Dual of L infinity is larger than L1 ($(L^{\infty }{)}^{\ast }⊋L^1$)

On ${\mathbb{R}}^n$, integration embeds $L^1$ isometrically into $(L^{\infty }{)}^{\ast }$, but this is not surjective. For example, the functional taking the limit at infinity on $\mathbb{C}1\oplus C_0({\mathbb{R}}^n)$ extends to $L^{\infty }$ by the Hahn-Banach theorem; it vanishes on compactly supported functions and is one on the constant function, so no $L^1$ density represents it.