 Solution

Weak-star convergence means pointwise convergence on the predual:For each $x\in X$, its image $J_{X}x\in X^{\prime \prime }$ under the canonical embedding into the bidual satisfies $(J_{X}x)(\Lambda )=\Lambda (x)$. Thus weak convergence implies weak-star convergence, and altogether

For the requested example, letThe Riemann-Lebesgue lemma gives $f_j\stackrel{\ast }{\rightharpoonup }0$ in $(L^1{)}^{\prime }$. Sincethe same lemma gives $f_j^2\stackrel{\ast }{\rightharpoonup }1/2$. Hence $f=0$ and $g=1/2\ne f^2$, exhibiting the failure of weak-star convergence to commute with squaring.