Consider linear functional on $L^p[0, 1]$, where $1 < p < ∞$:
$l(f) = \int_{1/3}^{1/2}f - \int_{3/4}^{1} f $. Definition of bounded linear funtional is $$||l|| = \sup\{|l(f)|:||f||_p=1\} <\infty$$
I know that $|l(f)| = |\int_{1/3}^{1/2}f - \int_{3/4}^{1} f| \le \int_0^1 |f|$. For $||f||_p=1$, we need $\int_0^1 |f|^p = 1$. I don't know how to proceed then.
I think it is unbounded. I am trying to find a function such that $\int_{0}^{1}|f|^p =1$ but somehow unbounded on interval $[1/3,1/2]$ or $[3/4,1]$.