Weak and Weak Star Topologies

Every displayed $U$ is a finite intersection of inverse images of open disks centered at $f_j(x)$, so is weakly open and contains $x$. Conversely, a finite subbasic intersection containing $x$ contains inverse images of disks of radii $r_j>0$ about $f_j(x)$; take $\epsilon ={\min }_j{r}_j$. With no conditions the intersection is $X$. Thus these sets form a neighborhood base.

Put $p(v)={\max }_j\mathrm{\mid }{f}_j(v)\mathrm{\mid }$, taking $p=0$ for an empty list. Scalar linearity and the triangle inequality give $p(v+w)\le p(v)+p(w)$ and $p(av)=\mathrm{\mid }a\mathrm{\mid }p(v)$. Consequently $\{p<\epsilon \}$ is balanced and real-convex. Moreover $p((v+w)-(v_0+w_0))<\epsilon$ whenever $p(v-v_0),p(w-w_0)<\epsilon \mathrm{/}2$. This proves continuity of addition at every pair.

At $(a_0,v_0)$ write $av-a_0{v}_0=a(v-v_0)+(a-a_0)v_0$. Require $\mathrm{\mid }a-a_0\mathrm{\mid }<\min (1,\epsilon \mathrm{/}(2(p(v_0)+1)))$ and $p(v-v_0)<\epsilon \mathrm{/}(2(\mathrm{\mid }{a}_0\mathrm{\mid }+1))$. Then $p(av-a_0{v}_0)<\epsilon$. These are product neighborhoods and work also at $a_0=0$ and $v_0=0$. Thus scalar multiplication is jointly continuous, and the convex zero-neighborhood base proves local convexity. $\mathrm{\square }$

Fix distinct $x,y\in X$. Apply dual norming to $v=x-y\mathrm{\ne }0$ to obtain $f\in X^{\ast }$ with $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }=\mathrm{\parallel }x-y\mathrm{\parallel }=:d>0$. This is the only use of HB; no simultaneous selection over pairs is needed.

Set $U=\{z:\mathrm{\mid }f(z-x)\mathrm{\mid }<d\mathrm{/}3\}$ and $V=\{z:\mathrm{\mid }f(z-y)\mathrm{\mid }<d\mathrm{/}3\}$. These are weak neighborhoods of $x$ and $y$. If $z$ belonged to both, the triangle inequality would give $d\le \mathrm{\mid }f(x-z)\mathrm{\mid }+\mathrm{\mid }f(z-y)\mathrm{\mid }<2d\mathrm{/}3$, impossible. Thus $U\cap V=\mathrm{\varnothing }$. Every distinct pair is separated; if $X=\{0\}$ there is no such pair. $\mathrm{\square }$

Each displayed set is a finite intersection of inverse scalar disks. Conversely, every finite intersection of subbasic sets containing $f$ contains such a set by shrinking each scalar open set to a disk and taking the smallest of the finitely many positive radii. The empty intersection needs no shrinking.

For a finite list set $p(h)={\max }_j\mathrm{\mid }h(x_j)\mathrm{\mid }$, with $p=0$ for an empty list. Linearity gives $p(h+k)\le p(h)+p(k)$ and $p(ah)=\mathrm{\mid }a\mathrm{\mid }p(h)$. Hence its open balls are balanced and real-convex; the bounds $p(h-h_0),p(k-k_0)<\epsilon \mathrm{/}2$ imply $p(h+k-h_0-k_0)<\epsilon$, proving addition is continuous.

To control scalar multiplication at $(a_0,h_0)$, use $p(ah-a_0{h}_0)\le \mathrm{\mid }a\mathrm{\mid }p(h-h_0)+\mathrm{\mid }a-a_0\mathrm{\mid }p(h_0)$. The requirements $\mathrm{\mid }a-a_0\mathrm{\mid }<\min (1,\epsilon \mathrm{/}(2(p(h_0)+1)))$ and $p(h-h_0)<\epsilon \mathrm{/}(2(\mathrm{\mid }{a}_0\mathrm{\mid }+1))$ make this less than $\epsilon$. Thus the topology is a locally convex vector topology.

If $f\mathrm{\ne }g$ as functions, some $x\in X$ has $f(x)\mathrm{\ne }g(x)$. Put $d=\mathrm{\mid }f(x)-g(x)\mathrm{\mid }>0$. The evaluation disks of radius $d\mathrm{/}3$ about these two values have disjoint inverse images containing $f$ and $g$, since a common member would give $d<2d\mathrm{/}3$. This proves Hausdorffness; on a singleton dual it is vacuous. No norming principle is involved. $\mathrm{\square }$

Continuity at zero supplies $f_1,\dots ,f_m\in X^{\ast }$ and $\epsilon >0$ such that $\mathrm{\mid }L(v)\mathrm{\mid }<1$ whenever ${\max }_j\mathrm{\mid }{f}_j(v)\mathrm{\mid }<\epsilon$. If every $f_j(v)=0$, every scalar multiple $tv$ is in this neighborhood. Then $\mathrm{\mid }tL(v)\mathrm{\mid }<1$ for all positive real $t$, forcing $L(v)=0$. For an empty list this already gives $L=0$.

Define $A:X\to {\mathbb{K}}^m$ by $Av=(f_1(v),\dots ,f_m(v))$. Step 1.1 makes $\mathrm{\ell }(Av)=L(v)$ a well-defined scalar-linear functional on $A(X)$. Choose a basis of this subspace and extend it to a basis of ${\mathbb{K}}^m$ by successively adding standard basis vectors when necessary; at most $m$ additions occur. Assign value zero on the added basis vectors. The resulting linear extension $\tilde{\mathrm{\ell }}$ has the form $\tilde{\mathrm{\ell }}(z)={\sum }_{j=1}^m{c}_j{z}_j$, where $c_j=\tilde{\mathrm{\ell }}(e_j)$. Only finite-dimensional basis choices occur.

By finite-dimensional boundedness $\mathrm{\ell }$ is bounded for the inherited norm, so factorization already gives norm boundedness of $L$. More explicitly $L={\sum }_j{c}_j{f}_j$ and $\mathrm{\mid }L(v)\mathrm{\mid }\le ({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\parallel }{f}_j\mathrm{\parallel })\mathrm{\parallel }v\mathrm{\parallel }$, hence $L\in X^{\ast }$. Conversely, for any $f\in X^{\ast }$ and any scalar disk around $f(x)$, its inverse image is a basic weak neighborhood, so $f$ is weakly continuous. Thus both inclusions hold. $\mathrm{\square }$

Continuity at zero gives points $x_1,\dots ,x_m$ and $\epsilon >0$ such that $\mathrm{\mid }L(f)\mathrm{\mid }<1$ when $\mathrm{\mid }f(x_j)\mathrm{\mid }<\epsilon$ for all $j$. If all evaluations vanish, the same bound holds for every $tf$, forcing $L(f)=0$. Thus $L$ vanishes on the kernel of $A(f)=(f(x_1),\dots ,f(x_m))$.

The rule $\mathrm{\ell }(Af)=L(f)$ is therefore well-defined and linear on $A(X^{\ast })\subseteq {\mathbb{K}}^m$. Take a finite basis of this image and extend it to a basis of ${\mathbb{K}}^m$, adding standard coordinate vectors successively. Extend $\mathrm{\ell }$ by zero on added basis vectors. Writing its coordinate coefficients as $c_j$ gives $L(f)={\sum }_j{c}_{j}f(x_j)=f({\sum }_j{c}_j{x}_j)$. Set $x={\sum }_j{c}_j{x}_j$. Empty coordinates give $L=0$ and $x=0$. This finite construction needs no choice axiom.

Every evaluation at a given $x$ is weak-star continuous. If $x,y$ give the same evaluation then $f(x-y)=0$ for every $f\in X^{\ast }$. Under HB point separation implies $x=y$. This includes the zero space and proves the asserted identification and its uniqueness. $\mathrm{\square }$

Since weak-open sets are norm open, weak-closed sets are norm closed, and ${\bar{C}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}\subseteq {\bar{C}}^w$. If $C=\mathrm{\varnothing }$, both closures are empty.

For nonempty $C$, put $K={\bar{C}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}$. It is convex: for $u,v\in K$ and $0<t<1$, approximate $u,v$ by points $a,b\in C$ within any positive $\delta$; then $ta+(1-t)b\in C$ and its distance from $tu+(1-t)v$ is less than $\delta$. The cases $t=0,1$ are just $v,u\in K$. Thus $K$ is nonempty, closed and convex. For each $x\mathrm{\notin }K$, separation gives $f\in X^{\ast }$ and a real level $a$ with $\mathrm{Re}f(z)<a<\mathrm{Re}f(x)$ for all $z\in K$.

The set $\{y:\mathrm{Re}f(y)>a\}$ is weakly open, contains $x$ and misses $C$. Therefore $x\mathrm{\notin }{\bar{C}}^w$, giving ${\bar{C}}^w\subseteq K$ and equality of closures. If $C$ is norm closed this equality makes it weakly closed; the reverse implication was step 1.1. $\mathrm{\square }$

The ball $B$ is convex by the triangle inequality and norm closed because $\mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }-\mathrm{\parallel }y\mathrm{\parallel }\mathrm{\mid }\le \mathrm{\parallel }x-y\mathrm{\parallel }$. Thus $B$ is weakly closed and contains $S$, giving ${\bar{S}}^w\subseteq B$.

Fix $x\in B$ and a weak neighborhood of $x$ containing the conditions $\mathrm{\mid }{f}_j(y-x)\mathrm{\mid }<\epsilon$, $1\le j\le m$. There is a nonzero $v$ with all $f_j(v)=0$: choose $m+1$ independent vectors in $X$ by finite induction; their images in ${\mathbb{K}}^m$ are dependent, so a nonzero linear combination of the original vectors lies in the common kernel. This also covers $m=0$.

If $\mathrm{\parallel }x\mathrm{\parallel }=1$, the point $x$ itself works. If $\mathrm{\parallel }x\mathrm{\parallel }<1$, put $T=(2+\mathrm{\parallel }x\mathrm{\parallel })\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }$. The real function $h(t)=\mathrm{\parallel }x+tv\mathrm{\parallel }$ on $[0,T]$ satisfies $\mathrm{\mid }h(t)-h(s)\mathrm{\mid }\le \mathrm{\mid }t-s\mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$, $h(0)<1$, and $h(T)\ge T\mathrm{\parallel }v\mathrm{\parallel }-\mathrm{\parallel }x\mathrm{\parallel }=2>1$. The intermediate value theorem gives $t\in [0,T]$ with $h(t)=1$. Then $y=x+tv\in S$ has $f_j(y-x)=0$ for every $j$, so lies in the given neighborhood. Every neighborhood of every $x\in B$ meets $S$, proving $B\subseteq {\bar{S}}^w$ and equality. $\mathrm{\square }$

For every $g\in X^{\ast }$, the scalar sequence $g(x_n)$ converges to $g(x)$, hence is bounded: a tail has modulus at most $\mathrm{\mid }g(x)\mathrm{\mid }+1$, and finitely many preceding moduli have a finite maximum. The maps $J_X{x}_n:X^{\ast }\to \mathbb{K}$ are therefore pointwise bounded bounded linear maps. Their domain $X^{\ast }=\mathcal{B}(X,\mathbb{K})$ is Banach since $\mathbb{K}$ is complete.

Sequential uniform boundedness applied to these maps gives ${\sup }_n\mathrm{\parallel }{J}_X{x}_n\mathrm{\parallel }<\mathrm{\infty }$. The HB isometry makes this ${\sup }_n\mathrm{\parallel }{x}_n\mathrm{\parallel }<\mathrm{\infty }$. Countable Choice is used exactly in F4; HB is used only in F2, and completeness of $X$ was not required.

For the second assertion, each scalar sequence $f_n(y)$ converges for fixed $y\in X$, so the same finite-head/tail estimate from step 1.1 gives pointwise boundedness. Apply F4 directly on the assumed Banach domain $X$ to obtain ${\sup }_n\mathrm{\parallel }{f}_n\mathrm{\parallel }<\mathrm{\infty }$. No bidual norming or HB is used in this case. If either domain is zero, all its operator norms are zero, so the same conclusions hold. $\mathrm{\square }$

Write $L={\sup }_{i_0}{\inf }_{i\ge i_0}\mathrm{\parallel }{x}_i\mathrm{\parallel }$. Every tail is nonempty because the index preorder is reflexive and nonempty, so its infimum exists in $[0,\mathrm{\infty })$, and their supremum exists in $[0,\mathrm{\infty }]$. For $f\in X^{\ast }$ with $\mathrm{\parallel }f\mathrm{\parallel }\le 1$ and any $\epsilon >0$, scalar convergence and $\mathrm{\mid }\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }a-b\mathrm{\mid }$ give eventually $\mathrm{\parallel }{x}_i\mathrm{\parallel }\ge \mathrm{\mid }f(x_i)\mathrm{\mid }>\mathrm{\mid }f(x)\mathrm{\mid }-\epsilon$. Thus one tail infimum is at least $\mathrm{\mid }f(x)\mathrm{\mid }-\epsilon$, whence $L\ge \mathrm{\mid }f(x)\mathrm{\mid }-\epsilon$.

If $L=\mathrm{\infty }$ the assertion holds. Otherwise letting the positive error decrease shows $L\ge \mathrm{\mid }f(x)\mathrm{\mid }$ for every dual unit-ball member: a positive gap is contradicted by half that gap. The HB norm formula yields $L\ge \mathrm{\parallel }x\mathrm{\parallel }$. At $x=0$ this follows already from $L\ge 0$; the zero functional ensures the dual unit ball is nonempty, even for the zero space. $\mathrm{\square }$

If $e_1,\dots ,e_d$ is a finite basis, its coordinate functionals $f_j$ are bounded by the boundedness of the inverse coordinate map. Moreover $\mathrm{\parallel }v\mathrm{\parallel }\le {\sum }_j\mathrm{\mid }{f}_j(v)\mathrm{\mid }\mathrm{\parallel }{e}_j\mathrm{\parallel }$. For $r>0$, the finite conditions $\mathrm{\mid }{f}_j(v)\mathrm{\mid }<r\mathrm{/}(1+{\sum }_j\mathrm{\parallel }{e}_j\mathrm{\parallel })$ imply $\mathrm{\parallel }v\mathrm{\parallel }<r$. Thus every norm ball about zero contains a weak neighborhood. Translating gives this at every point; weak-open sets are already norm open because their defining functionals are bounded. The topologies coincide. For $d=0$, $X$ is a singleton and the assertion holds directly.

Suppose instead $X$ is infinite dimensional. Given finitely many $f_1,\dots ,f_m$, choose $m+1$ independent vectors by finite induction. Their images in ${\mathbb{K}}^m$ are dependent; the resulting nontrivial combination gives a nonzero common-kernel vector $v$. Every real multiple $tv$ satisfies every zero-centered finite disk condition, and $\mathrm{\parallel }tv\mathrm{\parallel }=\mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$ is unbounded. By F1 every weak zero-neighborhood is therefore unbounded. It cannot lie in the norm unit ball, whereas equality of the topologies would make that ball a weak neighborhood. This excludes equality in infinite dimension and completes the equivalence. $\mathrm{\square }$

Suppose $(U_n)$ is a countable weak local base at zero. For each $n$ the set of finite lists of functionals and positive radii defining a basic neighborhood $V_n\subseteq U_n$ is nonempty. Apply Countable Choice once to these sets, fixing such finite lists. Enumerate their entries by pairs of natural numbers, padding each finite list with zero functionals. This gives a sequence $(h_k)$ containing every chosen functional.

Fix $f\in X^{\ast }$. It is weakly continuous by F2. Some $U_n$ lies inside $\{\mathrm{\mid }f\mathrm{\mid }<1\}$, so $V_n$ lies there too. Scaling shows that $f$ vanishes on the common kernel of its finite defining list. For its coordinate map $A$, the rule $\mathrm{\ell }(Ax)=f(x)$ is well-defined by this kernel inclusion. Extend a finite basis of $A(X)$ to one of ${\mathbb{K}}^m$, and set $\mathrm{\ell }=0$ on added basis vectors. If $c_j$ are the values of the extension on standard coordinate vectors, then $f(x)={\sum }_j{c}_j(Ax{)}_j$. Hence $f$ is a finite linear combination of the defining list. Thus the algebraic span of $(h_k)$ is all of $X^{\ast }$.

Scan $(h_k)$ in order, retaining an entry precisely when it is outside the span of preceding retained entries. This deterministic rule gives either a finite basis or an infinite subsequence forming a countably infinite Hamel basis of $X^{\ast }$: every discarded entry is in the earlier retained span, while each retained entry preserves independence. The infinite outcome is impossible because $X^{\ast }$ is Banach and F4 is choice-free. Therefore $X^{\ast }$ has a finite basis $g_1,\dots ,g_d$.

The map $x\mapsto (g_1(x),\dots ,g_d(x))$ is injective: a kernel vector is killed by every member of their span $X^{\ast }$ and is zero by HB separation. But $d+1$ independent vectors in $X$ would have dependent images in ${\mathbb{K}}^d$, contradicting injectivity. Such vectors exist by finite induction from infinite dimension. This contradiction excludes the supplied countable local base. Countable Choice was used only in step 1.1 and HB only in this step. $\mathrm{\square }$

By F1, $M^{\perp }$ is the intersection over $m\in M$ of the kernels of $f\mapsto f(m)$, and ${}^{\perp }N$ is the intersection over $f\in N$ of $\ker f$. Each kernel is a linear subspace and closed in the corresponding topology by F2 and closedness of $\{0\}\subseteq \mathbb{K}$. Intersections preserve both properties; empty intersections give the whole ambient spaces.

For linear $N$, F3 yields $({}^{\perp }N{)}^{\perp }={\bar{N}}^{w^{\ast }}$ with no primal norming used. For linear $M$, every functional vanishing on $M$ also vanishes on its norm closure, by norm continuity. Thus ${\bar{M}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}\subseteq {}^{\perp }(M^{\perp })$. Step 1.1 also implies ${\bar{M}}^w\subseteq {}^{\perp }(M^{\perp })$.

Assume HB and fix $x\mathrm{\notin }K={\bar{M}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}$. The set $K$ is a nonempty closed linear subspace: addition and scalar multiplication preserve closure by their norm estimates. By F4 there is $f\in X^{\ast }$ whose real part is bounded above on $K$ and strictly larger at $x$. Since $K$ is a real linear subspace, scaling forces $\mathrm{Re}f=0$ on $K$. In the complex case $ik\in K$ forces $\mathrm{Im}f(k)=0$ too. Hence $f\in M^{\perp }$ and $f(x)\mathrm{\ne }0$, excluding $x$ from ${}^{\perp }(M^{\perp })$. The open set $\{y:\mathrm{\mid }f(y)\mathrm{\mid }>\mathrm{\mid }f(x)\mathrm{\mid }\mathrm{/}2\}$ also excludes $x$ from ${\bar{M}}^w$. Norm closure is contained in weak closure because weak-open sets are norm open; all three sets therefore coincide. $\mathrm{\square }$

For $g\in Y^{\ast }$, $g\circ T\in X^{\ast }$ and $\mathrm{\mid }(g\circ T)(x)\mathrm{\mid }\le \mathrm{\parallel }g\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. The inverse under $T$ of a weak subbasic set $g^{-1}(V)$ is $(g\circ T{)}^{-1}(V)$, which is weakly open in $X$. Inverse images preserve unions and finite intersections, proving continuity of $T$. Equivalently, F2 gives $g(T{x}_i)\to g(Tx)$ for every weakly convergent net.

Taking the supremum over $\mathrm{\parallel }x\mathrm{\parallel }\le 1$ in the bound of step 1.1 gives $\mathrm{\parallel }{T}^{\ast }g\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }g\mathrm{\parallel }$, so $T^{\ast }$ is bounded. Apply the step 1.1 argument to this bounded map: for every $\mathrm{\Phi }\in X^{\ast \ast }$, $\mathrm{\Phi }\circ T^{\ast }\in Y^{\ast \ast }$ since $\mathrm{\mid }\mathrm{\Phi }(T^{\ast }g)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }g\mathrm{\parallel }$. Thus inverse weak subbasic sets are weakly open, proving weak-to-weak continuity of $T^{\ast }$. The estimates remain valid for zero maps and zero spaces. $\mathrm{\square }$

For bounded $T$, $\mathrm{\mid }(T^{\ast }f)(x)\mathrm{\mid }=\mathrm{\mid }f(Tx)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. Thus $T^{\ast }f\in X^{\ast }$ and $T^{\ast }$ is bounded and scalar-linear. For every $x\in X$, the composite of $T^{\ast }$ with evaluation at $x$ is evaluation at $Tx$. Its inverse scalar-open sets are weak-star open by the defining evaluation topology in F1. Thus $T^{\ast }$ is weak-star continuous.

For the converse, fix $x\in X$. The map $f\mapsto (Sf)(x)$ is weak-star continuous and scalar-linear, since $S$ is and evaluation at $x$ is. F1 gives a unique point $Tx\in Y$ with $f(Tx)=(Sf)(x)$ for all $f\in Y^{\ast }$. Unique specification defines $T$ on all $X$ without choosing from a family of non-singleton sets. For scalars $a,b$, evaluation gives $f(T(ax+bz))=(Sf)(ax+bz)=af(Tx)+bf(Tz)$ for every $f$. Point separation makes $T(ax+bz)=aTx+bTz$.

By F2 and boundedness of $S$, $\mathrm{\parallel }Tx\mathrm{\parallel }={\sup }_{\mathrm{\parallel }f\mathrm{\parallel }\le 1}\mathrm{\mid }(Sf)(x)\mathrm{\mid }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. Hence $T$ is bounded, and its defining identity says $S=T^{\ast }$. Any other preadjoint has the same evaluations at each $x$ and therefore equals $T$ by point separation. Zero spaces and $S=0$ satisfy the same formulas, with $T=0$. $\mathrm{\square }$

If $\mathrm{\parallel }{T}_i-T\mathrm{\parallel }\to 0$, then for each fixed $x$, $\mathrm{\parallel }(T_i-T)x\mathrm{\parallel }\le \mathrm{\parallel }{T}_i-T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\to 0$. Explicitly for $\epsilon >0$ it suffices that $\mathrm{\parallel }{T}_i-T\mathrm{\parallel }<\epsilon \mathrm{/}(1+\mathrm{\parallel }x\mathrm{\parallel })$. This is SOT convergence, also at $x=0$.

If $T_i\to T$ in SOT, then for every fixed $x\in X$ and $f\in Y^{\ast }$, $\mathrm{\mid }f((T_i-T)x)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }(T_i-T)x\mathrm{\parallel }\to 0$. The bound $\mathrm{\parallel }(T_i-T)x\mathrm{\parallel }<\epsilon \mathrm{/}(1+\mathrm{\parallel }f\mathrm{\parallel })$ suffices, including $f=0$. This is WOT convergence; combined with step 1.1 it proves the hierarchy. $\mathrm{\square }$