Weak and Weak Star Topologies — Examples

Put $b_k=F(e_k)$. For a finite initial segment let $v_k={\bar{b}}_k\mathrm{\mid }{b}_k{\mathrm{\mid }}^{q-2}$ if $b_k\mathrm{\ne }0$, and $v_k=0$ if $b_k=0$; set other coordinates zero. For real scalars conjugation fixes $b_k$. Then $b_k{v}_k=\mathrm{\mid }{b}_k{\mathrm{\mid }}^q$ and $\mathrm{\mid }{v}_k{\mathrm{\mid }}^p=\mathrm{\mid }{b}_k{\mathrm{\mid }}^{(q-1)p}=\mathrm{\mid }{b}_k{\mathrm{\mid }}^q$. If $A_N={\sum }_{k<N}\mathrm{\mid }{b}_k{\mathrm{\mid }}^q$, linearity and boundedness give $A_N=F(v)\le \mathrm{\parallel }F\mathrm{\parallel }{A}_N^{1\mathrm{/}p}$. If $A_N=0$ there is nothing to divide; otherwise $A_N^{1\mathrm{/}q}\le \mathrm{\parallel }F\mathrm{\parallel }$.

The nondecreasing partial sums $A_N$ are bounded by $\mathrm{\parallel }F{\mathrm{\parallel }}^q$, so their nonnegative series converges and $b_k\to 0$. Indeed infinitely many $\mathrm{\mid }{b}_k\mathrm{\mid }\ge \epsilon >0$ would make arbitrarily large finite sums exceed that bound. For completeness finite Hölder bounds ${\sum }_{k<N}\mathrm{\mid }{x}_k{b}_k\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_p\mathrm{\parallel }b{\mathrm{\parallel }}_q$, so the coefficient pairing is absolutely convergent, consistently over both fields. In particular $F(e_n)=b_n\to 0$ for every $F$, giving weak convergence. Directly $\mathrm{\parallel }{e}_n{\mathrm{\parallel }}_p=(1^p{)}^{1\mathrm{/}p}=1$, so there is no norm convergence to zero. $\mathrm{\square }$

Define $F(x)={\sum }_{k=0}^{\mathrm{\infty }}{x}_k$. Absolute convergence gives a scalar sum, linearity by limits of finite sums, and $\mathrm{\mid }F(x)\mathrm{\mid }\le {\sum }_k\mathrm{\mid }{x}_k\mathrm{\mid }=\mathrm{\parallel }x{\mathrm{\parallel }}_1$. Thus $F\in ({\mathrm{\ell }}^1{)}^{\ast }$, but $F(e_n)=1$ for every $n$, whereas $F(0)=0$. Therefore $e_n$ cannot converge weakly to zero.

More generally coordinate evaluation $P_k(x)=x_k$ is bounded since $\mathrm{\mid }{x}_k\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_1$. If $e_n\rightharpoonup x$, then $x_k={\lim }_n{P}_k(e_n)=0$ for each fixed $k$, so $x=0$, already excluded by step 1.1. Thus there is no weak limit. $\mathrm{\square }$

The complex coefficient identification is F1. For real scalars, a bounded sequence $b$ defines $h_b(x)={\sum }_k{b}_k{x}_k$ with $\mathrm{\mid }{h}_b(x)\mathrm{\mid }\le \mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x{\mathrm{\parallel }}_1$. Conversely if $h$ is bounded, $b_k=h(e_k)$ satisfies $\mathrm{\mid }{b}_k\mathrm{\mid }\le \mathrm{\parallel }h\mathrm{\parallel }$; finite truncations of $x$ converge in norm because their error is the absolute series tail, so $h(x)={\sum }_k{b}_k{x}_k$. Testing $e_k$ gives $\mathrm{\parallel }{h}_b\mathrm{\parallel }\ge {\sup }_k\mathrm{\mid }{b}_k\mathrm{\mid }$, proving isometry and uniqueness. Thus the identification holds over either field.

For every $x\in {\mathrm{\ell }}^1$, convergence of ${\sum }_k\mathrm{\mid }{x}_k\mathrm{\mid }$ forces $x_n\to 0$. Hence $E_n(x)\to 0$ for each fixed $x$, proving weak-star convergence. The inequality $\mathrm{\mid }{E}_n(x)\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_1$ and equality $E_n(e_n)=1$ give $\mathrm{\parallel }{E}_n\mathrm{\parallel }=1$ exactly. $\mathrm{\square }$

For every $x\in c_0$, $e_n(x)=x_n\to 0$ by the definition of $c_0$. Hence $e_n$ converges to zero for $\sigma ({\mathrm{\ell }}^1,c_0)$.

The scalar-linear map $H(a)={\sum }_k{a}_k$ on ${\mathrm{\ell }}^1$ is well-defined by absolute convergence and satisfies $\mathrm{\mid }H(a)\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_1$. Thus it is a weakly continuous functional on ${\mathrm{\ell }}^1$, but $H(e_n)=1$ for all $n$. The sequence is not weakly null. Equal topologies would have the same convergent sequences and limits, so the weak and the specified weak-star topologies differ. $\mathrm{\square }$

For a bounded real functional $F$, put $b_n=F(e_n)$. Test $F$ on $v_N={\sum }_{n=1}^N{b}_n{e}_n$. Then $B_N={\sum }_{n=1}^N{b}_n^2=F(v_N)\le \mathrm{\parallel }F\mathrm{\parallel }\sqrt{B_N}$, so $B_N\le \mathrm{\parallel }F{\mathrm{\parallel }}^2$, including $B_N=0$. Thus $(b_n)$ is square summable. Finite Hölder and passage to increasing finite sums show ${\sum }_n\mathrm{\mid }{x}_n{b}_n\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\parallel }b{\mathrm{\parallel }}_2$, consistent with these tests.

Fix a basic weak neighborhood of zero given by $F_1,\dots ,F_m$ and $\epsilon >0$. If it missed $A$, then for each $n\ge 1$ some $j$ would have $\sqrt{n}\mathrm{\mid }{F}_j(e_n)\mathrm{\mid }\ge \epsilon$. Therefore ${\sum }_j\mathrm{\mid }{F}_j(e_n){\mathrm{\mid }}^2\ge {\epsilon }^2\mathrm{/}n$. Summing over $n$ contradicts the finite sum of square-summability bounds from step 1.1: the harmonic partial sums are unbounded since each block $2^r\le n<2^{r+1}$ contributes at least $1\mathrm{/}2$. For $m=0$ the neighborhood is the whole space and already meets $A$. Thus every weak zero-neighborhood meets $A$, so $0\in {\bar{A}}^w$, while every member of $A$ has norm $\sqrt{n}>0$.

If a sequence in $A$ converges weakly, F3 bounds its norms by some finite $C$. Its indices therefore satisfy $n\le C^2$, so its range lies in a finite subset of $A$. A finite set is closed in a Hausdorff space: each singleton is closed because every other point has a disjoint neighborhood, and finite unions are closed. The weak limit lies in this finite set, hence in $A$. Thus $A$ is sequentially weakly closed but not weakly closed. $\mathrm{\square }$

Put $b_k=F(e_k)$ and test $F$ on $v={\sum }_{k<N}{\bar{b}}_k{e}_k$. By F3, $B_N=F(v)={\sum }_{k<N}\mathrm{\mid }{b}_k{\mathrm{\mid }}^2\le \mathrm{\parallel }F\mathrm{\parallel }\sqrt{B_N}$, giving $B_N\le \mathrm{\parallel }F{\mathrm{\parallel }}^2$, also when $B_N=0$. Thus $b\in {\mathrm{\ell }}^2$. Truncations converge in the square-sum norm, so $F(x)={\lim }_N{\sum }_{k<N}{x}_k{b}_k={\sum }_k{x}_k{b}_k$, with absolute convergence: F2 bounds every finite sum ${\sum }_{k<N}\mathrm{\mid }{x}_k{b}_k\mathrm{\mid }$ by $\mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\parallel }b{\mathrm{\parallel }}_2$, and the nonnegative partial sums converge to their finite supremum. The same finite inequality applied to tails gives the infinite tail bound used below.

Consequently $F(R^{n}x)={\sum }_{k\ge 0}{x}_k{b}_{k+n}$ and $\mathrm{\mid }F(R^{n}x)\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2({\sum }_{j\ge n}\mathrm{\mid }{b}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\to 0$. This proves WOT convergence. But $\mathrm{\parallel }{R}^{n}x{\mathrm{\parallel }}_2^2={\sum }_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so at $x=e_0$ the norms remain one. Thus SOT convergence to zero fails. $\mathrm{\square }$

The shift and its powers are scalar-linear and $\mathrm{\parallel }{L}^{n}x{\mathrm{\parallel }}_2^2={\sum }_{k\ge 0}\mathrm{\mid }{x}_{k+n}{\mathrm{\mid }}^2={\sum }_{j\ge n}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$. These tails tend to zero for every fixed $x$ by convergence of its nonnegative series. Therefore each $L^n$ is bounded and $L^n\to 0$ in SOT.

Step 1.1 gives $\mathrm{\parallel }{L}^n\mathrm{\parallel }\le 1$. But $L^n{e}_n=e_0$ and both coordinate vectors have norm one, so F2 gives $\mathrm{\parallel }{L}^n\mathrm{\parallel }\ge 1$. Thus the operator norms remain exactly one and cannot converge to zero. The witness varies with $n$, which is compatible with convergence on every fixed vector. $\mathrm{\square }$

For $n\ge 1$, the map $T_n$ is scalar-linear and $\mathrm{\mid }{T}_n(x)\mathrm{\mid }\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, with equality at the unit vector $e_n$. Thus $\mathrm{\parallel }{T}_n\mathrm{\parallel }=n$ for $n\ge 1$, while $\mathrm{\parallel }{T}_0\mathrm{\parallel }=0$. For any fixed finitely supported $x$, $T_n(x)=0$ past its last nonzero index; hence the sequence $(T_n{)}_{n\in \mathbb{N}}$ is pointwise bounded and even converges weak-star to the zero functional in $X^{\ast }$.

To check incompleteness, let $a_k^{(m)}=1\mathrm{/}k$ for $1\le k\le m$ and zero otherwise. For $m>r$, $\mathrm{\parallel }{a}^{(m)}-a^{(r)}{\mathrm{\parallel }}_{\mathrm{\infty }}=1\mathrm{/}(r+1)$, so these form a Cauchy sequence. A norm limit would have coordinate $1\mathrm{/}k$ for each fixed $k$, since coordinate evaluation has norm at most one. That sequence is not finitely supported, so no limit exists in $X$. The pointwise convergence in step 1.1 therefore gives no uniform norm bound on this incomplete domain. $\mathrm{\square }$

Let $D$ consist of all triples $(U,n,x)$ where $U$ is a weak zero-neighborhood, $n\ge 1$ an integer, $x\in U$, and $\mathrm{\parallel }x\mathrm{\parallel }\ge n$. Define $(U,n,x)⪯(V,m,y)$ if $V\subseteq U$ and $m\ge n$. This is reflexive and transitive. It is nonempty by F2. For two triples, $U\cap V$ is a weak zero-neighborhood and hence contains some $z$ with $\mathrm{\parallel }z\mathrm{\parallel }\ge \max (n,m)$. The triple $(U\cap V,\max (n,m),z)$ is a common upper bound. Thus $D$ is directed; antisymmetry is unnecessary.

Define the net by $a_{(U,n,x)}=x$, the point already carried in the index. For a weak zero-neighborhood $W$, F2 gives at least one index $(W,1,z)$. Every later index has neighborhood contained in $W$, so its carried point lies in $W$. Hence $a_d\rightharpoonup 0$. For any real $R>0$, take an integer $n>R$ and any index with integer coordinate $n$, whose existence follows from F2. Every later index has carried-point norm at least $n>R$. Thus $\mathrm{\parallel }{a}_d\mathrm{\parallel }\to \mathrm{\infty }$, and no tail is bounded. No choice function assigning one point to every neighborhood was used: all admissible points are included in the index set. $\mathrm{\square }$