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

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 $L^2[0,1]$?

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