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 $L^2[0,1]$?

Lev Bahn
  • 2,892
  • 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 09 '18 at 18:48

1 Answers1

1

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. $f(x) = x^{-1/2}$.

Robert Israel
  • 448,999