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Currently I am studying Behrend's $M$-structure and Banach-Stone Theorem. He introduced the following notation.

Notation: Consider a Banach space $X.$ Fix $x\in X$ and $r\geq 0.$ Consider the set $$K(x,r)=\{(x^*,x^*(x)+r)\in X^*\times \mathbb{R}:x^*\in B_{X^*}\}.$$

It is not hard to show that $K(x,r)$ is a compact convex set.

However, the author quoted the following at page $45.$

convex hull of $\bigcup_{i=1}^nK(x_i,r_i)$ is compact because convex hull of finite unions of compact convex sets is compact.

I fail to prove the above statement. Any hint is appreciated.

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The convex hull of a set is the set of convex combinations of elements of that set. I claim that the convex hull of an union of (non-empty) convex sets $C_1\cup \cdots \cup C_n$ is the set of convex combinations in the form $t_1a_1+\cdots +t_na_n$, where $a_i\in C_i$ for all $i$. In fact, consider $x=\sum_{i,j} t_{i,j} a_{i,j}$, where $a_{i,j}\in C_i$ and $t_{i,j}>0$ for all $i,j$, plus $\sum_{i,j} t_{i,j}=1$. Then, call $a_i=\sum_{j} \frac{t_{i,j}}{\sum_{j} t_{i,j}}a_{i,j}$. This is a convex combination of elements of $C_i$, and thus $a_i\in C_i$. Moreover, $$x=\sum_i \left(\sum_{j} t_{i,j}\right)a_i,$$ which is indeed a convex combination with at most one element from each $C_i$.

Now, the same argument you used for $\Bbb R^d$ - be it sequential compactness, or $\operatorname{co}\left(\bigcup_{i=1}^n C_i\right)$ being image of a closed subset of $[0,1]^n\times \prod_{i=1}^n C_i$ by a continuous function - carries on here.

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  • $\begingroup$ Do we require all $t_i$'s have sums equal to $1$ in your first claim? $\endgroup$
    – Idonknow
    Commented May 14, 2018 at 12:42
  • $\begingroup$ @idonknow Yes, and to be non-negative. That's what convex combination means. $\endgroup$
    – user228113
    Commented May 14, 2018 at 12:47
  • $\begingroup$ To continue your proof, we do not invoke Caratheodory theorem right? Because in our case, $C_i$'s may not be of finite dimensional. I proceed by showing that $\bigcup_{I=1}^n C_i$ is a continuous image of a compact set. $\endgroup$
    – Idonknow
    Commented May 14, 2018 at 14:53
  • $\begingroup$ @idonknow No, that theorem is not necessary. In the context where the OP uses it, it's needed to arrive at the same point where my hint already is. $\endgroup$
    – user228113
    Commented May 14, 2018 at 15:07

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