An example of bounded linear functional on $L^3[0,1]$ that is not on $L^2[0,1]$.

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Here, our measure is Lebesgue measure.



Is there a bounded linear functional on $L^3[0,1]$ that is not the restriction to $L^3[0,1]$ of a bounded linear functional on $L^2[0,1]$?




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    It is the same to select an element of $L^3/2$ which is not an element of $L^2$, in view of Riesz. I can think of a few of those...
    – Ian
    Aug 9 at 18:48















up vote
1
down vote

favorite
1












Here, our measure is Lebesgue measure.



Is there a bounded linear functional on $L^3[0,1]$ that is not the restriction to $L^3[0,1]$ of a bounded linear functional on $L^2[0,1]$?




lp-spaces




share|cite|improve this question
















  • 2




    It is the same to select an element of $L^3/2$ which is not an element of $L^2$, in view of Riesz. I can think of a few of those...
    – Ian
    Aug 9 at 18:48













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Here, our measure is Lebesgue measure.



Is there a bounded linear functional on $L^3[0,1]$ that is not the restriction to $L^3[0,1]$ of a bounded linear functional on $L^2[0,1]$?




lp-spaces




share|cite|improve this question












Here, our measure is Lebesgue measure.



Is there a bounded linear functional on $L^3[0,1]$ that is not the restriction to $L^3[0,1]$ of a bounded linear functional on $L^2[0,1]$?





lp-spaces





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 9 at 18:44









Lev Ban

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  • 2




    It is the same to select an element of $L^3/2$ which is not an element of $L^2$, in view of Riesz. I can think of a few of those...
    – Ian
    Aug 9 at 18:48













  • 2




    It is the same to select an element of $L^3/2$ which is not an element of $L^2$, in view of Riesz. I can think of a few of those...
    – Ian
    Aug 9 at 18:48








2




2




It is the same to select an element of $L^3/2$ which is not an element of $L^2$, in view of Riesz. I can think of a few of those...
– Ian
Aug 9 at 18:48





It is the same to select an element of $L^3/2$ which is not an element of $L^2$, in view of Riesz. I can think of a few of those...
– Ian
Aug 9 at 18:48











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The bounded linear functionals on $L^3[0,1]$ correspond to members of $L^3/2[0,1]$ (since $1/3 + 2/3 = 1$). But there are members of $L^3/2(0,1)$ that are not in $L^2[0,1]$, e.g. $f(x) = x^-1/2$.






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    The bounded linear functionals on $L^3[0,1]$ correspond to members of $L^3/2[0,1]$ (since $1/3 + 2/3 = 1$). But there are members of $L^3/2(0,1)$ that are not in $L^2[0,1]$, e.g. $f(x) = x^-1/2$.






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      up vote
      1
      down vote



      accepted










      The bounded linear functionals on $L^3[0,1]$ correspond to members of $L^3/2[0,1]$ (since $1/3 + 2/3 = 1$). But there are members of $L^3/2(0,1)$ that are not in $L^2[0,1]$, e.g. $f(x) = x^-1/2$.






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        The bounded linear functionals on $L^3[0,1]$ correspond to members of $L^3/2[0,1]$ (since $1/3 + 2/3 = 1$). But there are members of $L^3/2(0,1)$ that are not in $L^2[0,1]$, e.g. $f(x) = x^-1/2$.






        share|cite|improve this answer












        The bounded linear functionals on $L^3[0,1]$ correspond to members of $L^3/2[0,1]$ (since $1/3 + 2/3 = 1$). But there are members of $L^3/2(0,1)$ that are not in $L^2[0,1]$, e.g. $f(x) = x^-1/2$.







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        share|cite|improve this answer



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        answered Aug 9 at 18:47









        Robert Israel

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