Is the following operator a compact operator?












1














I have to choose whether the following operator



$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$



is a compact operator. I have tried to use the Cauchy Schwarz inequality



$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$



but how can I identify the compactness of an operator?






functional-analysis operator-theory cauchy-schwarz-inequality







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  • 2




    Any continuous linear map into a finite dimensional vector space is necessarily compact.
    – Omnomnomnom
    Nov 20 at 15:36
















1














I have to choose whether the following operator



$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$



is a compact operator. I have tried to use the Cauchy Schwarz inequality



$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$



but how can I identify the compactness of an operator?






functional-analysis operator-theory cauchy-schwarz-inequality







share|cite|improve this question


















  • 2




    Any continuous linear map into a finite dimensional vector space is necessarily compact.
    – Omnomnomnom
    Nov 20 at 15:36














1












1








1







I have to choose whether the following operator



$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$



is a compact operator. I have tried to use the Cauchy Schwarz inequality



$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$



but how can I identify the compactness of an operator?






functional-analysis operator-theory cauchy-schwarz-inequality







share|cite|improve this question













I have to choose whether the following operator



$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$



is a compact operator. I have tried to use the Cauchy Schwarz inequality



$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$



but how can I identify the compactness of an operator?






functional-analysis operator-theory cauchy-schwarz-inequality




functional-analysis operator-theory cauchy-schwarz-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 15:26









MathCracky

445212




445212








  • 2




    Any continuous linear map into a finite dimensional vector space is necessarily compact.
    – Omnomnomnom
    Nov 20 at 15:36














  • 2




    Any continuous linear map into a finite dimensional vector space is necessarily compact.
    – Omnomnomnom
    Nov 20 at 15:36








2




2




Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36




Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36










1 Answer
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$T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.






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    1 Answer
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    active

    oldest

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    active

    oldest

    votes









    3














    $T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.






    share|cite|improve this answer


























      3














      $T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.






      share|cite|improve this answer
























        3












        3








        3






        $T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.






        share|cite|improve this answer












        $T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 15:31









        Federico

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