Let $V$ be a (real or complex) vector space. (I am not assuming that $V$ is finite-dimensional.)
Is there a term for a set $S \subset V$ with the property that for all $x \in S$, if $x$ is not an extreme point of the convex hull of $S$ then there exist $x_1,x_2 \in S \setminus \{x\}$ and $\lambda \in (0,1)$ such that $\lambda x_1 + (1-\lambda) x_2 = x$?
Does this property have any other interesting equivalent characterisations?
(Terms: The convex hull of $S$ is the smallest convex set containing $S$; it is constructed by taking all the finite convex combinations of elements of $S$. An extreme point of a convex set $C$ is a point $x \in C$ for which there do not exist $x_1,x_2 \in C \setminus \{x\}$ and $\lambda \in (0,1)$ such that $\lambda x_1 + (1-\lambda) x_2 = x$.)
Examples: Any convex set has this property. An arbitrary union of open line segments has this property. A set consisting of the three corners of a triangle has this property. But a set consisting of the three corners of a triangle plus one interior point of the triangle does not have this property. A set consisting of the boundary of a triangle has this property, and so does a set consisting of the boundary of a triangle plus any subset of the interior of the triangle.
I realise I should say a little bit more about the motivation behind this question. Namely, it concerns another class of examples I've come up with, that's somewhat less trivial and more interesting than the ones listed above.
Motivation:
First note that the set of probability measures on a measurable space $(X,\mathcal{X})$ is a convex set. (Specifically, it is a convex subset of the vector space of real-valued functions on $\mathcal{X}$.)
Given a measurable space $(X,\mathcal{X})$ and a measurable map $f \colon X \to X$, a probability measure $\mu$ on $X$ is called $f$-invariant if $\mu(f^{-1}(\cdot))=\mu$.
Note that the set of $f$-invariant probability measures is convex. A probability measure $\mu$ on $X$ is called $f$-ergodic if the following equivalent statements hold:
- $\mu$ is an extreme point of the convex set of $f$-invariant measures;
- $\mu$ is $f$-invariant and for any measurable set $A$ with $f^{-1}(A)=A$, we have $\mu(A) \in \{0,1\}$;
- $\mu$ is $f$-invariant and for any measurable set $A$ with $\mu(f^{-1}(A) \triangle A)=0$, we have $\mu(A) \in \{0,1\}$.
(Outline of the proof:
The equivalence of the second and third statement is quite quick to prove: If $\mu(f^{-1}(A) \triangle A)=0$ and $\mu$ is $f$-invariant, then the set $\tilde{A}:=\{x \in X : f^i(x) \in A \textrm{ for infinitely many } i \geq 0\,\}$ has $f^{-1}(\tilde{A})=\tilde{A}$ and $\mu(\tilde{A} \triangle A)=0$.
The equivalence of the latter two statements ("ergodicity in terms of triviality of invariant sets") to the first statement ("ergodicity in terms of extreme points") requires a little bit more work: One first proves that ergodicity under the latter characterisations implies that the only $f$-invariant probability measure that is absolutely continuous with respect to $\mu$ is $\mu$ itself. [In fact, this gives another characterisation of ergodicity.] The desired result is then pretty much immediate.)
Now I have looked at extending the equivalence of the above characterisations of ergodicity from the setting of measurable maps $f \colon X \to X$ to the setting of "Markov operators".
Given a probability space $(X,\mathcal{X},\mu)$, we write $L^\infty(\mu)$ for the set of $\mu$-essentially bounded measurable functions $g \colon X \to \mathbb{R}$ identified up to $\mu$-almost sure equality. A Markov operator on $L^\infty(\mu)$ is a linear function $P \colon L^\infty(\mu) \to L^\infty(\mu)$ with the additional properties that
- $g \geq \mathbf{0} \Rightarrow Pg \geq \mathbf{0}\,$,
- $P\mathbf{1} = \mathbf{1}$,
- $\int_X Pg \, d\mu \ = \ \int_X g \, d\mu$.
The "trivial case" of Markov operators on $L^\infty(\mu)$ are precisely the operators of the form $g \mapsto g \circ f$, where $f \colon X \to X$ is a map for which $\mu$ is $f$-invariant.
Accordingly, I will say that a Markov operator $P$ on $L^\infty(\mu)$ is ergodic if for any measurable set $A \subset X$ with $P\mathbf{1}_A=\mathbf{1}_A$ (modulo $\mu$-null sets), we have $\mu(A) \in \{0,1\}$.
Of course, formulating an extension of the extreme-point characterisation of ergodicity to Markov operators isn't a trivial issue, as a Markov operator is already defined with reference to the "invariant measure". But I have come up with a version that is good enough for my purposes:
First, define an absolutely continuous invariant measure (ACIM) of a Markov operator $P$ on $L^\infty(\mu)$ to be a probability measure $\nu$ on $X$ that is absolutely continuous with respect to $\mu$ and has the property that $$ \int_X P\mathbf{1}_A \, d\nu \ = \ \nu(A) $$ for all measurable $A \subset X$. (This equality is well-defined, since $\nu$ is absolutely continuous with respect to $\mu$.)
It is not extremely difficult to show (by analogy to the proof for maps $f \colon X \to X$) that $P$ is ergodic if and only if the only ACIM of $P$ is $\mu$ itself. A consequence of this is the following:
Proposition (Ergodicity in terms of extreme points). Fix a measurable space $(X,\mathcal{X})$. Suppose we have a set $C$ of probability measures on $(X,\mathcal{X})$ and a $C$-indexed family $(P_\mu)_{\mu \in C}$ of Markov operators $P_\mu$ on $L^1(\mu)$ with the following property: for every $\mu \in C$, for every probability measure $\nu$ on $X$ that is absolutely continuous with respect to $\mu$, $\,\nu \in C$ if and only if $\nu$ is an ACIM of $P_\mu$. Then
- the set $C$ has exactly the property asked about in this Math.SE question;
- for each $\mu \in C$, $P_\mu$ is ergodic if and only if $\mu$ is an extreme point of the convex hull of $C$.
The point is that $C$ need not be convex: e.g. it could consist of two distinct Dirac masses, $C=\{\delta_{x_1},\delta_{x_2}\}$, with $P_{\delta_{x_i}}$ being the unique Markov operator (namely the identity function) on $L^\infty(\delta_{x_i})$.