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Consider the sequence $e_n = (0,0,\ldots,0,1,0,\ldots)$ in $\ell^1$ which is weakly convergent to zero in $\ell^p$. for all $1\leq p \leq \infty.$ It is then obvious, from the theorem that sequences completely characterises weak convergence, that $\{e_n\}\cup\{ 0\}$ is weakly compact.

However, I remember somehow that this non-trivial theorem is not needed, and there is a more obvious answer (of the proof that $\{e_n\}\cup\{ 0\}$ is weakly compact). Does anyone have the solution that uses the weak topology directly without using sequential characterisations?

Ma Joad
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2 Answers2

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$\ell_1$ has the Scur property so a sequence converges weakly iff it converges in norm. Hence the claim does not hold for $p=1$.

For $p \in (1,\infty)$ the dual of $\ell_p$ is $\ell_q$, where $p$ is the conjugate of $q$. So given any $ a = (a(i))_{i\ge 1} \in \ell_p^* = \ell_q $ you have that $$ |\langle a, e_n \rangle| = \big | \sum_{i \ge 1} a(i) e_n(i) \big | = a(n) \xrightarrow{n \to \infty } 0 $$ since $\sum_{n \ge 1} |a(n)|^q < \infty$.

For $p=\infty$ the claim still holds but the above proof does not apply. One can show that $\ell_\infty $ is isometrically isomoprhic to $C(K)$ where $K$ is a compact space (in fact, $K = \beta \mathbb N$). Now the dual of $C(K)$ is the space $\mathcal M(K)$ of all regular complex measures (or regular finite signed measures in the case where your field is $\mathbb R$). Α $μ \in \mathcal M(K)$ acts on a function $f \in C(K)$ by way of $$ \langle μ,f \rangle = \int_K f dμ$$

Hence a bounded sequence $(f_n) \subset C(K)$ converges weakly to $0$ if and only $f_n \to 0$ pointwise.

(For the $\implies$ direction consider for each $x \in K$ the dirac measure $δ_x \in \mathcal M(K)$ where $δ_x(f) = f(x)$ and for the $\impliedby$ direction apply the dominated convergence theorem.)

As $e_n \to 0$ pointwise you get that $e_n \to 0$ weakly in $\ell_\infty$.

Evangelopoulos Foivos
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Let $S=\{e_n\}_{n=1}^\infty$. Then $S$ is weakly closed and weakly non-compact in $\ell^1$.

Let $F_n\in(\ell^1)^*$ be the functional $F_n(x)=x_n$ and $F_\infty\in(\ell^1)^*$ be the functional $F_\infty(x)=\sum_{k=1}^\infty x_k$. They are all bounded $$ \begin{align} |F_n(x)|&=|x_n|\le\sum_{k=1}^\infty|x_k|=\|x\|_1\\ |F_\infty(x)|&=\left|\sum_{k=1}^\infty x_k\right|\le\sum_{k=1}^\infty|x_k|=\|x\|_1 \end{align} $$

Since $(\ell^1)^*$ separates points on $\ell_1$, the weak topology on $\ell^1$ is $\text{T}_1$ (i.e. every point is weakly-closed.)

Let $a\in\ell^1$.

If $a\ne 0$ there is some $m$ such that $a_m\ne 0$. The set $U=F_m^{-1}(\Bbb{C}\setminus\{0\})$ is weakly-open in $\ell^1$, $a\in U$ and for all $n\ne m$, $e_n\not\in U$ because $F_m(e_n)=0$. Therefore $U\cap S\subseteq \{e_m\}$.

If $a=e_m$ then $U\cap S=\{a\}=\{e_m\}$ showing that $a=e_m$ is a weakly isolated point of $S$.

If $a\ne e_m$ then $V=U\cap(\ell^1\setminus\{e_m\})$ is a weak open neighborhood of $a$ such that $V\cap S=\emptyset$ showing that $a$ is not a weak limit point of $S$.

If $a=0$ the set $U=F_\infty^{-1}(\{z\in\Bbb{C}\colon|z|<1\})$ is weakly-open in $\ell^1$, $0\in U$ and $e_n\not\in U$ for all $n$ since $F_\infty(e_n)=1$ for all $n$. Hence $U\cap S=\emptyset$ showing that $0$ is not a weak limit point of $S$.

This covers all cases so $S$ has no weak limit points in $\ell^1$ and is closed.

Let $U_n=F_n^{-1}(\Bbb{C}\setminus\{0\})$. The collection $\{U_n\}_{n=1}^\infty$ is a weak open cover of $S$ and $U_n\cap S=\{e_n\}$ for all $n$. Therefore there is no finite sub-cover. This shows $S$ is not weakly-compact.

Here's an elementary proof that $e_n\to 0$ weakly as $n\to\infty$ in $\ell^\infty$. Let $\phi\in(\ell^\infty)^*$ and $a_n:=\phi(e_n)$. We'll show that $\sum_{k=1}^\infty |a_k|<+\infty$. The rest of the proof is as in $\ell^p$ for $p\in(1,\infty)$. $a_n$ must converge to $0$ as $n\to\infty$, so for all $\epsilon>0$ there is an $m$ such that for all $n>m$ $|a_n|=|\phi(e_n)|<\epsilon$.

Let $z_n$ be complex numbers with $|z_n|=1$ such that $z_n a_n\ge 0$. Then $z_n a_n=|a_n|$. For an integer $N$ let $b=\sum_{k=1}^N z_k e_k$. Note that $\|b\|_\infty = \max_{k=1}^N |z_k|=1$. You have $$ \phi(b)=\sum_{k=1}^N z_k \phi(e_k)=\sum_{k=1}^N z_k a_k=\sum_{k=1}^N |a_k|\ge 0 $$ and also $\phi(b)=|\phi(b)|\le \|\phi\|\|b\|_\infty=\|\phi\|$. Therefore $$ \sum_{k=1}^N |a_k|\le \|\phi\| $$ for all $N$. Hence $\sum_{k=1}^\infty |a_k|<+\infty$ as claimed.

Chad K
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