Erdos-Turan Inequality

By using harmonic analysis to detect intervals, equidistribution of a sequence modulo one can is equivalent to the following:

Weyl’s criterion: A sequence of real numbers x_{1}, x_{2}, \ldots, x_n, \ldots is uniformly distributed mod one if and only if for every integer h \neq 0 we have

\displaystyle \left|\sum_{n \leq N} e\left(h x_{n}\right)\right|=o_{h}(N) \quad \text { as } N \rightarrow \infty .

That is this criterion gives equidistribution

\displaystyle \left\{n \le N: \alpha \le \{x_{n}\| \le \beta\right\} \sim(\beta-\alpha) N \quad \text { as } N \rightarrow \infty .

What is the rate of convergence? Can we make everything more effective in terms of the interval and the N.?

One way to do it is the following inequality by Erdos-Turan:

\displaystyle \left|\frac{1}{N} \left| \left\{n \leq N: \alpha<\left\{x_{n}\right\} \leq \beta \right\}\right|-(\beta-\alpha)\right| \leq \frac{1}{H+1}+3 \sum_{h=1}^{H} \frac{1}{h}\left|\frac{1}{N} \sum_{n \leq N} e\left(h x_{n}\right)\right|

This inequality is a specialization of the following: Let \mu be a probability measure on the unit circle Then we have

\displaystyle |\mu(A)-|A|| \leq C\left(\frac{1}{n}+\sum{k=1}^{n} \frac{|\hat{\mu}(k)|}{k}\right)

where

\displaystyle \hat{\mu}(k)=\int \exp (2 \pi i k \theta) d \mu(\theta)

Previously mentioned inequality is when the measure is equal delta masses at the points x_1, x_2, \cdots x_N.

The rate of convergence is the measure of discrepancy, that is how far the sequence is from the uniform distribution.

Erdos-Turan needed this inequality to prove the discrepancy of the angles of the roots of a polynomial of degree n. In fact, they prove

For a polynomial P(z) , with roots \displaystyle z_j =r_j e^{i \phi_j} , we have

\displaystyle \left|N(\alpha, \beta)-\frac{\beta-\alpha}{2 \pi} n\right| \leq 16 \sqrt{n \log \frac{|P|}{\sqrt{\left|a_{0} a_{n}\right|}}}

where \displaystyle N(\alpha, \beta) =|\{j: \alpha \le \{phi_j\} \le \beta\}| and \displaystyle |P| = \max _{|z|=1}|P(z)| .

Selberg Polynomials: To prove Erdos-Turan, we need good approximations of the characteristic function with trigonometric polynomials of degree at most H.

\displaystyle S_{H}^{-}(x) \leq \chi_{[\alpha, \beta]}(x) \leq S_{H}^{+}(x)

and

\displaystyle \int_{\pi} S_{H}^{\pm}(x) d x=\beta-\alpha \pm \frac{1}{H+1}

Plugging these approximations, and using

\displaystyle \left|\widehat{\chi}_{[\alpha, \beta]}(h)\right|=\left|\frac{\sin \pi h(\beta-\alpha)}{\pi h}\right| \leq \min \left(\beta-\alpha, \frac{1}{\pi|h|}\right)

we get the following estimate for the discrepancy

\displaystyle |D(N ; \alpha, \beta)| \leq \frac{1}{H+1}+2 \sum_{h=1}^{H}\left(\frac{1}{H+1}+\min \left(\beta-\alpha, \frac{1}{\pi h}\right)\right)\frac{1}{N}\left|\sum_{n=1}^{N} e\left(h u_{n}\right)\right|

Note that this above estimate is slightly stronger than the Erdos-Turan mentioned before.

Selberg, Vaaler, Beurling:

Consider the Fejer Kernel

\displaystyle \Delta_{H}(x)=\sum_{-H}^{H}\left(1-\frac{|h|}{H}\right) e(h x)=\frac{1}{H}\left(\frac{\sin \pi H x}{\sin \pi x}\right)^{2}

The Vaaler functions are given by

\displaystyle V_{H}(x)= \frac{1}{H+1} \sum_{h=1}^{H}\left(\frac{h}{H+1}-\frac{1}{2}\right) \Delta_{H+1}\left(x-\frac{h}{H+1}\right)
\displaystyle \quad \quad \quad +\frac{1}{2 \pi(H+1)} \sin 2 \pi(H+1) x-\frac{1}{2 \pi} \Delta_{H+1}(x) \sin 2 \pi x

Now the Beurling polynomial is

\displaystyle B_{H}(x)=V_{H}(x)+\frac{1}{2(H+1)} \Delta_{H+1}(x)

Finally the Selberg polynomials are

\displaystyle S_{H}^{+}(x)=\beta-\alpha+B_{H}(x-\beta)+B_{H}(\alpha-x)
\displaystyle S_{H}^{-}(x)=\beta-\alpha-B_{H}(\beta-x)-B_{H}(x-\alpha) &fg-000000

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