^2\leq |x|$$ This implies that $|x|\leq 1 \iff x \in [-1,1]$. Thus $\partial f(0) \subseteq [-1,1]$, which is the particular case of the inclusion you wanted to show. To finish the proof, take $v\in \mathbb{R}^n$ such that $v\in[-1,1]$, i.e. $|v|\leq 1$, then $\forall x \in \mathbb{R}^n$ we have: $$ \langle v, x-0\rangle = \langle v,x \rangle \leq |v||x|\quad \forall x\in\mathbb{R}^n$$ Since $|v|\leq 1$, we have that: $$ \langle v,x \rangle \leq |x| = f(x)-f(0)$$ Thus $v\in \partial f(0) \implies [-1,1] \subseteq \partial f(0)$, and we have equality of sets.
Consider $|x| = \text{max}\{x,-x\}$ at $x=0$ to see why you need the convex hull.