How would one prove that $L^3$ is reflexive?

Question

I have no idea where to even start with this. I think the idea should be that we need an isomorphism from $L^3\to (L^{3/2})^*$ and then an isomorphism from $(L^{3/2})^*\to (L^3)^{**}$, but I don't see how to do this.

Answer 1

Using the duality properties of $L^p$ spaces: $$ (L^p)^{**}= (L^{p'})^* = L^p,\qquad \frac1p + \frac 1{p'}= 1,\qquad 1<p<\infty. $$

EDIT: more details.

For $g\in L^{p'}$ we define $$k_p(g): f\longmapsto\int fg.$$ Can be proved that $k_p(g)\in (L^p)^*$ (easy) and that $$k_p:L^{p'}\longrightarrow (L^p)^*$$ is an isometric isomorphism (less easy). Finally, the composition $$(k_p^{-1})^*\circ k_{p'}: L^p\longrightarrow (L^{p'})^*\longrightarrow (L^p)^{**}$$ is an isometric isomorphism and coincides with the canonical embedding in the bidual.