Formalize answer to $\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }$

Question

I was trying to understand the accepted answer here this answer Calculate $\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\ \{n\sqrt{2}\} }$ , in particular the statement of the accepted answer:

Since $\{ n\sqrt{2} \}$ is equidistributed modulo $1$, the limit could be rewritten as the limit of the expected value of the geometric average of $n$ uniform random variables.

I find the statement intuitive, but do not understand how it can be formalized, hence this separate question.

Note 1: I am aware that the problem can be solved using Weyl's theorem as here: limit of arithmetic mean of $|\sin n|$, $n\in\mathbb{N}$ , but this I am not sure directly justifies the reported statement (even if it provides a solution to the problem).

OLDER VERSION OF THE QUESTION:

Since the statement involves random variables I thought to use https://en.wikipedia.org/wiki/Ergodic_theory . The only way I see here to introduce a measure space is to define $\Omega$ as the unit interval $[0,1]$ with the Lesbesgue measure. Than define $T :\Omega \rightarrow \Omega : x \rightarrow \{ x+\sqrt{2} \} $, with brackets indicating fractional part. We have to check that the transformation is ergodic to apply Birkhoff's theorem but, even after doing that, what we can conclude ? Birkhoff's theorem would say that the relation:

$$\frac{1}{N} \sum_{k=0}^{N}f(\{ x+k\sqrt{2}\}) \rightarrow \int_0^1 f(y)dy$$

would hold for almost any $x \in \Omega$. But to apply this to our problem we would need that the relation holds for $x=0$, that is a specific element, and I am not sure therefore that Birkhoff can be applied here. Further, Birkhoff would not justify the statement even if we introduced some probability concepts.

So I have these two questions:

1. can Birkhoff's theorem or some ergodic theory based on r.v. be used to formalize the accepted answer ?

2. if not, how can the reported statement be formalized? (maybe it is already formal enough but I do not see why)

Answer 1

I’ll be rather vague in using the term nice function, and for simplicity we can assume the term means Riemann inegrable (or improper integrable).

Here I explain why the method, or rather interpretation, described by @Varun Vejalla in his answer to the the problem here works, which is indeed a very clever reduction of the problem.

Suppose $\{x_n\}$ is an equicontinuous sequence in $(0,1)$. Then, by Weyl's equidistribution theorem, for any nice function $f$ \begin{align} \frac1n\sum^n_{k=1}f(x_k)\xrightarrow{n\rightarrow\infty}\int^1_0f(t)\,dt\tag{1}\label{one} \end{align} The right has side being the expectation of the random variable $f(U)$ where $U$ is uniformly distributed in $(0,1)$

Now, suppose $\phi:(0,1)\rightarrow(0,\infty)$ is a nice function such that $\log\circ \phi$ is also nice, then applying \eqref{one} to $f=\log\circ\phi$ yields \begin{align} \alpha_n&:=\sqrt[n]{\phi(x_1)\cdot\ldots \phi(x_n)}\\&=\exp\Big(\frac1n\sum^n_{k=1}\log\circ\phi(x_k)\Big)\xrightarrow{n\rightarrow\infty}\exp\Big(\int^1_0\log\circ\phi(t)\,du\Big)\tag{2}\label{two} \end{align}

On the other hand, the expectation of $\Big(\phi(U_1)\cdot\ldots\phi(U_n)\Big)^{1/n}$ where $U_1,\ldots, U_n$ are i.i.d uniformly distributed random variables in $(0,1)$ is \begin{align} \beta_n:=\mathbb{E}\left[\Big(\phi(U_1)\cdot\ldots\phi(U_n)\Big)^{1/n}\right]=\Big(\int^1_0 \phi^{1/n}(t)\,dt\Big)^n\tag{3}\label{three} \end{align} If for example $\phi(t)=t^\alpha$ then \eqref{two} yields $a_n\xrightarrow{n\rightarrow\infty}e^{-\alpha}$ while \eqref{three} yields $\beta_n\xrightarrow{n\rightarrow\infty}\lim_n\Big(1+\tfrac{\alpha}{n}\Big)^{-n}=e^{-\alpha}$ (observe that the original problem has $\phi(t)=t$).

The question then is whether for all nice functions $\phi$ $$\lim_{n\rightarrow\infty}\Big(\int^1_0\phi^{1/n}(t)\,dt\Big)^n=\exp\Big(\int^1_0\log\circ\phi(t)\,dt\Big)$$ This is in fact the case for any measurable function $\phi\geq0$ such that $\|\phi\|_p<\infty$ for some $0<p<\infty$.

That is the interpretation by Verun works, that is, for the purposes the the problem at hand, the sequence $(x_n:n\in\mathbb{N})$ can be considered as an i.i.d sample from the uniform distribution $U$ in $(0,1)$ to justify the application of the law of large numbers. This however, is just an interpretation of the original problem in terms of probabilistic elements. There is nothing random about the sequence $(\{\sqrt{2}n\}:n\in\mathbb{N}\}$.

Answer 2

I imagine this is overkill, but we can show $\{n\sqrt{2}\}_n$ is equidistributed with the (topological aspects of) ergodic theory. I haven't reviewed my notes in a while, so please forgive any logical omissions - trying to piece the whole argument together took a while, as I wanted this post to include all the relevant chains of reasoning.

A topological dynamic system $(K;\varphi)$ is a compact Hausdorff space $K$ with a distinguished $\varphi:K\to K$ a continuous mapping. Let all references to a 'measure' refer to a complex Baire probability measure. A measure $\mu$ on $K$ is said to be invariant if it is and $\mu\circ\varphi^{-1}=\mu$ identically. We say $(K;\varphi)$ is uniquely ergodic if there is a unique invariant measure for this system. It can be shown that this implies the unique invariant measure $\mu$ makes $(K,\mu;\varphi)$ an ergodic system. It can further be shown that this implies the Koopman operator $T:C(K)\to C(K),f\mapsto f\circ\varphi$ is mean ergodic, i.e. that the limit: $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^nT^j$$Exists as an operator in $C(K)$, called the mean ergodic projection $P_T$. In particular, this convergence is uniform on every element $f\in C(K)$. This also implies the fixed space of $T$ is just the space of constant functions. It can be shown that any mean ergodic projection is a projection onto the fixed space of $T$, meaning $TP_T=P_T$ identically (where $P_T$ exists). Thus $P_T$ is always a constant in this case. From the formula (coming from measure-preservation): $$\int_KP_T(f)\,\mathrm{d}\mu=\int_K f\,\mathrm{d}\mu$$We conclude: $$P_T(f)=\Bbb E[f]$$

A group rotation system is a pair $(\mathcal{G};\alpha)$ where $\mathcal{G}$ is a compact Hausdorff topological group, $\alpha\in\mathcal{G}$. Such are interpreted as topological dynamical systems via the action $\varphi:g\mapsto\alpha g$. Such systems come equipped with a unique Haar measure $\mathfrak{m}$. If it so happens that $(\mathcal{G};\alpha)$ is minimal, meaning any subset of $\mathcal{G}$ that is both topologically closed and closed under the dynamic is either empty or $\mathcal{G}$ itself, then the system is uniquely ergodic with unique invariant $\mathfrak{m}$.

In particular, the group system $([0,1);\alpha)$ (the action being addition modulo $1$) where $\alpha\in[0,1)$ is irrational and $[0,1)$ is topologised by identification with the unit circle under $x\mapsto\exp(2\pi ix)$ is well known to be minimal; this is essentially because the orbit of $\alpha$ is dense, if I remember correctly. Thus it is also uniquely ergodic and the associated $T$ is mean ergodic on $C(K)$.

Let's prove that $T$ is even mean ergodic on $R[0,1]$ the space of $1$-periodic bounded Riemann integrable functions with the supremum norm. Let $A_n:=\frac{1}{n}\sum_{j=1}^nT^j$ denote the Cesaro mean operator, $n\in\Bbb N$, acting on $R[0,1]$. For any characteristic function $\chi$ of a subinterval of $[0,1]$ and any $\epsilon>0$ we may find continuous functions $g_\epsilon,h_\epsilon$ on $[0,1]$ with: $$g_\epsilon\le\chi\le h_\epsilon$$And: $$\int_0^1(h_\epsilon-g_\epsilon)\,\mathrm{d}t<\epsilon$$We know $A_n g_\epsilon$ converges uniformly to the constant function $\Bbb E[g_\epsilon]$, likewise for $h_\epsilon$, and from the inequality: $$A_n g_\epsilon\le A_n\chi\le A_nh_\epsilon$$For all $n$, we get: $$\int_0^1g_\epsilon(t)\,\mathrm{d}t-\epsilon\le A_n\chi\le\int_0^1h_\epsilon(t)\,\mathrm{d}t+\epsilon$$For large $n$. Combining this with the defining choice of $g,h$, we get: $$\|A_n\chi-\int_0^1\chi(t)\,\mathrm{d}t\|_\infty\le2\epsilon$$For these large $n$, concluding that $P_T\chi$ exists and equals $\int_0^1\chi$. This extends to the $1$-periodic step functions immediately. By similarly approximating any $f\in R[0,1]$ in integral by step functions - with the same argument - we have $P_T(f)$ exists and equals $\int_0^1 f$.

Finally, denote $\alpha:=n\alpha\bmod1$ for all $n$ and choose $f=\begin{cases}\chi_{[a,b]}&b<1\\\chi_{\{0\}\cup[a,b]}&b=1\end{cases}$ for any $[a,b]\subseteq(0,1]$, so that $f\in R[0,1]$. We have: $$A_nf(0)=\frac{1}{n}\sum_{j=0}^{n-1}T^jf(0)=\frac{1}{n}\sum_{j=0}^{n-1}f(\alpha_j)=\frac{1}{n}\operatorname{card}\{j:0\le j\le n,\alpha_j\in[a,b]\}$$This converges uniformly to $\int_0^1f=b-a$, as desired. Note that the zero point is taken for convenience, it does not actually matter. The convergence holds uniformly and everywhere - the topological side of the theory doesn't have to worry too much about almost everywhere pathologies.

This proof can be adapted to show any sequence $(\alpha_n)_n\subseteq[0,1)$ is equidistributed iff. $$\frac{1}{n}\sum_{j=0}^{n-1}f(\alpha_j)=\int_0^1f$$For all bounded Riemann integrable $f:[0,1]\to\Bbb C$, or equivalently just for every continuous function. The long preamble was to explain why we knew this held for $\alpha_n:=n\alpha\bmod1$, for irrational $\alpha$.

The details can all be found in this book, with a focus to chapter $10$.