Past exam of the functional analysis course of the University of Oxford / 2011 / 2

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ making every continuous linear functional $f\in X^{\ast }$ continuous. A basic neighborhood of $x_0$ is specified by finitely many inequalities $|f_j(x-x_0)|<ϵ$.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology on $X^{\ast }$ making every evaluation $f\mapsto f(x)$, for $x\in X$, continuous. Its basic neighborhoods use finitely many such evaluations. Nets converge in these topologies exactly when all the corresponding scalar evaluations converge.

Let $J_X:X\to X^{\ast \ast }$ be the canonical isometric embedding, $J_{X}x(f)=f(x)$. Goldstine theorem statesHere the balls are closed unit balls. Thus any bidual unit vector can be approximated in finitely many dual evaluations by unit vectors from $X$; this does not claim norm density or sequential density in complete generality.

The weak topology is coarser than the norm topology, so norm convergence implies weak convergence, andFor the reverse inclusion, let $x\notin K={\overline{C}}^{\Vert \cdot \Vert }$. The set $K$ is closed and convex. The Hahn-Banach separation theorem supplies $f\in X^{\ast }$ and $a\in \mathbb{R}$ such that $f(x)>a\ge {\sup }_{y\in K}f(y)$. The set $\{z:f(z)>a\}$ is weakly open, contains $x$ and misses $C$. Hence $x$ is outside the weak closure as well. If $C$ is empty the equality is immediate. ThereforeThe separating functional explains the role of convexity: an arbitrary norm-closed subset need not be weakly closed.

A reflexive Banach space is one for which $J_X:X\to X^{\ast \ast }$ is onto. If $X$ is reflexive, every $\theta \in X^{\ast \ast }$ has the form $\theta (f)=f(x)$ for some $x\in X$. Thus the functionals defining $\sigma (X^{\ast },X^{\ast \ast })$ and $\sigma (X^{\ast },X)$ are precisely the same, proving equality of the topologies.

Conversely, assume those topologies coincide. For any $\theta \in X^{\ast \ast }$, the linear functional $f\mapsto \theta (f)$ on $X^{\ast }$ is then weak-star continuous. Its continuity at zero gives finitely many $x_1,\dots ,x_m\in X$ and $ϵ>0$ such thatIf all these evaluations are zero, arbitrary scalar multiples of $f$ still satisfy the inequalities, forcing $\theta (f)=0$. Hence $\theta$ factors through the finite-dimensional evaluation map $f\mapsto (f(x_1),\dots ,f(x_m))$. A linear functional on its image extends to ${\mathbb{R}}^m$, so for suitable scalars $a_j$,Thus $\theta$ lies in $J_X(X)$. Since it was arbitrary, $X$ is reflexive. We have proved

The domain carries $\sigma (Z^{\ast \ast },Z^{\ast })$, and the target carries $\sigma (X^{\ast },X)$. For each $x\in X\subseteq Z^{\ast }$,which is one of the defining weak-star continuous evaluations on $Z^{\ast \ast }$. Thus $R$ is weak-star continuous.

Restriction cannot increase a functional's norm, so $R(B_{Z^{\ast \ast }})\subseteq B_{X^{\ast }}$. Conversely, any $f\in B_{X^{\ast }}$ is a bounded linear functional on the subspace $X$ of $Z^{\ast }$. The Hahn-Banach theorem extends it to a functional $\theta \in (Z^{\ast }{)}^{\ast }=Z^{\ast \ast }$ with the same norm. Then $R\theta =f$ and $\Vert \theta \Vert \le 1$. ConsequentlyThis proves exact surjectivity on closed balls, not just density of the image.

Put $T=R\circ J_Z:Z\to X^{\ast }$. Then $(Tz)(x)=x(z)$ and $\Vert Tz\Vert \le \Vert z\Vert$. We must prove $T$ onto.

Take $f\in B_{X^{\ast }}$. Part (iv) supplies $\theta \in B_{Z^{\ast \ast }}$ with $R\theta =f$. Goldstine's theorem gives a net $z_{\alpha }\in B_Z$ with $J_Z{z}_{\alpha }\to \theta$ weak-star. By continuity of $R$, $T{z}_{\alpha }\to f$ in $\sigma (X^{\ast },X)$. Reflexivity of $X$ makes this the weak topology on $X^{\ast }$ by part (iii). Thus $B_{X^{\ast }}$ is contained in the weak closure of $T(B_Z)$.

The reverse containment holds because $T(B_Z)\subseteq B_{X^{\ast }}$, and the latter ball is weakly closed: it is the intersection of the closed evaluation constraints $|g(x)|\le 1$ for $x\in B_X$. The set $T(B_Z)$ is convex, so part (ii), now on the Banach space $X^{\ast }$, givesDensity of the ball gives exact surjectivity by a convergent correction argument. For nonzero $f\in X^{\ast }$, choose $z_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and $\Vert f-T{z}_1\Vert \le \Vert f\Vert /2$. Inductively approximate the current residual $r_{n-1}$ by $T{z}_n$ with $\Vert z_n\Vert \le \Vert r_{n-1}\Vert$ and $\Vert r_{n-1}-T{z}_n\Vert \le \Vert r_{n-1}\Vert /2$. If a residual is zero, stop.

The residual norms are bounded by $2^{-n}\Vert f\Vert$, while ${\sum }_n\Vert z_n\Vert \le 2\Vert f\Vert$. Completeness of $Z$ gives $z={\sum }_n{z}_n\in Z$, and continuity of $T$ gives $Tz=f$. For $f=0$ use $z=0$. ThereforeThis is evaluation onto the dual of a reflexive dual subspace. The correction argument is needed: a dense image alone would not imply surjectivity.