${α⋅ \log(n)}$ is not uniformly distributed mod1 in $[0,1]$

equidistributionergodic-theoryharmonic-analysisuniform distribution

$\qquad \qquad \bbox[15px,border:2px solid red] { x_n:=\text{\{α$\cdot$ log(n)\}}_{n\in \mathbb N}}$

I want to show that the sequence $x_n$ is not uniformly distributed mod1 in $[0, 1]$ for any $α\in \mathbb R$.

Note:

1)$\qquad \qquad \qquad \qquad$ Euler summation Formula:

$ \qquad \qquad \bbox[15px,border:2px solid red] { \sum_{n=1}^Nf(n)=\int_1^Nf(t)dt+\frac{1}{2}(f(1)+f(N)) +\int_1^N(\text{\{t\}-$\frac{1}{2})$ }f'(t)dt }$

2) $\qquad \qquad \qquad \qquad$ Weyl's equidistributed criterion:

$\qquad \qquad \qquad \qquad \quad \qquad$ The following are equivalent
$\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {x_n \quad \text{is equivalent modulo 1} }$
$\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {\forall \text{continuous & 1-peridic f:}\quad \frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f }$
$\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {\forall k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 }$

I've already proved it by using (1) & (2) , is there any other way to approach this problem?

Best Answer

Yea, just think about what's going on. For $n \approx e^k$, $\alpha\log(n) \approx k\alpha$ and for $n \approx e^{k+\frac{1}{2\alpha}}$, $\alpha\log n \approx k\alpha+\frac{1}{2}$. So $\{\alpha \log n\}$ is in a particular interval of size $\frac{1}{2}$ for $n$ between $e^k$ and $e^{k+\frac{1}{2\alpha}}$. And for large $k$ ($\alpha$ is fixed), nearly all positive integers less than $e^{k+\frac{1}{2\alpha}}$ are greater than $e^k$. We conclude that nearly all $n \le e^{k+\frac{1}{2\alpha}}$ (you can put a floor function if you want) have $\{\alpha \log(n)\} \in (k\alpha,k\alpha+\frac{1}{2}) \pmod{1}$, clearly violating uniform distribution.

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