Mixing
A variant of Kronecker's approximation theorem?
$begingroup$
Let $tau,sigmain(0,infty)$ with $frac{tau}{sigma}notinmathbb Q$. By Kronecker's approximation theorem, we know:
(1) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+ntau-msigma|<epsilon$.
In other words, if you keep adding $tau$ to $x$, you will eventually come arbitratily close to the set $sigmamathbb N$. But what happens if you keep adding values that are just approximately $tau$?
To make this a precise question, let $(tau_n)_{ninmathbb N_0}subset (0,infty)$ with
$$ tau_{n+1}-tau_n xrightarrow{ntoinfty}tau.$$
At first glance, the following seems plausible:
(2) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+tau_n-msigma|<epsilon$.
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!
real-analysis sequences-and-series approximation real-numbers diophantine-approximation
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
$endgroup$
add a comment |
$begingroup$
Let $tau,sigmain(0,infty)$ with $frac{tau}{sigma}notinmathbb Q$. By Kronecker's approximation theorem, we know:
(1) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+ntau-msigma|<epsilon$.
In other words, if you keep adding $tau$ to $x$, you will eventually come arbitratily close to the set $sigmamathbb N$. But what happens if you keep adding values that are just approximately $tau$?
To make this a precise question, let $(tau_n)_{ninmathbb N_0}subset (0,infty)$ with
$$ tau_{n+1}-tau_n xrightarrow{ntoinfty}tau.$$
At first glance, the following seems plausible:
(2) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+tau_n-msigma|<epsilon$.
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!
real-analysis sequences-and-series approximation real-numbers diophantine-approximation
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
$endgroup$
add a comment |
$begingroup$
Let $tau,sigmain(0,infty)$ with $frac{tau}{sigma}notinmathbb Q$. By Kronecker's approximation theorem, we know:
(1) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+ntau-msigma|<epsilon$.
In other words, if you keep adding $tau$ to $x$, you will eventually come arbitratily close to the set $sigmamathbb N$. But what happens if you keep adding values that are just approximately $tau$?
To make this a precise question, let $(tau_n)_{ninmathbb N_0}subset (0,infty)$ with
$$ tau_{n+1}-tau_n xrightarrow{ntoinfty}tau.$$
At first glance, the following seems plausible:
(2) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+tau_n-msigma|<epsilon$.
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!
real-analysis sequences-and-series approximation real-numbers diophantine-approximation
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
$endgroup$
Let $tau,sigmain(0,infty)$ with $frac{tau}{sigma}notinmathbb Q$. By Kronecker's approximation theorem, we know:
(1) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+ntau-msigma|<epsilon$.
In other words, if you keep adding $tau$ to $x$, you will eventually come arbitratily close to the set $sigmamathbb N$. But what happens if you keep adding values that are just approximately $tau$?
To make this a precise question, let $(tau_n)_{ninmathbb N_0}subset (0,infty)$ with
$$ tau_{n+1}-tau_n xrightarrow{ntoinfty}tau.$$
At first glance, the following seems plausible:
(2) For each $xin mathbb R$ and $epsilon>0$, there are $m,ninmathbb N$ such that $|x+tau_n-msigma|<epsilon$.
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!
real-analysis sequences-and-series approximation real-numbers diophantine-approximation
real-analysis sequences-and-series approximation real-numbers diophantine-approximation
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
edited Feb 6 at 17:33
Mars Plastic
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
asked Feb 3 at 12:59
Mars PlasticMars Plastic
1,465122
1,465122
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $|x+ntau-msigma| lt epsilon$.
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 $n,min {mathbb N}$, let
$$A_{n,m}= bigglbrace Xin {mathbb R} bigg| |X+ntau-m| ltepsilonbiggrbrace.tag{3}$$
Then, Kronecker's usual theorem says that whenever $tau$ is irrational, there are nonnegative integers $n(x),m(x)$ with $xin A_{n(x),m(x)}$.
Then $bigcup_{xin [0,1]} A_{n(x),m(x)}$ is an open covering of $[0,1]$. Since $[0,1]$ is compact, there is a finite subset $Isubseteq [0,1]$ such that $bigcup_{xin I} A_{n(x),m(x)}$ 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 $xin [0,1]$, we can find $n,m$ with $0 leq n,m leq M$ such that $$(*) : quad |x+ntau-m| leq epsilon.$$
Now, if $xgeq 1$, 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
$$
{mathbb R}^+ subseteq bigcup_{0 leq n leq M, mgeq 0}
A_{n,m}. tag{4'}
$$
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 $kgeq k_0$.
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.
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x+tau_0ge 0$ and
$$ |(tau_{k}-tau_{k-1})-tau|<frac{epsilon}{2M(epsilon/2)} quad text{for all $kinmathbb N$.}$$
Hence for all $nin{0,ldots,M(epsilon/2)}$ and $minmathbb N$ we have
begin{align*}
|x+tau_n-msigma|&=left|x+tau_0+sum_{k=1}^{n}((tau_{k}-tau_{k-1})-tau)+ntau-msigmaright| \
&le sum_{k=1}^{M(epsilon/2)}|(tau_{k}-tau_{k-1})-tau| + |x+tau_0+ntau-msigma| \
&le frac{epsilon}{2} + |x+tau_0+ntau-msigma|.
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).
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
$endgroup$
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
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
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 $|x+ntau-msigma| lt epsilon$.
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 $n,min {mathbb N}$, let
$$A_{n,m}= bigglbrace Xin {mathbb R} bigg| |X+ntau-m| ltepsilonbiggrbrace.tag{3}$$
Then, Kronecker's usual theorem says that whenever $tau$ is irrational, there are nonnegative integers $n(x),m(x)$ with $xin A_{n(x),m(x)}$.
Then $bigcup_{xin [0,1]} A_{n(x),m(x)}$ is an open covering of $[0,1]$. Since $[0,1]$ is compact, there is a finite subset $Isubseteq [0,1]$ such that $bigcup_{xin I} A_{n(x),m(x)}$ 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 $xin [0,1]$, we can find $n,m$ with $0 leq n,m leq M$ such that $$(*) : quad |x+ntau-m| leq epsilon.$$
Now, if $xgeq 1$, 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
$$
{mathbb R}^+ subseteq bigcup_{0 leq n leq M, mgeq 0}
A_{n,m}. tag{4'}
$$
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 $kgeq k_0$.
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.
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $|x+ntau-msigma| lt epsilon$.
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 $n,min {mathbb N}$, let
$$A_{n,m}= bigglbrace Xin {mathbb R} bigg| |X+ntau-m| ltepsilonbiggrbrace.tag{3}$$
Then, Kronecker's usual theorem says that whenever $tau$ is irrational, there are nonnegative integers $n(x),m(x)$ with $xin A_{n(x),m(x)}$.
Then $bigcup_{xin [0,1]} A_{n(x),m(x)}$ is an open covering of $[0,1]$. Since $[0,1]$ is compact, there is a finite subset $Isubseteq [0,1]$ such that $bigcup_{xin I} A_{n(x),m(x)}$ 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 $xin [0,1]$, we can find $n,m$ with $0 leq n,m leq M$ such that $$(*) : quad |x+ntau-m| leq epsilon.$$
Now, if $xgeq 1$, 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
$$
{mathbb R}^+ subseteq bigcup_{0 leq n leq M, mgeq 0}
A_{n,m}. tag{4'}
$$
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 $kgeq k_0$.
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.
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $|x+ntau-msigma| lt epsilon$.
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 $n,min {mathbb N}$, let
$$A_{n,m}= bigglbrace Xin {mathbb R} bigg| |X+ntau-m| ltepsilonbiggrbrace.tag{3}$$
Then, Kronecker's usual theorem says that whenever $tau$ is irrational, there are nonnegative integers $n(x),m(x)$ with $xin A_{n(x),m(x)}$.
Then $bigcup_{xin [0,1]} A_{n(x),m(x)}$ is an open covering of $[0,1]$. Since $[0,1]$ is compact, there is a finite subset $Isubseteq [0,1]$ such that $bigcup_{xin I} A_{n(x),m(x)}$ 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 $xin [0,1]$, we can find $n,m$ with $0 leq n,m leq M$ such that $$(*) : quad |x+ntau-m| leq epsilon.$$
Now, if $xgeq 1$, 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
$$
{mathbb R}^+ subseteq bigcup_{0 leq n leq M, mgeq 0}
A_{n,m}. tag{4'}
$$
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 $kgeq k_0$.
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.
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
$endgroup$
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 $|x+ntau-msigma| lt epsilon$.
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 $n,min {mathbb N}$, let
$$A_{n,m}= bigglbrace Xin {mathbb R} bigg| |X+ntau-m| ltepsilonbiggrbrace.tag{3}$$
Then, Kronecker's usual theorem says that whenever $tau$ is irrational, there are nonnegative integers $n(x),m(x)$ with $xin A_{n(x),m(x)}$.
Then $bigcup_{xin [0,1]} A_{n(x),m(x)}$ is an open covering of $[0,1]$. Since $[0,1]$ is compact, there is a finite subset $Isubseteq [0,1]$ such that $bigcup_{xin I} A_{n(x),m(x)}$ 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 $xin [0,1]$, we can find $n,m$ with $0 leq n,m leq M$ such that $$(*) : quad |x+ntau-m| leq epsilon.$$
Now, if $xgeq 1$, 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
$$
{mathbb R}^+ subseteq bigcup_{0 leq n leq M, mgeq 0}
A_{n,m}. tag{4'}
$$
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 $kgeq k_0$.
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.
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
edited Feb 9 at 11:24
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
answered Feb 5 at 18:23
Ewan DelanoyEwan Delanoy
42.1k443104
42.1k443104
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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
add a comment |
$begingroup$
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 $x+tau_0ge 0$ and
$$ |(tau_{k}-tau_{k-1})-tau|<frac{epsilon}{2M(epsilon/2)} quad text{for all $kinmathbb N$.}$$
Hence for all $nin{0,ldots,M(epsilon/2)}$ and $minmathbb N$ we have
begin{align*}
|x+tau_n-msigma|&=left|x+tau_0+sum_{k=1}^{n}((tau_{k}-tau_{k-1})-tau)+ntau-msigmaright| \
&le sum_{k=1}^{M(epsilon/2)}|(tau_{k}-tau_{k-1})-tau| + |x+tau_0+ntau-msigma| \
&le frac{epsilon}{2} + |x+tau_0+ntau-msigma|.
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).
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
$endgroup$
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
add a comment |
$begingroup$
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 $x+tau_0ge 0$ and
$$ |(tau_{k}-tau_{k-1})-tau|<frac{epsilon}{2M(epsilon/2)} quad text{for all $kinmathbb N$.}$$
Hence for all $nin{0,ldots,M(epsilon/2)}$ and $minmathbb N$ we have
begin{align*}
|x+tau_n-msigma|&=left|x+tau_0+sum_{k=1}^{n}((tau_{k}-tau_{k-1})-tau)+ntau-msigmaright| \
&le sum_{k=1}^{M(epsilon/2)}|(tau_{k}-tau_{k-1})-tau| + |x+tau_0+ntau-msigma| \
&le frac{epsilon}{2} + |x+tau_0+ntau-msigma|.
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).
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
$endgroup$
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
add a comment |
$begingroup$
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 $x+tau_0ge 0$ and
$$ |(tau_{k}-tau_{k-1})-tau|<frac{epsilon}{2M(epsilon/2)} quad text{for all $kinmathbb N$.}$$
Hence for all $nin{0,ldots,M(epsilon/2)}$ and $minmathbb N$ we have
begin{align*}
|x+tau_n-msigma|&=left|x+tau_0+sum_{k=1}^{n}((tau_{k}-tau_{k-1})-tau)+ntau-msigmaright| \
&le sum_{k=1}^{M(epsilon/2)}|(tau_{k}-tau_{k-1})-tau| + |x+tau_0+ntau-msigma| \
&le frac{epsilon}{2} + |x+tau_0+ntau-msigma|.
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).
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
$endgroup$
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 $x+tau_0ge 0$ and
$$ |(tau_{k}-tau_{k-1})-tau|<frac{epsilon}{2M(epsilon/2)} quad text{for all $kinmathbb N$.}$$
Hence for all $nin{0,ldots,M(epsilon/2)}$ and $minmathbb N$ we have
begin{align*}
|x+tau_n-msigma|&=left|x+tau_0+sum_{k=1}^{n}((tau_{k}-tau_{k-1})-tau)+ntau-msigmaright| \
&le sum_{k=1}^{M(epsilon/2)}|(tau_{k}-tau_{k-1})-tau| + |x+tau_0+ntau-msigma| \
&le frac{epsilon}{2} + |x+tau_0+ntau-msigma|.
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).
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
edited Feb 7 at 17:54
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
answered Feb 6 at 16:48
Mars PlasticMars Plastic
1,465122
1,465122
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
add a comment |
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
1
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
$begingroup$
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
add a comment |
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