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I was studying Krein-Milman Theorem in the aspect of infinite dimensional Banach spaces. We know that the closed unit ball $B_{X^*}=\{f\in X^*:\|f\|\leq 1\}$ of dual space $X^*$ of an infinite dimensional Banach space $X$ is convex and compact under weak-star topology. Then by applying Krein-Milman Theorem we have $B_{X^*}=\overline{\operatorname{conv}}^{weak^*}(E_{B_{X^*}})$, where $E_{B_{X^*}}$ is the collection of all extreme points of $B_{X^*}$.

Now, I am struggling with the following ideas:

$\textbf{(1)}$ What can be the formulation of a sequence in $\operatorname{conv}(E_{B_{X^*}})$?

$\textbf{(1.1)}$ Is it $\{\displaystyle{\sum_{i=1}^k} \lambda_i^{n}f_i\}_{n\in\mathbb{N}}$, where $f_i\in E_{B_{X^*}}$?

$\textbf{(1.2)}$ If so, then for any $f\in \overline{\operatorname{conv}}^{weak^*}(E_{B_{X^*}})$ there exists $\left\{\displaystyle{\sum_{i=1}^k} \lambda_i^{n}f_i\right\}_{n\in\mathbb{N}}$ such that $\displaystyle{\sum_{i=1}^k} \lambda_i^{n}f_i\xrightarrow{weak^*} f$, where $\lambda_i^{n}\in [0, 1]$ and $\displaystyle{\sum_{i=1}^k} \lambda_i^{n}=1$. Is this correct?

$\textbf{(2)}$ If it happens either way then what can be the form of the limit point $f$?

$\textbf{(2.1)}$ I am not getting the idea of limit of a sequence from convex hull of a set. Is it an infinite convex combination of the extreme points of $B_{X^*}$? (Even I do not know if there is any concept of infinite convex combination)

Please help me with an elaborate explanation with a possible example. Any kind of help is appreciated. Thank you in advance.

Tuh
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1 Answers1

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In general you cannot write the limit point as an infinite convex combination of extreme points. A first hint to this is that if $X$ is not separable, in a general a sequence of elements in $\operatorname{conv}(E_{X^*})$ might not be enough to get the limit points, you might need nets.

A nice example for extreme points and the Krein-Milman theorem is $X=C[0,1]$ (or more generally $C(K)$ for a compact Hausdorff space $X$). Then $X^*$ is the space of regular complex (or signed if you only work over $\Bbb R$) Borel measures thanks to the Riesz representation theorem. The extreme points of $B_{X^*}$ are precisely the delta measures $\delta_x$ with $x\in [0,1]$ and their multiples $c\delta_x$ with $|c|=1$. The Krein-Milman theorem says $B_{X^*}=\overline{\operatorname{conv}}^{w^*}{E_{B_X^*}}$. As a concrete example for an element of $B_{X^*}$ take the Lebesgue measure $\lambda$. Then a sequence of convex combinations of extreme points converging to $\lambda$ in the $w^*$ topology is given by $\frac1n\sum_{i=0}^{n-1}\delta_{i/n}$. Indeed, for $f\in C[0,1]$ we have $\int fd(\frac1n\sum_{i=0}^{n-1}\delta_{i/n})=\frac1n\sum_{i=0}^{n-1}f(i/n)\to\int fd\lambda$. But it is not difficult to see that one cannot write $\lambda$ as an infinite convex combination ($w^*$ convergent infinite sum) of delta measures.

But let me mention Choquet's Theorem which states that we may write elements in the convex set as a kind of continuous convex combination of extreme points (i.e. weak integral with respect to a probability measure).

leoli1
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  • Thank you very much for your answer. Is there any feasible example of a limit point to a net is of infinite convex combination? Does the $\lambda$ in your answer is a finite convex combination of the extreme points $\delta_x$? Please explain a little. – Tuh Sep 23 '24 at 04:45
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    Well, you can just start with an infinite convex combination, for example $\sum_{n=1}^\infty 2^{-n}\delta_{1/n}$. But $\lambda$ is not a convex combination (finite or infinite) of the extreme points – leoli1 Sep 23 '24 at 15:35
  • It is now clear to me. Thanks a lot. Could you please look into the construction of the net (please ignore the sequence word) mentioned in 1.2 above? is the construction correct? – Tuh Sep 23 '24 at 16:50
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    There is a problem with it: the $k$ and the $f_i$ might depend on $n$, so it should be something like $\sum_{i=1}^{k_n}\lambda^n_if^n_i$. For example in the answer with the Lebesgue measure we have $\lambda^n_i = \frac1n, k_n=n$ and $f^n_i = \delta_{(i-1)/n}$ – leoli1 Sep 23 '24 at 17:26
  • Got it now. Many thanks to you. – Tuh Sep 24 '24 at 03:45