# Ergodic splitting in L_p

I have a curiosity on the Ergodic decomposition given by the von Neumann's theorem:

$$L^2(X,\Sigma,\mu)=L^2(X,\Sigma_T,\mu)\oplus\overline{\{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu)\}},$$

that occurs for a measure-preserving map $T$ of a probability space $(X,\Sigma,\mu)$, $\Sigma_T$ being the sub-σ-algebra of all $T$-invariant measurable sets, and the (orthogonal) projector being the conditional expectation $E(\cdot|\Sigma_T)$. For all $1\leq p \leq\infty$, the conditional expectation is well-defined as a linear projector of norm 1 on $\textstyle L^p(X,\Sigma,\mu)$, with range the closed subspace $L^p(X,\Sigma_T,\mu) \subset L^p(X,\Sigma,\mu)$. Therefore it's quite natural to consider the analogue decomposition of the $L^p$ spaces given by the $L^p$ projector , that ''should'' be:

$$L^p(X,\Sigma,\mu)=L^p(X,\Sigma_T,\mu)\oplus\ {\overline{\{ f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)}\} }^{L^p}.$$

Now if $1\leq p\leq 2$, this splitting actually holds true, and it is easily obtained with a Lp-closure starting from the $L^2$ splitting. For $2\leq p \leq\infty$, a splitting is obtained by restriction to $L^p(X,\Sigma,\mu)$. And here it comes the problem: this way I get the right first factor (the range of the projector) $L^p(X,\Sigma_T,\mu)=L^2(X,\Sigma_T,\mu)\cap L^p(X,\Sigma,\mu)$, but I can't see why the kernel of the projector,

$$\overline{ \{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu) \} }^{L^2} \cap\ L^p(X,\Sigma,\mu)$$

should be equal to (and not larger than)

$$\overline{\{f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)\}}^{L^p}.$$

Maybe it's not a fundamental point (it does not enter in the proof of the main ergodic theorems) but I think that if a complete analogy holds true, it would be nice to state it, and if it doesn't, one would like to know what goes wrong. I checked the main texts of ergodic theory on this point, and found nothing.

Summarizing:

is there a (hopefully quick) way to see whether for $2\leq p \leq\infty$ there is an inclusion (hence equality)

$$\overline{ \{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu) \} }^{L^2} \cap\ L^p(X,\Sigma,\mu)\ \subset \ \overline{\{f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)\}}^{L^p}$$

$$\mathbf{?}$$

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TeX note: sorry, I could no way obtain the braces "{ }" in the 4 sets below the sign of "\overline" of closure. What's on? – Pietro Majer Jun 3 '10 at 22:45
If p>2 you have $L^{\infty}$ is dense in $L^p$. Also for the braces, try \\{ – Gjergji Zaimi Jun 3 '10 at 23:19
Thank you very much Gjergji, the \\} works well. And $L^\infty$ is dense in $L^p$, sure... But how does it imply the equality of the 2 spaces? – Pietro Majer Jun 4 '10 at 0:12
You can also put backticks around the whole dollar sign assembly, and this will fix the problem in general. – Harry Gindi Jun 4 '10 at 0:13

The mean ergodic theorem on L^p spaces is due to F. Riesz (1938) and S. Kakutani (1938). For p in $[1,\infty[$, this is theorem 1.2 ff in the book of Krengel, "Ergodic theorems".

If $p=\infty$, you get the splitting only on the set of functions for which the Birkhoff sums are converging, which, I think is not everything in general. In general Banach spaces, you always have the splitting in restriction to the space of vectors with converging averages (see Krengel, theorem 1.3).

The decomposition is more or less explicit. For all measurable function g, we can write

$g - {1\over n} S_n(g) = g_n - g_n\circ T$

with $S_n(g)=\Sigma_0^{n-1}g\circ T^k$ and $g_n={1\over n}\ \Sigma_0^{n-1} S_k(g)$.

If g is in the $L^2$ closure of coboundaries, we know that its conditional expectation w.r.t the invariant $\sigma$-algebra is zero. If moreover g is in $L^p$, $1\leq p < \infty$, we know that ${1\over n} S_n(g)$ goes to zero in $L^p$ norm, using the $L^p$ ergodic theorem. Also, the functions $g_n$ are in $L^p$, from their very definition. So g is in the closure of $L^p$ coboundaries.

I am not sure that the result holds for $p=\infty$.

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Thank you. I'm aware of Kakutani, Riesz, Wiener &c. But my concern is exactly on the direct sum decomposition in L_p, which is obtained by the orthogonal decomposition in L_2: precisely, by a closure, for p<2; by restriction, for p>2. While the first factor is always clear, the second one is not completely clear to me in the case p>2. – Pietro Majer Jun 4 '10 at 7:07
I edited my answer, so it should (almost) answer the question now. – coudy Jun 4 '10 at 8:11