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, 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?


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.


Let me first mention that actually Bogolyubov himself proved "the Krylov-Bogolyubov" theorem for actions of arbitrary amenable groups in 1939 (2 years after the Krylov-Bogolyubov theorem was published). Unfortunately, it was published in Ukrainian in a rather obscure place, and remained virtually unknown (see Anosov's article MR1311227 in Russian Math. Surveys 1994 for more details). Currently its Russian translation is available in various volumes of Bogolyubov's collected works, but as far as I know no English translation exists. Bogolyubov's argument is close to your "first proof", except for he works directly with means instead of Følner sequences (which had not been defined yet at the time).

Now, returning to your question about the "third proof". One way to do it (with a bit of cheating) consists in trying to replace invariance with respect to all elements of an amenable group with invariance just with respect to a single operator. This can, indeed, be done by using the fact that for any amenable group there is a probability measure $\mu$ on the group such that the sequence of its convolutions strongly converges to invariance (see the answer to the question characterizations of amenable groups which use the space $\ell_1(G)$ and convolution). Therefore, if a measure on the action space is invariant with respect to the convolution operator $P_\mu$ associated with the measure $\mu$, then it is necessarily group invariant. Now, for proving existence of $P_\mu$-invariant measures one can apply your argument. Of course, the same trick works for the "first" and for the "second" proofs as well.

PS The term "amenable action" already has several well established (and contradictory) meanings, and it is misleading to use it just for actions of amenable groups as you do in the last paragraph of your question.

  • $\begingroup$ Thanks. I have rephrased to avoid use of the words "amenable action". $\endgroup$ – Ian Morris Jul 31 '11 at 7:46

Please check Krasnosel'skii, M.A.; Lifshits, Je.A.; Sobolev, A.V. (1990), Positive Linear Systems: The method of positive operators, Sigma Series in Applied Mathematics,

Krasnoselskii told me that he presented his argument at Bogolubov's seminar in 1949?

  • $\begingroup$ Thanks very much, I will look at this book when I can get hold of it (according to COPAC, there are only seven copies of this book in the entire country!). I am not surprised that the argument is not new, but it does seem to be quite rare in the English-language literature. $\endgroup$ – Ian Morris Jan 18 '12 at 11:36

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