$l_2(S)$is a hilbert space where S is a subset












1












$begingroup$



Let S be a non-empty set and $l_2(S)$ be the set of all complex functions $f$ defined on $S$ with the following two properties:



$(1) {s:f(s)ne0}$ is empty or countable. $(2) sum{|f(s)|}^2 < +infty$



Then show that :



(a) $l_2(S)$ forms a complex linear space with respect to pointwise addition and scalar multiplication.



(b) If norm and inner product is defined as, $||f||=(sum{|f(s)|}^2)^{frac{1}{2}}$ and $ <f,g>=sum f(s)bar{g(s)}$ respectively, then $l_2(S)$ is actually a Hilbert space.




My attempt :



(a) Easily done.



(b) Clearly a complex linear space and then assuming it to be Banach, considering for each $f in l_2(S)$ , the form $f(s)=f_1(s)+if_2(s)$ i.e. considering its Real and Imaginary parts, computed and proved that the 'parallelogram law' holds for $l_2(S)$. But stuck on how to show that the space is actually complete.



Thanks in advance for help.






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Actually, $ell^p(S)$ is complete for all $1leq pleqinfty$ and any set $S$. See math.stackexchange.com/questions/486786/… for example.
    $endgroup$
    – SmileyCraft
    Jan 22 at 13:46










  • $begingroup$
    Besides if You are interested in the analogous case of $p$-Schatten classes You might have a look at Theorem 15 here:fernuni-hagen.de/analysis/download/diplomarbeit_melech.pdf
    $endgroup$
    – Peter Melech
    Jan 22 at 13:59
















1












$begingroup$



Let S be a non-empty set and $l_2(S)$ be the set of all complex functions $f$ defined on $S$ with the following two properties:



$(1) {s:f(s)ne0}$ is empty or countable. $(2) sum{|f(s)|}^2 < +infty$



Then show that :



(a) $l_2(S)$ forms a complex linear space with respect to pointwise addition and scalar multiplication.



(b) If norm and inner product is defined as, $||f||=(sum{|f(s)|}^2)^{frac{1}{2}}$ and $ <f,g>=sum f(s)bar{g(s)}$ respectively, then $l_2(S)$ is actually a Hilbert space.




My attempt :



(a) Easily done.



(b) Clearly a complex linear space and then assuming it to be Banach, considering for each $f in l_2(S)$ , the form $f(s)=f_1(s)+if_2(s)$ i.e. considering its Real and Imaginary parts, computed and proved that the 'parallelogram law' holds for $l_2(S)$. But stuck on how to show that the space is actually complete.



Thanks in advance for help.






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Actually, $ell^p(S)$ is complete for all $1leq pleqinfty$ and any set $S$. See math.stackexchange.com/questions/486786/… for example.
    $endgroup$
    – SmileyCraft
    Jan 22 at 13:46










  • $begingroup$
    Besides if You are interested in the analogous case of $p$-Schatten classes You might have a look at Theorem 15 here:fernuni-hagen.de/analysis/download/diplomarbeit_melech.pdf
    $endgroup$
    – Peter Melech
    Jan 22 at 13:59














1












1








1





$begingroup$



Let S be a non-empty set and $l_2(S)$ be the set of all complex functions $f$ defined on $S$ with the following two properties:



$(1) {s:f(s)ne0}$ is empty or countable. $(2) sum{|f(s)|}^2 < +infty$



Then show that :



(a) $l_2(S)$ forms a complex linear space with respect to pointwise addition and scalar multiplication.



(b) If norm and inner product is defined as, $||f||=(sum{|f(s)|}^2)^{frac{1}{2}}$ and $ <f,g>=sum f(s)bar{g(s)}$ respectively, then $l_2(S)$ is actually a Hilbert space.




My attempt :



(a) Easily done.



(b) Clearly a complex linear space and then assuming it to be Banach, considering for each $f in l_2(S)$ , the form $f(s)=f_1(s)+if_2(s)$ i.e. considering its Real and Imaginary parts, computed and proved that the 'parallelogram law' holds for $l_2(S)$. But stuck on how to show that the space is actually complete.



Thanks in advance for help.






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$





Let S be a non-empty set and $l_2(S)$ be the set of all complex functions $f$ defined on $S$ with the following two properties:



$(1) {s:f(s)ne0}$ is empty or countable. $(2) sum{|f(s)|}^2 < +infty$



Then show that :



(a) $l_2(S)$ forms a complex linear space with respect to pointwise addition and scalar multiplication.



(b) If norm and inner product is defined as, $||f||=(sum{|f(s)|}^2)^{frac{1}{2}}$ and $ <f,g>=sum f(s)bar{g(s)}$ respectively, then $l_2(S)$ is actually a Hilbert space.




My attempt :



(a) Easily done.



(b) Clearly a complex linear space and then assuming it to be Banach, considering for each $f in l_2(S)$ , the form $f(s)=f_1(s)+if_2(s)$ i.e. considering its Real and Imaginary parts, computed and proved that the 'parallelogram law' holds for $l_2(S)$. But stuck on how to show that the space is actually complete.



Thanks in advance for help.






functional-analysis operator-theory hilbert-spaces




functional-analysis operator-theory hilbert-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 22 at 13:41









CoherentCoherent

1,191523




1,191523








  • 1




    $begingroup$
    Actually, $ell^p(S)$ is complete for all $1leq pleqinfty$ and any set $S$. See math.stackexchange.com/questions/486786/… for example.
    $endgroup$
    – SmileyCraft
    Jan 22 at 13:46










  • $begingroup$
    Besides if You are interested in the analogous case of $p$-Schatten classes You might have a look at Theorem 15 here:fernuni-hagen.de/analysis/download/diplomarbeit_melech.pdf
    $endgroup$
    – Peter Melech
    Jan 22 at 13:59














  • 1




    $begingroup$
    Actually, $ell^p(S)$ is complete for all $1leq pleqinfty$ and any set $S$. See math.stackexchange.com/questions/486786/… for example.
    $endgroup$
    – SmileyCraft
    Jan 22 at 13:46










  • $begingroup$
    Besides if You are interested in the analogous case of $p$-Schatten classes You might have a look at Theorem 15 here:fernuni-hagen.de/analysis/download/diplomarbeit_melech.pdf
    $endgroup$
    – Peter Melech
    Jan 22 at 13:59








1




1




$begingroup$
Actually, $ell^p(S)$ is complete for all $1leq pleqinfty$ and any set $S$. See math.stackexchange.com/questions/486786/… for example.
$endgroup$
– SmileyCraft
Jan 22 at 13:46




$begingroup$
Actually, $ell^p(S)$ is complete for all $1leq pleqinfty$ and any set $S$. See math.stackexchange.com/questions/486786/… for example.
$endgroup$
– SmileyCraft
Jan 22 at 13:46












$begingroup$
Besides if You are interested in the analogous case of $p$-Schatten classes You might have a look at Theorem 15 here:fernuni-hagen.de/analysis/download/diplomarbeit_melech.pdf
$endgroup$
– Peter Melech
Jan 22 at 13:59




$begingroup$
Besides if You are interested in the analogous case of $p$-Schatten classes You might have a look at Theorem 15 here:fernuni-hagen.de/analysis/download/diplomarbeit_melech.pdf
$endgroup$
– Peter Melech
Jan 22 at 13:59










1 Answer
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$begingroup$

You can reduce to the case that $l_2$ is complete:



Assume you have a cauchy sequence $(f_n)$ in $l_2(S)$. Set $$N=bigcup_{nin mathbb{N} } {s:f_n(s)neq 0}.$$ Then you identify $l_2(N)$ as a subspace of $l_2(S)$ and under this identification your sequence lives in $l_2(N)$. As $N$ is countable $l_2(N)cong l_2$ or $l_2(N)cong mathbb{C}^N$ depending on whether $N$ is infinite or not, so your sequence converges.



A proof for the completeness of $l_2$ can be found here (just replace $mathbb R$ by $mathbb C$): Understanding the proof of $l_2$ being complete.






share|cite|improve this answer











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    $begingroup$

    You can reduce to the case that $l_2$ is complete:



    Assume you have a cauchy sequence $(f_n)$ in $l_2(S)$. Set $$N=bigcup_{nin mathbb{N} } {s:f_n(s)neq 0}.$$ Then you identify $l_2(N)$ as a subspace of $l_2(S)$ and under this identification your sequence lives in $l_2(N)$. As $N$ is countable $l_2(N)cong l_2$ or $l_2(N)cong mathbb{C}^N$ depending on whether $N$ is infinite or not, so your sequence converges.



    A proof for the completeness of $l_2$ can be found here (just replace $mathbb R$ by $mathbb C$): Understanding the proof of $l_2$ being complete.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You can reduce to the case that $l_2$ is complete:



      Assume you have a cauchy sequence $(f_n)$ in $l_2(S)$. Set $$N=bigcup_{nin mathbb{N} } {s:f_n(s)neq 0}.$$ Then you identify $l_2(N)$ as a subspace of $l_2(S)$ and under this identification your sequence lives in $l_2(N)$. As $N$ is countable $l_2(N)cong l_2$ or $l_2(N)cong mathbb{C}^N$ depending on whether $N$ is infinite or not, so your sequence converges.



      A proof for the completeness of $l_2$ can be found here (just replace $mathbb R$ by $mathbb C$): Understanding the proof of $l_2$ being complete.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You can reduce to the case that $l_2$ is complete:



        Assume you have a cauchy sequence $(f_n)$ in $l_2(S)$. Set $$N=bigcup_{nin mathbb{N} } {s:f_n(s)neq 0}.$$ Then you identify $l_2(N)$ as a subspace of $l_2(S)$ and under this identification your sequence lives in $l_2(N)$. As $N$ is countable $l_2(N)cong l_2$ or $l_2(N)cong mathbb{C}^N$ depending on whether $N$ is infinite or not, so your sequence converges.



        A proof for the completeness of $l_2$ can be found here (just replace $mathbb R$ by $mathbb C$): Understanding the proof of $l_2$ being complete.






        share|cite|improve this answer











        $endgroup$



        You can reduce to the case that $l_2$ is complete:



        Assume you have a cauchy sequence $(f_n)$ in $l_2(S)$. Set $$N=bigcup_{nin mathbb{N} } {s:f_n(s)neq 0}.$$ Then you identify $l_2(N)$ as a subspace of $l_2(S)$ and under this identification your sequence lives in $l_2(N)$. As $N$ is countable $l_2(N)cong l_2$ or $l_2(N)cong mathbb{C}^N$ depending on whether $N$ is infinite or not, so your sequence converges.



        A proof for the completeness of $l_2$ can be found here (just replace $mathbb R$ by $mathbb C$): Understanding the proof of $l_2$ being complete.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 28 at 22:17

























        answered Jan 22 at 13:55









        triitrii

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