# coboundary in the slow mixing systems

given dynamical system $$(X, T, \mu)$$, $$\mu$$ is probability, $$\mu \circ T =\mu$$, $$T$$'s transfer operator $$P$$ is defined by following relation: $$\int (P a) \cdot b d\mu= \int a \cdot (b \circ T) d\mu$$ for any $$a \in L^1, b \in L^{\infty}$$.

assume $$\phi_1 \in L^{\infty}(X), \int \phi_1 d\mu =0, \int |P^n \phi_1| d\mu \le \frac{1}{n^{1+\epsilon}}, \epsilon \in [0,1)$$ for any $$n \ge 1$$. And assume coboundary condition: $$0=\int \phi_1^2+2\cdot \sum_{i \ge 1} \phi_1\cdot P^i \phi_1 d\mu$$.

can we show: there is $$\xi\in L^1$$, such that $$\phi_1=\xi-\xi \circ T$$?

the present result I know so far is: when $$\epsilon \ge 1$$, we can use coboudary condition to prove $$\sup_n \int (\sum_{i \le n} \phi_1 \circ T^i)^2 d\mu \le n^{1-\epsilon} \le 1$$, then $$\sum_{i \le n} \phi_1 \circ T^i \to \xi$$ weakly in $$L^2$$, and we can show that $$\phi_1=\xi-\xi \circ T$$ by pass the limit to $$\xi \in L^2$$. But do not know how to deal with the case of slower mixing. I guess $$\xi = -\sum_{i \ge 0} P^i \phi_1 \in L^1$$?