$E=L^1(\Bbb R) \Rightarrow E'=L^{\infty}(\Bbb R) \Rightarrow E''\ne L^1(\Bbb R) $

Question

As an example of a banach space that is not reflexive we have $E=L^1(\Bbb R)$

As a proof I had this: $E=L^1(\Bbb R) \Rightarrow E'=L^{\infty}(\Bbb R) \Rightarrow E''\ne L^1(\Bbb R)$

I don't understand how the topological dual of $E$ is $L^{\infty}(\Bbb R)$ and why $E''\ne L^1(\Bbb R)$.

Thank you for your help.

Answer 1

I will partially answer the question.

We know that $(L^p(\Bbb R))'$ is isometric to $L^{p'}(\Bbb R)$ where $1/p+1/p'=1$ so indeed $(L^p(\Bbb R))'=L^\infty(\Bbb R)$