Weak convergence implies lower semicontinuity of the norm

Write $L={\sup }_{i_0}{\inf }_{i\ge i_0}\mathrm{\parallel }{x}_i\mathrm{\parallel }$. Every tail is nonempty because the index preorder is reflexive and nonempty, so its infimum exists in $[0,\mathrm{\infty })$, and their supremum exists in $[0,\mathrm{\infty }]$. For $f\in X^{\ast }$ with $\mathrm{\parallel }f\mathrm{\parallel }\le 1$ and any $\epsilon >0$, scalar convergence and $\mathrm{\mid }\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }a-b\mathrm{\mid }$ give eventually $\mathrm{\parallel }{x}_i\mathrm{\parallel }\ge \mathrm{\mid }f(x_i)\mathrm{\mid }>\mathrm{\mid }f(x)\mathrm{\mid }-\epsilon$. Thus one tail infimum is at least $\mathrm{\mid }f(x)\mathrm{\mid }-\epsilon$, whence $L\ge \mathrm{\mid }f(x)\mathrm{\mid }-\epsilon$.

If $L=\mathrm{\infty }$ the assertion holds. Otherwise letting the positive error decrease shows $L\ge \mathrm{\mid }f(x)\mathrm{\mid }$ for every dual unit-ball member: a positive gap is contradicted by half that gap. The HB norm formula yields $L\ge \mathrm{\parallel }x\mathrm{\parallel }$. At $x=0$ this follows already from $L\ge 0$; the zero functional ensures the dual unit ball is nonempty, even for the zero space. $\mathrm{\square }$