The question is already indicated as in the title, is it possible to find a normed space $X$ such that the continuous dual $X^{\ast}$ is isomorphic to the Euclidean Lebesgue space $L^{1}(\mathbb{R}^{n})$?
Having discussed with a professor, he said, if it were true, then it will contradict the Banach-Alaoglu Theorem: The closed unit ball $B_{1}(0)$ of $L^{1}(\mathbb{R}^{n})$ is weak$^{\ast}$-compact.
I still fail to see why there is such a contradiction.
Best Answer
@Ruy is right. $L^1(\mathbb R^n)$, with the usual Lebesgue measure, is not linearly homeomorphic to the dual of any Banach space. For example, one reason is that it is a separable space that fails the Radon-Nikodym property.
A nice reference on this material is
Diestel, J.; Uhl, J. J. jun., Vector measures, Mathematical Surveys. No. 15. Providence, R.I.: American Mathematical Society (AMS). XIII, 322 p. $ 35.60 (1977). ZBL0369.46039.