# How to show the weak and weak* topology on the dual space of $c_{00}$ are not the same?

Equip $c_{00}$ with $\|\cdot\|_2$. I want to show the weak and weak* topology on the dual space of $c_{00}$ are not the same.

I know I can use the fact that a normed space $X$ is reflexive if and only if the weak and weak* topology on $X^*$ are the same. Since $c_{00}$ is not reflexive, the weak and weak* topology on the dual space of $c_{00}$ are not the same.

But my professor mentioned I should directly show the weak and weak* topology on the dual space of $c_{00}$ are not the same rather than using the simple fact.

So my question is how to construct the weak and weak* topology on the dual space of $c_{00}$ and show they are not the same? Thank you!

• "So my question is how to construct the weak and weak* topology on the dual space of $c_{00}$" Do you not know the definitions of weak and weak* topology? – mathworker21 Jul 1 '18 at 19:09
• Which topology are you endowing $c_{00}$ with? – Daniel Fischer Jul 1 '18 at 19:21
• @DanielFischer♦ Sorry I forget to mention. Equip $c_{00}$ with $\|\cdot\|_2$. – Answer Lee Jul 1 '18 at 19:24
• So what are the dual and the bidual? – Daniel Fischer Jul 1 '18 at 19:28
• @mathworker21 Yes I know. But I don't know what ${c_{00}}^*$ and ${c_{00}}^{**}$ look like. – Answer Lee Jul 1 '18 at 19:28

An easy way to show that the topologies are different is to find a sequence which converges in one topology but not the other.

Let $(e_n)_n$ be the canonical basis for $\ell^2 = (c_{00})^*$ and consider $(ne_n)_n$.

For an arbitrary $x = (x_n)_n \in c_{00}$ there exists $n_0 \in \mathbb{N}$ such that $x_n = 0, \forall n \ge n_0$. Therefore for all $n \ge m_0$ we have $$(ne_n)(x) = nx_n = 0$$ so $(ne_n)(x) \xrightarrow{n\to\infty} 0$.

Hence $ne_n \xrightarrow{w^*} 0$.

However, $\|ne_n\| = n$ so $(ne_n)_n$ is not a bounded sequence. Therefore, it cannot be weakly convergent in $\ell^2$.

• Thank you so much for you help! You saved my life! And based on your answer I add some details and post an answer below. Can you tell a look and see whether it is ok? – Answer Lee Jul 14 '18 at 20:14
• Sorry to bother you again. This proof is my homework. And my professor wanted me to directly show $(f_{y_i})$ below does not convergence weakly in ${c_{00}}^*$. But it is really hard for me to do that without using every weakly convergent sequence is norm bounded. Can you give me a little hint? Thank you so much! – Answer Lee Jul 16 '18 at 21:24
• @AnswerLee Sure, it's easy: consider the functional $f\in(\ell^2)^{*} = \ell^2$ represented by the sequence $\left(\frac1{n^{3/4}}\right)_n$. Then $$f(ne_n) = n \cdot \frac1{n^{3/4}} = n^{1/4} \xrightarrow{n\to\infty} +\infty$$ so $(ne_n)_n$ does not converge weakly in $\ell^2$. – mechanodroid Jul 16 '18 at 22:15
• Thanks for your answer. But to be more accurate, I think we need to find $g\in {c_{00}}^*$ such that $g(f_{y_i})$ does not converge to 0. ($(f_{y_i})$ is the sequence we mentioned below.) – Answer Lee Jul 17 '18 at 3:52
To show $\sigma(X^*,X)\neq\sigma(X^*,X^{**})$ for $X=(c_{00},\|\cdot\|_2)$, we only need to find a weak* convergent sequence in ${c_{00}}^*$ that does not converge weakly in ${c_{00}}^*$.
Let $y=(y_n)\in\ell_2$. Then $f_y(x)=\sum\limits_{n=1}^{\infty}x_ny_n$ ($x=(x_n)\in c_{00}$) defines a bounded linear functional on $c_{00}$. In fact, $\phi:\ell_2\rightarrow {c_{00}}^*,y\mapsto f_y$ is an isometric isomorphism. Let $y_i=(y_n^i)_{n\in\mathbb{N}}$ ($i\in\mathbb{N}$) where $(y_n^i)_{n\in\mathbb{N}}$ is the sequence in $\mathbb{F}$ whose $i$th element is $i$ and other elements are $0$. Then $(y_i)_{i\in\mathbb{N}}$ is a sequence in $\ell_2$. Thus $(f_{y_i})_{i\in\mathbb{N}}$ is a sequence in ${c_{00}}^*$. Let $x=(x_n)\in c_{00}$. Then there exists $N_0\in\mathbb{N}$ such that $x_n=0$ for all $n\geqslant N_0$. Thus for all $i\geqslant N_0$, $f_{y_i}(x)=\sum\limits_{n=1}^{\infty}x_ny_n^i=ix_i=0$. It follows that $\lim\limits_{i\rightarrow\infty}f_{y_i}(x)=0$. Hence, $f_{y_i}\overset{wk^*}{\longrightarrow}0$. Let $i\in\mathbb{N}$. Then $\|y_i\|_2=(\sum\limits_{n=1}^{\infty}|y_n^i|^2)^{\frac{1}{2}}=i$. Since $\ell_2\cong {c_{00}}^*$, $\lim\limits_{i\rightarrow\infty}\|f_{y_i}\|=\|y_i\|_2=\infty$. Thus $(y_i)_{i\in\mathbb{N}}$ is not norm bounded in ${c_{00}}^*$. Hence, $(y_i)_{i\in\mathbb{N}}$ does not converge weakly in ${c_{00}}^*$.
We can also show $(f_{y_i})_{i\in\mathbb{N}}$ does not converge weakly to $0$ in ${c_{00}}^*$ directly:
Since $\ell_2\cong {c_{00}}^*$, for each $f\in {c_{00}}^*$, there exists $y\in\ell_2$ such that $f=f_y$, and thus we can use $f_y$ to present the element in ${c_{00}}^*$. Define $g:{c_{00}}^*\rightarrow\mathbb{F}$ by $g(f_y)=\sum\limits_{n=1}^{\infty}\frac{1}{n}y_n$ ($y=(y_n)\in\ell_2$). Let $\alpha\in\mathbb{F}$ and let $f_x,f_y\in{c_{00}}^*$ ($x=(x_n)\in\ell_2$ and $y=(y_n)\in\ell_2$). Then $$g(f_x+\alpha f_y)=\sum\limits_{n=1}^{\infty}\frac{1}{n}(x_n+\alpha y_n)=\sum\limits_{n=1}^{\infty}\frac{1}{n}x_n+\alpha\sum\limits_{n=1}^{\infty}\frac{1}{n}y_n=g(f_x)+\alpha g(f_y).$$ Hence, $g$ is linear. Let $f_y\in{c_{00}}^*$ ($y=(y_n)\in\ell_2$). Then $$|g(f_y)|=|\sum\limits_{n=1}^{\infty}\frac{1}{n}y_n|\leqslant\sum\limits_{n=1}^{\infty}\frac{1}{n}|y_n|\leqslant\sum\limits_{n=1}^{\infty}\frac{1}{n}\|f_y\|=\frac{\pi^2}{6}\|f_y\|.$$ Thus $g$ is a linear bounded functional on ${c_{00}}^*$; that is, $g\in{c_{00}}^{**}$. For all $i\in\mathbb{N}$, $g(f_{y_i})=\sum\limits_{n=1}^{\infty}\frac{1}{n}y_n^i=1$. Hence, $(f_{y_i})_{i\in\mathbb{N}}$ does not converge weakly to $0$ in ${c_{00}}^*$.