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My solution concerning a problem about Ergodic Recurrence requires me to prove that $\|P_T 1_B\| > 0$.

Where $P_T$ is the projection onto the space $I := \{f \in L^2 : f \circ T = f\}$, $T$ is a measure preserving mapping (so for all $B$ measurable $\mu(T^{-1} B) = \mu(B)$) and $B$ is of positive measure.

Can someone hint my why $\|P_T 1_B\|$ should be strictly positive? If $1_B$ would be in $I$ then I would have that $T^{-1} B = B$ which is probably not the case.

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Can you describe $P_T 1_B$ explicitly? –  Rasmus Feb 21 '11 at 21:12
    
@Rasmus: That didn't look too easy, should I try harder? –  Jonas Teuwen Feb 21 '11 at 21:14
    
I don't know either. I was just bothering. ;) –  Rasmus Feb 21 '11 at 21:17
    
I deleted my flawed solution. Maybe it could be repaired, but Theo's is the way to go. –  Rasmus Feb 22 '11 at 7:46

1 Answer 1

up vote 5 down vote accepted

It doesn't matter whether $B$ is invariant or not, it only has to be of positive measure, so that its characteristic function $[B]$ is nonzero in $L^{2}$. I assume that $X$ is a finite measure space (so $[X] \in L^{2}$) and that $T: X \to X$ is a measure-preserving transformation.

Clearly, $P_{T}[X] = [X]$, so \[ 0 < \mu(B) = \langle [B],[X] \rangle = \langle [B], P_{T}[X] \rangle = \langle P_{T}[B],[X]\rangle \] using $P_{T}^{\ast} = P_{T}$ in the last equation, hence $P_{T}[B] \neq 0$ in $L^{2}$.

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