Editing Banach–Alaoglu theorem (section)

===Proof involving duality theory===

{{math proof|drop=hidden|proof=
Denote by the underlying field of <math>X</math> by <math>\mathbb{K},</math> which is either the [[real number]]s <math>\R</math> or [[complex number]]s <math>\Complex.</math> 
This proof will use some of the basic properties that are listed in the articles: [[polar set]], [[dual system]], and [[continuous linear operator]]. 

To start the proof, some definitions and readily verified results are recalled. When <math>X^{\#}</math> is endowed with the [[weak-* topology]] <math>\sigma\left(X^{\#}, X\right),</math> then this [[Hausdorff space|Hausdorff]] [[Locally convex topological vector space|locally convex]] topological vector space is denoted by <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right).</math> 
The space <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> is always a [[Complete topological vector space|complete TVS]]; however, <math>\left(X^{\prime}, \sigma\left(X^{\prime}, X\right)\right)</math> may fail to be a complete space, which is the reason why this proof involves the space <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right).</math> 
Specifically, this proof will use the fact that a subset of a complete Hausdorff space is compact if (and only if) it is closed and [[Totally bounded (functional analysis)|totally bounded]]. 
Importantly, the [[subspace topology]] that <math>X^{\prime}</math> inherits from <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> is equal to <math>\sigma\left(X^{\prime}, X\right).</math> This can be readily verified by showing that given any <math>f \in X^{\prime},</math> a [[Net (mathematics)|net]] in <math>X^{\prime}</math> converges to <math>f</math> in one of these topologies if and only if it also converges to <math>f</math> in the other topology (the conclusion follows because two topologies are equal if and only if they have the exact same convergent nets). 

The triple <math>\left\langle X, X^{\prime} \right\rangle</math> is a [[Dual system|dual pairing]] although unlike <math>\left\langle X, X^{\#} \right\rangle,</math> it is in general not guaranteed to be a dual system. 
Throughout, unless stated otherwise, all polar sets will be taken with respect to the canonical [[Dual system|pairing]] <math>\left\langle X, X^{\prime} \right\rangle.</math> 

Let <math>U</math>  be a neighborhood of the origin in <math>X</math> and let: 
* <math>U^{\circ} = \left\{f \in X^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\right\}</math> be the polar of <math>U</math> with respect to the canonical pairing <math>\left\langle X, X^{\prime} \right\rangle</math>;
* <math>U^{\circ\circ} = \left\{x \in X ~:~ \sup_{f \in U^{\circ}} |f(x)|\leq 1 \right\}</math> be the [[Polar set|bipolar of <math>U</math>]] with respect to <math>\left\langle X, X^{\prime} \right\rangle</math>;
* <math>U^{\#} = \left\{f \in X^{\#} ~:~ \sup_{u \in U} |f(u)| \leq 1\right\}</math> be the polar of <math>U</math> with respect to the canonical dual system <math>\left\langle X, X^{\#} \right\rangle.</math> Note that <math>U^{\circ} = U^{\#} \cap X^{\prime}.</math>
A well known fact about polar sets is that <math>U^{\circ\circ\circ} \subseteq U^{\circ}.</math> 

# Show that <math>U^{\#}</math> is a <math>\sigma\left(X^{\#}, X\right)</math>-closed subset of <math>X^{\#}:</math> Let <math>f \in X^{\#}</math> and suppose that <math>f_{\bull} = \left(f_i\right)_{i \in I}</math> is a net in <math>U^{\#}</math> that converges to <math>f</math> in <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right).</math> To conclude that <math>f \in U^{\#},</math> it is sufficient (and necessary) to show that <math>|f(u)| \leq 1</math> for every <math>u \in U.</math> Because <math>f_i(u) \to f(u)</math> in the scalar field <math>\mathbb{K}</math> and every value <math>f_i(u)</math> belongs to the closed (in <math>\mathbb{K}</math>) subset <math>\left\{ s \in \mathbb{K} : |s| \leq 1 \right\},</math> so too must this net's limit <math>f(u)</math> belong to this set. Thus <math>|f(u)| \leq 1.</math> 
# Show that <math>U^{\#} = U^{\circ}</math> and then conclude that <math>U^{\circ}</math> is a closed subset of both <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> and <math>\left(X^{\prime}, \sigma\left(X^{\prime}, X\right)\right):</math> The inclusion <math>U^{\circ} \subseteq U^{\#}</math> holds because every continuous linear functional is (in particular) a linear functional. For the reverse inclusion <math>\,U^{\#} \subseteq U^{\circ},\,</math> let <math>f \in U^{\#}</math> so that <math>\;\sup_{u \in U} |f(u)| \leq 1,\,</math> which states exactly that the linear functional <math>f</math> is bounded on the neighborhood <math>U</math>; thus <math>f</math> is a [[Continuous linear operator|continuous linear functional]] (that is, <math>f \in X^{\prime}</math>) and so <math>f \in U^{\circ},</math> as desired. Using (1) and the fact that the intersection <math>U^{\#} \cap X^{\prime} = U^{\circ} \cap X^{\prime} = U^{\circ}</math> is closed in the subspace topology on <math>X^{\prime},</math> the claim about <math>U^{\circ}</math> being closed follows. 
# Show that <math>U^{\circ}</math> is a <math>\sigma\left(X^{\prime}, X\right)</math>-[[Totally bounded (functional analysis)|totally bounded]] subset of <math>X^{\prime}:</math> By the [[bipolar theorem]], <math>U \subseteq U^{\circ\circ}</math> where because the neighborhood <math>U</math> is an [[Absorbing set|absorbing subset]] of <math>X,</math> the same must be true of the set <math>U^{\circ\circ};</math> it is possible to prove that this implies that <math>U^{\circ}</math> is a <math>\sigma\left(X^{\prime}, X\right)</math>-[[Bounded set (topological vector space)|bounded subset]] of <math>X^{\prime}.</math> Because <math>X</math> [[dual system|distinguishes points]] of <math>X^{\prime},</math> a subset of <math>X^{\prime}</math> is <math>\sigma\left(X^{\prime}, X\right)</math>-bounded if and only if it is <math>\sigma\left(X^{\prime}, X\right)</math>-[[Totally bounded (functional analysis)|totally bounded]]. So in particular, <math>U^{\circ}</math> is also <math>\sigma\left(X^{\prime}, X\right)</math>-totally bounded. 
# Conclude that <math>U^{\circ}</math> is also a <math>\sigma\left(X^{\#}, X\right)</math>-totally bounded subset of <math>X^{\#}:</math> Recall that the <math>\sigma\left(X^{\prime}, X\right)</math> topology on <math>X^{\prime}</math> is identical to the subspace topology that <math>X^{\prime}</math> inherits from <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right).</math> This fact, together with (3) and the definition of "totally bounded", implies that <math>U^{\circ}</math> is a <math>\sigma\left(X^{\#}, X\right)</math>-totally bounded subset of <math>X^{\#}.</math> 
# Finally, deduce that <math>U^{\circ}</math> is a <math>\sigma\left(X^{\prime}, X\right)</math>-compact subset of <math>X^{\prime}:</math> Because <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> is a [[Complete topological vector space|complete TVS]] and <math>U^{\circ}</math> is a closed (by (2)) and totally bounded (by (4)) subset of <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right),</math> it follows that <math>U^{\circ}</math> is compact. [[Q.E.D.|<math>\blacksquare</math>]]
}}

If <math>X</math> is a [[normed vector space]], then the polar of a neighborhood is closed and norm-bounded in the dual space. 
In particular, if <math>U</math>  is the open (or closed) unit ball in <math>X</math> then the polar of <math>U</math> is the closed unit ball in the continuous dual space <math>X^{\prime}</math> of <math>X</math> (with the [[Operator norm|usual dual norm]]). 
Consequently, this theorem can be specialized to:

{{math theorem|name=Banach–Alaoglu theorem|math_statement= If <math>X</math> is a normed space then the closed unit ball in the continuous dual space <math>X^{\prime}</math> (endowed with its usual [[operator norm]]) is compact with respect to the [[weak-* topology]]. }}

When the continuous dual space <math>X^{\prime}</math> of <math>X</math> is an infinite dimensional normed space then it is {{em|impossible}} for the closed unit ball in <math>X^{\prime}</math> to be a compact subset when <math>X^{\prime}</math> has its usual norm topology. 
This is because the unit ball in the norm topology is compact if and only if the space is finite-dimensional (cf. [[F. Riesz theorem]]). 
This theorem is one example of the utility of having different topologies on the same vector space.

It should be cautioned that despite appearances, the Banach–Alaoglu theorem does {{em|not}} imply that the weak-* topology is [[locally compact]]. 
This is because the closed unit ball is only a neighborhood of the origin in the [[strong topology]], but is usually not a neighborhood of the origin in the weak-* topology, as it has empty interior in the weak* topology, unless the space is finite-dimensional. 
In fact, it is a result of [[André Weil|Weil]] that all [[locally compact]] [[Hausdorff space|Hausdorff]] topological vector spaces must be finite-dimensional.