Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading a paper in which the following argument is made (in the proof of Theorem 7). I will try to provide just the essentials necessary to ask my question, in particular omitting the computability & computable analysis parts. If I've not been fully specific after edit number $n$, and the argument as it stands is verifiably false, I will (in edit number $n+1$) refine the argument so it's closer to what the authors actually said.

Let $(X, \mu, \Sigma)$ be a probability space. Let $T \colon X \to X$ be measurable, measure-preserving (i.e. $\mu \circ T^{-1} = \mu$) and ergodic (i.e. $\mu(A \Delta T^{-1} A)=0 \Longrightarrow \mu A =0$ or $\mu A = 1$). If $B$ is measurable, then by von Neumann's mean ergodic theorem

$$a_{n} :=\frac{1}{n+1}\sum_{i\leq n} \chi_{T^{-i}B} \to \mu(B),$$

where convergence is with respect to the $L^{2}(X)$ norm. If $U\subset X$ is measurable, then they say thatand this is where my confusion liesby the Cauchy-Schwarz inequality

$$\label{conv:1}\langle \chi_{T^{-n}U}, a_{n} \rangle \to \mu(U)\cdot\mu(B).\tag{1}$$

I'm confused because a direct appliction of the Cauchy-Schwarz inequality implies less:

$$ \begin{align*} \langle \chi_{T^{-n}U}, a_{n} \rangle &\leq \lVert \chi_{T^{-n}U}\rVert_{L^{2}(X)} \cdot \lVert a_{n} \rVert_{L^{2}(x)} \\ &= \mu(T^{-n}(U)) \cdot \lVert a_{n} \rVert_{L^{2}(x)} \\ &= \mu(U) \cdot \lVert a_{n} \rVert_{L^{2}(x)} \quad \text{(Since } T \text{ is measure-preserving.)}\\ &\to \mu(U) \mu(B). \end{align*} $$

In the given setting, does the convergence at (1) hold? Why?

share|cite|improve this question
up vote 2 down vote accepted

Cauchy-Schwarz inequality gives $\langle \chi_{T^{-n}U}, a_{n}-\mu(B) \rangle^2\le\lVert \chi_{T^{-n}U}\rVert_{L^{2}(X)}^2 \cdot \lVert a_{n}-\mu(B) \rVert_{L^{2}(X)}^2$.

The first term on the RHS is at most $1$ and the second term goes to $0$ hence the LHS goes to $0$.

Now use the facts that $\langle \chi_{T^{-n}U}, \mu(B) \rangle=\mu(B)\cdot\langle \chi_{T^{-n}U}, 1 \rangle=\mu(B)\cdot\mu(T^{-n}U)$ and that $\mu(T^{-n}U)=\mu(U)$ to conclude.

share|cite|improve this answer
Boo ya! Thanks. – Quinn Culver Aug 11 '11 at 0:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.