# coboundary in Dynamical system

given measurable dynamic system $( X, \mathcal{B}, T, \mu)$, $\mu \circ T^{-1}=\mu$ ergodic.

$\phi \in L^{\infty}$ is coboundary with $\int \phi d\mu =0$, means the decomposition: $\exists \xi$, s.t $\phi=\xi \circ T-\xi$

$\phi$ is not coboundary, means the decomposition: $\exists \xi$, non zero function $\beta$, s.t $\phi=\beta+ \xi \circ T-\xi$

My question is what space this decomposition lies in? $L^1$ or $L^2$? that means $\xi, \beta$ are in $L^1$ or $L^2$?

I know some authors use $L^2$, is this the original definition?

Thanks!

• What you have described as a non-coboundary seems wrong. It’s just anything that cannot be written as a coboundary (including functions with a non-zero integral). There is s famous decomposition used in the proof of the von Neumann ergodic theorem, where it is shown that $L^2$ is the direct dum of the closure of the $T$-invariant functions and the closure of the coboundaries. Commented Jul 16, 2018 at 5:19
• Thanks! I updated my question. Yes, Von Neuman's decomposition is in $L^2$. Thanks for reminding me! Commented Jul 16, 2018 at 18:44
• your definition of non-coboundary seems absurd (take $\xi=0$, $\beta=\phi$)
– YCor
Commented Jul 16, 2018 at 22:22
• The regularity of $\xi$ depends on the regularity of $\phi$, but I think you will need additional hypotheses for $T$ (e.g. Anosov, topologically transitive). Such results are known as Livsic's theorems (I was told some confusion exists due to inconsistent romanizations of the surname Livsic), although a websearch for the $L^\infty$ case doesn't show much. Note that uniqueness and regularity are the main issues here, as we can use the formula $\xi (T(x))= \phi(x) + \xi(x)$ to define $\xi(x)$ recursively and get existence with a bit of care. Commented Aug 14, 2018 at 19:57