I want to prove that $l^\infty$ is not reflexive, however I do not want to check why canonical inclusion of $l^\infty$ to $(l^{\infty})^{**}$ is not an isometric isomorphism. I know that if a normed space is reflexive then every bounded sequence of its elements has weakly convergent subsequence. In the case of $l^\infty$ which sequence should I take to break this property?
Thank you.
Let $\phi$ be a Banach limit. Define $x_n$ as $$ x_n = (1,0,1,0,\dots, 1,0,0, \dots) = \sum_{k=1}^n e_{2n-1}. $$ Since only finitely many entries of $x_n$ are non-zero, $\phi(x_n) = 0$. The sequence $(x_n)$ converges weak-star in $l^\infty=(l^1)^*$ to the periodic sequence $x=(1,0,1,0,\dots)$. Now due to the properties of the Banach limit, $\phi(x)=\frac12 \ne \phi(0) = \lim_{n\to\infty} \phi(x_n)$. Hence $x_n \not\rightharpoonup x$ in $l^\infty$, and there can be no other weak limit, as the weak-star limit $x$ is uniquely determined.