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In page 45 of Peter Walters' "Introduction to Ergodic theory" there is a theorem which states that:

Theorem: $T:X \to X$ is weak-mixing if and only if for all $f,g \in L_2(X)$ we have that $$ \lim_{n\to \infty} \frac{1}{n} \sum^{n-1}_{i=0} \Big| \int U_T^n(f) \cdot g \: dm - \int f \: dm \int g \: dm \Big| = 0 $$ where $U_T(f)$ is defined by $U_T (f)(x) = (f\circ T)(x)$

The only if part of this proof follows trivially (apparently) by taking $f = \chi_A, \: g = \chi_B $ such that \begin{align} 0 & = \lim_{n \to \infty} \frac{1}{n} \sum_{i=0}^{n-1} \Big| \int|\chi_A \circ T^n \cdot \chi_B| dm - \int \chi_A dm \int \chi_B dm \Big|\\ & = \lim_{n \to \infty} \frac{1}{n} \sum^{n-1}_{i=0} \Big|m( T^nA \cap B) - m(A)m(B) \Big| \end{align}

However this final line does not actually imply weak mixing as far as I can tell since the weak-mixing condition is

$$ \lim_{n \to \infty} \frac{1}{n} \sum^{n-1}_{i=0} \Big|m( T^{-n}A \cap B) - m(A)m(B) \Big|. $$

That is, we need the statement to hold for $T^{-n}$ not $T^n$. If anyone could help me see why the difference doesn't matter, or has an alternative proof it would be very helpful.

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    $\begingroup$ You just made a mistake in writing $\chi_A\circ T^{n}$ as $\chi_{T^{n}(A)}$. $\endgroup$ – Kavi Rama Murthy Apr 28 '19 at 5:19
  • $\begingroup$ Yep I see. Thanks a lot. $\endgroup$ – Debreu Apr 28 '19 at 8:58
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I'll answer my own question. The expression with $T^n$ is wrong. Notice $$ U_T (\chi_B) = \chi_{T^{-1}B}$$

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