Show that the norm-closed unit ball of $c_0$ is not weakly compact; recall that $c_0^*=\ell_1$.

  • $\begingroup$ Did you mean $C_0$ instead of $c_0$? $\endgroup$ May 19, 2012 at 8:11
  • $\begingroup$ @Zev: What is $C_0$? I'm pretty sure $c_0$ here means the space of sequences converging to $0$, and that is standard notation. On the other hand, "$C_0$" is often followed by a topological space to denote the space of continuous complex-valued functions on that space that vanish at infinity. E.g. $C_0(\mathbb N)=c_0$ (when $\mathbb N$ is given the discrete topology). $\endgroup$ May 19, 2012 at 21:15

1 Answer 1


Hint: Let $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$. Suppose $z\in c_0$ is a weak cluster point of $(x_n)$. By considering the action of the standard unit vectors of $\ell_1$ on the $x_n$, obtain a contradiction by showing that we must have $z=(1,1,\ldots)$.

  • $\begingroup$ In general, $B(X)$ is weakly compact if and only if $X$ is reflexive. Additionally, Eberlein-Smulian Theorem tells us that in this case $B(X)$ is weakly sequentially compact. $\endgroup$
    – Theo
    May 18, 2012 at 23:35
  • $\begingroup$ @DavidMitra I am sorry I am new to analysis. Can you give a detailed answer. Thank you! $\endgroup$
    – Answer Lee
    Mar 18, 2018 at 23:35
  • $\begingroup$ @AnswerLee write $z=(z_n)$, take $e_m=(0,...,0,1,0,...)$ and note for every subsequence $x_{n_k}$ we must have $$z_n = \langle z,e_m\rangle = \lim_n\langle x_{n_k},e_m\rangle=1$$ Then, $z_n\not\in c_0$ $\endgroup$ Mar 18, 2018 at 23:44
  • $\begingroup$ @VeridianDynamics I am sorry I don't get that why $$z_n = \langle z,e_m\rangle = \lim_n\langle x_{n_k},e_m\rangle=1?$$ Can you be more specific? Thank you!! $\endgroup$
    – Answer Lee
    Mar 19, 2018 at 0:22
  • 1
    $\begingroup$ for sequences $(x_n)\in c_0$ and $(y_n)\in(c_0)^*=l^1$, the value of the functional $(y_n)$ on the point $(x_n)$ is given by: $$\langle (x_n),(y_n)\rangle = \sum_{n=1}^\infty x_ny_n$$ So, for every $n_k$ larger than $m$, we have $\langle x_{n_k},e_m\rangle = 1$. This gives $\lim_n\langle x_{n_k},e_m\rangle=1$. $\langle z,e_m\rangle = \lim_n\langle x_{n_k},e_m\rangle$ follows from the continuity of $z$ $\endgroup$ Mar 19, 2018 at 0:39

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