Prove the inequality $int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right)...












2












$begingroup$


I want to prove the inequality



begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}



Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!






integration functional-analysis analysis







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  • 5




    $begingroup$
    Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
    $endgroup$
    – Michael Burr
    Dec 15 '18 at 13:17








  • 1




    $begingroup$
    $ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
    $endgroup$
    – user587192
    Dec 16 '18 at 14:21
















2












$begingroup$


I want to prove the inequality



begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}



Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!






integration functional-analysis analysis







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
    $endgroup$
    – Michael Burr
    Dec 15 '18 at 13:17








  • 1




    $begingroup$
    $ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
    $endgroup$
    – user587192
    Dec 16 '18 at 14:21














2












2








2


1



$begingroup$


I want to prove the inequality



begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}



Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!






integration functional-analysis analysis







share|cite|improve this question









$endgroup$




I want to prove the inequality



begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}



Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!






integration functional-analysis analysis




integration functional-analysis analysis






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Dec 15 '18 at 13:15









Omojola MichealOmojola Micheal

1,853324




1,853324








  • 5




    $begingroup$
    Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
    $endgroup$
    – Michael Burr
    Dec 15 '18 at 13:17








  • 1




    $begingroup$
    $ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
    $endgroup$
    – user587192
    Dec 16 '18 at 14:21














  • 5




    $begingroup$
    Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
    $endgroup$
    – Michael Burr
    Dec 15 '18 at 13:17








  • 1




    $begingroup$
    $ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
    $endgroup$
    – user587192
    Dec 16 '18 at 14:21








5




5




$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17






$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17






1




1




$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21




$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21










1 Answer
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$begingroup$

Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
$$
mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
0 &;xne Omega.
end{cases}
$$
Here you have $Omega=[-3,3]$. Can you finish the proof?






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

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    3












    $begingroup$

    Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
    $$
    mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
    0 &;xne Omega.
    end{cases}
    $$
    Here you have $Omega=[-3,3]$. Can you finish the proof?






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
      $$
      mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
      0 &;xne Omega.
      end{cases}
      $$
      Here you have $Omega=[-3,3]$. Can you finish the proof?






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
        $$
        mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
        0 &;xne Omega.
        end{cases}
        $$
        Here you have $Omega=[-3,3]$. Can you finish the proof?






        share|cite|improve this answer









        $endgroup$



        Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
        $$
        mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
        0 &;xne Omega.
        end{cases}
        $$
        Here you have $Omega=[-3,3]$. Can you finish the proof?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 14:15









        BigbearZzzBigbearZzz

        8,69121652




        8,69121652






























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