Why couldn't Baez-Duarte prove the Riemann Hypothesis?
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Define
begin{equation} I_n=int_{0}^{1/n} |U s_{n}(x)|^2 mathrm{d}x
end{equation} where $Us_{n}(x)=frac{1}{x}sum_{j=1}^{n} frac{mu(j)}{j}rho(jx), mu$ denotes the Mobius function and $rho(y)$ is the fractional part of $y$. We make three crucial observations: Firstly, since $0leq rho(jx)leq jx$ for every $jgeq 1, xgeq 0$, note that $lim_{xrightarrow 0^+} Big(frac{rho(jx)}{jx}Big)<infty$, thus the integrand of $I_n$ is well-defined for all $xgeq 0$. Secondly, the integral $I_n$ is defined over a finite range $(0, 1/n)$. Thirdly, by $2.14$ of Baez-Duarte we have $lim _{nrightarrow infty} Us_{n}(x)=-frac{sin 2pi x}{pi x}$ hence $|Us_{n}(x)|<c/x$ for all $n$, where $c$ is some positive constant. Notice that these observations collectively imply that $I_n leq C$ for every positive integer $n$ where $C$ is some positive constant, or equivalently,
begin{equation}
Big(sum_{j=1}^n mu(j)Big)^2 = O(n),
end{equation} by identity 2.12 of Baez-Duarte that
begin{equation}
I_n = frac{1}{n}Big(sum_{j=1}^n mu(j)Big)^{2}.
end{equation}Since it is known that the RH is equivalent to the statement that $Big(sum_{j=1}^n mu(j)Big)^2 = O(n^{1+epsilon})$ for any $epsilon>0$, couldn't Baez-Duarte conclude this way that the RH is true ?
number-theory analytic-number-theory riemann-zeta
edited Dec 12 '18 at 10:00
Klangen
1,76211334
$endgroup$
|
show 11 more comments
$begingroup$
Define
begin{equation} I_n=int_{0}^{1/n} |U s_{n}(x)|^2 mathrm{d}x
end{equation} where $Us_{n}(x)=frac{1}{x}sum_{j=1}^{n} frac{mu(j)}{j}rho(jx), mu$ denotes the Mobius function and $rho(y)$ is the fractional part of $y$. We make three crucial observations: Firstly, since $0leq rho(jx)leq jx$ for every $jgeq 1, xgeq 0$, note that $lim_{xrightarrow 0^+} Big(frac{rho(jx)}{jx}Big)<infty$, thus the integrand of $I_n$ is well-defined for all $xgeq 0$. Secondly, the integral $I_n$ is defined over a finite range $(0, 1/n)$. Thirdly, by $2.14$ of Baez-Duarte we have $lim _{nrightarrow infty} Us_{n}(x)=-frac{sin 2pi x}{pi x}$ hence $|Us_{n}(x)|<c/x$ for all $n$, where $c$ is some positive constant. Notice that these observations collectively imply that $I_n leq C$ for every positive integer $n$ where $C$ is some positive constant, or equivalently,
begin{equation}
Big(sum_{j=1}^n mu(j)Big)^2 = O(n),
end{equation} by identity 2.12 of Baez-Duarte that
begin{equation}
I_n = frac{1}{n}Big(sum_{j=1}^n mu(j)Big)^{2}.
end{equation}Since it is known that the RH is equivalent to the statement that $Big(sum_{j=1}^n mu(j)Big)^2 = O(n^{1+epsilon})$ for any $epsilon>0$, couldn't Baez-Duarte conclude this way that the RH is true ?
number-theory analytic-number-theory riemann-zeta
edited Dec 12 '18 at 10:00
Klangen
1,76211334
$endgroup$
3
$begingroup$
Because you are mixing things you didn't look at carefully. For $x in (0,1/n), m le n$ then $rho(mx) = mx$. So what ? The complicated part of $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ is for $x in (1/n,1)$. Everything is explained in the paper and in several places on books, MSE and the web.
$endgroup$
– reuns
Dec 6 '18 at 13:14
2
$begingroup$
$int_{0}^{1/n} |Us_{n}(x)|^2 mathrm{d}x=OBigg(int_{0}^{1/n} frac{sin^2(2pi x)}{pi ^2 x^2} mathrm{d}x Bigg)$ doesn't make any sense. You are confusing $L^2$ and pointwise convergence. All you know is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin (pi x)/pi$ in $L^2([0,1])$ and under the RH it converges in $L^2([0,1], x^{-1+epsilon})$
$endgroup$
– reuns
Dec 6 '18 at 13:32
2
$begingroup$
The paper is exactly about that, why not read it, as well as the many explications you'll find on the web.
$endgroup$
– reuns
Dec 6 '18 at 13:35
2
$begingroup$
$|Us_{n}(x)|^2 leq frac{Csin^2 (2pi x)}{x^2}$ is wrong. All you know (from the Fourier series of $rho$ and $sum_{d | n} mu(d) = 1_{n=1}$ and say $sum_n mu(n)/n=0$ equivalent to the PNT) is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin(pi x)/pi$ in $L^2((0,1])$.
$endgroup$
– reuns
Dec 6 '18 at 13:55
1
$begingroup$
You are claiming that in his paper on some criterion of the RH Baez is mentioning/proving that $frac{1}{N} (sum_{n=1}^N mu(n))^2 = O(1)$ without noticing it ? Understanding the exact mode of convergence of $sum_{n=1}^{N} frac{mu(n)}{n} rho(nx)$ is complicated. Showing it converges uniformly to $-sin(2pi x)/pi$ is academic.oup.com/qjmath/article-abstract/os-8/1/313/… The only thing that is +/- obvious is to relate "zeta has no zeros for $Re(s) > sigma$" with $sup_N int_0^1 |sum_{n=1}^N mu(n)frac{rho( nx)}{n}x^{sigma-3/2}|^2 dx < infty$
$endgroup$
– reuns
Dec 6 '18 at 20:02
|
show 11 more comments
$begingroup$
Define
begin{equation} I_n=int_{0}^{1/n} |U s_{n}(x)|^2 mathrm{d}x
end{equation} where $Us_{n}(x)=frac{1}{x}sum_{j=1}^{n} frac{mu(j)}{j}rho(jx), mu$ denotes the Mobius function and $rho(y)$ is the fractional part of $y$. We make three crucial observations: Firstly, since $0leq rho(jx)leq jx$ for every $jgeq 1, xgeq 0$, note that $lim_{xrightarrow 0^+} Big(frac{rho(jx)}{jx}Big)<infty$, thus the integrand of $I_n$ is well-defined for all $xgeq 0$. Secondly, the integral $I_n$ is defined over a finite range $(0, 1/n)$. Thirdly, by $2.14$ of Baez-Duarte we have $lim _{nrightarrow infty} Us_{n}(x)=-frac{sin 2pi x}{pi x}$ hence $|Us_{n}(x)|<c/x$ for all $n$, where $c$ is some positive constant. Notice that these observations collectively imply that $I_n leq C$ for every positive integer $n$ where $C$ is some positive constant, or equivalently,
begin{equation}
Big(sum_{j=1}^n mu(j)Big)^2 = O(n),
end{equation} by identity 2.12 of Baez-Duarte that
begin{equation}
I_n = frac{1}{n}Big(sum_{j=1}^n mu(j)Big)^{2}.
end{equation}Since it is known that the RH is equivalent to the statement that $Big(sum_{j=1}^n mu(j)Big)^2 = O(n^{1+epsilon})$ for any $epsilon>0$, couldn't Baez-Duarte conclude this way that the RH is true ?
number-theory analytic-number-theory riemann-zeta
edited Dec 12 '18 at 10:00
Klangen
1,76211334
$endgroup$
Define
begin{equation} I_n=int_{0}^{1/n} |U s_{n}(x)|^2 mathrm{d}x
end{equation} where $Us_{n}(x)=frac{1}{x}sum_{j=1}^{n} frac{mu(j)}{j}rho(jx), mu$ denotes the Mobius function and $rho(y)$ is the fractional part of $y$. We make three crucial observations: Firstly, since $0leq rho(jx)leq jx$ for every $jgeq 1, xgeq 0$, note that $lim_{xrightarrow 0^+} Big(frac{rho(jx)}{jx}Big)<infty$, thus the integrand of $I_n$ is well-defined for all $xgeq 0$. Secondly, the integral $I_n$ is defined over a finite range $(0, 1/n)$. Thirdly, by $2.14$ of Baez-Duarte we have $lim _{nrightarrow infty} Us_{n}(x)=-frac{sin 2pi x}{pi x}$ hence $|Us_{n}(x)|<c/x$ for all $n$, where $c$ is some positive constant. Notice that these observations collectively imply that $I_n leq C$ for every positive integer $n$ where $C$ is some positive constant, or equivalently,
begin{equation}
Big(sum_{j=1}^n mu(j)Big)^2 = O(n),
end{equation} by identity 2.12 of Baez-Duarte that
begin{equation}
I_n = frac{1}{n}Big(sum_{j=1}^n mu(j)Big)^{2}.
end{equation}Since it is known that the RH is equivalent to the statement that $Big(sum_{j=1}^n mu(j)Big)^2 = O(n^{1+epsilon})$ for any $epsilon>0$, couldn't Baez-Duarte conclude this way that the RH is true ?
number-theory analytic-number-theory riemann-zeta
number-theory analytic-number-theory riemann-zeta
edited Dec 12 '18 at 10:00
Klangen
1,76211334
edited Dec 12 '18 at 10:00
Klangen
1,76211334
edited Dec 12 '18 at 10:00
Klangen
1,76211334
edited Dec 12 '18 at 10:00
Klangen
1,76211334
edited Dec 12 '18 at 10:00
Klangen
1,76211334
1,76211334
asked Dec 6 '18 at 12:01
user507152
3
$begingroup$
Because you are mixing things you didn't look at carefully. For $x in (0,1/n), m le n$ then $rho(mx) = mx$. So what ? The complicated part of $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ is for $x in (1/n,1)$. Everything is explained in the paper and in several places on books, MSE and the web.
$endgroup$
– reuns
Dec 6 '18 at 13:14
2
$begingroup$
$int_{0}^{1/n} |Us_{n}(x)|^2 mathrm{d}x=OBigg(int_{0}^{1/n} frac{sin^2(2pi x)}{pi ^2 x^2} mathrm{d}x Bigg)$ doesn't make any sense. You are confusing $L^2$ and pointwise convergence. All you know is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin (pi x)/pi$ in $L^2([0,1])$ and under the RH it converges in $L^2([0,1], x^{-1+epsilon})$
$endgroup$
– reuns
Dec 6 '18 at 13:32
2
$begingroup$
The paper is exactly about that, why not read it, as well as the many explications you'll find on the web.
$endgroup$
– reuns
Dec 6 '18 at 13:35
2
$begingroup$
$|Us_{n}(x)|^2 leq frac{Csin^2 (2pi x)}{x^2}$ is wrong. All you know (from the Fourier series of $rho$ and $sum_{d | n} mu(d) = 1_{n=1}$ and say $sum_n mu(n)/n=0$ equivalent to the PNT) is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin(pi x)/pi$ in $L^2((0,1])$.
$endgroup$
– reuns
Dec 6 '18 at 13:55
1
$begingroup$
You are claiming that in his paper on some criterion of the RH Baez is mentioning/proving that $frac{1}{N} (sum_{n=1}^N mu(n))^2 = O(1)$ without noticing it ? Understanding the exact mode of convergence of $sum_{n=1}^{N} frac{mu(n)}{n} rho(nx)$ is complicated. Showing it converges uniformly to $-sin(2pi x)/pi$ is academic.oup.com/qjmath/article-abstract/os-8/1/313/… The only thing that is +/- obvious is to relate "zeta has no zeros for $Re(s) > sigma$" with $sup_N int_0^1 |sum_{n=1}^N mu(n)frac{rho( nx)}{n}x^{sigma-3/2}|^2 dx < infty$
$endgroup$
– reuns
Dec 6 '18 at 20:02
|
show 11 more comments
3
$begingroup$
Because you are mixing things you didn't look at carefully. For $x in (0,1/n), m le n$ then $rho(mx) = mx$. So what ? The complicated part of $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ is for $x in (1/n,1)$. Everything is explained in the paper and in several places on books, MSE and the web.
$endgroup$
– reuns
Dec 6 '18 at 13:14
2
$begingroup$
$int_{0}^{1/n} |Us_{n}(x)|^2 mathrm{d}x=OBigg(int_{0}^{1/n} frac{sin^2(2pi x)}{pi ^2 x^2} mathrm{d}x Bigg)$ doesn't make any sense. You are confusing $L^2$ and pointwise convergence. All you know is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin (pi x)/pi$ in $L^2([0,1])$ and under the RH it converges in $L^2([0,1], x^{-1+epsilon})$
$endgroup$
– reuns
Dec 6 '18 at 13:32
2
$begingroup$
The paper is exactly about that, why not read it, as well as the many explications you'll find on the web.
$endgroup$
– reuns
Dec 6 '18 at 13:35
2
$begingroup$
$|Us_{n}(x)|^2 leq frac{Csin^2 (2pi x)}{x^2}$ is wrong. All you know (from the Fourier series of $rho$ and $sum_{d | n} mu(d) = 1_{n=1}$ and say $sum_n mu(n)/n=0$ equivalent to the PNT) is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin(pi x)/pi$ in $L^2((0,1])$.
$endgroup$
– reuns
Dec 6 '18 at 13:55
1
$begingroup$
You are claiming that in his paper on some criterion of the RH Baez is mentioning/proving that $frac{1}{N} (sum_{n=1}^N mu(n))^2 = O(1)$ without noticing it ? Understanding the exact mode of convergence of $sum_{n=1}^{N} frac{mu(n)}{n} rho(nx)$ is complicated. Showing it converges uniformly to $-sin(2pi x)/pi$ is academic.oup.com/qjmath/article-abstract/os-8/1/313/… The only thing that is +/- obvious is to relate "zeta has no zeros for $Re(s) > sigma$" with $sup_N int_0^1 |sum_{n=1}^N mu(n)frac{rho( nx)}{n}x^{sigma-3/2}|^2 dx < infty$
$endgroup$
– reuns
Dec 6 '18 at 20:02
3
3
$begingroup$
Because you are mixing things you didn't look at carefully. For $x in (0,1/n), m le n$ then $rho(mx) = mx$. So what ? The complicated part of $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ is for $x in (1/n,1)$. Everything is explained in the paper and in several places on books, MSE and the web.
$endgroup$
– reuns
Dec 6 '18 at 13:14
$begingroup$
Because you are mixing things you didn't look at carefully. For $x in (0,1/n), m le n$ then $rho(mx) = mx$. So what ? The complicated part of $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ is for $x in (1/n,1)$. Everything is explained in the paper and in several places on books, MSE and the web.
$endgroup$
– reuns
Dec 6 '18 at 13:14
2
2
$begingroup$
$int_{0}^{1/n} |Us_{n}(x)|^2 mathrm{d}x=OBigg(int_{0}^{1/n} frac{sin^2(2pi x)}{pi ^2 x^2} mathrm{d}x Bigg)$ doesn't make any sense. You are confusing $L^2$ and pointwise convergence. All you know is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin (pi x)/pi$ in $L^2([0,1])$ and under the RH it converges in $L^2([0,1], x^{-1+epsilon})$
$endgroup$
– reuns
Dec 6 '18 at 13:32
$begingroup$
$int_{0}^{1/n} |Us_{n}(x)|^2 mathrm{d}x=OBigg(int_{0}^{1/n} frac{sin^2(2pi x)}{pi ^2 x^2} mathrm{d}x Bigg)$ doesn't make any sense. You are confusing $L^2$ and pointwise convergence. All you know is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin (pi x)/pi$ in $L^2([0,1])$ and under the RH it converges in $L^2([0,1], x^{-1+epsilon})$
$endgroup$
– reuns
Dec 6 '18 at 13:32
2
2
$begingroup$
The paper is exactly about that, why not read it, as well as the many explications you'll find on the web.
$endgroup$
– reuns
Dec 6 '18 at 13:35
$begingroup$
The paper is exactly about that, why not read it, as well as the many explications you'll find on the web.
$endgroup$
– reuns
Dec 6 '18 at 13:35
2
2
$begingroup$
$|Us_{n}(x)|^2 leq frac{Csin^2 (2pi x)}{x^2}$ is wrong. All you know (from the Fourier series of $rho$ and $sum_{d | n} mu(d) = 1_{n=1}$ and say $sum_n mu(n)/n=0$ equivalent to the PNT) is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin(pi x)/pi$ in $L^2((0,1])$.
$endgroup$
– reuns
Dec 6 '18 at 13:55
$begingroup$
$|Us_{n}(x)|^2 leq frac{Csin^2 (2pi x)}{x^2}$ is wrong. All you know (from the Fourier series of $rho$ and $sum_{d | n} mu(d) = 1_{n=1}$ and say $sum_n mu(n)/n=0$ equivalent to the PNT) is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin(pi x)/pi$ in $L^2((0,1])$.
$endgroup$
– reuns
Dec 6 '18 at 13:55
1
1
$begingroup$
You are claiming that in his paper on some criterion of the RH Baez is mentioning/proving that $frac{1}{N} (sum_{n=1}^N mu(n))^2 = O(1)$ without noticing it ? Understanding the exact mode of convergence of $sum_{n=1}^{N} frac{mu(n)}{n} rho(nx)$ is complicated. Showing it converges uniformly to $-sin(2pi x)/pi$ is academic.oup.com/qjmath/article-abstract/os-8/1/313/… The only thing that is +/- obvious is to relate "zeta has no zeros for $Re(s) > sigma$" with $sup_N int_0^1 |sum_{n=1}^N mu(n)frac{rho( nx)}{n}x^{sigma-3/2}|^2 dx < infty$
$endgroup$
– reuns
Dec 6 '18 at 20:02
$begingroup$
You are claiming that in his paper on some criterion of the RH Baez is mentioning/proving that $frac{1}{N} (sum_{n=1}^N mu(n))^2 = O(1)$ without noticing it ? Understanding the exact mode of convergence of $sum_{n=1}^{N} frac{mu(n)}{n} rho(nx)$ is complicated. Showing it converges uniformly to $-sin(2pi x)/pi$ is academic.oup.com/qjmath/article-abstract/os-8/1/313/… The only thing that is +/- obvious is to relate "zeta has no zeros for $Re(s) > sigma$" with $sup_N int_0^1 |sum_{n=1}^N mu(n)frac{rho( nx)}{n}x^{sigma-3/2}|^2 dx < infty$
$endgroup$
– reuns
Dec 6 '18 at 20:02
|
show 11 more comments
1 Answer
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Your argument for "This collectively implies that $I_nle C$ for every positive integer $n$ where $C$ is some positive constant" is based on the following reasoning: If the functions $f_n$ converge pointwise to an integrable function $f$, then $int_{Omega_n} f_n$ is bounded by some function of $int_{Omega} f$ for some $Omega$ which contains all $Omega_n$.
This is simply false. For example the functions $f_n(x)=frac{1}{nx}$ converge pointwise on $(0,1)$ to $f(x)=0$. Yet $int_0^{1/n} f_n=infty$.
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
$endgroup$
$begingroup$
Your purported counter-example isn't valid because it doesn't satisfy all the outlined three observations upon which the argument is based. Note that $lim_{xrightarrow 0} f_{n}(x)$ doesn't exist.
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– user507152
Dec 9 '18 at 1:47
$begingroup$
@FunIsMath. Where in your argument are you making use of continuity at zero?
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 1:48
1
$begingroup$
@FunIsMath. Let me know if you would like to add fourth or fifth or sixth conditions, :)
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 2:35
2
$begingroup$
@FunIsMath. We are again at the starting point. "$|Us_{n}(x)|<c/x$ for every $n$ and $x in [0,1/n]$" doesn't imply anything about $int_0^1 |Us_{n}(x)|^2dx$ which can be $infty$. Why are you not capable to state clearly that you want $Us_n$ to converge uniformly to $-frac{sin 2pi x}{pi x}$, which is what suggests reading too fast Baez paper right after 2.11 (reading it more slowly shows they meant $x Us_n$ converges uniformly to $-frac{sin 2pi x}{pi }$, which is already a complicated theorem proven in Davenport, that I'd like to find a copy of)
$endgroup$
– reuns
Dec 9 '18 at 5:39
1
$begingroup$
@reuns, my bad, you're right. The main issue is i had overlooked the fact that $ sum_{j=1}^{n} frac{mu(j)}{j}rho(jx)$ is simply $sum_{j=1}^{n} mu(j)$ for $xin(0, 1/n)$, which made all of my subsequent arguments erroneous. Thanks !
$endgroup$
– user507152
Dec 9 '18 at 16:27
|
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1 Answer
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Your argument for "This collectively implies that $I_nle C$ for every positive integer $n$ where $C$ is some positive constant" is based on the following reasoning: If the functions $f_n$ converge pointwise to an integrable function $f$, then $int_{Omega_n} f_n$ is bounded by some function of $int_{Omega} f$ for some $Omega$ which contains all $Omega_n$.
This is simply false. For example the functions $f_n(x)=frac{1}{nx}$ converge pointwise on $(0,1)$ to $f(x)=0$. Yet $int_0^{1/n} f_n=infty$.
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
$endgroup$
$begingroup$
Your purported counter-example isn't valid because it doesn't satisfy all the outlined three observations upon which the argument is based. Note that $lim_{xrightarrow 0} f_{n}(x)$ doesn't exist.
$endgroup$
– user507152
Dec 9 '18 at 1:47
$begingroup$
@FunIsMath. Where in your argument are you making use of continuity at zero?
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 1:48
1
$begingroup$
@FunIsMath. Let me know if you would like to add fourth or fifth or sixth conditions, :)
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 2:35
2
$begingroup$
@FunIsMath. We are again at the starting point. "$|Us_{n}(x)|<c/x$ for every $n$ and $x in [0,1/n]$" doesn't imply anything about $int_0^1 |Us_{n}(x)|^2dx$ which can be $infty$. Why are you not capable to state clearly that you want $Us_n$ to converge uniformly to $-frac{sin 2pi x}{pi x}$, which is what suggests reading too fast Baez paper right after 2.11 (reading it more slowly shows they meant $x Us_n$ converges uniformly to $-frac{sin 2pi x}{pi }$, which is already a complicated theorem proven in Davenport, that I'd like to find a copy of)
$endgroup$
– reuns
Dec 9 '18 at 5:39
1
$begingroup$
@reuns, my bad, you're right. The main issue is i had overlooked the fact that $ sum_{j=1}^{n} frac{mu(j)}{j}rho(jx)$ is simply $sum_{j=1}^{n} mu(j)$ for $xin(0, 1/n)$, which made all of my subsequent arguments erroneous. Thanks !
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– user507152
Dec 9 '18 at 16:27
|
show 26 more comments
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Your argument for "This collectively implies that $I_nle C$ for every positive integer $n$ where $C$ is some positive constant" is based on the following reasoning: If the functions $f_n$ converge pointwise to an integrable function $f$, then $int_{Omega_n} f_n$ is bounded by some function of $int_{Omega} f$ for some $Omega$ which contains all $Omega_n$.
This is simply false. For example the functions $f_n(x)=frac{1}{nx}$ converge pointwise on $(0,1)$ to $f(x)=0$. Yet $int_0^{1/n} f_n=infty$.
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
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Your purported counter-example isn't valid because it doesn't satisfy all the outlined three observations upon which the argument is based. Note that $lim_{xrightarrow 0} f_{n}(x)$ doesn't exist.
$endgroup$
– user507152
Dec 9 '18 at 1:47
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@FunIsMath. Where in your argument are you making use of continuity at zero?
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 1:48
1
$begingroup$
@FunIsMath. Let me know if you would like to add fourth or fifth or sixth conditions, :)
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 2:35
2
$begingroup$
@FunIsMath. We are again at the starting point. "$|Us_{n}(x)|<c/x$ for every $n$ and $x in [0,1/n]$" doesn't imply anything about $int_0^1 |Us_{n}(x)|^2dx$ which can be $infty$. Why are you not capable to state clearly that you want $Us_n$ to converge uniformly to $-frac{sin 2pi x}{pi x}$, which is what suggests reading too fast Baez paper right after 2.11 (reading it more slowly shows they meant $x Us_n$ converges uniformly to $-frac{sin 2pi x}{pi }$, which is already a complicated theorem proven in Davenport, that I'd like to find a copy of)
$endgroup$
– reuns
Dec 9 '18 at 5:39
1
$begingroup$
@reuns, my bad, you're right. The main issue is i had overlooked the fact that $ sum_{j=1}^{n} frac{mu(j)}{j}rho(jx)$ is simply $sum_{j=1}^{n} mu(j)$ for $xin(0, 1/n)$, which made all of my subsequent arguments erroneous. Thanks !
$endgroup$
– user507152
Dec 9 '18 at 16:27
|
show 26 more comments
$begingroup$
Your argument for "This collectively implies that $I_nle C$ for every positive integer $n$ where $C$ is some positive constant" is based on the following reasoning: If the functions $f_n$ converge pointwise to an integrable function $f$, then $int_{Omega_n} f_n$ is bounded by some function of $int_{Omega} f$ for some $Omega$ which contains all $Omega_n$.
This is simply false. For example the functions $f_n(x)=frac{1}{nx}$ converge pointwise on $(0,1)$ to $f(x)=0$. Yet $int_0^{1/n} f_n=infty$.
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
$endgroup$
Your argument for "This collectively implies that $I_nle C$ for every positive integer $n$ where $C$ is some positive constant" is based on the following reasoning: If the functions $f_n$ converge pointwise to an integrable function $f$, then $int_{Omega_n} f_n$ is bounded by some function of $int_{Omega} f$ for some $Omega$ which contains all $Omega_n$.
This is simply false. For example the functions $f_n(x)=frac{1}{nx}$ converge pointwise on $(0,1)$ to $f(x)=0$. Yet $int_0^{1/n} f_n=infty$.
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
answered Dec 9 '18 at 1:34
Gjergji ZaimiGjergji Zaimi
1,234911
1,234911
$begingroup$
Your purported counter-example isn't valid because it doesn't satisfy all the outlined three observations upon which the argument is based. Note that $lim_{xrightarrow 0} f_{n}(x)$ doesn't exist.
$endgroup$
– user507152
Dec 9 '18 at 1:47
$begingroup$
@FunIsMath. Where in your argument are you making use of continuity at zero?
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 1:48
1
$begingroup$
@FunIsMath. Let me know if you would like to add fourth or fifth or sixth conditions, :)
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 2:35
2
$begingroup$
@FunIsMath. We are again at the starting point. "$|Us_{n}(x)|<c/x$ for every $n$ and $x in [0,1/n]$" doesn't imply anything about $int_0^1 |Us_{n}(x)|^2dx$ which can be $infty$. Why are you not capable to state clearly that you want $Us_n$ to converge uniformly to $-frac{sin 2pi x}{pi x}$, which is what suggests reading too fast Baez paper right after 2.11 (reading it more slowly shows they meant $x Us_n$ converges uniformly to $-frac{sin 2pi x}{pi }$, which is already a complicated theorem proven in Davenport, that I'd like to find a copy of)
$endgroup$
– reuns
Dec 9 '18 at 5:39
1
$begingroup$
@reuns, my bad, you're right. The main issue is i had overlooked the fact that $ sum_{j=1}^{n} frac{mu(j)}{j}rho(jx)$ is simply $sum_{j=1}^{n} mu(j)$ for $xin(0, 1/n)$, which made all of my subsequent arguments erroneous. Thanks !
$endgroup$
– user507152
Dec 9 '18 at 16:27
|
show 26 more comments
$begingroup$
Your purported counter-example isn't valid because it doesn't satisfy all the outlined three observations upon which the argument is based. Note that $lim_{xrightarrow 0} f_{n}(x)$ doesn't exist.
$endgroup$
– user507152
Dec 9 '18 at 1:47
$begingroup$
@FunIsMath. Where in your argument are you making use of continuity at zero?
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 1:48
1
$begingroup$
@FunIsMath. Let me know if you would like to add fourth or fifth or sixth conditions, :)
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 2:35
2
$begingroup$
@FunIsMath. We are again at the starting point. "$|Us_{n}(x)|<c/x$ for every $n$ and $x in [0,1/n]$" doesn't imply anything about $int_0^1 |Us_{n}(x)|^2dx$ which can be $infty$. Why are you not capable to state clearly that you want $Us_n$ to converge uniformly to $-frac{sin 2pi x}{pi x}$, which is what suggests reading too fast Baez paper right after 2.11 (reading it more slowly shows they meant $x Us_n$ converges uniformly to $-frac{sin 2pi x}{pi }$, which is already a complicated theorem proven in Davenport, that I'd like to find a copy of)
$endgroup$
– reuns
Dec 9 '18 at 5:39
1
$begingroup$
@reuns, my bad, you're right. The main issue is i had overlooked the fact that $ sum_{j=1}^{n} frac{mu(j)}{j}rho(jx)$ is simply $sum_{j=1}^{n} mu(j)$ for $xin(0, 1/n)$, which made all of my subsequent arguments erroneous. Thanks !
$endgroup$
– user507152
Dec 9 '18 at 16:27
$begingroup$
Your purported counter-example isn't valid because it doesn't satisfy all the outlined three observations upon which the argument is based. Note that $lim_{xrightarrow 0} f_{n}(x)$ doesn't exist.
$endgroup$
– user507152
Dec 9 '18 at 1:47
$begingroup$
Your purported counter-example isn't valid because it doesn't satisfy all the outlined three observations upon which the argument is based. Note that $lim_{xrightarrow 0} f_{n}(x)$ doesn't exist.
$endgroup$
– user507152
Dec 9 '18 at 1:47
$begingroup$
@FunIsMath. Where in your argument are you making use of continuity at zero?
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 1:48
$begingroup$
@FunIsMath. Where in your argument are you making use of continuity at zero?
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 1:48
1
1
$begingroup$
@FunIsMath. Let me know if you would like to add fourth or fifth or sixth conditions, :)
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 2:35
$begingroup$
@FunIsMath. Let me know if you would like to add fourth or fifth or sixth conditions, :)
$endgroup$
– Gjergji Zaimi
Dec 9 '18 at 2:35
2
2
$begingroup$
@FunIsMath. We are again at the starting point. "$|Us_{n}(x)|<c/x$ for every $n$ and $x in [0,1/n]$" doesn't imply anything about $int_0^1 |Us_{n}(x)|^2dx$ which can be $infty$. Why are you not capable to state clearly that you want $Us_n$ to converge uniformly to $-frac{sin 2pi x}{pi x}$, which is what suggests reading too fast Baez paper right after 2.11 (reading it more slowly shows they meant $x Us_n$ converges uniformly to $-frac{sin 2pi x}{pi }$, which is already a complicated theorem proven in Davenport, that I'd like to find a copy of)
$endgroup$
– reuns
Dec 9 '18 at 5:39
$begingroup$
@FunIsMath. We are again at the starting point. "$|Us_{n}(x)|<c/x$ for every $n$ and $x in [0,1/n]$" doesn't imply anything about $int_0^1 |Us_{n}(x)|^2dx$ which can be $infty$. Why are you not capable to state clearly that you want $Us_n$ to converge uniformly to $-frac{sin 2pi x}{pi x}$, which is what suggests reading too fast Baez paper right after 2.11 (reading it more slowly shows they meant $x Us_n$ converges uniformly to $-frac{sin 2pi x}{pi }$, which is already a complicated theorem proven in Davenport, that I'd like to find a copy of)
$endgroup$
– reuns
Dec 9 '18 at 5:39
1
1
$begingroup$
@reuns, my bad, you're right. The main issue is i had overlooked the fact that $ sum_{j=1}^{n} frac{mu(j)}{j}rho(jx)$ is simply $sum_{j=1}^{n} mu(j)$ for $xin(0, 1/n)$, which made all of my subsequent arguments erroneous. Thanks !
$endgroup$
– user507152
Dec 9 '18 at 16:27
$begingroup$
@reuns, my bad, you're right. The main issue is i had overlooked the fact that $ sum_{j=1}^{n} frac{mu(j)}{j}rho(jx)$ is simply $sum_{j=1}^{n} mu(j)$ for $xin(0, 1/n)$, which made all of my subsequent arguments erroneous. Thanks !
$endgroup$
– user507152
Dec 9 '18 at 16:27
|
show 26 more comments
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Because you are mixing things you didn't look at carefully. For $x in (0,1/n), m le n$ then $rho(mx) = mx$. So what ? The complicated part of $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ is for $x in (1/n,1)$. Everything is explained in the paper and in several places on books, MSE and the web.
$endgroup$
– reuns
Dec 6 '18 at 13:14
2
$begingroup$
$int_{0}^{1/n} |Us_{n}(x)|^2 mathrm{d}x=OBigg(int_{0}^{1/n} frac{sin^2(2pi x)}{pi ^2 x^2} mathrm{d}x Bigg)$ doesn't make any sense. You are confusing $L^2$ and pointwise convergence. All you know is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin (pi x)/pi$ in $L^2([0,1])$ and under the RH it converges in $L^2([0,1], x^{-1+epsilon})$
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– reuns
Dec 6 '18 at 13:32
2
$begingroup$
The paper is exactly about that, why not read it, as well as the many explications you'll find on the web.
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– reuns
Dec 6 '18 at 13:35
2
$begingroup$
$|Us_{n}(x)|^2 leq frac{Csin^2 (2pi x)}{x^2}$ is wrong. All you know (from the Fourier series of $rho$ and $sum_{d | n} mu(d) = 1_{n=1}$ and say $sum_n mu(n)/n=0$ equivalent to the PNT) is that $sum_{m=1}^{n} frac{mu(m)}{m} rho(mx)$ converges to $-sin(pi x)/pi$ in $L^2((0,1])$.
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– reuns
Dec 6 '18 at 13:55
1
$begingroup$
You are claiming that in his paper on some criterion of the RH Baez is mentioning/proving that $frac{1}{N} (sum_{n=1}^N mu(n))^2 = O(1)$ without noticing it ? Understanding the exact mode of convergence of $sum_{n=1}^{N} frac{mu(n)}{n} rho(nx)$ is complicated. Showing it converges uniformly to $-sin(2pi x)/pi$ is academic.oup.com/qjmath/article-abstract/os-8/1/313/… The only thing that is +/- obvious is to relate "zeta has no zeros for $Re(s) > sigma$" with $sup_N int_0^1 |sum_{n=1}^N mu(n)frac{rho( nx)}{n}x^{sigma-3/2}|^2 dx < infty$
$endgroup$
– reuns
Dec 6 '18 at 20:02