The Krylov-Bogolioubov theorem is a fundamental result in the ergodic theory of dynamical systems which is typically stated as follows: if $T$ is a continuous transformation of a nonempty compact metric space $X$, then there exists a Borel probability measure $\mu$ on $X$ which is invariant under $T$ in the sense that $\mu(A)=\mu(T^{-1}A)$ for all Borel sets $A \subseteq X$, or equivalently $\int f d\mu = \int (f \circ T)d\mu$ for every continuous function $f \colon X \to \mathbb{R}$. This question is concerned with proofs of that theorem.


The most popular proof of the Krylov-Bogolioubov theorem operates as follows. Let $\mathcal{M}$ denote the set of all Borel probability measures on $X$, and equip $\mathcal{M}$ with the coarsest topology such that for every continuous $f \colon X \to \mathbb{R}$, the function from $\mathcal{M}$ to $\mathbb{R}$ defined by $\mu \mapsto \int f d\mu$ is continuous. In this topology $\mathcal{M}$ is compact and metrisable, and a sequence $(\mu\_n)$ of elements of $\mathcal{M}$ converges to a limit $\mu$ if and only if $\int fd\mu\_n \to \int fd\mu$ for every continuous $f \colon X \to \mathbb{R}$. 
Now let $x \in X$ be arbitrary, and define a sequence of elements of $\mathcal{M}$ by $\mu\_n:=(1/n)\sum\_{i=0}^{n-1}\delta\_{T^ix}$, where $\delta\_z$ denotes the Dirac probability measure concentrated at $z$. Using the sequential compactness of $\mathcal{M}$ we may extract an accumulation point $\mu$ which is invariant under $T$ by an easy calculation. This proof, together with minor variations thereupon, is fairly ubiquitous in ergodic theory textbooks. In the answers to [this question](https://mathoverflow.net/questions/66669/proof-of-krylov-bogoliubov-theorem), Vaughn Climenhaga notes the following alternative proof: the map taking the measure $\mu$ to the measure $T\_\*\mu$ defined by $(T\_\*\mu)(A):=\mu(T^{-1}A)$ is a continuous transformation of the compact convex set $\mathcal{M}$, and hence has a fixed point by the Schauder-Tychonoff theorem. A couple of years ago I thought of another proof, given below. The first part of this question is: has the following proof ever been published?


This third proof is as follows. Clearly it suffices to show that there exists a *finite* Borel measure on $X$ which is invariant under $T$, since we may normalise this measure to produce a probability measure. By the Hahn decomposition theorem it follows that it suffices to find a nonzero finite *signed* measure on $X$ which is invariant under $T$. By the Riesz representation theorem for measures this is equivalent to the statement that there exists a nonzero continuous linear functional $L \colon C(X) \to \mathbb{R}$ such that $L(f \circ T)=f$ for all continuous functions $f \colon X \to \mathbb{R}$. Let $B(X)$ be the closed subspace of $C(X)$ which is equal to the closure of the set of all continuous functions which take the form $g \circ T - g$ for some continuous $g$. Clearly a continuous linear functional $L \colon C(X) \to \mathbb{R}$ satisfies $L(f \circ T)=f$ if and only if it vanishes on $B(X)$, so to construct an invariant measure it suffices to show that the dual of $C(X)/B(X)$ is nontrivial. A consequence of the Hahn-Banach theorem is that the dual of $C(X)/B(X)$ is nontrivial as long as $C(X)/B(X)$ is itself nontrivial, so to prove the theorem it is sufficient to show that the complement of $B(X)$ in $C(X)$ is nonempty. But the constant function $h(x):=1$ is not in $B(X)$ because if $|(g \circ T - g)(x) - 1|<1/2$ for all $x \in X$, then $g(Tx)>g(x)+1/2$ for all $x \in X$ and hence $g(T^nx)>g(x)+n/2$ for all $n \geq 1$ and $x \in X$. This is impossible since $g$ is continuous and $X$ is compact. We conclude that $B(X)$ is a proper Banach subspace of $C(X)$ and the desired functional exists.


The second part of the question deals with the Krylov-Bogolioubov theorem for measures invariant under amenable groups of transformations. Let $\Gamma = \{T\_\gamma\}$ be a countable amenable group of continuous transformations of the compact metric space $X$. We shall say that $\mu \in \mathcal{M}$ is invariant under $\Gamma$ if $(T\_\gamma)\_\*\mu = \mu$ for all $T\_\gamma$. I believe that by using Følner sequences one may generalise the first proof of the Krylov-Bogolioubov theorem to show that every such amenable group has an invariant Borel probability measure. I seem to recall that the second proof also generalises to this scenario (in Glasner's book, perhaps?). Despite a certain amount of thought I have not been able to see how the third proof might generalise to this situation, even when $\Gamma$ is generated by just two commuting elements. So, the second part of this question is: can anyone see how the third proof generalises to the amenable case?



Alternative proofs of the Krylov-Bogolioubov theorem
Your proof can be modified a bit so that it works for a general countable, amenable group $\Gamma$. In this case, we can take $B(X)$ to be the closure of $\mbox{span} \{ f \circ T\_{\gamma} - f : f \in C(X), \gamma \in \Gamma \}$ and we show that $B(X)$ doesn't contain the constant functions on $X$ as follows:


I claim that every function $h \in \mbox{span} \{ f \circ T\_{\gamma} - f : f \in C(X), \gamma \in \Gamma \}$ is non-negative at some point in $X$, namely $\max\_{x \in X} h(x) \ge 0$. This is sometimes called the "translate property" for amenable groups, if I'm not mistaken. Anyway, after establishing that, the conclusion follows easily: it follows that one cannot uniformly approximate a negative constant (say $-7$) by a function of the above form.


The proof of the translate property I'm familiar with goes like this:


Assume for contradiction that $\max\_{x \in X} h(x) = \delta < 0$ and let $\varepsilon > 0$. Let $h=\sum\_{i=1}^{n}f\_{i}\circ T\_{\gamma\_{i}}-f\_{i}$. Since $\Gamma$ is amenable we can find a finite set $U \subset \Gamma$ such that $\frac{\left|\gamma\_{i}U\triangle U\right|}{\left|U\right|}<\varepsilon$ for all $i=1,\dots,n$ (the Følner condition). Now consider the function $F:=\frac{1}{\left|U\right|}\sum\_{u\in U}h\circ T\_{u}$. Choose some point $x\_0 \in X$. On the one hand, by our assumption, $F\left(x\_{0}\right) = \frac{1}{\left|U\right|}\sum\_{u\in U} h \left(T\_{u}x\_{0}\right)\le\delta<0$. On the other hand, 


$\left|F\left(x\_{0}\right)\right|=\left|\frac{1}{\left|U\right|}\sum\_{u\in U}\sum\_{i=1}^{n}f\_{i}\left(T\_{\gamma\_{i}}T\_{u}x\_{0}\right)-f\_{i}\left(T\_{u}x\_{0}\right)\right|$


By the triangle inequality, changing order of summation and using the property of U, you see that the above is at most $2n \varepsilon \max\_{x\in X,1\le i\le n}\left|f\_{i}\left(x\right)\right|$. So if you choose $\varepsilon$ small enough, you get a contradiction.


