Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 106 / 5

Use the canonical embedding into the bidual $J_{X}x(f)=f(x)$. It is an isometry by the Hahn-Banach theorem. The weak-star topology $\sigma (X^{\ast \ast },X^{\ast })$, restricted to $J_{X}X$, is exactly the weak topology $\sigma (X,X^{\ast })$: both are determined by the same scalar evaluations $f(x)$. Therefore if $K\subseteq X$ is a weakly compact set, $J_{X}K$ is compact in the Hausdorff weak-star topology of $X^{\ast \ast }$, and hence is closed for the weak-star topology. This explains the first assertion without treating the whole embedded space $J_{X}X$ as weak-star closed.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint isFor each $x\in X$, the scalar evaluation $y^{\ast }\mapsto (T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$ is weak-star continuous on $Y^{\ast }$. These are exactly the defining evaluations of the weak-star topology on $X^{\ast }$, so $T^{\ast }$ is weak-star-to-weak-star continuous. Equivalently, if $y_{\alpha }^{\ast }\stackrel{w^{\ast }}{\to }{y}^{\ast }$, then $(T^{\ast }{y}_{\alpha }^{\ast })(x)\to (T^{\ast }{y}^{\ast })(x)$ for every $x$.

The Goldstine theorem states that $J_X(B_X)$ is dense in $B_{X^{\ast \ast }}$ for $\sigma (X^{\ast \ast },X^{\ast })$, where the balls are closed unit balls. The Banach-Alaoglu theorem states that for any normed vector space $E$, the closed dual unit ball $B_{E^{\ast }}$ is compact for $\sigma (E^{\ast },E)$. Neither assertion claims norm compactness or, in general, sequential compactness.

For the weakly compact operator equivalences below, identify $X$ and $Y$ with their canonical images only when stated. In particular, the range condition means $T^{\ast \ast }(X^{\ast \ast })\subseteq J_{Y}Y$. We shall prove $(i)⟺(ii)$, $(ii)⟹(iii)⟹(iv)⟹(ii)$. The key identities, obtained by evaluating on $y^{\ast }\in Y^{\ast }$, are

Weak compactness forces the second adjoint to take values in $Y$. Suppose $C={\overline{T(B_X)}}^{\Vert \cdot \Vert }$ is weakly compact in $Y$. By the first observation, $J_{Y}C$ is closed for the weak-star topology in $Y^{\ast \ast }$. The second adjoint $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star-to-weak-star continuous, by the adjoint-continuity argument applied to $T^{\ast }$.

Thus $(T^{\ast \ast }{)}^{-1}(J_{Y}C)$ is closed for the weak-star topology in $X^{\ast \ast }$. The identity $T^{\ast \ast }{J}_X=J_{Y}T$ makes it contain $J_X(B_X)$. The Goldstine theorem therefore givesScaling any $F\in X^{\ast \ast }$ into the closed unit ball provesThis is the forward direction of the bidual characterization of weakly compact operators.

The bidual range condition implies both weak compactness and the stronger continuity of $T^{\ast }$. First assume $T^{\ast \ast }(X^{\ast \ast })\subseteq J_{Y}Y$. The Banach-Alaoglu theorem makes $B_{X^{\ast \ast }}$ weak-star compact, and $T^{\ast \ast }$ is weak-star continuous. Hence its image is compact for the weak-star topology in $Y^{\ast \ast }$. Because it lies in $J_{Y}Y$, it corresponds to a weakly compact subset $C\subseteq Y$ under $J_Y^{-1}$.

The identity $T^{\ast \ast }{J}_X=J_{Y}T$ gives $T(B_X)\subseteq C$. A weakly compact set in a Hausdorff space is closed for the weak topology, so $C$ is also norm-closed and contains the norm closure of $T(B_X)$. That closure is a convex set closed in the norm topology, hence weakly closed by the Mazur theorem. It is therefore a weakly closed subset of the weakly compact $C$, and is a weakly compact set. This proves $(ii)⟹(i)$.

For $(ii)⟹(iii)$, fix $F\in X^{\ast \ast }$. By the range assumption there is $y_F\in Y$ with $T^{\ast \ast }F=J_Y{y}_F$. For every $y^{\ast }\in Y^{\ast }$,The right side is a defining weak-star continuous evaluation on $Y^{\ast }$. Since the weak topology on $X^{\ast }$ is determined by all $F\in X^{\ast \ast }$, every scalar coordinate of $T^{\ast }$ for that topology is weak-star continuous. Therefore

Weak-star-to-weak continuity of the adjoint implies its weak compactness. By the Banach-Alaoglu theorem, $B_{Y^{\ast }}$ is compact for the weak-star topology. Under the assumed continuous map $T^{\ast }:(Y^{\ast },w^{\ast })\to (X^{\ast },w)$, its image $T^{\ast }(B_{Y^{\ast }})$ is weakly compact in $X^{\ast }$.

This image is closed for the weak topology, and hence for the norm topology, because the weak topology is Hausdorff and weaker than the norm topology. Thus its norm closure is the same weakly compact set. By the definition of a weakly compact operator, $T^{\ast }$ is weakly compact, proving $(iii)⟹(iv)$.

Weak compactness of the adjoint forces weak compactness of the original operator. Assume $T^{\ast }:Y^{\ast }\to X^{\ast }$ is weakly compact. Apply the already proved $(i)⟹(ii)$ to this operator, rather than assuming the adjoint equivalence in advance. It givesTake $\psi \in Y^{\ast \ast \ast }$ annihilating $J_{Y}Y$, so $\psi (J_{Y}y)=0$ for every $y\in Y$. Write $T^{\ast \ast \ast }\psi =J_{X^{\ast }}f$ for some $f\in X^{\ast }$. For every $x\in X$,Hence $f=0$ and $T^{\ast \ast \ast }\psi =0$. It follows thatfor every $\psi$ annihilating $J_{Y}Y$.

The vector subspace $J_{Y}Y$ is closed in the norm topology in $Y^{\ast \ast }$ because $Y$ is complete and $J_Y$ is an isometry. The Hahn-Banach theorem says that any point outside a closed linear subspace can be separated from it by a bounded linear functional vanishing on that subspace. Applying this in $Y^{\ast \ast }$ shows that an element annihilated by all such $\psi \in Y^{\ast \ast \ast }$ must belong to $J_{Y}Y$. Therefore $T^{\ast \ast }F\in J_{Y}Y$ for every $F$, proving $(iv)⟹(ii)$. The reverse implications above now establish all four equivalences, including weak compactness of an operator and its adjoint:

Finally, if $X$ is a reflexive Banach space, every $F\in X^{\ast \ast }$ is $J_{X}x$ for some $x$, and $T^{\ast \ast }F=J_{Y}Tx\in J_{Y}Y$. If $Y$ is a reflexive Banach space, $Y^{\ast \ast }=J_{Y}Y$, so the same range condition holds automatically. In either case, $(ii)⟹(i)$ proves