 Solution

The weak topology on a normed vector space $X$ is $\sigma (X,X^{\ast })$, the coarsest topology making every bounded linear functional continuous. A neighbourhood base at $x$ is given by finitely many inequalities $|f_j(y-x)|<\epsilon$, with $f_j\in X^{\ast }$. In the complex case, separation uses real parts of bounded linear functionals.

Mazur theorem states that the weak closure of a convex set equals its closure in the norm topology. The weak topology is coarser than the norm topology, so the norm closure is contained in the weak closure. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and $a\in \mathbb{R}$ with $\mathrm{Re}f(x)>a\ge {\sup }_{c\in C}\mathrm{Re}f(c)$. The corresponding weak topology neighbourhood of $x$ misses $C$, so $x$ is outside its weak closure. This proves Mazur theorem, including the empty-set case. In particular, if $x_n$ converges weakly to $x$, then $x$ is in the weak closure of every tail and therefore in the norm closure of its convex hull. Choosing a finite convex combination of the $n$th tail within $1/n$ of $x$ proves the usual Mazur lemma formulation as well.

The weak-star topology on the continuous dual space $X^{\ast }$ is $\sigma (X^{\ast },X)$: convergence means pointwise convergence on $X$, and a neighbourhood base prescribes finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this weak-star topology, even if $X$ is incomplete. Embed this closed unit ball intowhere $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Each factor is compact, so $P$ is compact by the Tychonoff theorem. Inside $P$, the equations $a_{x+y}=a_x+a_y$ and $a_{\lambda x}=\lambda a_x$ define a closed set. Every such point defines a linear functional satisfying $|a_x|\le \Vert x\Vert$, hence belongs to the closed unit ball of $X^{\ast }$. Thus this image is closed in $P$. The product topology on it is exactly the weak-star topology, proving Banach-Alaoglu theorem. Evaluations also separate its points, so the weak-star topology is Hausdorff.

For the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, we have $(J_{X}x)(f)=f(x)$. Restricting all evaluations at $f\in X^{\ast }$ therefore gives exactly $\sigma (X,X^{\ast })$. Thus the induced subspace topology is the weak topology on $X$.

Now identify $X$ with $J_{X}X$. Let $C$ be a bounded convex set, write $D$ for its norm closure in $X$, and let $K$ be its weak-star topology closure in $X^{\ast \ast }$. Boundedness places $K$ in a multiple of the closed unit ball of $X^{\ast \ast }$, so $K$ is compact by Banach-Alaoglu theorem. The preceding subspace topology identification and Mazur theorem giveIf $D$ is a weakly compact set, its image in the Hausdorff weak-star topology is compact and therefore closed. It contains $C$, so $K\subseteq D\subseteq X$. Conversely, if $K\subseteq X$, the displayed identity gives $K=D$, and its compactness is precisely weak compactness in $X$. Hence $\boxed{D\text{ is weakly compact}⟺K\subseteq X}$. This argument also covers $C=\varnothing$.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint is $T^{\ast }:Y^{\ast }\to X^{\ast }$, defined by $(T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$. Evaluation at any fixed $x$ is thus evaluation at $Tx$ after applying $T^{\ast }$. Each is continuous in the relevant weak-star topology, proving that $T^{\ast }$ is weak-star continuous. Applying the same result to $T^{\ast }$ shows that $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star continuous. Direct evaluation gives

Use $B_X$ for the closed unit ball; using the open ball gives the same norm closure of $T(B_X)$. Goldstine theorem says that $J_X{B}_X$ is weak-star dense in $B_{X^{\ast \ast }}$. Put $K=T^{\ast \ast }(B_{X^{\ast \ast }})$. It is compact and closed in the weak-star topology by Banach-Alaoglu theorem and the established continuity. It contains $J_{Y}T(B_X)$. Conversely, Goldstine theorem gives, for every $x^{\ast \ast }\in B_{X^{\ast \ast }}$, a net $(J_X{x}_{\alpha })$ from $J_X{B}_X$ converging weak-star to $x^{\ast \ast }$; its image converges weak-star to $T^{\ast \ast }{x}^{\ast \ast }$. ConsequentlyApply the preceding bounded convex set criterion in $Y$, and then scale the closed unit ball. We obtain the bidual characterization of weakly compact operators:The canonical embedding into the bidual on the right specifies exactly which copy of $Y$ is intended.