What can be said about the distribution of the set {p_n x}, where {x} denotes the fractional part, and p_n is the nth prime number? Can it be said that:

(a) For any $y$ in [0,1], for any positive $\epsilon$, and for any irrational $x$, there is an $n$ such that $|\{p_{n}x\}-y|<\epsilon$?
(b) The set running through all natural numbers $n$ of $\{p_{n}x\}$ is equidistributed in [0,1]?

What I wanted to add earlier but didn't have time to: I've been able to prove,

and, the subsequent generalization:

for a monotonically increasing sequence of natural numbers $r_n$, such that $\sum 1/r_n$ diverges, then for almost all $x$, if $\sum _{n=1}^N\{r_{n}x{\}}^m)/N$ tends to a limit, it tends to $1/(m+1)$ for some natural number $m$.

Obviously, $p_n$ satisfies the requirements for $r_n$. What gets me is proving that $\sum _{n=1}^N\{r_{n}x{\}}^m)/N$ tends to a limit. I don't see how to do this.
Brad

In the link , I gave both the result and the proof of the result:

for almost all $x$. We make a generalization of it. Write

$f_{m,N}(x)=\sum _{n=1}^N(\{r_{n}x{\}}^m-\frac{1}{m+1})/r_n$,

for any integer $m$. We wish to prove:

Theorem 1: $f_{m,N}(x)=O(1)$ for almost all $x$ (for any integer $m$).

We prove it by induction. Note that if we define

${\displaystyle C_n(x)=\{\begin{array}{c}\sum _k\sin (2\pi kx)/k^n\text{if}n\text{odd}\\ \sum _k\cos (2\pi kx)/k^n\text{if}n\text{even}\end{array},}$

then for a given $n$, there are constants $a_k$ such that

$a_n\{x{\}}^n+a_{n-1}\{x{\}}^{n-1}+\dots +a_1\{x\}+a_0=C_n(x)$   (*) [ $a_k$ will not remain the same from $C_n$ to $C_{n+1}$. In other words, $a_0$ might be something different when $n=1$ than when $n=2$] This is easily verified by integration of the Fourier series given in the link (and thus this holds only for non-integer $x$). For ease later on, denote that polynomial in $\{x\}$ that equals $C_n(x)$ as $P_n(x)$. Consider for odd $m$,

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{g_N(x)}& =& \sum _{n=1}^N{P}_m(r_{n}x)/r_n\\ \multicolumn{1}{c}{}& =& \sum _{k=1}^{\infty }\sum _{n=1}^N\sin (2\pi r_n\mathrm{kx})/(r_n{k}^m)\\ \multicolumn{1}{c}{}& =& \sum _{h=1}^{\infty }{\varphi }_N(h)\sin (2\pi h\backslash x),\end{array}}$

where ${\varphi }_N(h)=\sum _{b_{n}k=h;n\le N}1/(r_n{k}^m)\le {\varphi }_N(h)/h$, for all irrational $x$. Upon integrating (keeping in mind that the fourier series representation holds for a set of measure 1), we have by parseval

${\int }_0^1{g}_N^2(x)\mathrm{dx}=\sum _{h=1}^{\infty }{\varphi }_N<\mathrm{quote}/{>}^2(h)\le \sum _{h=1}^{\infty }{\varphi }_n^2(h)/h^2=O(1)$.

Thus,

Lemma 1: $g_N(x)$ is bounded for almost all $x$ (for all $m$)

We will use this later. Integrating (*) from 0 to 1, we have

$a_m/(m+1)+a_{m-1}/m+...+a_1/2+a_0=0$,

which implies that

$P_m(x)=a_n(\{x{\}}^n-\frac{1}{n+1})+a_{n-1}(\{x{\}}^{n-1}-\frac{1}{n})+\dots +a_1(x-\frac{1}{2})$.

Assume that Theorem 1 is proven for all $j<m$. Then by the Lemma

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{O(1)}& =& \sum _{n=1}^N{P}_m(r_{n}x)/r_n\\ \multicolumn{1}{c}{}& =& a_m\sum _{n=1}^N(\{r_{n}x{\}}^m-\frac{1}{m+1})/r_n+a_{m-1}\sum _{n=1}^N(\{r_{n}x{\}}^{m-1}-\frac{1}{m})/r_n+\dots +a_1\sum _{n=1}^N(\{r_{n}x\}-\frac{1}{2})/r_n\\ \multicolumn{1}{c}{}& =& (a_m\sum _{n=1}^N(\{r_{n}x{\}}^m-\frac{1}{m+1})/r_n)+O(1)\end{array}}$

by inductive hypothesis for almost all $x$ (as the measure of the intersection of a finite number of sets of measure 1 is still 1). As we know theorem 1 is true for $m=1$, it is true for all integer $m$ by induction.

The obvious corollary of Theorem 1 is

Corollary 1: $\sum _{n=1}^{N}p(\{r_{n}x\})-{\int }_0^{1}p(t)\mathrm{dt}=O(1)$ for almost all $x$, where $p$ is any finite polynomial.

We need only Theorem 1 to prove the first result I gave above. Suppose that $x$ is one of älmost all $x$" that satisfy the theorem and that $(\sum _{n=1}^N\{r_{n}x{\}}^m)/N$ converges to $x$. We then have

$\Lambda (N)=\sum _{n=1}^N\{r_{n}x{\}}^m=\mathrm{xN}+o(N)$     (1),

and

$\sum _{n=1}^N\{r_{n}x{\}}^m/r_n=O(1)+\frac{1}{m+1}\sum _{n=1}^{N}1/r_n$ .    (2)

Using the Abel summation formula on (2) and then substituting (1) in, we are given

$O(1)+\frac{1}{m+1}\sum _{n=1}^{N}1/r_n=\sum _{n=1}^{N-1}\Lambda (n)(1/r_n-1/r_{n+1})+O(1)=\sum _{n=1}^N\mathrm{xN}.(1/r_n-1/r_{n+1})+o(\sum _{n=1}^{N}N.(1/r_n-1/r_{n+1}))$,

or $\frac{1}{m+1}\sum _{n=1}^{N}1/r_n~\sum _{n=1}^N\mathrm{xN}.(1/r_n-1/r_{n+1})=\sum _{n=1}^{N}x/r_n+O(1)$,

upon performing the Abel summation formula again. Immediately we are given $x=1/(m+1)$. (Here we have assumed that $\sum 1/r_n$ is divergent throughout)
I'm in a rush again right now, this time to a dentist's appointment. I'll get around to typing up the proof of my second result later tonight - it's just a simple exercise with corollary 1.

Brad

Here's the rest. First off, I've made a typo with the corollary; it should read:

We use this to prove the second result I gave before Michael's post. Suppose we are given an $\alpha$. Construct a polynomial such that ${\int }_0^{1}p(x)\mathrm{dx}=0$, and such that $p(z)<-\mu$ for $|z-\alpha |>=\epsilon$ for some $\epsilon$. (All $\alpha$, $\mu$, and $\epsilon$ are positive, and $\alpha$ is in (0,1).) [You may check that a quadratic that satisfies these conditions may be constructed] Then, assume that there are not an infinitude of solutions to $|\{r_{n}x\}-\alpha |<\epsilon$ for some given $x$ in the set of almost all $x$ in the corollary. We then have

$O(1)=\sum _{n=1}^{N}p(\{r_{n}x\})/r_n<O(1)+A(N)$,

where $A(N)<-\mu \sum _{n=1}^{N}p(\{r_{n}x\})/r_n$ and thus diverges when $\sum 1/r_n$ diverges. This is a contradiction for all $r_n$ such that $\sum 1/r_n$ does indeed diverge, so we have:

Theorem 2: For a monotonically increasing sequence of integers $r_n$ such that $\sum 1/r_n$ diverges, the sequence $\{r_{n}x\}$ is dense in [0,1) for almost all $x$.

$p_n$ is such a sequence.

Brad

Thank you Brad; I'm about half way through reading it, though it may be a while before I can comment intelligibly!