Is it $L^2$ function ?












1












$begingroup$


For $f ∈ L^2
([0, 1))$, define $Vf(x) = int_{0}^{x}f(t) dt$.
Is $Vf$ continuous function?
Is it in $L^2$?



Thanks for any help






real-analysis integration continuity







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$endgroup$








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 2 '18 at 16:11






  • 1




    $begingroup$
    Please add your thoughts. Do you see why if the answer to the first question is yes, then you can answer the second question?
    $endgroup$
    – qbert
    Dec 2 '18 at 16:45
















1












$begingroup$


For $f ∈ L^2
([0, 1))$, define $Vf(x) = int_{0}^{x}f(t) dt$.
Is $Vf$ continuous function?
Is it in $L^2$?



Thanks for any help






real-analysis integration continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 2 '18 at 16:11






  • 1




    $begingroup$
    Please add your thoughts. Do you see why if the answer to the first question is yes, then you can answer the second question?
    $endgroup$
    – qbert
    Dec 2 '18 at 16:45














1












1








1





$begingroup$


For $f ∈ L^2
([0, 1))$, define $Vf(x) = int_{0}^{x}f(t) dt$.
Is $Vf$ continuous function?
Is it in $L^2$?



Thanks for any help






real-analysis integration continuity







share|cite|improve this question









$endgroup$




For $f ∈ L^2
([0, 1))$, define $Vf(x) = int_{0}^{x}f(t) dt$.
Is $Vf$ continuous function?
Is it in $L^2$?



Thanks for any help






real-analysis integration continuity




real-analysis integration continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 16:10









PolPol

1268




1268








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 2 '18 at 16:11






  • 1




    $begingroup$
    Please add your thoughts. Do you see why if the answer to the first question is yes, then you can answer the second question?
    $endgroup$
    – qbert
    Dec 2 '18 at 16:45














  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – MisterRiemann
    Dec 2 '18 at 16:11






  • 1




    $begingroup$
    Please add your thoughts. Do you see why if the answer to the first question is yes, then you can answer the second question?
    $endgroup$
    – qbert
    Dec 2 '18 at 16:45








1




1




$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 2 '18 at 16:11




$begingroup$
What have you tried?
$endgroup$
– MisterRiemann
Dec 2 '18 at 16:11




1




1




$begingroup$
Please add your thoughts. Do you see why if the answer to the first question is yes, then you can answer the second question?
$endgroup$
– qbert
Dec 2 '18 at 16:45




$begingroup$
Please add your thoughts. Do you see why if the answer to the first question is yes, then you can answer the second question?
$endgroup$
– qbert
Dec 2 '18 at 16:45










1 Answer
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$begingroup$

The answer is yes to both questions: Let $x,yin [0,1)$ such that $y<x$. Then
$$
|Vf(x)-Vf(y)|=|int_y^xf(t)dt|leqint_y^x|f(t)|dt=int_0^1|f(t)|chi_{[y,x]}(t)dtleq ||f||_2|x-y|^{1/2}
$$
where in the last inequality we use the Cauchy-Schwartz inequality. So $Vf$ is continuous. On the other hand, by the previous step
$$
int_0^1|int_0^xf(t)dt|^2dxleqint_0^1||f||_2^2xdx=frac{||f||_2^2}{2}
$$
So $Vf$ is in $L^2[0,1)$.






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    $begingroup$

    The answer is yes to both questions: Let $x,yin [0,1)$ such that $y<x$. Then
    $$
    |Vf(x)-Vf(y)|=|int_y^xf(t)dt|leqint_y^x|f(t)|dt=int_0^1|f(t)|chi_{[y,x]}(t)dtleq ||f||_2|x-y|^{1/2}
    $$
    where in the last inequality we use the Cauchy-Schwartz inequality. So $Vf$ is continuous. On the other hand, by the previous step
    $$
    int_0^1|int_0^xf(t)dt|^2dxleqint_0^1||f||_2^2xdx=frac{||f||_2^2}{2}
    $$
    So $Vf$ is in $L^2[0,1)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The answer is yes to both questions: Let $x,yin [0,1)$ such that $y<x$. Then
      $$
      |Vf(x)-Vf(y)|=|int_y^xf(t)dt|leqint_y^x|f(t)|dt=int_0^1|f(t)|chi_{[y,x]}(t)dtleq ||f||_2|x-y|^{1/2}
      $$
      where in the last inequality we use the Cauchy-Schwartz inequality. So $Vf$ is continuous. On the other hand, by the previous step
      $$
      int_0^1|int_0^xf(t)dt|^2dxleqint_0^1||f||_2^2xdx=frac{||f||_2^2}{2}
      $$
      So $Vf$ is in $L^2[0,1)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The answer is yes to both questions: Let $x,yin [0,1)$ such that $y<x$. Then
        $$
        |Vf(x)-Vf(y)|=|int_y^xf(t)dt|leqint_y^x|f(t)|dt=int_0^1|f(t)|chi_{[y,x]}(t)dtleq ||f||_2|x-y|^{1/2}
        $$
        where in the last inequality we use the Cauchy-Schwartz inequality. So $Vf$ is continuous. On the other hand, by the previous step
        $$
        int_0^1|int_0^xf(t)dt|^2dxleqint_0^1||f||_2^2xdx=frac{||f||_2^2}{2}
        $$
        So $Vf$ is in $L^2[0,1)$.






        share|cite|improve this answer









        $endgroup$



        The answer is yes to both questions: Let $x,yin [0,1)$ such that $y<x$. Then
        $$
        |Vf(x)-Vf(y)|=|int_y^xf(t)dt|leqint_y^x|f(t)|dt=int_0^1|f(t)|chi_{[y,x]}(t)dtleq ||f||_2|x-y|^{1/2}
        $$
        where in the last inequality we use the Cauchy-Schwartz inequality. So $Vf$ is continuous. On the other hand, by the previous step
        $$
        int_0^1|int_0^xf(t)dt|^2dxleqint_0^1||f||_2^2xdx=frac{||f||_2^2}{2}
        $$
        So $Vf$ is in $L^2[0,1)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 16:59









        Dante GrevinoDante Grevino

        1,100112




        1,100112






























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