$\{(\log n)^\alpha\}$ not equidistributed if $0<\alpha\leq 1$, so how is it distributed?

Question

The brackets denote the fractional part function. It is well known that the distribution (defined as the limit of the empirical distribution) is $F(x)=(e^x - 1)/(e-1)$, with $x\in [0, 1]$, if $\alpha=1$, see here. Also, $\{n^\alpha\}$ is equidistributed if $\alpha>0$ is not an integer, that is, the distribution is uniform on $[0, 1]$, in other words $F(x)=x$. I suspect $\{(\log n)^\alpha\}$ is equidistributed if $\alpha>1$.

For slow growing functions, we don't have equidistribution, and one might even argue that there is no limiting distribution due to lack of convergence. Here I propose a simple framework to make the limiting distribution exist and be well and uniquely defined. My questions are:

• What is the distribution of $\{(\log n)^\alpha\}$ when $0<\alpha\leq 1$?
• Is $\{(\log n)^\alpha\}$ indeed equidistributed when $\alpha>1$?
• What is the "slowest growth" function that guarantees equidistribution? For instance, is $\{(\log n)(\log \log n)\}$ equidistributed, given that $\{\log n\}$ is not, and $\{(\log n)^2\}$ seems to be?

Potential approach to this problem

Here I summarize my research. It allows you to quickly find the limiting distribution for $\{h(n)\}$ (or a least provides a sure way to find it), whether equidistributed or not, for strictly increasing sequences $h(n)$ satisfying both $h(n)=o(n)$ and $h(n)\rightarrow \infty$. With this methodology, it becomes trivial to prove that $\{ \sqrt{n}\}$ is equidistributed, and that $\{\log_b n\}$ has limiting distribution $(b^x-1)/(b-1)$. Thus my question focuses on less trivial examples.

To find the limiting distribution whether uniform on not (if uniform on $[0, 1]$ it implies equidistribution), proceed as follows:

• Find a large $m$ such that $h(m) = n$ is an (obviously large) integer. Thus $m = h^{-1}(n)$. Find $m'$ such that $h(m')=n+1$, thus $m'=h^{-1}(n+1)$.
• We are interested in the distribution of $h(m+k)$ between $m$ and $m'$, with $k=0,1,2,\dots, m'-m$. Let $x=k/(m'-m)$ so that $0\leq x\leq 1$.
• Let $g_n(x) = h(m+k)-h(m) = h(m+x(m'-m))-h(m)$, and $k(n)=h^{-1}(n)$. Thus $g_n(x) = -n + h(k(n) + [k(n+1)-k(n)]\cdot x)$.
• Define $g(x) =\lim_{n\rightarrow\infty} g_n(x)$. Then the distribution of interest is $F(x)=g^{-1}(x)$, with $0\leq x\leq 1$.

I used this to find that $\{(\log_2 n)^2\}$ is equidistributed ($F(x)=x$) and $\{ \log_2 n\}$ has distribution $F(x) = 2^x -1$. I am wondering if Mathematica would have found it, the computation of the limiting function $g_n(x)$, in the end, being purely mechanical.

I haven't computed it for the general case $\{(\log_2 n)^\alpha\}$, and that's part of my question. In that case, we have

$$g_n(x)=\Big[\log_2(2^{n^{1/\alpha}} + (2^{(n+1)^{1/\alpha}} - 2^{n^{1/\alpha}})\cdot x)\Big]^\alpha -n .$$

The next step is to compute $\lim_{n\rightarrow\infty} g_n(x)$.

Update on 10/28/2021

I computed the above limit $g(x)=\lim_{n\rightarrow \infty} g_n(x)$, with Mathematica, see here for $\alpha=\frac{1}{2}$. It confirms the answer provided below by Iosif. If $\alpha<1$ then $g(x) = 1$ if $x>0$, thus $g$ can not be inverted, so there is no limiting distribution of any kind. If $\alpha = 1$ then $g(x)=\log(1+(e-1)\cdot x)$ which is correct, yielding $F(x)=g^{-1}(x)=(e^x-1)/(e-1)$. And if $\alpha>1$ then $g(x)=x$, thus $F(x)=x$ is the uniform distribution. It is possible that regardless of $h(n)$ satisfying my requirements, these are the only three options.

Using Fejer's criterium mentioned in the comments, we also have equidistribution for $\{(\log n) (\log \log n)^\alpha\}$ if $\alpha>0$.

Below is the output from Mathematica:

Answer 1

There is no limit distribution at all for $a:=\alpha\in(0,1)$: for each $x\in(0,1)$, the relative frequency \begin{equation*} f_n:=f_n(x):=\frac1n\,\sum_{j=1}^{n-1}1(x_j<x) \end{equation*} will be forever oscillating between $0$ and $1$ as $n\to\infty$, where \begin{equation*} x_j:=\{\ln^a j\}. \end{equation*} (The oscillations will be very slow for large $n$, because the function $\ln$ is varying slowly.)

Indeed, take any $a\in(0,1)$ and any $x\in(0,1)$. If $k$ and $j$ are natural numbers and $k\le\ln^a j<k+1$, then $x_j<x\iff e^{k^p}\le j<e^{(k+x)^p}$, where $p:=1/a>1$.

So, with $u\wedge v:=\min(u,v)$, \begin{equation*} \begin{aligned} S&:=\sum_{j=1}^{n-1}1(x_j<x) \\ &=\sum_{j=1}^{n-1}\sum_{k=1}^\infty 1(e^{k^p}\le j<e^{(k+x)^p}) \\ &=\sum_{k=1}^\infty\sum_{j=1}^{n-1} 1(e^{k^p}\le j<e^{(k+x)^p}) \\ &=\sum_{k=1}^\infty\sum_{j=1}^\infty 1(e^{k^p}\le j<n\wedge e^{(k+x)^p}) \\ &=\sum_{1\le k<\ln^a n}\;\sum_{j=1}^\infty 1(e^{k^p}\le j<n\wedge e^{(k+x)^p}) \\ &=\sum_{1\le k<\ln^a n}\;(n\wedge e^{(k+x)^p}-e^{k^p}+O(1)) \\ &=S_1+S_2+O(\ln^a n)=S_1+S_2+o(n), \end{aligned} \end{equation*} where \begin{equation*} S_1:=\sum_{k=1}^{k_n}(e^{(k+x)^p}-e^{k^p}),\quad S_2:=\sum_{k_n+1\le k<\ln^a n}(n-e^{k^p}), \end{equation*} \begin{equation*} k_n:=k_{x,n}:=\lceil \ln^a n-x\rceil-1, \end{equation*} so that \begin{equation*} \ln^a n-x\le k_n+1<\ln^a n-x+1. \tag{1} \end{equation*} Since $p>1$ and $x\in(0,1)$, for $k\to\infty$ and natural $l<k$ we have $e^{(k+x)^p}>>e^{k^p}>>e^{(l+x)^p}$, where $A>>B$ means $A/B\to\infty$, so that $e^{(k+x)^p}-e^{k^p}\sim e^{(k+x)^p}>> e^{(l+x)^p}-e^{l^p}$. So, \begin{equation*} S_1\sim e^{(k_n+x)^p}. \end{equation*} Next, \begin{equation*} S_2=1(k_n+1<\ln^a n)(n-e^{(k_n+1)^p}) \end{equation*} and \begin{equation*} 1(k_n+1<\ln^a n)=1(\ln^a n-x\le k_n+1<\ln^a n). \end{equation*}

Take any $y\in(x,1+x)$. Note that $\ln^a(n+1)-\ln^a n\sim a\ln^a n/(n\ln n)$. So, there is a sequence $(n_k)=(n_{y;k})$ of integers such that $\ln^a n<k+y\le (\ln^a n)(1+1/(n\ln n))=\ln^a n+o(1)$ for $n=n_k$ and all $k$. So, by (1), eventually (that is, for all large enough natural $k$) we have $k_n=k$ if $n=n_k$, and hence \begin{equation*} (k_n+y)^p =(\ln n)[1+O(1)/(n\ln n)]^p=\ln n+O(1/n)=\ln n+o(1), \end{equation*} so that \begin{equation*} n\sim e^{(k_n+y)^p}>>e^{(k_n+x)^p}\sim S_1, \end{equation*} whereas $\ln^a n<k_n+y<k_n+1$ if $y\in(x,1)$, so that $S_2=0$, and thus $f_n=(S+o(1))/n\to0$.

However, if $y\in(1,1+x)$, then eventually we have $\ln^a n=k_n+y+o(1)>k_n+1$, so that \begin{equation*} S_2=n-e^{(k_n+1)^p}=n-o(e^{(k_n+y)^p})=n-o(n), \end{equation*} which implies $f_n\ge(S_2+o(n))/n\to1$, whence $f_n\to1$.

Thus, as $k\to\infty$, we have $f_{n_{y;k}}\to0$ if $y\in(x,1)$ and $f_{n_{y;k}}\to1$ if $y\in(1,1+x)$.

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As an illustration, here are the connected graphs $\{(n,f_n)\colon1\le n\le 10^2\}$ and $\{(n,f_n)\colon10^2\le n\le 10^6\}$ for $a=5/10$ and $x=4/10$:

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In the case $a=1$, one can use reasoning mostly quite similar to the above reasoning for $a\in(0,1)$, but the conclusion here is a bit different. Namely, here, for $y\in(x,1+x)$ (as above) and $n=n_k=n_{y;k}$, \begin{equation} S_1\sim\frac{e^x-1}{e-1}\,e^{1-y}n, \end{equation} and also $S_2=0$ eventually if $y\in(x,1)$, so that \begin{equation} f_{n_{y;k}}\to\frac{e^x-1}{e-1}\,e^{1-y} \end{equation} if $y\in(x,1)$, so that $\lim_k f_{n_{y;k}}$ depends on the choice of $y\in(x,1)$. So, there is no limit distribution for $a=1$ either.

The case $a>1$ is covered in comments. In this case, the limit distribution is uniform.

Thus, the borderline case is $a=1$, in terms of the existence of a limit distribution.