# Try to find a contradiction in showing a unit ball is not weakly compact.

Let $c_{00}$ be the linear space of all sequences in $\mathbb{F}$ that are eventually zero. Then $c_{00}$ is a linear subspace of $\ell_2$ and hence $(c_{00},\|\cdot\|_2)$ is a normed space.

Let $B$ be the norm closed unit ball of $c_{00}$. I want to use contradiction to show $B$ is not weakly compact. Let $x=(\frac{1}{2^n})_{n\in\mathbb{N}}$. Then $x\in\ell_2-c_{00}$. Let $x_i=\{\frac{1}{2},\cdots,\frac{1}{2^i},0\cdots\}$ $(i\in\mathbb{N})$. Then $(x_i)_{i\in\mathbb{N}}$ is a sequence in $\text{ball$c_{00}$}$. We have $\|x-x_{i}\|_2=(\sum\limits_{n=i+1}^{\infty}|\frac{1}{2^n}|^2)^{\frac{1}{2}}\rightarrow0$. Thus $x_{i}\overset{\|\cdot\|_2}{\longrightarrow} x$ in $\ell_2$. Hence, $f(x_i)\rightarrow f(x)$ for all $f\in {\ell_2}^*$; that is, $x_{i}\overset{wk}{\longrightarrow} x$ in $\ell_2$. Thus every subnet of $(x_{i})$ weakly converges to $x$ in $\ell_2$. Now assume that $B$ is weakly compact. Then there exists a subnet $(x_{i'})$ of $(x_{i})$ such that $x_{i'}\overset{wk}{\longrightarrow} y$ in $c_{00}$ for some $y\in B$.

And I am stuck here because I realized the two weakly convergent sequences above are not weakly convergent in the same space. So I am wondering:

1. Can we get $(x_{i'})$ has no weakly limit point in $c_{00}$ from every subnet of $(x_{i})$ weakly converges to $x$ in $\ell_2$?

2. Or can we get $x_{i'}\overset{wk}{\longrightarrow} y$ in $\ell_2$?

Are there any better contradiction we can find? Thank you in advance!

You need this lemma:

Let $M$ be a subspace of a normed space $X$ and let $(a_n)_n$ be a sequence in $M$ which converges weakly in $M$ to an element $a \in M$. Then $(a_n)_n$ also converges to $a$ weakly in $X$.

Proof:

Let $f \in X^*$ be a bounded linear functional. Then the restriction $f|_{M}$ is a bounded linear functional on $M$ so from the assumption $a_n \xrightarrow{\text{weakly in } M} a$ it follows $f|_M(a_n) \xrightarrow{n\to\infty} f|_M(a)$. But this means precisely $f(a_n) \xrightarrow{n\to\infty} f(a)$.

Since $f$ was arbitrary, we conclude $a_n \xrightarrow{\text{weakly in } X} a$.

Again, you are proving that your sequence has no weakly convergent subsequences. Weak limit points are not necessarily limits of weakly convergent subsequences.

• Thanks for your answer. But why $f|_M(a_n) \xrightarrow{n\to\infty} f|_M(a)$ means precisely $f(a_n) \xrightarrow{n\to\infty} f(a)$? – Answer Lee Jul 12 '18 at 22:25
• @AnswerLee $a_n$ and $a$ are elements of $M$ so $f|_M(a_n) = f(a_n)$ and $f|_M(a) = f(a)$. – mechanodroid Jul 12 '18 at 22:26
• Thank you so much for you help again. I have another question troubles me for a long time. Because you are familiar with my content. Can we find an example of a sequence in ${c_{00}}^*$ that is weak* convergent but not weakly convergent in ${c_{00}}^*$? I can open a new question if you think it is necessary. Appreciate!! – Answer Lee Jul 13 '18 at 16:13
• @AnswerLee Yes, try the sequence $(ne_n)_n$ in $\ell^1 = (c_{00})^*$ where $e_n$ is the $n$-th canonical vector. Then $ne_n \xrightarrow{w^*} 0$ in $\ell^1$ but it doesn't converge weakly because it is not bounded. – mechanodroid Jul 13 '18 at 16:16
• Equip $c_{00}$ with $\|\cdot\|_2$. Does this sequence still have the property we need? – Answer Lee Jul 13 '18 at 17:54