5
$\begingroup$

Let $\tau,\sigma\in(0,\infty)$ with $\frac{\tau}{\sigma}\notin\mathbb Q$. By Kronecker's approximation theorem, we know:

(1) For each $x\in \mathbb R$ and $\epsilon>0$, there are $m,n\in\mathbb N$ such that $|x+n\tau-m\sigma|<\epsilon$.

In other words, if you keep adding $\tau$ to $x$, you will eventually come arbitratily close to the set $\sigma\mathbb N$. But what happens if you keep adding values that are just approximately $\tau$?

To make this a precise question, let $(\tau_n)_{n\in\mathbb N_0}\subset (0,\infty)$ with

$$ \tau_{n+1}-\tau_n \xrightarrow{n\to\infty}\tau.$$

Then, the according conjecture is:

(2) For each $x\in \mathbb R$ and $\epsilon>0$, there are $m,n\in\mathbb N$ such that $|x+\tau_n-m\sigma|<\epsilon$.

If one assumes that

$$ \sum_{n=0}^\infty \left((\tau_{n+1}-\tau_n)-\tau \right) \text{ converges in $\mathbb R$,}$$

it is indeed relatively easy to deduce (2) from (1).

QUESTION: If $\sum_{n=0}^\infty \left((\tau_{n+1}-\tau_n)-\tau \right)$ diverges, does (2) still hold?

My ad hoc ideas didn't quite work out and before I start to think deeper about it, I thought I might ask if anyone here knows of any result in this direction.

Thanks a lot in advance!

$\endgroup$
4
+50
$\begingroup$

Your claim is true, and here is why.

Summary of proof. A compactness argument allows one to use a strengthened version of Kronecker's theorem that is "more uniform" in $x$ namely :

Main lemma. There is a constant $M$ (depending only on $\sigma,\tau$ and $\epsilon$ and not on $x$) such that for any $x \geq 0$, there are integers $(n,m)\in[0,M]\times {\mathbb N}$ with $|x+n\tau-m\sigma| \lt \epsilon$.

Detailed proof. Replacing $(x,\tau,\sigma,\epsilon)$ with $(\frac{x}{\sigma},\frac{\tau}{\sigma},1,\frac{\epsilon}{\sigma})$, we may assume without loss that $\sigma=1$.

For $n,m\in {\mathbb N}$, let

$$A_{n,m}= \bigg\lbrace X\in {\mathbb R} \bigg| |X+n\tau-m| \lt\epsilon\bigg\rbrace.\tag{3}$$

Then, Kronecker's usual theorem says that whenever $\tau$ is irrational, there are nonnegative integers $n(x),m(x)$ with $x\in A_{n(x),m(x)}$.

Then $\bigcup_{x\in [0,1]} A_{n(x),m(x)}$ is an open covering of $[0,1]$. Since $[0,1]$ is compact, there is a finite subset $I\subseteq [0,1]$ such that $\bigcup_{x\in I} A_{n(x),m(x)}$ is still a covering of $[0,1]$. Denote by $M$ the maximum value of $n(x)$ or $m(x)$ when $x$ varies in the finite set $I$. We have then that

$$ [0,1] \subseteq \bigcup_{0 \leq n,m \leq M} A_{n,m}. \tag{4} $$

(4) means that for any $x\in [0,1]$, we can find $n,m$ with $0 \leq n,m \leq M$ such that $$(*) : \quad |x+n\tau-m| \leq \epsilon.$$ Now, if $x\geq 1$, and we put $x'=x-\lfloor x \rfloor$ (the fractional part of $x$), then $x'\in [0,1]$ so that $|x'+n'\tau-m'| \leq \epsilon$ for some $(n',m')=(n(x'),m(x'))$. But then ($*$) holds also for $(n',m'+\lfloor x \rfloor)$ in place of $(n,m)$. We deduce that

$$ {\mathbb R}^+ \subseteq \bigcup_{0 \leq n \leq M, m\geq 0} A_{n,m}. \tag{4'} $$

This concludes the proof of the main lemma. Let us now prove (2). Using $\frac{\epsilon}{2}$ instead of $\epsilon$ in the main lemma, there is a $M>0$ such that for any $y \geq 0$, there are integers $(n(y),m(y))\in[0,M]\times {\mathbb N}$ with

$$|y+n(y)\tau-m(y)| \lt \frac{\epsilon}{2}.\tag{5}$$

Let $\delta >0$ be a positive constant whose value is to be decided later. By hypothesis, there is a $k_0$ such that $x+\tau_1+\sum_{k=1}^{k_0-1}\tau_{k+1}-\tau_k \geq 0$ and $|\tau_{k+1}-\tau_k-\tau| \leq \delta$ for any $k\geq k_0$.

Let $y=x+\tau_1+\sum_{k=1}^{k_0-1}\tau_{k+1}-\tau_k=x+\tau_{k_0}$ ; we know that $y$ is nonnegative. By (5),

$$\bigg|x+\tau_{k_0}+n(y)\tau-m(y)\bigg| \lt \frac{\epsilon}{2}.\tag{6}$$

On the other hand, we have

$$ \bigg| \sum_{k=k_0}^{k_0+n(y)-1} \tau_{k+1}-\tau_k-\tau \bigg| \leq n(y)\delta \leq \delta M. \tag{7}$$

Adding (6) and (7) and using the triangle inequality, we obtain

$$ \bigg|x+\tau_{k_0+n(y)}-m(y)\bigg|=\bigg|x+\sum_{k=1}^{k_0+n(y)-1}(\tau_{k+1}-\tau_k)-m(y)\bigg| \lt \frac{\epsilon}{2}+\delta M. $$

Taking $\delta=\frac{\epsilon}{2M}$, we are done.

$\endgroup$
  • $\begingroup$ Thanks a lot for this immensely helpful answer! Please see my own answer below (which was too long for the comment section) for some additional discussion. $\endgroup$ – Mars Plastic Feb 6 at 16:50
  • $\begingroup$ By the way: Is your Lemma a well-known variant of Kronecker's Theorem? Can you give me a reference? $\endgroup$ – Mars Plastic Feb 6 at 16:57
  • 1
    $\begingroup$ I do not know a reference for this lemma, but there probably is. Someone else will perhaps be able to come with one $\endgroup$ – Ewan Delanoy Feb 6 at 17:28
1
$\begingroup$

Thank you so very, very much, Ewan Delanoy! Your Lemma is exactly what I needed. Let me say that I'm only posting this as an answer, because it is too long for a comment.

Your Lemma basically says that the number $n$ in (1) can in fact always be taken from the set $\{0,\ldots,M=M(\epsilon)\}$. This is indeed all I need to prove (2) (since this allows to argue in the same manner, as if the series were convergent). Your proof of (2) is a bit off though, as you seem to demand that $\tau_k$ converges to $\tau$, which is not what I assume. For the sake of completeness and clarity, let me rework that part as follows:

Let $\epsilon>0$. Without loss of generality we can assume that $x+\tau_0\ge 0$ and $$ |(\tau_{k}-\tau_{k-1})-\tau|<\frac{\epsilon}{2M(\epsilon/2)} \quad \text{for all $k\in\mathbb N$.}$$ Hence for all $n\in\{0,\ldots,M(\epsilon/2)\}$ and $m\in\mathbb N$ we have \begin{align*} |x+\tau_n-m\sigma|&=\left|x+\tau_0+\sum_{k=1}^{n}((\tau_{k}-\tau_{k-1})-\tau)+n\tau-m\sigma\right| \\ &\le \sum_{k=1}^{M(\epsilon/2)}|(\tau_{k}-\tau_{k-1})-\tau| + |x+\tau_0+n\tau-m\sigma| \\ &\le \frac{\epsilon}{2} + |x+\tau_0+n\tau-m\sigma|. \end{align*} Now, thanks to your Lemma, we can choose $n$ and $m$ such that the second summand is also smaller than $\frac{\epsilon}{2}$.

Once again thank you very much for providing me with this essential ingredient (and its neat proof).

$\endgroup$
  • 1
    $\begingroup$ In fact, I misread the OP and what I called $\tau_k$ in my old version was really $\tau_{k+1}-\tau_k$. It's corrected now $\endgroup$ – Ewan Delanoy Feb 6 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.