Completion of inner product space












1












$begingroup$


Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.



I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.



Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?






linear-algebra functional-analysis inner-product-space complete-spaces







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$endgroup$

















    1












    $begingroup$


    Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.



    I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.



    Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?






    linear-algebra functional-analysis inner-product-space complete-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.



      I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.



      Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?






      linear-algebra functional-analysis inner-product-space complete-spaces







      share|cite|improve this question









      $endgroup$




      Let $(X,langlecdot,cdotrangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.



      I know that in the case $X$ is complete, the linear isometry we can use is $Phi_X:Xto X'$ defined by $Phi_X(x)=langlecdot,xrangle$, and use the Fréchet-Riesz representation theorem.



      Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $overline{Phi_X(X)}=X'$. The first "$subseteq$" follows easily, but how do i show "$supseteq$"?






      linear-algebra functional-analysis inner-product-space complete-spaces




      linear-algebra functional-analysis inner-product-space complete-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 23 at 20:14









      JamesJames

      916318




      916318






















          1 Answer
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          $begingroup$

          The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But what I want to show that $Phi_X$ is a completion.
            $endgroup$
            – James
            Jan 23 at 20:29










          • $begingroup$
            Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:32










          • $begingroup$
            Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
            $endgroup$
            – James
            Jan 23 at 20:33










          • $begingroup$
            Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:44






          • 1




            $begingroup$
            Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
            $endgroup$
            – Martin Argerami
            Jan 23 at 23:05











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          1 Answer
          1






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          active

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          0












          $begingroup$

          The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But what I want to show that $Phi_X$ is a completion.
            $endgroup$
            – James
            Jan 23 at 20:29










          • $begingroup$
            Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:32










          • $begingroup$
            Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
            $endgroup$
            – James
            Jan 23 at 20:33










          • $begingroup$
            Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:44






          • 1




            $begingroup$
            Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
            $endgroup$
            – Martin Argerami
            Jan 23 at 23:05
















          0












          $begingroup$

          The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But what I want to show that $Phi_X$ is a completion.
            $endgroup$
            – James
            Jan 23 at 20:29










          • $begingroup$
            Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:32










          • $begingroup$
            Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
            $endgroup$
            – James
            Jan 23 at 20:33










          • $begingroup$
            Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:44






          • 1




            $begingroup$
            Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
            $endgroup$
            – Martin Argerami
            Jan 23 at 23:05














          0












          0








          0





          $begingroup$

          The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.






          share|cite|improve this answer









          $endgroup$



          The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $Phi_X(X)$ is dense in $X'$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 20:27









          Martin ArgeramiMartin Argerami

          128k1183183




          128k1183183












          • $begingroup$
            But what I want to show that $Phi_X$ is a completion.
            $endgroup$
            – James
            Jan 23 at 20:29










          • $begingroup$
            Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:32










          • $begingroup$
            Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
            $endgroup$
            – James
            Jan 23 at 20:33










          • $begingroup$
            Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:44






          • 1




            $begingroup$
            Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
            $endgroup$
            – Martin Argerami
            Jan 23 at 23:05


















          • $begingroup$
            But what I want to show that $Phi_X$ is a completion.
            $endgroup$
            – James
            Jan 23 at 20:29










          • $begingroup$
            Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:32










          • $begingroup$
            Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
            $endgroup$
            – James
            Jan 23 at 20:33










          • $begingroup$
            Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
            $endgroup$
            – Martin Argerami
            Jan 23 at 20:44






          • 1




            $begingroup$
            Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
            $endgroup$
            – Martin Argerami
            Jan 23 at 23:05
















          $begingroup$
          But what I want to show that $Phi_X$ is a completion.
          $endgroup$
          – James
          Jan 23 at 20:29




          $begingroup$
          But what I want to show that $Phi_X$ is a completion.
          $endgroup$
          – James
          Jan 23 at 20:29












          $begingroup$
          Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
          $endgroup$
          – Martin Argerami
          Jan 23 at 20:32




          $begingroup$
          Yes, the closure of $Phi_X(X)$ is $X'$. Not sure what else you expect.
          $endgroup$
          – Martin Argerami
          Jan 23 at 20:32












          $begingroup$
          Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
          $endgroup$
          – James
          Jan 23 at 20:33




          $begingroup$
          Well, I have to show of course... and I have trouble showing that $X'subseteqoverline{Phi_X(X)}$
          $endgroup$
          – James
          Jan 23 at 20:33












          $begingroup$
          Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
          $endgroup$
          – Martin Argerami
          Jan 23 at 20:44




          $begingroup$
          Sorry, I fail to see what else you want. You work on the completion of $X$, then as you mentioned you get that $overline{phi_X(overline X)}=X'$, but $Phi_X$ is an isometry so $phi_X(overline X)=overline{phi_X(X)}$.
          $endgroup$
          – Martin Argerami
          Jan 23 at 20:44




          1




          1




          $begingroup$
          Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
          $endgroup$
          – Martin Argerami
          Jan 23 at 23:05




          $begingroup$
          Yes, that's what I was saying all the time. You have $$overline{T(X)}=overline{Phi_X(iota(X))}=Phi_X(overline{iota(X)})=Phi_X(overline X)=X'.$$
          $endgroup$
          – Martin Argerami
          Jan 23 at 23:05


















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