Alaoglu's Theorem

Alaoglu's Theorem

Recall from the Separable Criterion for the Compactness and Sequential Compactness of the Closed Unit Ball of X* in the Weak* Topology page that if $X$ is a normed linear space and if $X$ is separable (that is, $X$ has a countable dense subset) then the closed unit ball $B_*$ of the topological dual $X^*$ is compact and sequentially compact with respect to the weak* topology.

This theorem is rather restrictive as it requires the separability property which many spaces do not possess. We would still like to know when the closed unit ball of the topological dual $X^*$ is compact and sequentially compact with respect to the weak* topology.

Fortunately, there's a very nice theorem called Alaoglu's theorem which tells us that if $X$ is a normed linear space then the closed unit ball of the topological dual $X^*$ is always compact with respect to the weak* topology - although, we are not guaranteed sequential compactness with respect to the weak* topology.

We state Alaoglu's theorem below but omit the proof (as it is rather technical).

Theorem 1 (Alaoglu's Theorem): Let $X$ be a normed linear space. Then the closed unit ball of the topological dual $X^*$ is compact with respect to the weak* topology.

It must be emphasized again that Alaoglu's theorem says NOTHING about sequential compactness of the closed unit ball $B_*$ in the topological dual $X^*$ with respect to the weak* Topology. The following example gives us a normed linear space $X$ where $B_*$ fails to be sequentially compact.

Example 1

Let $\ell^{\infty}$ denote the normed linear space of all bounded sequences of complex numbers with the norm $\| \cdot \| : \ell^{\infty} \to [0, \infty)$ defined for all $(x_n)_{n=1}^{\infty} \in \ell^{\infty}$ by:

(1)
\begin{align} \quad \| (x_n)_{n=1}^{\infty} \| = \sup_{n \in \mathbb{N}} | x_n | \end{align}

We exhibit a sequence of linear functionals on the closed unit ball of $(\ell^{\infty})^*$ that has no convergent subsequence. For each $n \in \mathbb{N}$ let $\varphi_n : \ell^{\infty} \to \mathbb{C}$ be defined by:

(2)
\begin{align} \quad \varphi_n((x_n)_{n=1}^{\infty}) = x_n \end{align}

Observe that for each $n \in \mathbb{N}$ we have that:

(3)
\begin{align} \quad \| \varphi_n ((x_n)_{n=1}^{\infty}) \| = 1 \| x_n \| \end{align}

And so for each $n \in \mathbb{N}$ we have that:

(4)
\begin{align} \quad \| \varphi_n \| \leq 1 \end{align}

So indeed, $(\varphi_n)_{n=1}^{\infty}$ is a sequence of continuous linear functionals on $X^*$. Now suppose that there exists a convergent subsequence, $(\varphi_{n_k})_{k=1}^{\infty}$. Then for each sequence $(x_m)_{m=1}^{\infty} \in \ell^{\infty}$ we have that $(\varphi_{n_k}((x_m)_{m=1}^{\infty})_{k=1}^{\infty} = (x_{n_k})_{k=1}^{\infty}$ converges. But this does not always happen. For example, we can let $(x_{n_k})_{k=1}^{\infty}$ be the sequence defined such that every odd term is $0$ and every even term is $1$.

Therefore the closed unit ball of the topological dual $(\ell^{\infty})^*$ is not sequentially compact in the weak* topology

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