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a question about the definition:

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!

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    $\begingroup$ 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. $\endgroup$
    – Anthony Quas
    Jul 16, 2018 at 5:19
  • $\begingroup$ Thanks! I updated my question. Yes, Von Neuman's decomposition is in $ L^2 $. Thanks for reminding me! $\endgroup$
    – jason
    Jul 16, 2018 at 18:44
  • $\begingroup$ your definition of non-coboundary seems absurd (take $\xi=0$, $\beta=\phi$) $\endgroup$
    – YCor
    Jul 16, 2018 at 22:22
  • $\begingroup$ 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. $\endgroup$
    – David Hughes
    Aug 14, 2018 at 19:57

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