Alaoglu's Theorem

# Alaoglu's Theorem

Recall from the Separable Criterion for the Weak-* Compactness and Weak-* Sequential Compactness of the Closed Unit Ball of X* 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 $B_{X^*}$ of the topological dual $X^*$ is weak-* compact and weak-* sequentially compact

Alaoglu's theorem tells us that if $X$ is a normed linear space then $B_{X^*}$ is always weak-* compact regardless of whether or not $X$ is separable. We will see on the The Closed Unit Ball of ℓ∞ is NOT Weak-* Sequentially Compact page that weak-* sequential compactness is NOT implied though.

 Theorem 1 (Alaoglu's Theorem): Let $X$ be a normed linear space. Then the closed unit ball of $X^*$ is weak-* compact.
• Proof: Let $B_{X^*}$ denote the closed unit ball of $X^*$, that is:
(1)
\begin{align} \quad B_{X^*} = \{ f \in X^* : \| f \| \leq 1 \} \end{align}
• Then if $f \in B_{X^*}$ we have that $\| f \| \leq 1$ and so for every $x \in X$, $|f(x)| \leq \| f \| \| x \| \leq \| x \|$. So for every $x \in X$ and every $f \in B_{X^*}$ we have that $f(x) \in [- \| x \|, \| x \|]$. For each $x \in X$ let $I_x = [- \| x \|, \| x \|]$ and let:
(2)
\begin{align} \quad P = \prod_{x \in X} I_x \end{align}
• Then:
(3)
\begin{align} \quad B_{X^*} \subset P \end{align}
• Each $I_x$ is a compact space and so by Tychonoff's Theorem, the space $P$ with the product topology is also compact. Note that by definition, the product topology on $P$ is the initial topology induced by the projection maps $\{ p_y : y \in X \}$ where $p_y : \prod_{x \in X} I_x \to I_y$ is defined for each $f \in p_y$ by $p_y(f) = f(y)$. In other words, the product topology on $P$ is the weakest topology which makes the maps $\{ p_y : y \in X \}$ continuous. The subspace topology on $B_{X^*}$ is thus the weakest topology which makes $\{ \hat{x} : x \in X \}$ continuous, i.e., the weak-* topology.
• Recall that a closed subspace of a compact topological space is also compact. Therefore, to show that $B_{X^*}$ wit the weak-* topology is weak-* compact, all we need to show is that $B_{X^*}$ is weak-* closed, that is, we want to show $B_{X^*} = \mathrm{cl} (B_{X^*})$ where $\mathrm{cl} (B_{X^*})$ denotes the weak-* closed of $B_{X^*}$.
• Clearly $B_{X^*} \subseteq \mathrm{cl} (B_{X^*})$. For the reverse inclusion, let $f \in \mathrm{cl} (B_{X^*})$. We aim to show that $f \in B_{X^*}$, that is, $f$ is linear and $f$ is bounded with $\| f \| \leq 1$.
• Linearity of $f$; Let $s, t \in X$ and let $a, b \in \mathbb{R}$. Let $u = as + bt$. Let $\epsilon > 0$ and let:
(4)
\begin{align} \quad N = \{ g \in P : |g(s) - f(s)| < \epsilon, |g(t) - f(t)| < \epsilon, |g(u) - f(u)| < \epsilon \} \end{align}
• Then $N$ is a neighbourhood of $f$. Furthermore, observe that:
(5)
\begin{align} \quad N = \prod_{x \in X \setminus \{ s, t, u \}} I_x \times \left [ (f(s) - \epsilon, f(s) + \epsilon) \cap I_s \right ] \times \left [ (f(t) - \epsilon, f(t) + \epsilon) \cap I_t \right ] \times \left [ (f(u) - \epsilon, f(u) + \epsilon) \cap I_u \right ] \end{align}
• By definition, a basis for the product topology on $P$ is $\{ \prod_{x \in X} U_x : U_x \: \mathrm{open \: in \:} I_x, U_x = I_x \: \mathrm{for \: all \: but \: finitely \: many \:} x \in X \}$. Therefore, $N$ is a basis element of the product topology on $P$ and so $N$ is an open neighbourhood of $f$.
• Since $f \in \mathrm{cl} (B_{X^*})$ and $N$ is an open neighbourhood of $f$ we have that:
(6)
\begin{align} \quad B_{X^*} \cap N \neq \emptyset \end{align}
• Let $g \in B_{X^*} \cap N$. Since $g \in B_{X^*}$ we have that $g$ is linear, and so $g(u) = g(as + bt) = ag(s) + bg(t)$. Thus:
(7)
\begin{align} \quad g(u) - ag(s) - bg(t) = 0 \quad (*) \end{align}
• Since $g \in N$ we also have that:
(8)
\begin{align} \quad |g(s) - f(s)| &< \epsilon \\ \quad |g(t) - f(t)| &< \epsilon \\ \quad |g(u) - f(t)| &< \epsilon \quad (**) \end{align}
• Therefore:
(9)
\begin{align} \quad |f(u) - af(s) - bf(t)| &= |f(u) - af(s) - bf(t) - 0| \\ & \overset{(*)}= |f(u) - af(s) - bf(t) - [g(u) - ag(s) - bg(t)] | \\ &= |[f(u) - g(u)] + a[g(s) - f(s)] + b[g(t) - f(t)]| \\ &< |f(u) - g(u)| + |a||g(s) - f(s)| + |b||g(t) - f(t)| \\ & \overset{(**)} < \epsilon + |a| \epsilon + |b| \epsilon \\ & < (1 + |a| + |b|) \epsilon \end{align}
• Since the inequality above is true for all $\epsilon > 0$ we see that $f(u) - af(s) - bf(t) = 0$. That is, $f(as + bt) = af(s) + bf(t)$. So $f$ is linear.
• Lastly we show that $f$ is bounded with $\| f \| \leq 1$. Since $f \in \mathrm{cl} (B_{X^*})$ there exists $(f_n) \subset B_{X^*}$ such that $(f_n) \to f$. For each $n \in \mathbb{N}$ since $f_n \in B_{X^*}$ we have that $|f_n(x)| \leq \| x \|$. Thus:
(10)
\begin{align} \quad |f(x)| = \left | \lim_{n \to \infty} f_n(x) \right | = \lim_{n \to \infty} |f_n(x)| \leq \| x \| \end{align}
• So $f$ is bounded and $\| f \| \leq 1$.
• Therefore $f \in B_{X^*}$ which shows that $B_{X^*} = \mathrm{cl} (B_{X^*})$. So $B_{X^*}$ is weak-* closed subspace of the compact space $P$, i.e., $B_{X^*}$ is weak-* compact. $\blacksquare$