Characterization of $ell^q$ sequences











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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!










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    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!










    share|cite|improve this question


























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      up vote
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      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!










      share|cite|improve this question















      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






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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 $sum_nx(u(n))=x(sum_nu(n))=infty$.






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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 $sum_nx(u(n))=x(sum_nu(n))=infty$.






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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 $sum_nx(u(n))=x(sum_nu(n))=infty$.






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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 $sum_nx(u(n))=x(sum_nu(n))=infty$.






                share|cite|improve this answer














                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 $sum_nx(u(n))=x(sum_nu(n))=infty$.







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