Coordinate evaluations converge weak star to zero in ell one star

The complex coefficient identification is F1. For real scalars, a bounded sequence $b$ defines $h_b(x)={\sum }_k{b}_k{x}_k$ with $\mathrm{\mid }{h}_b(x)\mathrm{\mid }\le \mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x{\mathrm{\parallel }}_1$. Conversely if $h$ is bounded, $b_k=h(e_k)$ satisfies $\mathrm{\mid }{b}_k\mathrm{\mid }\le \mathrm{\parallel }h\mathrm{\parallel }$; finite truncations of $x$ converge in norm because their error is the absolute series tail, so $h(x)={\sum }_k{b}_k{x}_k$. Testing $e_k$ gives $\mathrm{\parallel }{h}_b\mathrm{\parallel }\ge {\sup }_k\mathrm{\mid }{b}_k\mathrm{\mid }$, proving isometry and uniqueness. Thus the identification holds over either field.

For every $x\in {\mathrm{\ell }}^1$, convergence of ${\sum }_k\mathrm{\mid }{x}_k\mathrm{\mid }$ forces $x_n\to 0$. Hence $E_n(x)\to 0$ for each fixed $x$, proving weak-star convergence. The inequality $\mathrm{\mid }{E}_n(x)\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_1$ and equality $E_n(e_n)=1$ give $\mathrm{\parallel }{E}_n\mathrm{\parallel }=1$ exactly. $\mathrm{\square }$