Why is $L^q$ congruent to $(L^p)^*$?

Question

Why is $L^q$ congruent to $(L^p)^*$ for $1\leq p < \infty$? Here $q=\frac{p}{p-1}$.

Answer 1

The inclusion $L^q \subset (L^p)^*$ is fairly obvious; for $g \in L^q$, define the functional $f\mapsto \int fg \,d\mu$. This is a bounded linear functional on $L^p$ by Holder's inequality.

The reverse inclusion uses the Radon-Nikodym theorem. Indeed, take $\phi \in (L^p)^*$ and define the complex measure $\nu(A) := \phi(\mathbf 1_A)$ for any measurable $A$. [It's not super obvious that this is $\nu$ is indeed a measure; you need to prove this.] Then $\nu \ll \mu$ where $\mu$ is the original measure. Thus, there is a a Radon-Nikodym derivative $g$ such that $$\phi(\mathbf 1_A) = \nu(A) = \int_A g \,d\mu = \int_X \mathbf 1_A g \, d\mu.$$ FInally, you need to prove that $g \in L^q$ and that $\phi(f) = \int_X fg \,d\mu$ for all $f \in L^p$.

As an aside, you can google $L^p$-$L^q$ duality and find this proof or a very similar one in the first few links.