functions whose average along orbits is zero or a constant - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T15:06:56Zhttp://mathoverflow.net/feeds/question/94813http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/94813/functions-whose-average-along-orbits-is-zero-or-a-constantfunctions whose average along orbits is zero or a constantJames Propp2012-04-22T06:12:38Z2012-10-13T01:37:07Z
<p>Is there some name in ergodic theory or integrable systems theory for a function whose average value on every orbit is zero? (Of course when I say "every orbit" in the context of ergodic theory I mean "modulo a set of measure zero".)</p>
<p>The space of $L^2$ functions with ergodic average zero is the orthocomplement of the space of invariant $L^2$ functions, under the dynamical inner product $\langle f,g \rangle = \lim_{n \rightarrow \infty} (1/n) \sum_{k=0}^{n-1} f(T^k x)\overline{g}(x)$, so it seems like a natural space for ergodic theorists to consider. And the space of functions that average to zero along orbits seems even more natural in the setting of integrable systems, where evolution laws and conserved quantities can switch places.</p>
<p>I'm also interested in knowing if there's a name for functions whose average value on every orbit is some orbit-independent constant. I don't want to invent terminology for such functions if satisfactory terminology already exists, and I suspect it does, though I haven't been able to find it on the web. I looked at an introductory article on cocycles and coboundaries in ergodic theory, but didn't find what I was looking for.</p>
http://mathoverflow.net/questions/94813/functions-whose-average-along-orbits-is-zero-or-a-constant/94826#94826Answer by waler for functions whose average along orbits is zero or a constantwaler2012-04-22T09:23:50Z2012-10-13T01:37:07Z<p>Second part: if your system is nontrivial, and function $f$ is "smooth", than condition
of the form
$$
\lim_{n\to\infty} \frac 1n\sum_{i=0}^{n-1} f(T^i x) =c
$$
for <em>all</em> $\ x$, should imply that
$$
f(x)= g(x)-g(Tx)+c
$$
for some function $g$. In this case, one says that $f$ is cohomologous to a constant.</p>