I am studying real analysis, actually, I need to prepare a seminar for my professor about convex analysis and I am using the Kenneth Davidson book called ''real analysis and application'', https://carma.newcastle.edu.au/jon/Preprints/Books/Other/TimesReal.pdf.
I need to prove the Theorem 16.7.13 at the page 595, I got the proof but there is a step that is confusing:
That is, why $\hspace{0.3cm}$ $\bigcup_{j \in J(a)} \partial f_j(a) = \operatorname{conv} \{ \partial f_j(a): j \in J(a)\} ?? $
If this information helps, I Know that $\partial f_j(a)$ is a convex set, for all $j$, and clearly that the right hand is subset of left hand, but how to show the otherwise?
The statement in the question is incorrect, however if you 'view' the sets through a linear functional $\phi$ then the result is true.
For any linear functional $\phi$ and any set $A$ we have $\sup_{x \in A} \phi(x) = \sup_{x \in \operatorname{co} A} \phi(x)$.
Take $A= \cup_k \partial f(a_k)$.
It is clear that $\sup_{x \in A} \phi(x) \le \sup_{x \in \operatorname{co} A} \phi(x)$, just from containment.
Suppose $x \in \operatorname{co} A$, then $x = \sum_k \lambda_k a_k$, with $a_k \in A$. Choose $k_0$ such that $\phi(a_{k_0}) \ge \phi(a_k)$, then $\phi(a_{k_0}) \ge \phi(x)$, and hence we have equality.