Mixing
Characterization of $ell^q$ sequences
I'm working on a solution for this one:
Let $x:=(x_n)_n subset mathbb{C}^mathbb{N}.$ Suppose that for any $yin ell^p, pin [1,+infty],$ the series $sum_{ninmathbb{N}}x_ny_n$ converges. Prove that $(x_n)_n in ell^q$ for $qin [1,+infty]$ satisfying $frac{1}{p}+frac{1}{q} = 1.$
I tried to prove the statement for the three cases in the following way, but don't come so far..
$p=infty :$ Let $(y_n)_nin ell^infty$, so we have $sup_{ninmathbb{N}}|y_n|<infty$ and that the series $sum_{ninmathbb{N}}x_ny_n$ converges. Now I have to show that $sum_{n=1}^{infty}|x_n|<infty$ to get $(x_n)_n in ell^1.$
$p=1:$ I got the hint to construct a sequence $(y_n)_n in ell^1$ such that the sequence $x_ny_n$ does not converge to $0.$
$pin(1,infty):$ here we should apply the Banach Steinhaus Theorem to a well chosen family of linear maps $T_N: ell^ptomathbb{C},Ninmathbb{N}.$
Thank you for your help!
functional-analysis
asked Nov 20 '18 at 18:47
hAM1t
33
add a comment |
I'm working on a solution for this one:
Let $x:=(x_n)_n subset mathbb{C}^mathbb{N}.$ Suppose that for any $yin ell^p, pin [1,+infty],$ the series $sum_{ninmathbb{N}}x_ny_n$ converges. Prove that $(x_n)_n in ell^q$ for $qin [1,+infty]$ satisfying $frac{1}{p}+frac{1}{q} = 1.$
I tried to prove the statement for the three cases in the following way, but don't come so far..
$p=infty :$ Let $(y_n)_nin ell^infty$, so we have $sup_{ninmathbb{N}}|y_n|<infty$ and that the series $sum_{ninmathbb{N}}x_ny_n$ converges. Now I have to show that $sum_{n=1}^{infty}|x_n|<infty$ to get $(x_n)_n in ell^1.$
$p=1:$ I got the hint to construct a sequence $(y_n)_n in ell^1$ such that the sequence $x_ny_n$ does not converge to $0.$
$pin(1,infty):$ here we should apply the Banach Steinhaus Theorem to a well chosen family of linear maps $T_N: ell^ptomathbb{C},Ninmathbb{N}.$
Thank you for your help!
functional-analysis
asked Nov 20 '18 at 18:47
hAM1t
33
add a comment |
I'm working on a solution for this one:
Let $x:=(x_n)_n subset mathbb{C}^mathbb{N}.$ Suppose that for any $yin ell^p, pin [1,+infty],$ the series $sum_{ninmathbb{N}}x_ny_n$ converges. Prove that $(x_n)_n in ell^q$ for $qin [1,+infty]$ satisfying $frac{1}{p}+frac{1}{q} = 1.$
I tried to prove the statement for the three cases in the following way, but don't come so far..
$p=infty :$ Let $(y_n)_nin ell^infty$, so we have $sup_{ninmathbb{N}}|y_n|<infty$ and that the series $sum_{ninmathbb{N}}x_ny_n$ converges. Now I have to show that $sum_{n=1}^{infty}|x_n|<infty$ to get $(x_n)_n in ell^1.$
$p=1:$ I got the hint to construct a sequence $(y_n)_n in ell^1$ such that the sequence $x_ny_n$ does not converge to $0.$
$pin(1,infty):$ here we should apply the Banach Steinhaus Theorem to a well chosen family of linear maps $T_N: ell^ptomathbb{C},Ninmathbb{N}.$
Thank you for your help!
functional-analysis
asked Nov 20 '18 at 18:47
hAM1t
33
I'm working on a solution for this one:
Let $x:=(x_n)_n subset mathbb{C}^mathbb{N}.$ Suppose that for any $yin ell^p, pin [1,+infty],$ the series $sum_{ninmathbb{N}}x_ny_n$ converges. Prove that $(x_n)_n in ell^q$ for $qin [1,+infty]$ satisfying $frac{1}{p}+frac{1}{q} = 1.$
I tried to prove the statement for the three cases in the following way, but don't come so far..
$p=infty :$ Let $(y_n)_nin ell^infty$, so we have $sup_{ninmathbb{N}}|y_n|<infty$ and that the series $sum_{ninmathbb{N}}x_ny_n$ converges. Now I have to show that $sum_{n=1}^{infty}|x_n|<infty$ to get $(x_n)_n in ell^1.$
$p=1:$ I got the hint to construct a sequence $(y_n)_n in ell^1$ such that the sequence $x_ny_n$ does not converge to $0.$
$pin(1,infty):$ here we should apply the Banach Steinhaus Theorem to a well chosen family of linear maps $T_N: ell^ptomathbb{C},Ninmathbb{N}.$
Thank you for your help!
functional-analysis
functional-analysis
asked Nov 20 '18 at 18:47
hAM1t
33
asked Nov 20 '18 at 18:47
hAM1t
33
edited Nov 20 '18 at 20:24
asked Nov 20 '18 at 18:47
hAM1t
33
asked Nov 20 '18 at 18:47
hAM1t
33
asked Nov 20 '18 at 18:47
hAM1t
33
33
add a comment |
add a comment |
1 Answer
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For $0ne zin Bbb C$ let $z'=|z|^{q-2}bar z,$ and let $0'=0.$ Note that $|z'|^p=zz'= |z|^q.$
Suppose $xnot in l_q.$ Then there is a strictly increasing sequence $(M_j)_{jin Bbb N}$ in $Bbb N$ such that $$F(n)=sum_{j=M_{2n-1}}^{M_{2n}} |x_j|^q>2^n$$ for each $nin Bbb N.$
For $nin Bbb N$ and $M_{2n-1}leq jleq M_{2n}$ let $y_j=(x_j)'/F(n).$ Let $y_j=0$ for all other $jin Bbb N.$
We have $$sum_{k=1}^{infty}|y_k|^p=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}|x_j|^q/F(n)^{p}=$$ $$=sum_{n=1}^{infty}F(n)^{1-p}<sum_{n=1}^{infty}2^{n(1-p)}<infty$$ because $1-p<0.$
We have $$sum_{k=1}^{infty}x_ky_k=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}x_j(x_j)'/F(n)=$$ $$=sum_{n=1}^{infty}1=infty.$$
Remarks. Treating $x$ as a function that sends each $yin l_p$ to $sum_jx_jy_j, $ we cannot assume $x$ is continuous. But let $u(n)=(u(n)_j)_jin l_p$ where $u(n)_j=y_j$ for $M_{2n-1}leq M_{2n}$ and $u(n)_j=0$ for all other $j$. Then $sum_nu(n)=(y_j)_jin l_p$ and also $sum_nx(u(n))=x(sum_nu(n))=infty$.
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
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1 Answer
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For $0ne zin Bbb C$ let $z'=|z|^{q-2}bar z,$ and let $0'=0.$ Note that $|z'|^p=zz'= |z|^q.$
Suppose $xnot in l_q.$ Then there is a strictly increasing sequence $(M_j)_{jin Bbb N}$ in $Bbb N$ such that $$F(n)=sum_{j=M_{2n-1}}^{M_{2n}} |x_j|^q>2^n$$ for each $nin Bbb N.$
For $nin Bbb N$ and $M_{2n-1}leq jleq M_{2n}$ let $y_j=(x_j)'/F(n).$ Let $y_j=0$ for all other $jin Bbb N.$
We have $$sum_{k=1}^{infty}|y_k|^p=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}|x_j|^q/F(n)^{p}=$$ $$=sum_{n=1}^{infty}F(n)^{1-p}<sum_{n=1}^{infty}2^{n(1-p)}<infty$$ because $1-p<0.$
We have $$sum_{k=1}^{infty}x_ky_k=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}x_j(x_j)'/F(n)=$$ $$=sum_{n=1}^{infty}1=infty.$$
Remarks. Treating $x$ as a function that sends each $yin l_p$ to $sum_jx_jy_j, $ we cannot assume $x$ is continuous. But let $u(n)=(u(n)_j)_jin l_p$ where $u(n)_j=y_j$ for $M_{2n-1}leq M_{2n}$ and $u(n)_j=0$ for all other $j$. Then $sum_nu(n)=(y_j)_jin l_p$ and also $sum_nx(u(n))=x(sum_nu(n))=infty$.
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
add a comment |
For $0ne zin Bbb C$ let $z'=|z|^{q-2}bar z,$ and let $0'=0.$ Note that $|z'|^p=zz'= |z|^q.$
Suppose $xnot in l_q.$ Then there is a strictly increasing sequence $(M_j)_{jin Bbb N}$ in $Bbb N$ such that $$F(n)=sum_{j=M_{2n-1}}^{M_{2n}} |x_j|^q>2^n$$ for each $nin Bbb N.$
For $nin Bbb N$ and $M_{2n-1}leq jleq M_{2n}$ let $y_j=(x_j)'/F(n).$ Let $y_j=0$ for all other $jin Bbb N.$
We have $$sum_{k=1}^{infty}|y_k|^p=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}|x_j|^q/F(n)^{p}=$$ $$=sum_{n=1}^{infty}F(n)^{1-p}<sum_{n=1}^{infty}2^{n(1-p)}<infty$$ because $1-p<0.$
We have $$sum_{k=1}^{infty}x_ky_k=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}x_j(x_j)'/F(n)=$$ $$=sum_{n=1}^{infty}1=infty.$$
Remarks. Treating $x$ as a function that sends each $yin l_p$ to $sum_jx_jy_j, $ we cannot assume $x$ is continuous. But let $u(n)=(u(n)_j)_jin l_p$ where $u(n)_j=y_j$ for $M_{2n-1}leq M_{2n}$ and $u(n)_j=0$ for all other $j$. Then $sum_nu(n)=(y_j)_jin l_p$ and also $sum_nx(u(n))=x(sum_nu(n))=infty$.
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
add a comment |
For $0ne zin Bbb C$ let $z'=|z|^{q-2}bar z,$ and let $0'=0.$ Note that $|z'|^p=zz'= |z|^q.$
Suppose $xnot in l_q.$ Then there is a strictly increasing sequence $(M_j)_{jin Bbb N}$ in $Bbb N$ such that $$F(n)=sum_{j=M_{2n-1}}^{M_{2n}} |x_j|^q>2^n$$ for each $nin Bbb N.$
For $nin Bbb N$ and $M_{2n-1}leq jleq M_{2n}$ let $y_j=(x_j)'/F(n).$ Let $y_j=0$ for all other $jin Bbb N.$
We have $$sum_{k=1}^{infty}|y_k|^p=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}|x_j|^q/F(n)^{p}=$$ $$=sum_{n=1}^{infty}F(n)^{1-p}<sum_{n=1}^{infty}2^{n(1-p)}<infty$$ because $1-p<0.$
We have $$sum_{k=1}^{infty}x_ky_k=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}x_j(x_j)'/F(n)=$$ $$=sum_{n=1}^{infty}1=infty.$$
Remarks. Treating $x$ as a function that sends each $yin l_p$ to $sum_jx_jy_j, $ we cannot assume $x$ is continuous. But let $u(n)=(u(n)_j)_jin l_p$ where $u(n)_j=y_j$ for $M_{2n-1}leq M_{2n}$ and $u(n)_j=0$ for all other $j$. Then $sum_nu(n)=(y_j)_jin l_p$ and also $sum_nx(u(n))=x(sum_nu(n))=infty$.
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
For $0ne zin Bbb C$ let $z'=|z|^{q-2}bar z,$ and let $0'=0.$ Note that $|z'|^p=zz'= |z|^q.$
Suppose $xnot in l_q.$ Then there is a strictly increasing sequence $(M_j)_{jin Bbb N}$ in $Bbb N$ such that $$F(n)=sum_{j=M_{2n-1}}^{M_{2n}} |x_j|^q>2^n$$ for each $nin Bbb N.$
For $nin Bbb N$ and $M_{2n-1}leq jleq M_{2n}$ let $y_j=(x_j)'/F(n).$ Let $y_j=0$ for all other $jin Bbb N.$
We have $$sum_{k=1}^{infty}|y_k|^p=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}|x_j|^q/F(n)^{p}=$$ $$=sum_{n=1}^{infty}F(n)^{1-p}<sum_{n=1}^{infty}2^{n(1-p)}<infty$$ because $1-p<0.$
We have $$sum_{k=1}^{infty}x_ky_k=sum_{n=1}^{infty}sum_{j=M_{2n-1}}^{M_{2n}}x_j(x_j)'/F(n)=$$ $$=sum_{n=1}^{infty}1=infty.$$
Remarks. Treating $x$ as a function that sends each $yin l_p$ to $sum_jx_jy_j, $ we cannot assume $x$ is continuous. But let $u(n)=(u(n)_j)_jin l_p$ where $u(n)_j=y_j$ for $M_{2n-1}leq M_{2n}$ and $u(n)_j=0$ for all other $j$. Then $sum_nu(n)=(y_j)_jin l_p$ and also $sum_nx(u(n))=x(sum_nu(n))=infty$.
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
edited Nov 21 '18 at 6:52
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
answered Nov 21 '18 at 6:21
DanielWainfleet
34.2k31647
34.2k31647
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