Consider $(\Omega ,A,\mu , T)$ a measure preserving system. Recall by Birkoff Ergodic Theorem for $f\in L_1(\mu)$, $\lim_{n\to\infty} \sum_{k=0}^{n-1} f(T^kx)=\tilde{f}(x)$ a.e. In the following cases identify $\tilde{f}$ and verify $||\tilde{f}||_{L_1}\leq ||f||_{L_1}$.
$(\Omega ,A,\mu , T)$ is Bernoulli Shift* and $f(\dots ,x_{-1},x_0,x_1, \dots )= 1_{ \{i \} }(x_0)$.
$Tx=x+\alpha (\mod 1)$, (where $X=[0,1)$.). and $\alpha$ is irrational and $f=1_{[0,1/2]}$.(**)
$X=\mathbb{R}$. $\mu$ is Lebesgue measure, $Tx=x+1$ and $f$ is some $L_1$ function.
Note: *For Bernoulli Shift see http://www.impan.pl/~gutman/The%20Theory%20of%20Bernoulli%20Shifts.pdf Page -3.
(**) I find $f(T^kx) =1$ if $0\leq x+k\alpha \leq 1/2$ or $0\leq x+k\alpha -1\leq 1/2$ or ... $0\leq x+k\alpha -k\leq 1/2$. $$0$$ o.w.
For 3. We take some $f\in L_1$, $f(T^kx)=f(x+k)$. If we choose $f=e^x$ and $\mu =1$, the inequality does not hold.
In question 2. $\sum_{k=0}^{n-1} f(T^kx) \leq n$ for $x\leq 1/2$. So the inequality should hold, but I could not find any form of $\tilde{f}$.
PLWASE HELP!!!
No. 2 is more likely doable. In No. 3, we have $\tilde{f} = \lim_{n\to\infty} \sum_{k=0}^{n-1} f(x+k)$. Then $$||\tilde{f}||_{L_1} =\int_{-\infty}^{\infty} |\lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^{n-1} f(x+k)|d\mu(x)$$ i.e. $||\tilde{f}||_{L_1} =\int_{-\infty}^{\infty} |\lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^{n-1} f(x+k)|dx \leq \int_{-\infty}^{\infty} \lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^{n-1} |f(x+k)|dx$. But I could not find out what to do next!!
If anyone has any clue please help.