Posted by: yglima | June 17, 2010

ERT13: Roth theorem

This post is nothing more than a complement of ERT12. I preferred to separate it to show its own importance, specially in the historic development of the achievements. With the help of the tools developed in ERT12, we prove the

Theorem 1 (Roth, 1953) If {A\subset\mathbb Z} has positive upper-Banach density, then it contains arithmetic progressions of length {3}.

His proof relied on a Fourier-analytic argument of energy increment for functions: one decomposes a function {f} as {g+b}, where {g} is good and {b} is bad in a specific sense. If the effect of {b} is large, it is possible to break it into good and bad parts again and so on. In each step, the “energy” of {b} increases a fixed amount. Being bounded, it must stop after a finite number of steps. At the end, {g} controls the behavior of {f} and for it the result is straightforward. This follows the same philosophy of Calderón-Zygmund theory of singular integrals in harmonic analysis. See this article of Alexander Arbieto, Carlos Matheus and Carlos Gustavo Moreira for further details.

This represented the first affirmative result in the direction of Szemerédi theorem, at that time known as the Erdös-Turán conjecture. Only 16 year later, in 1969, Szemerédi proved the existence of arithmetic progressions of length 4 and, finally, in 1975 proved the conjecture in its full generality.

By Furstenberg correspondence principle (see ERT4), Theorem 1 will follow from

Theorem 2 (Roth, quantitative version) Let {f\in L^2(\mu)\backslash\{0\}} be such that {f\ge 0}. Then

\displaystyle \liminf_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\int_X fT^nfT^{2n}f>0\,.


In order to prove this, we apply Koopman-von Neumann theorem obtained in ERT12 to write

\displaystyle f=f_{\rm c}+f_{\rm wm}\,,\ \text{ where } f_{\rm c}\in\mathcal H_{\rm c} \text{ and } f_{\rm wm}\in\mathcal H_{\rm wm}.

By linearity, \displaystyle \dfrac{1}{N}\sum_{n=1}^N\int_X fT^nfT^{2n}fd\mu =\dfrac{1}{N}\sum_{n=1}^N\int_X f_{\rm c}T^nf_{\rm c}T^{2n}f_{\rm c}d\mu+\text{other terms}\,, \ \ \ \ \ (1)

where the other terms are the sum of 7 expressions of the type

\displaystyle \dfrac{1}{N}\sum_{n=1}^N\int_X fT^ngT^{2n}hd\mu\,,

where at least one of {f,g,h} is weak mixing. It turns out that the appearance of at least one weak mixing function vanishes the above limit. This is the content of the

Proposition 3 If at least one of {f,g,h\in L^2(\mu)} is weak mixing, then\displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\int_X fT^ngT^{2n}hd\mu=0.

Proof: We assume {\mu} is ergodic. The general result follows by ergodic decomposition. We will prove that if {g} or {h} is weak mixing, then \displaystyle \lim_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N T^ngT^{2n}h=0. \ \ \ \ \ (2)

We can also assume this because

\displaystyle \int fT^ngT^{2n}hd\mu=\int hT^{-n}gT^{-2n}fd\mu

and so the positions of {f,g,h} can be switched. To prove (2), we make use of van der Corput trick. Defining {x_n=T^ngT^{2n}h}, we have

\displaystyle \begin{array}{rcl} (x_{n+m},x_n)&=&\displaystyle\int_X\left(T^{n+m}gT^{2n+2m}h\right)\cdot\left(T^n\overline{g}T^{2n}\overline{h}\right)d\mu\\ &&\\ &=&\displaystyle\int_X\left(T^{m}gT^{n+2m}h\right)\cdot\left(\overline{g}T^{n}\overline{h}\right)d\mu\\ &&\\ &=&\displaystyle\int_X\left(\overline{g}T^mg\right)\cdot T^n\left(\overline{h}T^{2m}h\right)d\mu\\ &&\\ &=&\left(\overline{g}T^mg,T^n\left(hT^{2m}\overline{h}\right)\right) \end{array}

and then

\displaystyle \begin{array}{rcl} y_m&\dot=&\displaystyle\lim_{N\rightarrow+\infty}\dfrac{1}{N}\displaystyle\sum_{n=1}^N(x_{n+m},x_n)\\ &&\\ &=&\left(\overline{g}T^{m}g,\displaystyle\lim_{N\rightarrow+\infty}\dfrac{1}{N}\displaystyle\sum_{n=1}^N T^n\left(hT^{2m}\overline{h}\right)\right)\\ &&\\ &=&\left(\displaystyle\int_X \overline{g}T^m gd\mu\right)\cdot\left(\displaystyle\int_X hT^{2m}\overline{h}d\mu\right)\, , \end{array}

where in the last equality we used the ergodicity of {\mu}. If {L=\max\{\|g\|^2,\|h\|^2\}},

\displaystyle |y_m|\le L\cdot\min\left\{|(g,T^mg)|^2,|(h,T^mh)|^2\right\}

which guarantees that

\displaystyle \left|\dfrac{1}{M}\sum_{m=1}^M y_m\right|\rightarrow 0\,.

\Box

Proposition 3 proves Theorem 2. In fact, by (1),

\displaystyle \liminf_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\int_X fT^nfT^{2n}fd\mu= \liminf_{N\rightarrow+\infty}\dfrac{1}{N}\sum_{n=1}^N\int_X f_{\rm c}T^nf_{\rm c}T^{2n}f_{\rm c}d\mu\,.

The component {f_{\rm c}} is non-negative, by Proposition 4 of ERT1. In addition,

\displaystyle \int f_{\rm c}d\mu=\int fd\mu>0\,,

because {f_{\rm wm}}, being weak mixing, is orthogonal to the constant functions, that is, has zero mean. This guarantees the non-negativity of the inferior limit. To prove it is positive, we proceed as in ERT10: given {\varepsilon>0}, the norm {\|T^nf_{\rm c}-f_{\rm c}\|<\varepsilon} for a syndetic set of iterates {n}. If {\varepsilon} is sufficiently small, to each of these iterates the integral {\int f_{\rm c}T^nf_{\rm c}T^{2n}f_{\rm c}d\mu} is positive. This concludes the proof of Theorem 2.

Previous posts: ERT0, ERT1, ERT2, ERT3, ERT4, ERT5, ERT6, ERT7, ERT8, ERT9, ERT10, ERT11, ERT12.


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