 Solution

For a Banach space $X$, weak convergence $u_k\rightharpoonup u$ meansFor its continuous dual space $X^{\ast }$, convergence in the weak-star topology meansWeak convergence on $X^{\ast }$ tests against all of $X^{\ast \ast }$, whereas weak-star convergence tests only against the canonical image of $X$. Thus weak convergence always implies weak-star convergence. If $X$ is a reflexive Banach space, that image is all of $X^{\ast \ast }$, so the weak and weak-star topologies on $X^{\ast }$ coincide. Weak-star topology is defined relative to a chosen predual; it is not an extra intrinsic topology on an arbitrary Banach space without such an identification.