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I recently tackled a problem and I arrived at something of the following form,

$$ \frac{1}{n^2} \sum_{i=0}^{n-1}\sum_{j=0}^{n-1} f(T^ix, T^jy), $$

where $T$ is a measure preserving transformation and I am interested in the limit as $n$ tends to infinity. In my case it is the shift operator in a probability space.

In the uni-variate case Birkhoff's ergodic theorem states that for a measure preserving transformation $T$,in a measurable space $(X, \mathscr{B})$, with a $\sigma$-finite measure $\mu$ $$\frac{1}{n} \sum_{i=0}^{n-1} f(T^ix)$$ converges a.e. to a function $f*\in L^1$, with $f^* = \frac{1}{\mu(X)}\int fd\mu$. Is there an equivalent result for the multivariate case? Will the first equation converge to it's average?

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  • $\begingroup$ I do not have a reference right now, but I would say that it is probably false and that the almost sure convergence only holds for functions in $L \log L (\Omega, \mu)$ (this is what happens for multi-parametric martingales) $\endgroup$ Commented May 25, 2019 at 10:32
  • $\begingroup$ Proving the result for $L \log L(\Omega)$ would not be difficult. You need to combine the Maximal ergodic inequality with Yanos extrapolation to get the $L \log L \to L_1$ bound for the ergodic maximal. After that you can use the same argument that is used for martingale sequences. $\endgroup$ Commented May 25, 2019 at 10:37
  • $\begingroup$ I think that I can move from here, based on your comments and the answer below. Although I first need to learn about Yanos extrapolation. Thank you for the input $\endgroup$ Commented May 27, 2019 at 11:12
  • $\begingroup$ No problem. I think i will write a more detailed answer because I think it is likely that the largest Orlicz class for which there is pointwise ergodic convergence is $L \log L(\Omega)$. $\endgroup$ Commented May 27, 2019 at 11:29

2 Answers 2

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Let me expand here the point made in the the comments. First, I misread slightly your question. I had in mind two-parametric ergodic means given by $$ A_{n,m}(f) = \sum_{i = 0}^{n - 1} \sum_{j = 0}^{m - 1} f(T^i x, T^j y). $$ Not the means $A_{n}(f) = A_{n,n}(f)$. For the first you need to be in $L \log L(\Omega)$. For the second I do not have a definitive answer. It may be enough to be in $L^1$, see the edit bellow.

First, the following result is known.

Theorem: For every $f \in L \log L(\Omega,\mu)$ , where $(\Omega, \mu) = (\Omega_1 \times \Omega_2, \mu_1 \otimes \mu_2)$ and ergodic transformation $T_i:X_i \to X_i$, the two parametric ergodic averages $$ A_{n,m}(f) = \sum_{i = 0}^{n - 1} \sum_{j = 0}^{m - 1} f(T^i x, T^j y) $$ converge almost everywhere $A_{n,m}(f) \to f$.

The proof uses the following steps

  • For each $i$ use the maximal weak-type $(1,1)$ maximal ergodic inequality.
  • Use real interpolation to get that: $$ \Big\| \sup_{n \geq 0} A^i_{n}(|f|) \Big\|_p \lesssim \max \Big\{ 1, \frac1{p - 1} \Big\} \, \| f \|_p $$
  • use Yano's extrapolation [Y] to obtain that the above maximal operator above maps $L \log L(\Omega)$ into $L^1(\Omega)$ for every $i$.

Then, copying the argument in [JMZ], you have that the map $f \mapsto f^\ast$ given by $$ f^\ast(x,y) = \limsup_{n, m \to \infty} A_{n,m}(|f|) $$ maps $L \log L(\Omega)$ into $L_1(\Omega)$. Just by composition, we have that $$ \limsup_{n, m \to \infty} A_{n,m}(|f|) \leq \limsup_{n \to \infty} A^1_{n} \bigg( \underbrace{\sup_{m \geq 0} A_m^2|f|}_{g} \bigg) $$ But the function $g$ is in $L^1(\Omega)$ as the function $f$ is in $L \log L$. Now, using that for every function in $L^1(\Omega)$ we have almost everywhere convergence (and therefore the limsup is exchangeable by the lim) we can conclude. The fact that $f^\ast$ is in $L^1$ gives almost everywhere convergence by the same argument that is used with maximal functions.

I will go further and conjecture that the following is true (probably known to the experts):

Open (to my knowledge) For every $\varphi$ with $\varphi \in o(x \log x)$ we have that there is an element $f \in L_\varphi(\Omega)$, such that $A_{n,m}(f) \not\to f$.

It is known, see [JMZ, Theorem 8]. That this holds in the case of differentiability of integrals. I will try to adapt the argument to the ergodic case. It is also likely to be true, since something similar holds for martingales [G].

[G] Gundy, R. F., On the class L log L, martingales, and singular integrals, Stud. Math. 33, 109-118 (1969). ZBL0181.44202.

[JMZ] Jessen, B.; Marcinkiewicz, J.; Zygmund, A., Note on the differentiability of multiple integrals., Fundamenta Math. 25, 217-234 (1935). ZBL61.0255.01.

[Y] Yano, Shigeki, Notes on Fourier analysis. XXIX. An extrapolation theorem, J. Math. Soc. Japan 3, 296-305 (1951). ZBL0045.17901.

P.D.: For your purposes, perhaps it will be enough if you take the usual ergodic averages with respect to $$ S = p (\mathrm{id} \otimes T) + q (T \otimes \mathrm{id}) $$ for $p + q = 1$. Its ergodic averages will be weighted summations concentrated as Gaussian around the diagonal $i = j$.

Edit I found a (almost) complete solution in $L^1$ in the following paper:

You need to interpret $(i,j) \mapsto T^i \otimes T^j$ as an action of $\mathbb{Z}^2$ and use that $[0,N]$ is a well-tempered Foelner sequence (in the sense of that paper). That would give you almost everywhere convergence for the means $A_{n,n}(f)$, for every $f \in L^1(\Omega)$.

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    $\begingroup$ Small remark on the edit: The case at hand is already covered by Emerson's pointwise ergodic theorem (The Pointwise Ergodic Theorem for Amenable Groups, American J. Math., 1974), and I would guess that the result for $\mathbb{Z}^d$-actions is even older. But for $\mathbb{Z}^d$-actions on noncommutative spaces, the case $p=1$ still seems to be open. $\endgroup$
    – MaoWao
    Commented Oct 4, 2019 at 9:21
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    $\begingroup$ Isn't the noncommutative case covered by this recent work on polynomial growth groups: arxiv.org/abs/1705.04851? $\endgroup$ Commented Oct 4, 2019 at 11:36
  • $\begingroup$ I haven't looked at it in detail, but it seems so. I'm still a little confused by their claim to treat more general operators than Junge--Xu, so it's perhaps not quite what I was thinking of. $\endgroup$
    – MaoWao
    Commented Oct 4, 2019 at 12:17
  • $\begingroup$ Which makes me realize that the theorems by Lindenstrauss and Emerson both do not cover the situation in the question, because they both only treat group actions, but the OP does not assume invertibility. One would really need a similar ergodic theorem for semigroup actions, such as this: mathoverflow.net/questions/152465/… $\endgroup$
    – MaoWao
    Commented Oct 4, 2019 at 12:24
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Rephrased in an other way, the question is about the convergence of $$ \frac 1{n^2}\sum_{i=1}^n\sum_{j=1}^nf\left(X_i,X_j\right), $$ where $\left(X_i\right)_{i\geqslant 1}$ is a strictly stationary sequence. In order to hope something, we have to assume that $f\left(X_i,X_j\right)$ is integrable for all $i,j$. The diagonal part $\sum_{i=1}^nf\left(X_i,X_i\right)$ has, by the classical ergodic theorem, a negligible contribution hence we are reduced to study the asymptotic behavior of $$ \frac 1{n^2}\sum_{1\leqslant i<j\leqslant n}f\left(X_i,X_j\right). $$ (the part with $j>i$ follows by using $g\colon (u,v)\mapsto f(v,u)$). The keyword is then $U$-statistics. A version of the ergodic theorem could be $$ \frac 1{n^2}\sum_{1\leqslant i<j\leqslant n}\left(f\left(X_i,X_j\right)-\mathbb E\left[f\left(X_i,X_j\right)\right]\right)\to 0 \mbox{ a.s.} $$ but the question seems to be open.

There are some results when the product of marginal distribution functions is continuous (see

Borovkova, S.; Burton, R.; Dehling, H. Consistency of the Takens estimator for the correlation dimension. Ann. Appl. Probab. 9 (1999), no. 2, 376--390. doi:10.1214/aoap/1029962747. https://projecteuclid.org/euclid.aoap/1029962747).

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