I'm reading page 72 of this pdf, and I can't understand the proof of the weighted sums property of the subdifferential $\partial f$. Weighted sums satisfies:
Given two convex functions $f$ and $g$ on $\mathbb{R}^n$ and $\lambda, \mu > 0$, and a function $h(x) = \lambda f(x) + \mu g(x)$, then: $$\partial h(x) = \lambda \partial f(x) + \mu \partial g(x) $$ For any $x \in \text{int Dom } h$
For the proof, author uses Corollary 1.2.1 and propositions 3.4.1:
Proposition 3.4.1 (page 69): $$f′_h(x) = max\{h^T d ; d \in \partial f(x)\}$$ with $h \in \mathbb{R}^n$ and $f$ is a convex function. In short, the directional derivative is the support function of the set $\partial f$.
Corollary 1.2.1 in this pdf:
Let $\psi_M(x) = \sup\{y^Tx; y \in M\}$.
$M$ is a convex set. Function $\psi_M(x)$ is called the support function of the set $M$.
Let $M1$ and $M2$ be two closed convex sets.
If for any $x \in \text{Dom }\psi_{M2}$ we have $\psi_{M1}(x) \leq \psi_{M2}(x)$ then $M1 \subset M2$
Let $\text{Dom }\psi_{M1} = \text{Dom }\psi_{M2}$ and for any $x \in \text{Dom }\psi_{M1}$ we have $\psi_{M1}(x) = \psi_{M2}(x)$. Then $M1 \equiv M2$
Proof of weighted sums:
Let $x \in \text{int Dom } f \cap \text{int Dom } g$. Then for any vector $h \in \mathbb{R}^n$ (notation in this pdf is weird because author uses the same letter for vectors and functions):
$$f'_h(x) = \lambda f'_h(x) + \mu g'_h(x)$$
I think here's a typo: the term to the left should be $h'_h(x)$:
$$h'_h(x) = \lambda f'_h(x) + \mu g'_h(x) $$
$$h'_h(x) = max\{\lambda h^Td_1 ; d_1 \in \partial f (x) \} + max \{ \mu h^Td_2 ; d_2 \in \partial g(x) \} $$ $$ = max\{h^T (\lambda d_1 + \mu d_2) ; d_1 \in \partial f (x) ; d_2 \in \partial g(x) \} $$ $$ = max\{h^T d ; d \in \lambda \partial f (x) + \mu \partial g(x) \}$$
from this point, author uses Corollary 1.2.1 to obtain the property of weighted sums, but doesn't say how. Was the author saying that function $f$ is equal to function $h$ because of this corollary? How do I obtain the equivalence with $\partial h(x)$?