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?










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    Your answer is correct and complete.
    – Kavi Rama Murthy
    Nov 30 at 10:22






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    complete and correct.
    – DisintegratingByParts
    Nov 30 at 10:38
















1















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?










share|cite|improve this question


















  • 1




    Your answer is correct and complete.
    – Kavi Rama Murthy
    Nov 30 at 10:22






  • 1




    complete and correct.
    – DisintegratingByParts
    Nov 30 at 10:38














1












1








1








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?










share|cite|improve this question














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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asked Nov 30 at 10:15









sound wave

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  • 1




    Your answer is correct and complete.
    – Kavi Rama Murthy
    Nov 30 at 10:22






  • 1




    complete and correct.
    – DisintegratingByParts
    Nov 30 at 10:38














  • 1




    Your answer is correct and complete.
    – Kavi Rama Murthy
    Nov 30 at 10:22






  • 1




    complete and correct.
    – DisintegratingByParts
    Nov 30 at 10:38








1




1




Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22




Your answer is correct and complete.
– Kavi Rama Murthy
Nov 30 at 10:22




1




1




complete and correct.
– DisintegratingByParts
Nov 30 at 10:38




complete and correct.
– DisintegratingByParts
Nov 30 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).






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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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      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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        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












        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 at 11:23









        астон вілла олоф мэллбэрг

        37.2k33376




        37.2k33376






























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