Given the subspace of $L^2$ made by constant functions, characterize its orthogonal complement












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Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.




I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.



$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$



$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$



So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.



Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?






functional-analysis hilbert-spaces







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  • 1




    $begingroup$
    Your answer is correct and complete.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 10:22






  • 1




    $begingroup$
    complete and correct.
    $endgroup$
    – DisintegratingByParts
    Nov 30 '18 at 10:38
















1












$begingroup$



Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.




I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.



$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$



$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$



So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.



Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?






functional-analysis hilbert-spaces







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your answer is correct and complete.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 10:22






  • 1




    $begingroup$
    complete and correct.
    $endgroup$
    – DisintegratingByParts
    Nov 30 '18 at 10:38














1












1








1





$begingroup$



Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.




I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.



$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$



$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$



So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.



Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?






functional-analysis hilbert-spaces







share|cite|improve this question









$endgroup$





Let $X={vin L^2([-1,1]):v text{ is constant a.e.}}$ be a subspace of $L^2([-1,1])$, characterize $X^{perp}$.




I don't really know what characterize means, I know that: $X^{perp}={fin L^2([-1,1]):langle f,v rangle=0, forall vin X}$.



$$langle f,v rangle = int^1_{-1}fv dx = vint^1_{-1}f,quad text{since $v$ is constant.}$$



$$vint^1_{-1}f=0quad text{for every $vin X$ iff}quad int^1_{-1}f=0.$$



So $X^{perp}={fin L^2([-1,1]):int^1_{-1}f=0}$.



Is this enough to characterize the set $X^{perp}$ or I need to say something on the orthogonal projections of functions in $L^2([-1,1])$ on $X$ too?






functional-analysis hilbert-spaces




functional-analysis hilbert-spaces






share|cite|improve this question













share|cite|improve this question











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asked Nov 30 '18 at 10:15









sound wavesound wave

1618




1618








  • 1




    $begingroup$
    Your answer is correct and complete.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 10:22






  • 1




    $begingroup$
    complete and correct.
    $endgroup$
    – DisintegratingByParts
    Nov 30 '18 at 10:38














  • 1




    $begingroup$
    Your answer is correct and complete.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 10:22






  • 1




    $begingroup$
    complete and correct.
    $endgroup$
    – DisintegratingByParts
    Nov 30 '18 at 10:38








1




1




$begingroup$
Your answer is correct and complete.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 10:22




$begingroup$
Your answer is correct and complete.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 10:22




1




1




$begingroup$
complete and correct.
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:38




$begingroup$
complete and correct.
$endgroup$
– DisintegratingByParts
Nov 30 '18 at 10:38










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While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.



Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).






share|cite|improve this answer









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    $begingroup$

    While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.



    Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.



      Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.



        Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).






        share|cite|improve this answer









        $endgroup$



        While the word "characterize" is somewhat ambiguous, since the characterization of a set may be given in many ways (enumeration of elements, "in words", via those satisfying a given proposition, or a union/intersection of some described sets), I am sure whoever has asked this question(person/text) will have expected the answer that you have given i.e. those with integral zero.



        Which also means that there is no need to say something on the orthogonal projections of $L^2[-1,1]$ on $X$ (although you may want to think about this : it is not too difficult either).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 11:23









        астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

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