View source for Banach–Alaoglu theorem

{{Short description|Theorem in functional analysis}}
In [[functional analysis]] and related branches of [[mathematics]], the '''Banach–Alaoglu theorem''' (also known as '''Alaoglu's theorem''') states that the [[Closed set|closed]] [[Ball (mathematics)|unit ball]] of the [[dual space]] of a [[normed vector space]] is [[Compact space|compact]] in the [[weak* topology]].<ref>{{harvnb|Rudin|1991}}, Theorem 3.15.</ref> 
A common proof identifies the unit ball with the weak-* topology as a closed subset of a [[Cartesian product|product]] of compact sets with the [[product topology]]. 
As a consequence of [[Tychonoff's theorem]], this product, and hence the unit ball within, is compact.

This theorem has applications in physics when one describes the set of states of an algebra of observables, namely that any state can be written as a convex linear combination of so-called pure states.

==History==

According to Lawrence Narici and Edward Beckenstein, the Alaoglu theorem is a “very important result—maybe {{em|the}} most important fact about the [[weak-* topology]]—[that] echos throughout functional analysis.”{{sfn|Narici|Beckenstein|2011|pp=235-240}} 
In 1912, Helly proved that the unit ball of the continuous dual space of <math>C([a, b])</math> is countably weak-* compact.{{sfn|Narici|Beckenstein|2011|pp=225-273}} 
In 1932, [[Stefan Banach]] proved that the closed unit ball in the continuous dual space of any [[Separable metric space|separable]] [[normed space]] is sequentially weak-* compact (Banach only considered [[sequential compactness]]).{{sfn|Narici|Beckenstein|2011|pp=225-273}} 
The proof for the general case was published in 1940 by the mathematician [[Leonidas Alaoglu]]. 
According to Pietsch [2007], there are at least twelve mathematicians who can lay claim to this theorem or an important predecessor to it.{{sfn|Narici|Beckenstein|2011|pp=235-240}} 

The '''Bourbaki–Alaoglu theorem''' is a generalization<ref>{{harvnb|Köthe|1983}}, Theorem (4) in §20.9.</ref><ref>{{harvnb|Meise|Vogt|1997}}, Theorem 23.5.</ref> of the original theorem by [[Nicolas Bourbaki|Bourbaki]] to [[dual topology|dual topologies]] on [[locally convex space]]s. 
This theorem is also called the '''Banach–Alaoglu theorem''' or the '''weak-* compactness theorem''' and it is commonly called simply the '''Alaoglu theorem'''.{{sfn|Narici|Beckenstein|2011|pp=235-240}}

==Statement==
{{See also|Topological vector space#Dual space|Dual system|Polar set}}

If <math>X</math> is a vector space over the field <math>\mathbb{K}</math> then <math>X^{\#}</math> will denote the [[algebraic dual space]] of <math>X</math> and these two spaces are henceforth associated with the [[Bilinear map|bilinear]] {{em|evaluation map}} <math>\left\langle \cdot, \cdot \right\rangle : X \times X^{\#} \to \mathbb{K}</math> defined by 
<math display=block>\left\langle x, f \right\rangle ~\stackrel{\scriptscriptstyle\text{def}}{=}~ f(x)</math>
where the triple <math>\left\langle X, X^{\#},\left\langle \cdot, \cdot \right\rangle \right\rangle</math> forms a [[dual system]] called the {{em|canonical dual system}}. 

If <math>X</math> is a [[topological vector space]] (TVS) then its [[continuous dual space]] will be denoted by <math>X^{\prime},</math> where <math>X^{\prime} \subseteq X^{\#}</math> always holds. 
Denote the [[weak-* topology]] on <math>X^{\#}</math> by <math>\sigma\left(X^{\#}, X\right)</math> and denote the weak-* topology on <math>X^{\prime}</math> by <math>\sigma\left(X^{\prime}, X\right).</math> 
The weak-* topology is also called the [[topology of pointwise convergence]] because given a map <math>f</math> and a [[Net (mathematics)|net]] of maps <math>f_{\bull} = \left(f_i\right)_{i \in I},</math> the net <math>f_{\bull}</math> converges to <math>f</math> in this topology if and only if for every point <math>x</math> in the domain, the net of values <math>\left(f_i(x)\right)_{i \in I}</math> converges to the value <math>f(x).</math> 

{{Math theorem|name=Alaoglu theorem{{sfn|Narici|Beckenstein|2011|pp=225-273}}|math_statement=
For any [[topological vector space]] (TVS) <math>X</math> ({{em|not}} necessarily [[Hausdorff space|Hausdorff]] or [[locally convex]]) with [[continuous dual space]] <math>X^{\prime},</math> the [[Polar set|polar]] 
<math display=block>U^{\circ} = \left\{f \in X^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\right\}</math> 
of any [[Neighborhood (topology)|neighborhood]] <math>U</math> of origin in <math>X</math> is compact in the [[weak topology (polar topology)|weak-* topology]]<ref group=note>Explicitly, a subset <math>B^{\prime} \subseteq X^{\prime}</math> is said to be "compact (resp. totally bounded, etc.) in the weak-* topology" if when <math>X^{\prime}</math> is given the [[weak-* topology]] and the subset <math>B^{\prime}</math> is given the [[subspace topology]] inherited from <math>\left(X^{\prime}, \sigma\left(X^{\prime}, X\right)\right),</math> then <math>B^{\prime}</math> is a [[Compact set|compact]] (resp. [[totally bounded]], etc.) space.</ref> <math>\sigma\left(X^{\prime}, X\right)</math> on <math>X^{\prime}.</math> Moreover, <math>U^{\circ}</math> is equal to the polar of <math>U</math>  with respect to the canonical system <math>\left\langle X, X^{\#} \right\rangle</math> and it is also a compact subset of <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right).</math>
}}

===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.

===Elementary proof===

The following [[elementary proof]] does not utilize [[duality theory]] and requires only basic concepts from set theory, topology, and functional analysis. 
What is needed from topology is a working knowledge of [[Net (mathematics)|net]] [[Net convergence|convergence]] in [[topological space]]s and familiarity with the fact that a [[Continuous linear functional|linear functional is continuous]] if and only if it is [[Continuous linear operator#bounded on a neighborhood of a point|bounded on a neighborhood]] of the origin (see the articles on [[continuous linear functional]]s and [[sublinear functional]]s for details). 
Also required is a proper understanding of the technical details of how the space <math>\mathbb{K}^X</math> of all functions of the form <math>X \to \mathbb{K}</math> is identified as the [[Cartesian product]] <math display=inline>\prod_{x \in X} \mathbb{K},</math> and the relationship between [[pointwise convergence]], the [[product topology]], and [[Subspace topology|subspace topologies]] they induce on subsets such as the [[algebraic dual space]] <math>X^{\#}</math> and products of subspaces such as <math display=inline>\prod_{x \in X} B_{r_x}.</math> 
An explanation of these details is now given for readers who are interested. 

{{collapse top|title=Primer on product/function spaces, nets, and pointwise convergence|left=true}}

For every real <math>r,</math> <math>B_r ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{c \in \mathbb{K} : |c| \leq r\}</math> will denote the closed ball of radius <math>r</math> centered at <math>0</math> and <math>r U ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{r u : u \in U\}</math> for any <math>U \subseteq X,</math>

'''Identification of functions with tuples'''

The Cartesian product <math display=inline>\prod_{x \in X} \mathbb{K}</math> is usually thought of as the set of all <math>X</math>-indexed [[tuple]]s <math>s_{\bull} = \left(s_x\right)_{x \in X}</math> but, since tuples are technically just functions from an indexing set, it can also be identified with the space <math>\mathbb{K}^X</math> of all functions having prototype <math>X \to \mathbb{K},</math> as is now described: 

* {{em|Function <math>\to</math> Tuple}}: A function <math>s : X \to \mathbb{K}</math> belonging to <math>\mathbb{K}^X</math> is identified with its (<math>X</math>-indexed) "{{em|tuple of values}}" <math>s_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ (s(x))_{x \in X}.</math>
* {{em|Tuple <math>\to</math> Function}}: A tuple <math>s_{\bull} = \left(s_x\right)_{x \in X}</math> in <math display=inline>\prod_{x \in X} \mathbb{K}</math> is identified with the function <math>s : X \to \mathbb{K}</math> defined by <math>s(x) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s_x</math>; this function's "tuple of values" is the original tuple <math>\left(s_x\right)_{x \in X}.</math>

This is the reason why many authors write, often without comment, the equality 
<math display=block>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math>
and why the Cartesian product <math display=inline>\prod_{x \in X} \mathbb{K}</math> is sometimes taken as the definition of the set of maps <math>\mathbb{K}^X</math> (or conversely). 
However, the Cartesian product, being the [[Product (category theory)|(categorical) product]] in the [[Category (mathematics)|category]] of [[Category of sets|sets]] (which is a type of [[inverse limit]]), also comes equipped with associated maps that are known as its (coordinate) {{em|projections}}. 

The {{visible anchor|canonical projection of the Cartesian product}} at a given point <math>z \in X</math> is the function
<math display=block>\Pr{}_z : \prod_{x \in X} \mathbb{K} \to \mathbb{K} \quad \text{ defined by } \quad s_{\bull} = \left(s_x\right)_{x \in X} \mapsto s_z</math>
where under the above identification, <math>\Pr{}_z</math> sends a function <math>s : X \to \mathbb{K}</math> to
<math display=block>\Pr{}_z(s) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s(z).</math> 
Stated in words, for a point <math>z</math> and function <math>s,</math> "plugging <math>z</math> into <math>s</math>" is the same as "plugging <math>s</math> into <math>\Pr{}_z</math>". 

In particular, suppose that <math>\left(r_x\right)_{x \in X}</math> are non-negative real numbers. 
Then <math>\prod_{x \in X} B_{r_x} \subseteq \prod_{x \in X} \mathbb{K} = \mathbb{K}^X,</math> where under the above identification of tuples with functions, <math>\prod_{x \in X} B_{r_x}</math> is the set of all functions <math>s \in \mathbb{K}^X</math> such that <math>s(x) \in B_{r_x}</math> for every <math>x \in X.</math> 

If a subset <math>U \subseteq X</math> [[Partition of a set|partitions]] <math>X</math> into <math>X = U \, \cup \,(X \setminus U)</math> then the linear bijection
<math display=block>\begin{alignat}{4}
H :\;&& \prod_{x \in X} \mathbb{K} &&\;\to    \;& \left(\prod_{u \in U} \mathbb{K}\right) \times \prod_{x \in X \setminus U} \mathbb{K} \\[0.3ex]
     && \left(f_x\right)_{x \in X} &&\;\mapsto\;& \left( \left(f_u\right)_{u \in U}, \; \left(f_x\right)_{x \in X \setminus U} \right) \\
\end{alignat}</math>
canonically identifies these two Cartesian products; moreover, this map is a [[homeomorphism]] when these products are endowed with their product topologies. 
In terms of function spaces, this bijection could be expressed as 
<math display=block>\begin{alignat}{4}
H :\;&& \mathbb{K}^X &&\;\to    \;& \mathbb{K}^U \times \mathbb{K}^{X \setminus U} \\[0.3ex]
     && f &&\;\mapsto\;& \left(f\big\vert_U, \; f\big\vert_{X \setminus U}\right) \\
\end{alignat}.</math>

'''Notation for nets and function composition with nets'''

A [[Net (mathematics)|net]] <math>x_{\bull} = \left(x_i\right)_{i \in I}</math> in <math>X</math> is by definition a function <math>x_{\bull} : I \to X</math> from a non-empty [[directed set]] <math>(I, \leq).</math> 
Every [[sequence]] in <math>X,</math> which by definition is just a function of the form <math>\N \to X,</math> is also a net. 
As with sequences, the value of a net <math>x_{\bull}</math> at an index <math>i \in I</math> is denoted by <math>x_i</math>; however, for this proof, this value <math>x_i</math> may also be denoted by the usual function parentheses notation <math>x_{\bull}(i).</math> 
Similarly for [[function composition]], if <math>F : X \to Y</math> is any function then the net (or sequence) that results from "plugging <math>x_{\bull}</math> into <math>F</math>" is just the function <math>F \circ x_{\bull} : I \to Y,</math> although this is typically denoted by <math>\left(F\left(x_i\right)\right)_{i \in I}</math> (or by <math>\left(F\left(x_i\right)\right)_{i=1}^{\infty}</math> if <math>x_{\bull}</math> is a sequence). 
In the proofs below, this resulting net may be denoted by any of the following notations 
<math display=block>F\left(x_{\bull}\right) = \left(F\left(x_i\right)\right)_{i \in I} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ F \circ x_{\bull},</math> 
depending on whichever notation is cleanest or most clearly communicates the intended information. 
In particular, if <math>F : X \to Y</math> is continuous and <math>x_{\bull} \to x</math> in <math>X,</math> then the conclusion commonly written as <math>\left(F\left(x_i\right)\right)_{i \in I} \to F(x)</math> may instead be written as <math>F\left(x_{\bull}\right) \to F(x)</math> or <math>F \circ x_{\bull} \to F(x).</math> 

'''Topology'''

The set <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math> is assumed to be endowed with the [[product topology]]. It is well known that the product topology is identical to the [[topology of pointwise convergence]]. 
This is because given <math>f</math> and a [[Net (mathematics)|net]] <math>\left(f_i\right)_{i \in I},</math> where <math>f</math> and every <math>f_i</math> is an element of <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K},</math> then the net <math>\left(f_i\right)_{i \in I} \to f</math> [[Convergent net|converges]] in the product topology if and only if

:for every <math>z \in X,</math> the net <math>\Pr{}_z\left(\left(f_i\right)_{i \in I}\right) \to \Pr{}_z(f)</math> converges in <math>\mathbb{K},</math> 

where because <math>\;\Pr{}_z(f) = f(z)\;</math> and 
<math display=inline>\Pr{}_z\left(\left(f_i\right)_{i \in I}\right) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(\Pr{}_z\left(f_i\right)\right)_{i \in I} = \left(f_i(z)\right)_{i \in I},</math> 
this happens if and only if 

:for every <math>z \in X,</math> the net <math>\left(f_i(z)\right)_{i \in I} \to f(z)</math> converges in <math>\mathbb{K},</math> 

Thus <math>\left(f_i\right)_{i \in I}</math> converges to <math>f</math> in the product topology if and only if it converges to <math>f</math> pointwise on <math>X.</math>

This proof will also use the fact that the topology of pointwise convergence is preserved when passing to [[topological subspace]]s. 
This means, for example, that if for every <math>x \in X,</math> <math>S_x \subseteq \mathbb{K}</math> is some [[Topological subspace|(topological) subspace]] of <math>\mathbb{K}</math> then the topology of pointwise convergence (or equivalently, the product topology) on <math display=inline>\prod_{x \in X} S_x</math> is equal to the [[subspace topology]] that the set <math display=inline>\prod_{x \in X} S_x</math> inherits from <math display=inline>\prod_{x \in X} \mathbb{K}.</math> 
And if <math>S_x</math> is closed in <math>\mathbb{K}</math> for every <math>x \in X,</math> then <math display=inline>\prod_{x \in X} S_x</math> is a closed subset of <math display=inline>\prod_{x \in X} \mathbb{K}.</math> 

'''Characterization of <math>\sup_{u \in U} |f(u)| \leq r</math>'''

An important fact used by the proof is that for any real <math>r,</math> 
<math display=block>\sup_{u \in U} |f(u)| \leq r \qquad \text{ if and only if } \qquad f(U) \subseteq B_r</math>
where <math>\,\sup\,</math> denotes the [[supremum]] and <math>f(U) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{f(u) : u \in U\}.</math> 
As a side note, this characterization does not hold if the closed ball <math>B_r</math> is replaced with the open ball <math>\{c \in \mathbb{K} : |c| < r\}</math> (and replacing <math>\;\sup_{u \in U} |f(u)| \leq r\;</math> with the strict inequality <math>\;\sup_{u \in U} |f(u)| < r\;</math> will not change this; for counter-examples, consider <math>X ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \mathbb{K}</math> and the [[identity map]] <math>f ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \operatorname{Id}</math> on <math>X</math>). 
{{collapse bottom}} 

The essence of the Banach–Alaoglu theorem can be found in the next proposition, from which the Banach–Alaoglu theorem follows. 
Unlike the Banach–Alaoglu theorem, this proposition does {{em|not}} require the vector space <math>X</math> to endowed with any topology. 

{{Math theorem|name=Proposition{{sfn|Narici|Beckenstein|2011|pp=225-273}}|math_statement=
Let <math>U</math> be a subset of a vector space <math>X</math> over the field <math>\mathbb{K}</math> (where <math>\mathbb{K} = \R \text{ or } \mathbb{K} = \Complex</math>) and for every real number <math>r,</math> endow the closed ball <math display=inline>B_r ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{s \in \mathbb{K} : |s| \leq r\}</math> with its [[Subspace topology|usual topology]] (<math>X</math> need not be endowed with any topology, but <math>\mathbb{K}</math> has its usual [[Euclidean topology]]). 
Define
<math display=block>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in X^{\#} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\}.</math>

If for every <math>x \in X,</math> <math>r_x > 0</math> is a real number such that <math>x \in r_x U,</math> then <math>U^{\#}</math> is a closed and [[Compact space|compact]] [[Topological subspace|subspace]] of the [[product space]] <math>\prod_{x \in X} B_{r_x}</math> (where because this [[product topology]] is identical to the [[topology of pointwise convergence]], which is also called the [[weak-* topology]] in functional analysis, this means that <math>U^{\#}</math> is compact in the weak-* topology or "weak-* compact" for short).
}}

Before proving the proposition above, it is first shown how the Banach–Alaoglu theorem follows from it (unlike the proposition, Banach–Alaoglu assumes that <math>X</math> is a [[topological vector space]] (TVS) and that <math>U</math> is a neighborhood of the origin). 

{{Math proof|drop=hidden|title=Proof that Banach–Alaoglu follows from the proposition above|proof=
Assume that <math>X</math> is a [[topological vector space]] with continuous dual space <math>X^{\prime}</math> and that <math>U</math> is a neighborhood of the origin. 
Because <math>U</math> is a neighborhood of the origin in <math>X,</math> it is also an [[Absorbing set|absorbing subset]] of <math>X,</math> so for every <math>x \in X,</math> there exists a real number <math>r_x > 0</math> such that <math>x \in r_x U.</math> 
Thus the hypotheses of the above proposition are satisfied, and so the set <math>U^{\#}</math> is therefore compact in the [[weak-* topology]].  
The proof of the Banach–Alaoglu theorem will be complete once it is shown that <math>U^{\#} = U^{\circ},</math><ref group=note>If <math>\tau</math> denotes the topology that <math>X</math> is (originally) endowed with, then the equality <math>U^{\circ} = U^{\#}</math> shows that the polar <math>U^{\circ} = \Big\{f \in X^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\}</math> of <math>U</math> is dependent {{em|only}} on <math>U</math> (and <math>X^{\#}</math>) and that the rest of the topology <math>\tau</math> can be ignored.  To clarify what is meant, suppose <math>\sigma</math> is any TVS topology on <math>X</math> such that the set <math>U</math> is (also) a neighborhood of the origin in <math>(X, \sigma).</math>  Denote the continuous dual space of <math>(X, \sigma)</math> by <math>(X, \sigma)^{\prime}</math> and denote the polar of <math>U</math> with respect to <math>(X, \sigma)</math> by 
<math display=block>U^{\circ, \sigma} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in (X, \sigma)^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\}</math>
so that <math>U^{\circ, \tau}</math> is just the set <math>U^{\circ}</math> from above. 
Then <math>U^{\circ, \tau} = U^{\circ, \sigma}</math> because both of these sets are equal to <math>U^{\#}.</math>  Said differently, the polar set <math>U^{\circ, \sigma}</math>'s defining "requirement" that <math>U^{\circ, \sigma}</math> be a subset of the '''{{em|continuous}}''' dual space <math>(X, \sigma)^{\prime}</math> is inconsequential and can be ignored because it does not have any effect on the resulting set of linear functionals. However, if <math>\nu</math> is a TVS topology on <math>X</math> such that <math>U</math> is {{em|not}} a neighborhood of the origin in <math>(X, \nu)</math> then the polar <math>U^{\circ, \nu}</math> of <math>U</math> with respect to <math>(X, \nu)</math> is not guaranteed to equal <math>U^{\#}</math> and so the topology <math>\nu</math> can not be ignored.</ref> 
where recall that <math>U^{\circ}</math> was defined as
<math display=block>U^{\circ} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in X^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\} ~=~ U^{\#} \cap X^{\prime}.</math>

'''Proof that <math>U^{\circ} = U^{\#}:</math>''' 
Because <math>U^{\circ} = U^{\#} \cap X^{\prime},</math> the conclusion is equivalent to <math>U^{\#} \subseteq X^{\prime}.</math> 
If <math>f \in U^{\#}</math> then <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 functional]] (that is, <math>f \in X^{\prime}</math>), as desired. 
<math>\blacksquare</math>
}}

{{Math proof|drop=hidden|title=Proof of Proposition|proof=
The [[Product space (topology)|product space]] <math display=inline>\prod_{x \in X} B_{r_x}</math> is compact by [[Tychonoff's theorem]] (since each closed ball <math>B_{r_x} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{s \in \mathbb{K} : |s| \leq r_x\}</math> is a [[Hausdorff space|Hausdorff]]<ref group=note>Because every <math>B_{r_x}</math> is also a [[Hausdorff space]], the conclusion that <math>\prod_{x \in X} B_{r_x}</math> is compact only requires the so-called "Tychonoff's theorem for compact Hausdorff spaces," which is equivalent to the [[ultrafilter lemma]] and strictly weaker than the [[axiom of choice]].</ref> [[compact space]]). Because a closed subset of a compact space is compact, the proof of the proposition will be complete once it is shown that
<math display=block>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in X^{\#} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\} ~=~ \left\{f \in X^{\#} ~:~ f(U) \subseteq B_1\right\}</math>
is a closed subset of <math display=inline>\prod_{x \in X} B_{r_x}.</math> 
The following statements guarantee this conclusion:
#<math>U^{\#} \subseteq \prod_{x \in X} B_{r_x}.</math>
#<math>U^{\#}</math> is a closed subset of the [[product space]] <math>\prod_{x \in X} \mathbb{K} = \mathbb{K}^X.</math>

'''Proof of (1)''': 

For any <math>z \in X,</math> let <math display=inline>\Pr{}_z : \prod_{x \in X} \mathbb{K} \to \mathbb{K}</math> denote the projection to the <math>z</math><sup>th</sup> coordinate ([[#canonical projection of the Cartesian product|as defined above]]). 
To prove that <math display=inline>U^{\#} \subseteq \prod_{x \in X} B_{r_x},</math> it is sufficient (and necessary) to show that <math>\Pr{}_x\left(U^{\#}\right) \subseteq B_{r_x}</math> for every <math>x \in X.</math> 
So fix <math>x \in X</math> and let <math>f \in U^{\#}.</math> 
Because <math>\Pr{}_x(f) \,=\, f(x),</math> it remains to show that <math>f(x) \in B_{r_x}.</math> 
Recall that <math>r_x > 0</math> was defined in the proposition's statement as being any positive real number that satisfies <math>x \in r_x U</math> (so for example, <math>r_u := 1</math> would be a valid choice for each <math>u \in U</math>), which implies <math>\,u_x ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \frac{1}{r_x} \, x \in U.\,</math> 
Because <math>f</math> is a [[Strict positive homogeneity|positive homogeneous]] function that satisfies <math>\;\sup_{u \in U} |f(u)| \leq 1,\,</math> 
<math display=block>\frac{1}{r_x}|f(x)|
= \left|\frac{1}{r_x} f(x)\right|
= \left|f\left(\frac{1}{r_x} x\right)\right|
= \left|f\left(u_x\right)\right| 
\leq \sup_{u \in U} |f(u)| 
\leq 1.</math> 

Thus <math>|f(x)| \leq r_x,</math> which shows that <math>f(x) \in B_{r_x},</math> as desired. 

'''Proof of (2)''': 

The [[algebraic dual space]] <math>X^{\#}</math> is always a closed subset of <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math> (this is proved in [[#Algebraic dual space is closed in the space of all functions|the lemma below]] for readers who are not familiar with this result). 
The set
<math display=block>\begin{alignat}{9}
U_{B_1} 
&\,\stackrel{\scriptscriptstyle\text{def}}{=}\, \Big\{ ~~\;~~\;~~\;~~ f\ \in \mathbb{K}^X  ~~\;~~ : \sup_{u \in U} |f(u)| \leq 1\Big\} \\
&= \big\{ ~~\;~~\;~~\;~~f \, \in \mathbb{K}^X  ~~\;~~ : f(u) \in B_1 \text{ for all } u \in U\big\} \\
&= \Big\{\left(f_x\right)_{x \in X} \in \prod_{x \in X} \mathbb{K} \,~:~ \; ~f_u~ \in B_1  \text{ for all } u \in U\Big\} \\
&= \prod_{x \in X} C_x \quad \text{ where } \quad C_x ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \begin{cases}
B_1        & \text{ if } x \in U \\
\mathbb{K} & \text{ if } x \not\in U \\
\end{cases} \\
\end{alignat}</math>
is closed in the [[product topology]] on <math>\prod_{x \in X} \mathbb{K} = \mathbb{K}^X</math> since it is a product of closed subsets of <math>\mathbb{K}.</math> 
Thus <math>U_{B_1} \cap X^{\#} = U^{\#}</math> is an intersection of two closed subsets of <math>\mathbb{K}^X,</math> which proves (2).<ref group=note>The conclusion <math>U_{B_1} = \prod_{x \in X} C_x</math> can be written as <math>U_{B_1} ~=~ \Big(\prod_{u \in U} B_1\Big) \times \prod_{x \in X \setminus U} \mathbb{K}.</math> The set <math>U^{\#}</math> may thus equivalently be defined by <math>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ X^{\#} \cap \left[\Big(\prod_{u \in U} B_1\Big) \times \prod_{x \in X \setminus U} \mathbb{K}\right].</math> Rewriting the definition in this way helps make it apparent that the set <math>U^{\#}</math> is closed in <math>\prod_{x \in X} \mathbb{K}</math> because [[#Algebraic dual space is closed in the space of all functions|this is true of <math>X^{\#}.</math>]]</ref>
<math>\blacksquare</math>
}}

The conclusion that the set <math>U_{B_1} = \left\{f \in \mathbb{K}^X : f(U) \subseteq B_1\right\}</math> is closed can also be reached by applying the following more general result, this time proved using nets, to the special case <math>Y := \mathbb{K}</math> and <math>B := B_1.</math><!-- NOTE: This added generality does not make the proof more difficult because the proof is nearly identical to the proof of the special space where Y = K and B = B_1. In fact, the generality simplified the proof slightly (compared to the older version from e.g. 26 November 2021) -->

:'''Observation''': If <math>U \subseteq X</math> is any set and if <math>B \subseteq Y</math> is a [[Closed set|closed]] subset of a topological space <math>Y,</math> then <math>U_B ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{f \in Y^X : f(U) \subseteq B\right\}</math> is a closed subset of <math>Y^X</math> in the topology of pointwise convergence. 

:'''Proof of observation''': Let <math>f \in Y^X</math> and suppose that <math>\left(f_i\right)_{i \in I}</math> is a net in <math>U_B</math> that converges pointwise to <math>f.</math> It remains to show that <math>f \in U_B,</math> which by definition means <math>f(U) \subseteq B.</math> For any <math>u \in U,</math> because <math>\left(f_i(u)\right)_{i \in I} \to f(u)</math> in <math>Y</math> and every value <math>f_i(u) \in f_i(U) \subseteq B</math> belongs to the closed (in <math>Y</math>) subset <math>B,</math> so too must this net's limit belong to this closed set; thus <math>f(u) \in B,</math> which completes the proof. <math>\blacksquare</math>

{{math theorem|name=Lemma (<math>X^{\#}</math> is closed in <math>\mathbb{K}^X</math>){{anchor|Algebraic dual space is closed in the space of all functions}}|note=|style=|math_statement= 
The [[algebraic dual space]] <math>X^{\#}</math> of any vector space <math>X</math> over a field <math>\mathbb{K}</math> (where <math>\mathbb{K}</math> is <math>\R</math> or <math>\Complex</math>) is a closed subset of <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math> in the topology of pointwise convergence. (The vector space <math>X</math> need not be endowed with any topology). 
}}

{{collapse top|title=Proof of lemma|left=true}}
Let <math>f \in \mathbb{K}^X</math> and suppose that <math>f_{\bull} = \left(f_i\right)_{i \in I}</math> is a net in <math>X^{\#}</math> the converges to <math>f</math> in <math>\mathbb{K}^X.</math> 
To conclude that <math>f \in X^{\#},</math> it must be shown that <math>f</math> is a linear functional. So let <math>s</math> be a scalar and let <math>x, y \in X.</math> 

For any <math>z \in X,</math> let <math>f_{\bull}(z) : I \to \mathbb{K}</math> denote {{em|<math>f_{\bull}</math>'s net of values at <math>z</math>}}
<math display=block>f_{\bull}(z) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(f_i(z)\right)_{i \in I}.</math>
Because <math>f_{\bull} \to f</math> in <math>\mathbb{K}^X,</math> which has the topology of pointwise convergence, <math>f_{\bull}(z) \to f(z)</math> in <math>\mathbb{K}</math> for every <math>z \in X.</math> 
By using <math>x, y, sx, \text{ and } x + y,</math> in place of <math>z,</math> it follows that each of the following nets of scalars converges in <math>\mathbb{K}:</math>
<math display=block>f_{\bull}(x) \to f(x), \quad f_{\bull}(y) \to f(y), \quad f_{\bull}(x + y) \to f(x + y), \quad \text{ and } \quad f_{\bull}(sx) \to f(sx).</math>


Proof that <math>f(s x) = s f(x):</math> 
Let <math>M : \mathbb{K} \to \mathbb{K}</math> be the "multiplication by <math>s</math>" map defined by <math>M(c) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s c.</math> 
Because <math>M</math> is continuous and <math>f_{\bull}(x) \to f(x)</math> in <math>\mathbb{K},</math> it follows that <math>M\left(f_{\bull}(x)\right) \to M(f(x))</math> where the right hand side is <math>M(f(x)) = s f(x)</math> and the left hand side is
<math display=block>\begin{alignat}{4}
M\left(f_{\bull}(x)\right) 
\stackrel{\scriptscriptstyle\text{def}}{=}&~ M \circ f_{\bull}(x)                            && \text{ by definition of notation } \\
=&~ \left(M\left(f_i(x)\right)\right)_{i \in I}  ~~~ && \text{ because } f_{\bull}(x) = \left(f_i(x)\right)_{i \in I} : I \to \mathbb{K} \\
=&~ \left(s f_i(x)\right)_{i \in I}                  && M\left(f_i(x)\right) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s f_i(x) \\
=&~ \left(f_i(s x)\right)_{i \in I}                  && \text{ by linearity of } f_i \\
=&~ f_{\bull}(sx)                                    && \text{ notation }
\end{alignat}</math>
which proves that <math>f_{\bull}(sx) \to s f(x).</math> Because also <math>f_{\bull}(sx) \to f(sx)</math> and limits in <math>\mathbb{K}</math> are unique, it follows that <math>s f(x) = f(s x),</math> as desired. 


Proof that <math>f(x + y) = f(x) + f(y):</math> 
Define a net <math>z_{\bull} = \left(z_i\right)_{i \in I} : I \to \mathbb{K} \times \mathbb{K}</math> by letting <math>z_i ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(f_i(x), f_i(y)\right)</math> for every <math>i \in I.</math> 
Because <math>f_{\bull}(x) = \left(f_i(x)\right)_{i \in I} \to f(x)</math> and <math>f_{\bull}(y) = \left(f_i(y)\right)_{i \in I} \to f(y),</math> it follows that <math>z_{\bull} \to ( f(x), f(y) )</math> in <math>\mathbb{K} \times \mathbb{K}.</math> 
Let <math>A : \mathbb{K} \times \mathbb{K} \to \mathbb{K}</math> be the addition map defined by <math>A(x, y) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ x + y.</math> 
The continuity of <math>A</math> implies that <math>A\left(z_{\bull}\right) \to A(f(x), f(y))</math> in <math>\mathbb{K}</math> where the right hand side is <math>A(f(x), f(y)) = f(x) + f(y)</math> and the left hand side is
<math display=block>A\left(z_{\bull}\right) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ A \circ z_{\bull} 
= \left(A\left(z_i\right)\right)_{i \in I} 
= \left(A\left(f_i(x), f_i(y)\right)\right)_{i \in I} 
= \left(f_i(x) + f_i(y)\right)_{i \in I} 
= \left(f_i(x + y)\right)_{i \in I} 
= f_{\bull}(x + y)
</math>
which proves that <math>f_{\bull}(x + y)  \to f(x) + f(y).</math> Because also <math>f_{\bull}(x + y) \to f(x + y),</math> it follows that <math>f(x + y) = f(x) + f(y),</math> as desired. 
<math>\blacksquare</math>

<!-- Removed information
Alternative proof outline: 
The proof above did not involve [[Cauchy net]]s or even anything related to the [[uniform structure]] (specifically, the [[product uniformity]]) that the product space <math>\prod_{x \in X} \mathbb{K} = \mathbb{K}^X</math> is endowed with. 
However, it is possible to prove this lemma by first showing that <math>X^{\#}</math> is a [[Complete topological vector space|complete TVS]] and then concluding that it is a closed subset of the Hausdorff space <math>\mathbb{K}^X</math> (because every complete subspace of a Hausdorff space is necessarily a closed subset). 
This involves dealing with [[Cauchy net]]s, where it can be shown that a net <math>f_{\bull}</math> is Cauchy in the topology of pointwise convergence if and only if for every point <math>x \in X,</math> the net of values <math>\left(f_i(x)\right)_{i \in I}</math> is Cauchy in <math>\mathbb{K}.</math> 
Showing that <math>X^{\#}</math> is a complete space requires showing that every Cauchy net <math>f_{\bull}</math> in <math>X^{\#}</math> converges to some element <math>f</math> of <math>X^{\#}.</math> 
However, such a proof would require additional steps such as constructing the function <math>f</math> from the net <math>f_{\bull}</math> and then proving that <math>f \in X^{\#}</math> and that <math>f_{\bull} \to f</math> in <math>X^{\#}.</math> 
--- End: Removed information -->
{{collapse bottom}}<!-- END OF PROOF OF LEMMA -->

[[#Algebraic dual space is closed in the space of all functions|The lemma above]] actually also follows from its corollary below since <math>\prod_{x \in X} \mathbb{K}</math> is a Hausdorff [[complete uniform space]] and any subset of such a space (in particular <math>X^{\#}</math>) is closed if and only if it is complete. 

{{math theorem|name=Corollary to lemma (<math>X^{\#}</math> is weak-* complete)|note=|style=|math_statement= 
When the [[algebraic dual space]] <math>X^{\#}</math> of a vector space <math>X</math> is equipped with the topology <math>\sigma\left(X^{\#}, X\right)</math> of pointwise convergence (also known as the weak-* topology) then the resulting [[topological space]] <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> is a [[Complete topological vector space|complete]] [[Hausdorff space|Hausdorff]] [[Locally convex topological vector space|locally convex]] [[topological vector space]]. 
}}

{{collapse top|title=Proof of corollary to lemma|left=true}}
<!-- {{math proof|title=Proof of corollary|proof=}} -->
Because the underlying field <math>\mathbb{K}</math> is a complete Hausdorff locally convex topological vector space, the same is true of the [[product space]] <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}.</math> 
A closed subset of a complete space is complete, so by the lemma, the space <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> is complete.
<math>\blacksquare</math> 
{{collapse bottom}}


The above elementary proof of the Banach–Alaoglu theorem actually shows that if <math>U \subseteq X</math> is any subset that satisfies <math>X = (0, \infty) U ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{r u : r > 0, u \in U\}</math> (such as any [[Absorbing set|absorbing subset]] of <math>X</math>), then <math>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{f \in X^{\#} : f(U) \subseteq B_1\right\}</math> is a [[Weak-* topology|weak-* compact]] subset of <math>X^{\#}.</math> 

As a side note, with the help of the above elementary proof, it may be shown (see this footnote)<ref group=proof>
<!-- It suffices to prove that <math>P = \left(\cap \operatorname{Box}_P\right) \cap X^{\#} </math> because then <math>P \subseteq X^{\prime} \subseteq X^{\#}</math> will imply <math>P = \left(\cap \operatorname{Box}_P\right) \cap X^{\prime}.</math> -->
For any non-empty subset <math>A \subseteq [0, \infty),</math> the equality <math>\cap \left\{B_a : a \in A\right\} = B_{\inf_{} A}</math> holds (the intersection on the left is a closed, rather than open, disk − possibly of radius <math>0</math> − because it is an intersection of closed subsets of <math>\mathbb{K}</math> and so must itself be closed). For every <math>x \in X,</math> let <math>m_x = \inf_{} \left\{ R_x : R_{\bull} \in T_P \right\}</math> so that the previous set equality implies <math>\cap \operatorname{Box}_P = \bigcap_{R_{\bull} \in T_P} \prod_{x \in X} B_{R_x} = \prod_{x \in X} \bigcap_{R_{\bull} \in T_P} B_{R_x} = \prod_{x \in X} B_{m_x}.</math> From <math>P \subseteq \cap \operatorname{Box}_P</math> it follows that <math>m_{\bull} \in T_P</math> and <math>\cap \operatorname{Box}_P \in \operatorname{Box}_P,</math> thereby making <math>\cap \operatorname{Box}_P</math> the [[least element]] of <math>\operatorname{Box}_P</math> with respect to <math>\,\subseteq.\,</math> (In fact, the [[Family of sets|family]] <math>\operatorname{Box}_P</math> is closed under (non-[[Nullary intersection|nullary]]) arbitrary intersections and also under finite unions of at least one set). The elementary proof showed that <math>T_P</math> and <math>\operatorname{Box}_P</math> are not empty and moreover, it also even showed that <math>T_P</math> has an element <math>\left(r_x\right)_{x \in X}</math> that satisfies <math>r_u = 1</math> for every <math>u \in U,</math> which implies that <math>m_u \leq 1</math> for every <math>u \in U.</math> The inclusion <math>P ~\subseteq~ \left(\cap \operatorname{Box}_P\right) \cap X^{\prime} ~\subseteq~ \left(\cap \operatorname{Box}_P\right) \cap X^{\#}</math> is immediate; to prove the reverse inclusion, let <math>f \in \left(\cap \operatorname{Box}_P\right) \cap X^{\#}.</math> By definition, <math>f \in P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ U^{\#}</math> if and only if <math>\sup_{u \in U} |f(u)| \leq 1,</math> so let <math>u \in U</math> and it remains to show that <math>|f(u)| \leq 1.</math> From <math>f \in \cap \operatorname{Box}_P = \prod B_{m_\bull},</math> it follows that <math>f(u) = \Pr{}_u (f) \in \Pr{}_u \left(\prod_{x \in X} B_{m_x}\right) = B_{m_u},</math> which implies that <math>|f(u)| \leq m_u \leq 1,</math> as desired. 
<math>\blacksquare</math>
</ref> 
that there exist <math>X</math>-indexed non-negative real numbers <math>m_{\bull} = \left(m_x\right)_{x \in X}</math> such that 
<math display=block>\begin{alignat}{4}
U^{\circ} 
&= U^{\#}     && \\
&= X^{\#}     && \cap \prod_{x \in X} B_{m_x} \\
&= X^{\prime} && \cap \prod_{x \in X} B_{m_x} \\
\end{alignat}</math>
where these real numbers <math>m_{\bull}</math> can also be chosen to be "minimal" in the following sense: 
using <math>P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ U^{\circ}</math> (so <math>P = U^{\#}</math> as in the proof) and defining the notation <math>\prod B_{R_\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \prod_{x \in X} B_{R_x}</math> for any <math>R_{\bull} = \left(R_x\right)_{x \in X} \in \R^X,</math> if
<math display=block>T_P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{R_{\bull} \in \R^X ~:~ P \subseteq \prod B_{R_\bull}\right\}</math> 
then <math>m_{\bull} \in T_P</math> and for every <math>x \in X,</math> <math>m_x = \inf \left\{ R_x : R_{\bull} \in T_P \right\},</math> 
which shows that these numbers <math>m_{\bull}</math> are unique; indeed, this [[infimum]] formula can be used to define them. 

In fact, if <math>\operatorname{Box}_P</math> denotes the set of all such products of closed balls containing the polar set <math>P,</math>
<math display=block>\operatorname{Box}_P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{ \prod B_{R_\bull} ~:~ R_{\bull} \in T_P \right\} ~=~ \left\{ \prod B_{R_\bull} ~:~ P \subseteq \prod B_{R_\bull} \right\},</math> 
then 
<math display=inline>\prod B_{m_\bull} = \cap \operatorname{Box}_P \in \operatorname{Box}_P</math> 
where <math display=inline>\bigcap \operatorname{Box}_P</math> denotes the intersection of all sets belonging to <math>\operatorname{Box}_P.</math>

This implies (among other things<ref group=note>This tuple <math>m_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(m_x\right)_{x \in X}</math> is the [[least element]] of <math>T_P</math> with respect to natural induced pointwise [[partial order]] defined by <math>R_{\bull} \leq S_{\bull}</math> if and only if <math>R_x \leq S_x</math> for every <math>x \in X.</math> Thus, every neighborhood <math>U</math> of the origin in <math>X</math> can be associated with this unique (minimum) function <math>m_{\bull} : X \to [0, \infty).</math> For any <math>x \in X,</math> if <math>r > 0</math> is such that <math>x \in r U</math> then <math>m_x \leq r</math> so that in particular, <math>m_0 = 0</math> and <math>m_u \leq 1</math> for every <math>u \in U.</math></ref>) 
that <math display=inline>\prod B_{m_\bull} = \prod_{x \in X} B_{m_x}</math> the unique [[least element]] of <math>\operatorname{Box}_P</math> with respect to <math>\,\subseteq;</math> this may be used as an alternative definition of this (necessarily [[Convex set|convex]] and [[Balanced set|balanced]]) set.  
The function <math>m_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(m_x\right)_{x \in X} : X \to [0, \infty)</math> is a [[seminorm]] and it is unchanged if <math>U</math> is replaced by the [[convex balanced hull]] of <math>U</math> (because <math>U^{\#} = [\operatorname{cobal} U]^{\#}</math>). 
Similarly, because <math>U^{\circ} = \left[\operatorname{cl}_X U\right]^{\circ},</math> <math>m_{\bull}</math> is also unchanged if <math>U</math> is replaced by its [[Closure (topology)|closure]] in <math>X.</math>

==Sequential Banach–Alaoglu theorem==

A special case of the Banach–Alaoglu theorem is the sequential version of the theorem, which asserts that the closed unit ball of the dual space of a [[separable metric space|separable]] normed vector space is [[sequentially compact]] in the weak-* topology. 
In fact, the weak* topology on the closed unit ball of the dual of a separable space is [[metrizable]], and thus compactness and sequential compactness are equivalent.

Specifically, let <math>X</math> be a separable normed space and <math>B</math> the closed unit ball in <math>X^{\prime}.</math> Since <math>X</math> is separable, let <math>x_{\bull} = \left(x_n\right)_{n=1}^{\infty}</math> be a countable dense subset. 
Then the following defines a metric, where for any <math>x, y \in B</math> 
<math display=block>\rho(x,y) = \sum_{n=1}^\infty \, 2^{-n} \, \frac{\left|\langle x - y, x_n \rangle\right|}{1 + \left|\langle x - y, x_n \rangle\right|}</math>
in which <math>\langle\cdot, \cdot\rangle</math> denotes the duality pairing of <math>X^{\prime}</math> with <math>X.</math> 
Sequential compactness of <math>B</math> in this metric can be shown by a [[diagonalization argument]] similar to the one employed in the proof of the [[Arzelà–Ascoli theorem]].

Due to the constructive nature of its proof (as opposed to the general case, which is based on the axiom of choice), the sequential Banach–Alaoglu theorem is often used in the field of [[partial differential equations]] to construct solutions to PDE or [[Calculus of variations|variational problems]]. 
For instance, if one wants to minimize a functional <math>F : X^{\prime} \to \R</math> on the dual of a separable normed vector space <math>X,</math> one common strategy is to first construct a minimizing sequence <math>x_1, x_2, \ldots \in X^{\prime}</math> which approaches the infimum of <math>F,</math> use the sequential Banach–Alaoglu theorem to extract a subsequence that converges in the weak* topology to a limit <math>x,</math> and then establish that <math>x</math> is a minimizer of <math>F.</math> 
The last step often requires <math>F</math> to obey a (sequential) [[Semi-continuity|lower semi-continuity]] property in the weak* topology.

When <math>X^{\prime}</math> is the space of finite Radon measures on the real line (so that <math>X = C_0(\R)</math> is the space of continuous functions vanishing at infinity, by the [[Riesz–Markov–Kakutani representation theorem|Riesz representation theorem]]), the sequential Banach–Alaoglu theorem is equivalent to the [[Helly selection theorem]].

{{math proof|drop=hidden|proof=
For every <math>x \in X,</math> let
<math display=block>D_x = \{c \in \Complex : |c| \leq \|x\|\}</math>
and let
<math display=block>D = \prod_{x \in X} D_x</math>
be endowed with the [[product topology]]. 
Because every <math>D_x</math> is a compact subset of the complex plane, [[Tychonoff's theorem]] guarantees that their product <math>D</math> is compact. 

The closed unit ball in <math>X^{\prime},</math> denoted by <math>B_1^{\,\prime},</math> can be identified as a subset of <math>D</math> in a natural way:
<math display=block>\begin{alignat}{4}
F :\;&& B_1^{\,\prime} &&\;\to    \;& D \\[0.3ex]
     && f              &&\;\mapsto\;& (f(x))_{x \in X}. \\
\end{alignat}</math>

This map is injective and it is continuous when <math>B_1^{\,\prime}</math> has the [[weak-* topology]]. 
This map's inverse, defined on its image, is also continuous. 

It will now be shown that the image of the above map is closed, which will complete the proof of the theorem. 
Given a point <math>\lambda_{\bull} = \left(\lambda_x\right)_{x \in X} \in D</math> and a net <math>\left(f_i(x)\right)_{x \in X}</math> in the image of <math>F</math> indexed by <math>i \in I</math> such that 
<math display=block>\lim_{i} \left(f_i(x)\right)_{x \in X} \to \lambda_{\bull} \quad \text{ in } D,</math>
the functional <math>g : X \to \Complex</math> defined by
<math display=block>g(x) = \lambda_x \qquad \text{ for every } x \in X,</math>
lies in <math>B_1^{\,\prime}</math> and <math>F(g) = \lambda_{\bull}.</math> 
<math>\blacksquare</math>
}}

==Consequences==

===Consequences for normed spaces===

Assume that <math>X</math> is a [[normed space]] and endow its continuous dual space <math>X^{\prime}</math> with the usual [[dual norm]]. 

<ul>
<li>The closed unit ball in <math>X^{\prime}</math> is weak-* compact.{{sfn|Narici|Beckenstein|2011|pp=225-273}}  So if <math>X^{\prime}</math> is infinite dimensional then its closed unit ball is necessarily {{em|not}} compact in the norm topology by [[F. Riesz's theorem]] (despite it being weak-* compact).
</li>
<li>A [[Banach space]] is [[Reflexive space|reflexive]] if and only if its closed unit ball is <math>\sigma\left(X, X^{\prime}\right)</math>-compact; this is known as [[James' theorem]].{{sfn|Narici|Beckenstein|2011|pp=225-273}}</li>
<li>If <math>X</math> is a [[reflexive Banach space]], then every bounded sequence in <math>X</math> has a weakly convergent subsequence. 
(This follows by applying the Banach–Alaoglu theorem to a weakly metrizable subspace of <math>X</math>; or, more succinctly, by applying the [[Eberlein–Šmulian theorem]].) 
For example, suppose that <math>X</math> is the space [[Lp space]] <math>L^p(\mu)</math> where <math>1 < p < \infty</math> and let <math>q</math> satisfy <math>\frac{1}{p} + \frac{1}{q} = 1.</math>
Let <math>f_1, f_2, \ldots</math> be a bounded sequence of functions in <math>X.</math> 
Then there exists a subsequence <math>\left(f_{n_k}\right)_{k=1}^{\infty}</math> and an <math>f \in X</math> such that
<math display=block>\int f_{n_k} g\,d\mu \to \int f g\,d\mu \qquad \text{ for all } g \in L^q(\mu) = X^{\prime}.</math>
The corresponding result for <math>p = 1</math> is not true, as <math>L^1(\mu)</math> is not reflexive.</li>
</ul>

===Consequences for Hilbert spaces===

<ul>
<li>In a Hilbert space, every bounded and closed set is weakly relatively compact, hence every bounded net has a weakly convergent subnet (Hilbert spaces are [[reflexive space|reflexive]]).</li>
<li>As norm-closed, convex sets are weakly closed ([[Hahn–Banach theorem]]), norm-closures of convex bounded sets in Hilbert spaces or reflexive Banach spaces are weakly compact.</li>
<li>Closed and bounded sets in <math>B(H)</math> are precompact with respect to the [[weak operator topology]] (the weak operator topology is weaker than the [[ultraweak topology]] which is in turn the weak-* topology with respect to the predual of <math>B(H),</math> the [[trace class]] operators). Hence bounded sequences of operators have a weak accumulation point. 
As a consequence, <math>B(H)</math> has the [[Heine–Borel theorem|Heine–Borel property]], if equipped with either the weak operator or the ultraweak topology.</li>
</ul>

==Relation to the axiom of choice and other statements==
{{See also|Krein–Milman theorem#Relation to other statements}}

The Banach–Alaoglu may be proven by using [[Tychonoff's theorem]], which under the [[Zermelo–Fraenkel set theory]] ('''ZF''') axiomatic framework is equivalent to the [[axiom of choice]]. 
Most mainstream functional analysis relies on '''ZF''' + the axiom of choice, which is often denoted by '''ZFC'''. 
However, the theorem does {{em|not}} rely upon the axiom of choice in the [[Separable space|separable]] case (see [[#Sequential Banach–Alaoglu theorem|above]]): in this case there actually exists a constructive proof. 
In the general case of an arbitrary normed space, the [[ultrafilter Lemma]], which is strictly weaker than the axiom of choice and equivalent to Tychonoff's theorem for compact {{em|Hausdorff}} spaces, suffices for the proof of the Banach–Alaoglu theorem, and is in fact equivalent to it.

{{anchor|Relation to the Hahn–Banach theorem}}The Banach–Alaoglu theorem is equivalent to the [[ultrafilter lemma]], which implies the [[Hahn–Banach theorem]] for [[real vector space]]s ('''HB''') but is not equivalent to it (said differently, Banach–Alaoglu is also strictly stronger than '''HB'''). 
However, the [[Hahn–Banach theorem]] is equivalent to the following weak version of the Banach–Alaoglu theorem for normed space<ref name=BellFremlin1972>{{cite journal|last1=Bell|first1=J.|last2=Fremlin|first2=David|title=A Geometric Form of the Axiom of Choice|journal=Fundamenta Mathematicae|date=1972|volume=77|issue=2|pages=167–170|doi=10.4064/fm-77-2-167-170|url=http://matwbn.icm.edu.pl/ksiazki/fm/fm77/fm77116.pdf|access-date=26 Dec 2021}}</ref> in which the conclusion of compactness (in the [[weak-* topology]] of the closed unit ball of the dual space) is replaced with the conclusion of {{em|quasicompactness}} (also sometimes called {{em|convex compactness}}); 

{{Math theorem|name={{visible anchor|Weak version of Alaoglu theorem}}<ref name=BellFremlin1972 />|math_statement=
Let <math>X</math> be a normed space and let <math>B</math> denote the closed unit ball of its [[continuous dual space]] <math>X^{\prime}.</math> Then <math>B</math> has the following property, which is called ([[weak-* topology|weak-*]]) {{em|{{visible anchor|quasicompactness}}}} or {{em|{{visible anchor|convex compactness}}}}: whenever <math>\mathcal{C}</math> is a cover of <math>B</math> by {{em|convex}} [[weak-* topology|weak-* closed]] subsets of <math>X^{\prime}</math> such that <math>\{B \cap C : C \in \mathcal{C}\}</math> has the [[finite intersection property]], then <math>B \cap \bigcap_{C \in \mathcal{C}} C</math> is not empty.
}}

[[Compact space|Compactness]] implies [[#convex compactness|convex compactness]] because a topological space is compact if and only if every [[Family of sets|family]] of closed subsets having the [[finite intersection property]] (FIP) has non-empty intersection. 
The [[#convex compactness|definition of convex compactness]] is similar to this characterization of [[compact space]]s in terms of the FIP, except that it only involves those closed subsets that are also [[Convex set|convex]] (rather than all closed subsets).

==See also==

* {{annotated link|Bishop–Phelps theorem}}
* {{annotated link|Banach–Mazur theorem}}
* {{annotated link|Delta-convergence|Delta-compactness theorem}}
* {{annotated link|Dixmier–Ng theorem}}
* {{annotated link|Eberlein–Šmulian theorem}}
* {{annotated link|Goldstine theorem}}
* {{annotated link|James' theorem}}
* {{annotated link|Krein-Milman theorem}}
* {{annotated link|Mazur's lemma}}
* {{annotated link|Topological vector space}}

==Notes==

{{reflist|group=note}}

'''Proofs'''

{{reflist|group=proof}}

==Citations==

{{reflist}}

==References==

{{refbegin}}
* {{Köthe Topological Vector Spaces I}} <!--{{sfn|Köthe|1969|p=}}-->
* {{cite book|last1=Meise|first1=Reinhold|last2=Vogt|first2=Dietmar|title=Introduction to Functional Analysis|publisher=Clarendon Press|location=Oxford, England|year=1997|isbn=0-19-851485-9|section=Theorem 23.5|page=264}}
* {{Narici Beckenstein Topological Vector Spaces|edition=2}} <!--{{sfn|Narici|Beckenstein|2011|p=}}-->
* {{Rudin Walter Functional Analysis|edition=2}} <!--{{sfn|Rudin|1991|p=}}--> See Theorem 3.15, p.&nbsp;68.
* {{Schaefer Wolff Topological Vector Spaces|edition=2}} <!--{{sfn|Schaefer|Wolff|1999|p=}}-->
* {{Schechter Handbook of Analysis and Its Foundations}} <!--{{sfn|Schechter|1996|p=}}-->
* {{Trèves François Topological vector spaces, distributions and kernels}} <!--{{sfn|Trèves|2006|p=}}-->
{{refend}}

==Further reading==

* {{Conway A Course in Functional Analysis|edition=2}} <!--{{sfn|Conway|1990|p=}}-->
* {{Lax Functional Analysis}} <!--{{sfn|Lax|2002|p=}}-->

{{Duality and spaces of linear maps}}
{{Functional analysis}}
{{Topological vector spaces}}

{{DEFAULTSORT:Banach-Alaoglu theorem}}

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[[Category:Theorems in functional analysis]]
[[Category:Topological vector spaces]]
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