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3

$begingroup$

I am trying to prove a theorem using an optimal solution. Given $(y_n)_n in mathbbN^*$ such that $exists delta > 0, forall n in mathbbN^* y_n+1 - y_n geq delta$ then for almost every $xi in mathbbR$ (Lebesgue) the sequence $(xi y_n)_n in mathbbN^*$ is equidistributed modulo $1$.

I found a proof considering an arbitrary interval $[a,b]$ but some people told me it's sufficient to prove the result for almost every $xi in [0,1]$ but I am not convinced. In fact, if we prove the result just for almost every $xi in [0,1]$ the sequence $xi y_n$ does not cover all the interval $xi in [0,1]$.

I am trying to find a counterexample, if we have a sequence $ frac116 leq y_n leq frac12$ then multiplying by $xi in [0,1]$ gives us $ 0 leq xi y_n leq frac12$ and if we take $xi in [1,2]$ we got $ 0 leq xi y_n leq 1$. Do you see what I mean? I think that the argument reducing to $[0,1]$ is not always true.

Thanks in advance !

real-analysis proof-verification equidistribution

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edited Mar 29 at 10:55

gt6989b

35.5k22557

asked Mar 29 at 10:54

bsmbsm

261

$endgroup$

add a comment | 

3

$begingroup$

I am trying to prove a theorem using an optimal solution. Given $(y_n)_n in mathbbN^*$ such that $exists delta > 0, forall n in mathbbN^* y_n+1 - y_n geq delta$ then for almost every $xi in mathbbR$ (Lebesgue) the sequence $(xi y_n)_n in mathbbN^*$ is equidistributed modulo $1$.

I found a proof considering an arbitrary interval $[a,b]$ but some people told me it's sufficient to prove the result for almost every $xi in [0,1]$ but I am not convinced. In fact, if we prove the result just for almost every $xi in [0,1]$ the sequence $xi y_n$ does not cover all the interval $xi in [0,1]$.

I am trying to find a counterexample, if we have a sequence $ frac116 leq y_n leq frac12$ then multiplying by $xi in [0,1]$ gives us $ 0 leq xi y_n leq frac12$ and if we take $xi in [1,2]$ we got $ 0 leq xi y_n leq 1$. Do you see what I mean? I think that the argument reducing to $[0,1]$ is not always true.

Thanks in advance !

real-analysis proof-verification equidistribution

share|cite|improve this question

edited Mar 29 at 10:55

gt6989b

35.5k22557

asked Mar 29 at 10:54

bsmbsm

261

$endgroup$

add a comment | 

$begingroup$

I am trying to prove a theorem using an optimal solution. Given $(y_n)_n in mathbbN^*$ such that $exists delta > 0, forall n in mathbbN^* y_n+1 - y_n geq delta$ then for almost every $xi in mathbbR$ (Lebesgue) the sequence $(xi y_n)_n in mathbbN^*$ is equidistributed modulo $1$.

I found a proof considering an arbitrary interval $[a,b]$ but some people told me it's sufficient to prove the result for almost every $xi in [0,1]$ but I am not convinced. In fact, if we prove the result just for almost every $xi in [0,1]$ the sequence $xi y_n$ does not cover all the interval $xi in [0,1]$.

I am trying to find a counterexample, if we have a sequence $ frac116 leq y_n leq frac12$ then multiplying by $xi in [0,1]$ gives us $ 0 leq xi y_n leq frac12$ and if we take $xi in [1,2]$ we got $ 0 leq xi y_n leq 1$. Do you see what I mean? I think that the argument reducing to $[0,1]$ is not always true.

Thanks in advance !

real-analysis proof-verification equidistribution

share|cite|improve this question

edited Mar 29 at 10:55

gt6989b

35.5k22557

asked Mar 29 at 10:54

bsmbsm

261

$endgroup$

share|cite|improve this question

edited Mar 29 at 10:55

gt6989b

35.5k22557

asked Mar 29 at 10:54

bsmbsm

261

share|cite|improve this question

edited Mar 29 at 10:55

gt6989b

35.5k22557

asked Mar 29 at 10:54

bsmbsm

261

edited Mar 29 at 10:55

gt6989b

35.5k22557

edited Mar 29 at 10:55

gt6989b

35.5k22557

asked Mar 29 at 10:54

bsmbsm

261

asked Mar 29 at 10:54

bsmbsm

261