# A variant of Kronecker's approximation theorem?

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.

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.

• 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. – Mars Plastic Feb 6 at 16:50
• By the way: Is your Lemma a well-known variant of Kronecker's Theorem? Can you give me a reference? – Mars Plastic Feb 6 at 16:57
• I do not know a reference for this lemma, but there probably is. Someone else will perhaps be able to come with one – Ewan Delanoy Feb 6 at 17:28

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).

• 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 – Ewan Delanoy Feb 6 at 17:34