Some subspaces are either closed or dense












2














a) Let $a_{n}rightarrow 0$ and $M={xin ell^{2}:sum a_{n}x_{n}=0}$. Show
that the subspace $M$ of $ell^{2}$ is closed or dense according as $%
{a_{n}}in ell^{2}$ or ${a_{n}}notin ell^{2}$.



b) Let $g$ be any positive measurable function on $[0,1]$. Let $$N=left{fin
L^{2}([0,1]):int_{0}^{1}f(x)g(x)dx=0right}$$ Prove that $N$ is closed
or dense in $L^{2}([0,1])$ according as $gin L^{2}([0,1])$ or $gnotin
L^{2}([0,1])$.




Several special cases of this have appeared here. I am trying to consolidate these. I have provided an answer but my answer is probably not the best. I would look forward to comments, alternate proofs and generalizations.






hilbert-spaces orthogonality







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    a) Let $a_{n}rightarrow 0$ and $M={xin ell^{2}:sum a_{n}x_{n}=0}$. Show
    that the subspace $M$ of $ell^{2}$ is closed or dense according as $%
    {a_{n}}in ell^{2}$     or ${a_{n}}notin ell^{2}$.



    b) Let $g$ be any positive measurable function on $[0,1]$. Let $$N=left{fin
    L^{2}([0,1]):int_{0}^{1}f(x)g(x)dx=0right}$$     Prove that $N$ is closed
    or dense in $L^{2}([0,1])$ according as $gin L^{2}([0,1])$ or $gnotin
    L^{2}([0,1])$    .




    Several special cases of this have appeared here. I am trying to consolidate these. I have provided an answer but my answer is probably not the best. I would look forward to comments, alternate proofs and generalizations.






    hilbert-spaces orthogonality







    share|cite|improve this question



























      2












      2








      2







      a) Let $a_{n}rightarrow 0$ and $M={xin ell^{2}:sum a_{n}x_{n}=0}$. Show
      that the subspace $M$ of $ell^{2}$ is closed or dense according as $%
      {a_{n}}in ell^{2}$       or ${a_{n}}notin ell^{2}$.



      b) Let $g$ be any positive measurable function on $[0,1]$. Let $$N=left{fin
      L^{2}([0,1]):int_{0}^{1}f(x)g(x)dx=0right}$$       Prove that $N$ is closed
      or dense in $L^{2}([0,1])$ according as $gin L^{2}([0,1])$ or $gnotin
      L^{2}([0,1])$      .




      Several special cases of this have appeared here. I am trying to consolidate these. I have provided an answer but my answer is probably not the best. I would look forward to comments, alternate proofs and generalizations.






      hilbert-spaces orthogonality







      share|cite|improve this question















      a) Let $a_{n}rightarrow 0$ and $M={xin ell^{2}:sum a_{n}x_{n}=0}$. Show
      that the subspace $M$ of $ell^{2}$ is closed or dense according as $%
      {a_{n}}in ell^{2}$       or ${a_{n}}notin ell^{2}$.



      b) Let $g$ be any positive measurable function on $[0,1]$. Let $$N=left{fin
      L^{2}([0,1]):int_{0}^{1}f(x)g(x)dx=0right}$$       Prove that $N$ is closed
      or dense in $L^{2}([0,1])$ according as $gin L^{2}([0,1])$ or $gnotin
      L^{2}([0,1])$      .




      Several special cases of this have appeared here. I am trying to consolidate these. I have provided an answer but my answer is probably not the best. I would look forward to comments, alternate proofs and generalizations.






      hilbert-spaces orthogonality




      hilbert-spaces orthogonality






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      edited Nov 29 at 8:44









      Chinnapparaj R

      5,2271826




      5,2271826










      asked Nov 29 at 8:31









      Kavi Rama Murthy

      48.9k31854




      48.9k31854






















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          For a) take $xin l^{0}$ (the space of finitely non-zero sequences).
          Consider $x-c_{n}x^{(n)}$ where $x_{j}^{(n)}=frac{a_{j}}{b_{n}}$ for $jleq
          n$           and $0$ for $j>n,c_{n}=-frac{sum_{j=1}^{n}a_{j}x_{j}}{%
          sum_{j=1}^{n}a_{j}^{2}}$           and $b_{n}=sqrt{sum_{j=1}^{n}a_{j}^{2}}$. Note that $leftVert x^{(n)}rightVert =1$
          for each $n$ and $c_{n}rightarrow 0$ if ${a_{n}}notin l^{2}$. Hence $M$
          is dense in this case. If ${a_{n}}in l^{2}$ then $M$ is the orthogonal
          complement of ${a_{n}}$. hence it is closed.



          For b) first observe that any $fin L^{2}([0,1])$ can be approximated by
          a $hin L^{2}([0,1])$ such that $int_{0}^{1}leftvert
          h(x)g(x)rightvert dx<infty $          . [ Take $h=fI_{{|fg|leq n}}$ ]. Now consider $%
          h-c_{n}frac{gI_{{|g|leq n}}}{b_{n}}$           where $b_{n}=sqrt{int
          g(x)^{2}I_{{|g|leq n}}(x)dx}$           and $c_{n}=-b_{n}frac{int_{0}^{1}h(x)g(x)dx}{int_{0}^{1}g^{2}I_{{|g|leq n}}dx}=%
          frac{int_{0}^{1}h(x)g(x)dx}{sqrt{int_{0}^{1}g^{2}I_{%
          {|g|leq n}}dx}}$          . Note that $c_{n}rightarrow 0$ if $gnotin L^{2}([0,1])$
          and that $frac{gI_{{|g|leq n}}}{b_{n}}$ has norm $1$ for each $n$. The
          proof is completed as in a).






          share|cite|improve this answer





















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            For a) take $xin l^{0}$ (the space of finitely non-zero sequences).
            Consider $x-c_{n}x^{(n)}$ where $x_{j}^{(n)}=frac{a_{j}}{b_{n}}$ for $jleq
            n$             and $0$ for $j>n,c_{n}=-frac{sum_{j=1}^{n}a_{j}x_{j}}{%
            sum_{j=1}^{n}a_{j}^{2}}$             and $b_{n}=sqrt{sum_{j=1}^{n}a_{j}^{2}}$. Note that $leftVert x^{(n)}rightVert =1$
            for each $n$ and $c_{n}rightarrow 0$ if ${a_{n}}notin l^{2}$. Hence $M$
            is dense in this case. If ${a_{n}}in l^{2}$ then $M$ is the orthogonal
            complement of ${a_{n}}$. hence it is closed.



            For b) first observe that any $fin L^{2}([0,1])$ can be approximated by
            a $hin L^{2}([0,1])$ such that $int_{0}^{1}leftvert
            h(x)g(x)rightvert dx<infty $            . [ Take $h=fI_{{|fg|leq n}}$ ]. Now consider $%
            h-c_{n}frac{gI_{{|g|leq n}}}{b_{n}}$             where $b_{n}=sqrt{int
            g(x)^{2}I_{{|g|leq n}}(x)dx}$             and $c_{n}=-b_{n}frac{int_{0}^{1}h(x)g(x)dx}{int_{0}^{1}g^{2}I_{{|g|leq n}}dx}=%
            frac{int_{0}^{1}h(x)g(x)dx}{sqrt{int_{0}^{1}g^{2}I_{%
            {|g|leq n}}dx}}$            . Note that $c_{n}rightarrow 0$ if $gnotin L^{2}([0,1])$
            and that $frac{gI_{{|g|leq n}}}{b_{n}}$ has norm $1$ for each $n$. The
            proof is completed as in a).






            share|cite|improve this answer


























              1














              For a) take $xin l^{0}$ (the space of finitely non-zero sequences).
              Consider $x-c_{n}x^{(n)}$ where $x_{j}^{(n)}=frac{a_{j}}{b_{n}}$ for $jleq
              n$               and $0$ for $j>n,c_{n}=-frac{sum_{j=1}^{n}a_{j}x_{j}}{%
              sum_{j=1}^{n}a_{j}^{2}}$               and $b_{n}=sqrt{sum_{j=1}^{n}a_{j}^{2}}$. Note that $leftVert x^{(n)}rightVert =1$
              for each $n$ and $c_{n}rightarrow 0$ if ${a_{n}}notin l^{2}$. Hence $M$
              is dense in this case. If ${a_{n}}in l^{2}$ then $M$ is the orthogonal
              complement of ${a_{n}}$. hence it is closed.



              For b) first observe that any $fin L^{2}([0,1])$ can be approximated by
              a $hin L^{2}([0,1])$ such that $int_{0}^{1}leftvert
              h(x)g(x)rightvert dx<infty $              . [ Take $h=fI_{{|fg|leq n}}$ ]. Now consider $%
              h-c_{n}frac{gI_{{|g|leq n}}}{b_{n}}$               where $b_{n}=sqrt{int
              g(x)^{2}I_{{|g|leq n}}(x)dx}$               and $c_{n}=-b_{n}frac{int_{0}^{1}h(x)g(x)dx}{int_{0}^{1}g^{2}I_{{|g|leq n}}dx}=%
              frac{int_{0}^{1}h(x)g(x)dx}{sqrt{int_{0}^{1}g^{2}I_{%
              {|g|leq n}}dx}}$              . Note that $c_{n}rightarrow 0$ if $gnotin L^{2}([0,1])$
              and that $frac{gI_{{|g|leq n}}}{b_{n}}$ has norm $1$ for each $n$. The
              proof is completed as in a).






              share|cite|improve this answer
























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                For a) take $xin l^{0}$ (the space of finitely non-zero sequences).
                Consider $x-c_{n}x^{(n)}$ where $x_{j}^{(n)}=frac{a_{j}}{b_{n}}$ for $jleq
                n$                 and $0$ for $j>n,c_{n}=-frac{sum_{j=1}^{n}a_{j}x_{j}}{%
                sum_{j=1}^{n}a_{j}^{2}}$                 and $b_{n}=sqrt{sum_{j=1}^{n}a_{j}^{2}}$. Note that $leftVert x^{(n)}rightVert =1$
                for each $n$ and $c_{n}rightarrow 0$ if ${a_{n}}notin l^{2}$. Hence $M$
                is dense in this case. If ${a_{n}}in l^{2}$ then $M$ is the orthogonal
                complement of ${a_{n}}$. hence it is closed.



                For b) first observe that any $fin L^{2}([0,1])$ can be approximated by
                a $hin L^{2}([0,1])$ such that $int_{0}^{1}leftvert
                h(x)g(x)rightvert dx<infty $                . [ Take $h=fI_{{|fg|leq n}}$ ]. Now consider $%
                h-c_{n}frac{gI_{{|g|leq n}}}{b_{n}}$                 where $b_{n}=sqrt{int
                g(x)^{2}I_{{|g|leq n}}(x)dx}$                 and $c_{n}=-b_{n}frac{int_{0}^{1}h(x)g(x)dx}{int_{0}^{1}g^{2}I_{{|g|leq n}}dx}=%
                frac{int_{0}^{1}h(x)g(x)dx}{sqrt{int_{0}^{1}g^{2}I_{%
                {|g|leq n}}dx}}$                . Note that $c_{n}rightarrow 0$ if $gnotin L^{2}([0,1])$
                and that $frac{gI_{{|g|leq n}}}{b_{n}}$ has norm $1$ for each $n$. The
                proof is completed as in a).






                share|cite|improve this answer












                For a) take $xin l^{0}$ (the space of finitely non-zero sequences).
                Consider $x-c_{n}x^{(n)}$ where $x_{j}^{(n)}=frac{a_{j}}{b_{n}}$ for $jleq
                n$                 and $0$ for $j>n,c_{n}=-frac{sum_{j=1}^{n}a_{j}x_{j}}{%
                sum_{j=1}^{n}a_{j}^{2}}$                 and $b_{n}=sqrt{sum_{j=1}^{n}a_{j}^{2}}$. Note that $leftVert x^{(n)}rightVert =1$
                for each $n$ and $c_{n}rightarrow 0$ if ${a_{n}}notin l^{2}$. Hence $M$
                is dense in this case. If ${a_{n}}in l^{2}$ then $M$ is the orthogonal
                complement of ${a_{n}}$. hence it is closed.



                For b) first observe that any $fin L^{2}([0,1])$ can be approximated by
                a $hin L^{2}([0,1])$ such that $int_{0}^{1}leftvert
                h(x)g(x)rightvert dx<infty $                . [ Take $h=fI_{{|fg|leq n}}$ ]. Now consider $%
                h-c_{n}frac{gI_{{|g|leq n}}}{b_{n}}$                 where $b_{n}=sqrt{int
                g(x)^{2}I_{{|g|leq n}}(x)dx}$                 and $c_{n}=-b_{n}frac{int_{0}^{1}h(x)g(x)dx}{int_{0}^{1}g^{2}I_{{|g|leq n}}dx}=%
                frac{int_{0}^{1}h(x)g(x)dx}{sqrt{int_{0}^{1}g^{2}I_{%
                {|g|leq n}}dx}}$                . Note that $c_{n}rightarrow 0$ if $gnotin L^{2}([0,1])$
                and that $frac{gI_{{|g|leq n}}}{b_{n}}$ has norm $1$ for each $n$. The
                proof is completed as in a).







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Nov 29 at 8:33









                Kavi Rama Murthy

                48.9k31854




                48.9k31854






























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