Posted by: yglima | March 2, 2010

ERT9: Weak Mixing implies Weak Mixing of all orders

Remember one of the characterizations we got in ERT8 of weak mixing: a mps {\mathbb X=(X,\mathcal A,\mu,T)} is weak mixing if and only if, for any {f,g\in L^2(\mu)},

\displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\left|(T^nf,g)-(f,1)\cdot(1,g)\right|=0\,,

that is, if and only if the sequence of functions {N^{-1}\sum_{n=1}^NT^nfg} converges in the {L^2(\mu)}-norm to {\int fd\mu\int gd\mu}.

For our interests, the notion of weak mixing will be useful only if the above property extends to multiple functions, because, as we saw in ERT4, the convergence of these sequence corresponds to multiple recurrence. This actually holds, as we will see below, and is called weak mixing implies weak mixing of all orders. Such property will follow from two main ingredients:

  1. The product characterization of weak mixing can be extended to multiple products.
  2. The van der Corput trick allows an inductive argument.

It is worth mentioning that this is the second time van der Corput trick appears in these lectures (the first one was in ERT2). The reader not used to it might think it is just a technical step in the proof, but it is actually an ingredient present in many situations of Ergodic Ramsey Theory when dealing with random components of mps. It has many versions, each of them to the purpose of particular notions of multiple recurrence. We first discuss the one we need.

1. The van der Corput trick

Theorem 1 (van der Corput trick) If {(u_n)_{n\in{\mathbb N}}} is a bounded sequence in a Hilbert space {\mathcal H} and if\displaystyle \lim_{H\rightarrow+\infty}\limsup_{N\rightarrow+\infty}\dfrac{1}{H}\sum_{h=1}^H\dfrac{1}{N}\sum_{n=1}^N (u_{n+h},u_n)=0\,,

then {\left\|\dfrac{1}{N}\sum_{n=1}^N u_n\right\|\rightarrow 0}.

Proof: Take {M>0} such that {\left\|u_n\right\|\le M}. Notice that, for a fixed {H\ge 1},

\displaystyle \begin{array}{rcl}&& \left\|\dfrac{1}{N}\sum_{n=1}^N u_n-\dfrac{1}{N}\dfrac{1}{H}\sum_{n=1}^N\sum_{h=1}^{H} u_{n+h}\right\|\\ & & \\ &=& \left\|\dfrac{1}{N}\dfrac{1}{H}\sum_{h=1}^H\left(\sum_{n=1}^Nu_n-\sum_{n=1}^Nu_{n+h}\right)\right\|\\ & & \\ &\le&\sup_{1\le h\le H}\dfrac{\left\|u_1+\cdots+u_h-u_{N+1}-\cdots-u_{N+h}\right\|}{N}\\ & & \\ &\le&\dfrac{2HM}{N} \end{array}

goes to zero as {N\rightarrow+\infty} and so the assertion is equivalent to

\displaystyle \lim_{H\rightarrow+\infty}\limsup_{N\rightarrow+\infty}\left\|\dfrac{1}{N}\dfrac{1}{H}\sum_{n=1}^N\sum_{h=1}^{H} u_{n+h}\right\|=0\,.

By the triangle inequality,

\displaystyle \begin{array}{rcl} & & \limsup_{N\rightarrow+\infty}\left\|\dfrac{1}{N}\dfrac{1}{H}\sum_{n=1}^N\sum_{h=1}^{H} u_{n+h}\right\|^2 \\ & & \\ &\le & \limsup_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\left\|\dfrac{1}{H}\sum_{h=1}^{H} u_{n+h}\right\|^2\\ & & \\ &=&\limsup_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\dfrac{1}{H^2}\sum_{h_1,h_2=1}^{H}(u_{n+h_1},u_{n+h_2})\\ & & \\&\le& \dfrac{M^2}{H}+\limsup_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\dfrac{1}{H^2}\sum_{h_1,h_2=1\atop{h_1\not=h_2}}^{H}(u_{n+h_1},u_{n+h_2})\\ & & \\&=&\dfrac{M^2}{H}, \end{array}

which goes to zero as {H\rightarrow+\infty}. \Box

The significance of this inequality is that it replaces the task of bounding a sum of coefficients {u_n} by that of bounding a sum of “differentiated” coefficients {u_{n+h}\overline{u_n}}. This trick is thus useful in “polynomial” type situations when the differentiated coefficients are often simpler (have smaller order) than the original coefficients.

The above theorem is written in a modern fashion. The original van der Corput trick is actually known as van der Corput difference theorem and comes from the theory of uniform distributions.

Definition 2 We say that a sequence {(x_n)\subset[0,1]} is uniformly distributed if, for every interval {[a,b]\subset[0,1]},\displaystyle \lim_{N\rightarrow+\infty}\dfrac{|\{1\le n\le N\,;\,x_n\in[a,b]\}|}{N}=b-a.

Exercise 1 Given a sequence {(x_n)\subset[0,1]}, prove that the following assertions are equivalent.

  1. {(x_n)} is uniformly distributed.
  2. For every continuous {f:[0,1]\rightarrow{\mathbb R}}, \displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N f(x_n)=\int fdm\,, where {m} is the Lebesgue measure.
  3. (Weyl criterion) For any {m\in{\mathbb Z}\backslash\{0\}}, \displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N e^{2\pi imx_n}=0\,.

Theorem 3 (van der Corput difference theorem) A sequence {(x_n)_n\subset[0,1]} is uniformly distributed if {(x_{n+h}-x_n)_n} is uniformly distributed for every {h\ge 1}.

Proof: By Weyl criterion and the assumption, {\dfrac{1}{N}\sum_{n=1}^N e^{2\pi im(x_{n+h}-x_n)}\rightarrow 0}. Fix {m\not=0}, {h\ge 1} and define {u_n=e^{2\pi imx_n}\in{\mathbb C}}. Then

\displaystyle \dfrac{1}{N}\sum_{n=1}^N (u_{n+h},u_n)=\dfrac{1}{N}\sum_{n=1}^N e^{2\pi im(x_{n+h}-x_n)}

converges to zero as {N\rightarrow+\infty}. By van der Corput trick,

\displaystyle \dfrac{1}{N}\sum_{n=1}^N u_n=\dfrac{1}{N}\sum_{n=1}^N e^{2\pi imx_n}

converges to zero which, again by Weyl criterion, proves that {(x_n)} is uniformly distributed. \Box

The previous result implies Weyl equidistribution theorem, which is a generalization of Kronecker theorem on the uniform distribution of {n\alpha\text{ mod }1}, for irrational {\alpha}.

Theorem 4 (Weyl) If {p(x)=a_dx^d+\cdots+a_1x+a_0} is a polynomial with at least one of its coefficients {a_0,\ldots,a_d} irrational, then {p(n)\text{ mod }1}, {n\in{\mathbb N}}, is uniformly distributed.

Proof: Follows from Kronecker theorem by {d-1} sucessive applications of Theorem 3. \Box

2. Weak mixing implies weak mixing of all orders

Remember that {\mathbb X=(X,\mathcal A,\mu,T)} is weak mixing if and only if {\mathbb X\times\mathbb Y} is ergodic whenever {\mathbb Y} is ergodic. For each {a\in{\mathbb Z}}, let {\mathbb X^a=(X,\mathcal A,\mu,T^a)}.

Proposition 5 If {\mathbb X} is weak mixing, then {\mathbb X^{a_1}\times \cdots\times\mathbb X^{a_k}} is ergodic, for every integers {a_1,\ldots,a_k}. In fact, it is weak mixing.

Proof: Note that {\mathbb X^a} is weak mixing, for any {a\in{\mathbb Z}}. To prove this, consider {A,B\in\mathcal A} and {E\subset{\mathbb Z}} of zero density such that

\displaystyle \lim_{n\rightarrow+\infty\atop{n\not\in E}}\mu(T^{-n}A\cap B)=\mu(A)\cdot\mu(B)\,.

Let {E_0=(E\cap a{\mathbb Z})/a}. This set has zero density, as

\displaystyle \dfrac{|E_0\cap\{1,\ldots,N\}|}{N}= a\cdot\dfrac{|(E\cap a{\mathbb Z})\cap\{1,\ldots,aN\}|}{aN}\le a\cdot\dfrac{|E\cap\{1,\ldots,aN\}|}{aN}

and the last fraction converges to zero.

We proceed by induction on {k}. The case {k=1} was proved in the last paragraph. Suppose that, for every integers {a_1,\ldots,a_k}, {\mathbb X^{a_1}\times\cdots\times\mathbb X^{a_k}} is ergodic. If {a_{k+1}\in{\mathbb Z}},

\displaystyle \mathbb X^{a_1}\times\cdots\times\mathbb X^{a_k}\times\mathbb X^{a_{k+1}}=(\mathbb X^{a_1}\times\cdots\times\mathbb X^{a_k})\times\mathbb X^{a_{k+1}}

is the product of an ergodic and a weak mixing system, which proves our assertion. \Box

The main result of this section is

Theorem 6 Let {\mathbb X=(X,\mathcal A,\mu,T)} be weak mixing. Then, for any {f_1,\ldots,f_k\in L^{\infty}(\mu)},\displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^NT^nf_1\cdot T^{2n}f_2\cdots T^{kn}f_k=\int_X f_1d\mu\cdots\int_X f_kd\mu

in the {L^2(\mu)}-norm.

In other words, this means that, for any {f_0,\ldots,f_{k}\in L^{\infty}(\mu)}, \displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\int_X f_0\cdot T^nf_1\cdots T^nf_kd\mu=\int_X f_0d\mu\int_X f_1d\mu\cdots\int_X f_kd\mu\,. \ \ \ (1)

The proof will consist in an inductive argument, with the use of van der Corput trick to reduce the case {k} to {k-1} and so on. By linearity of the expression, we assume that {\int_X f_1d\mu=0}. In this setting, we want to prove that

\displaystyle \left\|\lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^NT^nf_1\cdots T^nf_k\right\|=0,

where {\left\|f\right\|=\sqrt{(f,f)}} represents the norm of {L^2(\mu)}.

Proof: The case {k=1} follows from the ergodicity of {T}:

\displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^{N}T^nf_1=\int_X f_1d\mu\,.

Suppose the result is true for {k-1} and take {f_1,\ldots,f_k\in L^{\infty}(\mu)}. Define

\displaystyle u_n=T^nf_1\cdot T^{2n}f_2\cdots T^{kn}f_k\,,\ n\ge 1.

Given {h\ge 0}, let {g_{hi}=f_i\cdot T^{ih}\overline{f_i}}. Then

\displaystyle \begin{array}{rcl} (u_n,u_{n+h})&=&\int_X T^nf_1\cdots T^{kn}f_k\cdot T^{n+h}\overline{f_1}\cdots T^{k(n+h)}\overline{f_k}d\mu\\ & & \\ &=&\int_X T^n\left(f_1\cdot T^h\overline{f_1}\right)T^{2n}\left(f_2\cdot T^{2h}\overline{f_2}\right) \cdots T^{kn}\left(f_k\cdot T^{kh}\overline{f_k}\right)d\mu\\ & & \\ &=&\int_X T^n g_{h1}\cdot T^{2n}g_{h2}\cdots T^{kn}g_{hk}d\mu\\ & & \\ &=&\int_X g_{h1}T^{n}g_{h2}\cdots T^{(k-1)n}g_{hk}d\mu\\ & & \\ &=&\left(T^{n}g_{h2}\cdots T^{(k-1)n}g_{hk},\overline{g_{h1}}\right). \end{array}

By hypothesis,

\displaystyle \begin{array}{rcl} r_h&=&\lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N (u_n,u_{n+h})\\ & & \\ &=&\int_X g_{h1}d\mu\cdots\int_X g_{hk}d\mu\,. \end{array}

Rewrite the above expression as

\displaystyle \int_{X^k} (f_1\otimes\cdots\otimes f_k)(T\times\cdots\times T^k)^h(\overline{f_1}\otimes\cdots\otimes \overline{f_k})d(\mu\times\cdots\times\mu),

where {X^k=X\times\cdots\times X} and {f_1\otimes\cdots\otimes f_k:X^k\rightarrow{\mathbb C}} is given by

\displaystyle (f_1\otimes\cdots\otimes f_k)(x_1,\ldots,x_k)=f_1(x_1)\cdots f_k(x_k).

As {\mathbb X\times\cdots\times\mathbb X^k} is ergodic,

\displaystyle \begin{array}{rcl} \lim_{H\rightarrow+\infty}\dfrac{1}{H}\sum_{h=1}^H r_h &=&\left(\int_{X^k} (f_1\otimes\cdots\otimes f_k)d(\mu\times\cdots\times\mu)\right)^2\\ & & \\ &=&\left(\int_X f_1d\mu\cdots\int_X f_kd\mu\right)^2\\ & & \\ &=&0\,. \end{array}

By the van der Corput trick,

\displaystyle \lim_{M\rightarrow+\infty}\left\|\dfrac{1}{N}\sum_{n=1}^{N}u_n\right\|=0,

which concludes the proof. \Box

Corollary 7 (Multiple Poincaré Recurrence for weak mixing mps) If {\mathbb X=(X,\mathcal A,\mu,T)} is weak mixing and {A\in\mathcal A}, then\displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\mu(A\cap T^{-n}A\cdots\cap T^{-kn}A)=\mu(A)^{k+1},

for every {k\ge 1}.

Proof: Just consider {f_0=\cdots=f_{k+1}=\chi_A} in (1). \Box

Previous posts: ERT0, ERT1, ERT2, ERT3, ERT4, ERT5, ERT6, ERT7, ERT8.

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Responses

  1. [...] most . Indeed, every weakly mixing system is in fact weakly mixing of all orders; see for instance this blog post of Carlos Matheus, or these lecture notes of myself. So the problem is to exclude the possibility of correlation [...]


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