The Closed Unit Ball of ℓ∞ is NOT Weak-* Sequentially Compact

Recall from the Alaoglu's Theorem page that Alaoglu's theorem states that if $X$ is a normed linear space then the closed unit ball of $X^*$ is weak-* compact.

Note that Alaoglu's theorem says nothing about weak-* sequential compactness of the closed unit ball of $X^*$. The following proposition tells us that for example, the closed unit ball of $\ell^{\infty*}$ is not weak-* sequentially compact.

Proposition: The closed unit ball of $\ell^{\infty*}$ is not weak-* sequentially compact.

• Proof: We need to find a sequence in $\ell^{\infty*}$ that has no weak-* convergent subsequence.

• For each $n \in \mathbb{N}$ let $f_n : \ell^{\infty} \to \mathbb{R}$ be defined for each $(x_k)_{k=1}^{\infty}$ by:

(1)

\begin{align} \quad f_n((x_k)) = x_n \end{align}

• Clearly each $f_n$ is a linear function. Furthermore, note that for each $n \in \mathbb{N}$:

(2)

\begin{align} \quad |f_n((x_k))| = |x_n| \leq \sup_{k \geq 1} |x_k| = \| (x_k) \|_{\infty} \end{align}

• Therefore $f_n$ is a bounded linear function with $\| f_n \| \leq 1$. So $(f_n)$ is in the closed unit ball of $\ell^{\infty*}$. Let $(f_{n_k})$ be any subsequence of $(f_n)$. Consider the sequence $(x_i)$ defined by:

(3)

\begin{align} \quad x_i = \left\{\begin{matrix} 1 & \mathrm{if} \: i = n_k, \: \mathrm{k} \: \mathrm{odd}\\ 0 & \mathrm{otherwise} \end{matrix}\right. \end{align}

• Note that $(x_i) \in \ell^{\infty}$ since $\sup_{i \geq 1} |x_i| = 1$. Also note that then $f_{n_k}((x_i))$ is equal to $1$ and $0$ infinitely often as $k \to \infty$. Therefore $\lim_{k \to \infty} f_{n_k} ((x_i))$ does not exist. So $(f_{n_k})$ does not weak-* converge.

• Hence $(f_n)$ has no weak-* convergent subsequence so the closed unit ball of $\ell^{\infty*}$ is not weak-* sequentially compact. $\blacksquare$