Banach–Alaoglu theorem

In functional analysis and related branches of mathematics, the Banach–Alaoglu theorem (also known as Alaoglu's theorem) states that the closed unit ball of the dual space of a normed vector space is compact in the weak* topology.
A common proof identifies the unit ball with the weak-* topology as a closed subset of a 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 most important fact about the weak-* topology—hatechos throughout functional analysis.”
In 1912, Helly proved that the unit ball of the continuous dual space of C( , b is countably weak-* compact.
In 1932, Stefan Banach proved that the closed unit ball in the continuous dual space of any separable normed space is sequentially weak-* compact (Banach only considered sequential compactness).
The proof for the general case was published in 1940 by the mathematician Leonidas Alaoglu.
According to Pietsch 007 there are at least twelve mathematicians who can lay claim to this theorem or an important predecessor to it.

The Bourbaki–Alaoglu theorem is a generalization, Theorem 23.5. of the original theorem by Bourbaki to dual topologies on locally convex spaces.
This theorem is also called the Banach–Alaoglu theorem or the weak-* compactness theorem and it is commonly called simply the Alaoglu theorem.

Statement

If $X$ is a vector space over the field $\mathbb$ then $X^$ will denote the algebraic dual space of $X$ and these two spaces are henceforth associated with the bilinear $\left\langle \cdot, \cdot \right\rangle : X \times X^ \to \mathbb$ defined by
$\left\langle x, f \right\rangle ~\stackrel~ f(x)$
where the triple $\left\langle X, X^,\left\langle \cdot, \cdot \right\rangle \right\rangle$ forms a dual system called the .

If $X$ is a topological vector space (TVS) then its continuous dual space will be denoted by $X^,$ where $X^ \subseteq X^$ always holds.
Denote the weak-* topology on $X^$ by $\sigma\left(X^, X\right)$ and denote the weak-* topology on $X^$ by $\sigma\left(X^, X\right).$
The weak-* topology is also called the topology of pointwise convergence because given a map $f$ and a net of maps $f_ = \left(f_i\right)_,$ the net $f_$ converges to $f$ in this topology if and only if for every point $x$ in the domain, the net of values $\left(f_i(x)\right)_$ converges to the value $f(x).$

Proof involving duality theory

If $X$ is a normed vector space, then the polar of a neighborhood is closed and norm-bounded in the dual space.
In particular, if $U$ is the open (or closed) unit ball in $X$ then the polar of $U$ is the closed unit ball in the continuous dual space $X^$ of $X$ (with the usual dual norm).
Consequently, this theorem can be specialized to:

When the continuous dual space $X^$ of $X$ is an infinite dimensional normed space then it is for the closed unit ball in $X^$ to be a compact subset when $X^$ 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 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 Weil that all locally compact 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 convergence in topological spaces and familiarity with the fact that a linear functional is continuous if and only if it is bounded on a neighborhood of the origin (see the articles on continuous linear functionals and sublinear functionals for details).
Also required is a proper understanding of the technical details of how the space $\mathbb^X$ of all functions of the form $X \to \mathbb$ is identified as the Cartesian product $\prod_ \mathbb,$ and the relationship between pointwise convergence, the product topology, and subspace topologies they induce on subsets such as the algebraic dual space $X^$ and products of subspaces such as $\prod_ B_.$
An explanation of these details is now given for readers who are interested.

For every real $r,$ $B_r ~\stackrel~ \$ will denote the closed ball of radius $r$ centered at $0$ and $r U ~\stackrel~ \$ for any $U \subseteq X,$

Identification of functions with tuples

The Cartesian product $\prod_ \mathbb$ is usually thought of as the set of all $X$-indexed tuples $s_ = \left(s_x\right)_$ but, since tuples are technically just functions from an indexing set, it can also be identified with the space $\mathbb^X$ of all functions having prototype $X \to \mathbb,$ as is now described:

* : A function $s : X \to \mathbb$ belonging to $\mathbb^X$ is identified with its ($X$-indexed) "" $s_ ~\stackrel~ (s(x))_.$
* : A tuple $s_ = \left(s_x\right)_$ in $\prod_ \mathbb$ is identified with the function $s : X \to \mathbb$ defined by $s(x) ~\stackrel~ s_x$; this function's "tuple of values" is the original tuple $\left(s_x\right)_.$

This is the reason why many authors write, often without comment, the equality
$\mathbb^X = \prod_ \mathbb$
and why the Cartesian product $\prod_ \mathbb$ is sometimes taken as the definition of the set of maps $\mathbb^X$ (or conversely).
However, the Cartesian product, being the (categorical) product in the category of sets (which is a type of inverse limit), also comes equipped with associated maps that are known as its (coordinate) .

The at a given point $z \in X$ is the function
$\Pr_z : \prod_ \mathbb \to \mathbb \quad \text \quad s_ = \left(s_x\right)_ \mapsto s_z$
where under the above identification, $\Pr_z$ sends a function $s : X \to \mathbb$ to
$\Pr_z(s) ~\stackrel~ s(z).$
Stated in words, for a point $z$ and function $s,$ "plugging $z$ into $s$" is the same as "plugging $s$ into $\Pr_z$".

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

If a subset $U \subseteq X$ partitions $X$ into $X = U \, \cup \,(X \setminus U)$ then the linear bijection
\begin
H :\;&& \prod_ \mathbb &&\;\to \;& \left(\prod_ \mathbb\right) \times \prod_ \mathbb \\ .3ex && \left(f_x\right)_ &&\;\mapsto\;& \left( \left(f_u\right)_, \; \left(f_x\right)_ \right) \\
\end
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
\begin
H :\;&& \mathbb^X &&\;\to \;& \mathbb^U \times \mathbb^ \\ .3ex && f &&\;\mapsto\;& \left(f\big\vert_U, \; f\big\vert_\right) \\
\end.

Notation for nets and function composition with nets

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

Topology

The set $\mathbb^X = \prod_ \mathbb$ 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 $f$ and a net $\left(f_i\right)_,$ where $f$ and every $f_i$ is an element of $\mathbb^X = \prod_ \mathbb,$ then the net $\left(f_i\right)_ \to f$ converges in the product topology if and only if

:for every $z \in X,$ the net $\Pr_z\left(\left(f_i\right)_\right) \to \Pr_z(f)$ converges in $\mathbb,$

where because $\;\Pr_z(f) = f(z)\;$ and
$\Pr_z\left(\left(f_i\right)_\right) ~\stackrel~ \left(\Pr_z\left(f_i\right)\right)_ = \left(f_i(z)\right)_,$
this happens if and only if

:for every $z \in X,$ the net $\left(f_i(z)\right)_ \to f(z)$ converges in $\mathbb,$

Thus $\left(f_i\right)_$ converges to $f$ in the product topology if and only if it converges to $f$ pointwise on $X.$

This proof will also use the fact that the topology of pointwise convergence is preserved when passing to topological subspaces.
This means, for example, that if for every $x \in X,$ $S_x \subseteq \mathbb$ is some (topological) subspace of $\mathbb$ then the topology of pointwise convergence (or equivalently, the product topology) on $\prod_ S_x$ is equal to the subspace topology that the set $\prod_ S_x$ inherits from $\prod_ \mathbb.$
And if $S_x$ is closed in $\mathbb$ for every $x \in X,$ then $\prod_ S_x$ is a closed subset of $\prod_ \mathbb.$

Characterization of $\sup_ , f(u), \leq r$

An important fact used by the proof is that for any real $r,$
$\sup_ , f(u), \leq r \qquad \text \qquad f(U) \subseteq B_r$
where $\,\sup\,$ denotes the supremum and $f(U) ~\stackrel~ \.$
As a side note, this characterization does not hold if the closed ball $B_r$ is replaced with the open ball $\$ (and replacing $\;\sup_ , f(u), \leq r\;$ with the strict inequality $\;\sup_ , f(u), < r\;$ will not change this; for counter-examples, consider $X ~\stackrel~ \mathbb$ and the identity map $f ~\stackrel~ \operatorname$ on $X$).

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 require the vector space $X$ to endowed with any topology.

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

The conclusion that the set $U_ = \left\$ is closed can also be reached by applying the following more general result, this time proved using nets, to the special case $Y := \mathbb$ and $B := B_1.$

:Observation: If $U \subseteq X$ is any set and if $B \subseteq Y$ is a closed subset of a topological space $Y,$ then $U_B ~\stackrel~ \left\$ is a closed subset of $Y^X$ in the topology of pointwise convergence.

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

Let $f \in \mathbb^X$ and suppose that $f_ = \left(f_i\right)_$ is a net in $X^$ the converges to $f$ in $\mathbb^X.$
To conclude that $f \in X^,$ it must be shown that $f$ is a linear functional. So let $s$ be a scalar and let $x, y \in X.$

For any $z \in X,$ let $f_(z) : I \to \mathbb$ denote
$f_(z) ~\stackrel~ \left(f_i(z)\right)_.$
Because $f_ \to f$ in $\mathbb^X,$ which has the topology of pointwise convergence, $f_(z) \to f(z)$ in $\mathbb$ for every $z \in X.$
By using $x, y, sx, \text x + y,$ in place of $z,$ it follows that each of the following nets of scalars converges in $\mathbb:$
$f_(x) \to f(x), \quad f_(y) \to f(y), \quad f_(x + y) \to f(x + y), \quad \text \quad f_(sx) \to f(sx).$

Proof that $f(s x) = s f(x):$
Let $M : \mathbb \to \mathbb$ be the "multiplication by $s$" map defined by $M(c) ~\stackrel~ s c.$
Because $M$ is continuous and $f_(x) \to f(x)$ in $\mathbb,$ it follows that $M\left(f_(x)\right) \to M(f(x))$ where the right hand side is $M(f(x)) = s f(x)$ and the left hand side is
$\begin M\left(f_(x)\right) \stackrel&~ M \circ f_(x) && \text \\ =&~ \left(M\left(f_i(x)\right)\right)_ ~~~ && \text f_(x) = \left(f_i(x)\right)_ : I \to \mathbb \\ =&~ \left(s f_i(x)\right)_ && M\left(f_i(x)\right) ~\stackrel~ s f_i(x) \\ =&~ \left(f_i(s x)\right)_ && \text f_i \\ =&~ f_(sx) && \text \end$
which proves that $f_(sx) \to s f(x).$ Because also $f_(sx) \to f(sx)$ and limits in $\mathbb$ are unique, it follows that $s f(x) = f(s x),$ as desired.

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

The lemma above actually also follows from its corollary below since $\prod_ \mathbb$ is a Hausdorff complete uniform space and any subset of such a space (in particular $X^$) is closed if and only if it is complete.

Because the underlying field $\mathbb$ is a complete Hausdorff locally convex topological vector space, the same is true of the product space $\mathbb^X = \prod_ \mathbb.$
A closed subset of a complete space is complete, so by the lemma, the space $\left(X^, \sigma\left(X^, X\right)\right)$ is complete.
$\blacksquare$

The above elementary proof of the Banach–Alaoglu theorem actually shows that if $U \subseteq X$ is any subset that satisfies $X = (0, \infty) U ~\stackrel~ \$ (such as any absorbing subset of $X$), then $U^ ~\stackrel~ \left\$ is a weak-* compact subset of $X^.$

As a side note, with the help of the above elementary proof, it may be shown (see this footnote)

For any non-empty subset $A \subseteq [0, \infty),$ the equality $\cap \left\ = B_$ holds (the intersection on the left is a closed, rather than open, disk − possibly of radius $0$ − because it is an intersection of closed subsets of $\mathbb$ and so must itself be closed). For every $x \in X,$ let $m_x = \inf_ \left\$ so that the previous set equality implies $\cap \operatorname_P = \bigcap_ \prod_ B_ = \prod_ \bigcap_ B_ = \prod_ B_.$ From $P \subseteq \cap \operatorname_P$ it follows that $m_ \in T_P$ and $\cap \operatorname_P \in \operatorname_P,$ thereby making $\cap \operatorname_P$ the least element of $\operatorname_P$ with respect to $\,\subseteq.\,$ (In fact, the family $\operatorname_P$ is closed under (non-Nullary intersection, nullary) arbitrary intersections and also under finite unions of at least one set). The elementary proof showed that $T_P$ and $\operatorname_P$ are not empty and moreover, it also even showed that $T_P$ has an element $\left(r_x\right)_$ that satisfies $r_u = 1$ for every $u \in U,$ which implies that $m_u \leq 1$ for every $u \in U.$ The inclusion $P ~\subseteq~ \left(\cap \operatorname_P\right) \cap X^ ~\subseteq~ \left(\cap \operatorname_P\right) \cap X^$ is immediate; to prove the reverse inclusion, let $f \in \left(\cap \operatorname_P\right) \cap X^.$ By definition, $f \in P ~\stackrel~ U^$ if and only if $\sup_ , f(u), \leq 1,$ so let $u \in U$ and it remains to show that $, f(u), \leq 1.$ From $f \in \cap \operatorname_P = \prod B_,$ it follows that $f(u) = \Pr_u (f) \in \Pr_u \left(\prod_ B_\right) = B_,$ which implies that $, f(u), \leq m_u \leq 1,$ as desired.
$\blacksquare$

that there exist $X$-indexed non-negative real numbers $m_ = \left(m_x\right)_$ such that
$\begin U^ &= U^ && \\ &= X^ && \cap \prod_ B_ \\ &= X^ && \cap \prod_ B_ \\ \end$
where these real numbers $m_$ can also be chosen to be "minimal" in the following sense:
using $P ~\stackrel~ U^$ (so $P = U^$ as in the proof) and defining the notation $\prod B_ ~\stackrel~ \prod_ B_$ for any $R_ = \left(R_x\right)_ \in \R^X,$ if
$T_P ~\stackrel~ \left\$
then $m_ \in T_P$ and for every $x \in X,$ $m_x = \inf \left\,$
which shows that these numbers $m_$ are unique; indeed, this infimum formula can be used to define them.

In fact, if $\operatorname_P$ denotes the set of all such products of closed balls containing the polar set $P,$
$\operatorname_P ~\stackrel~ \left\ ~=~ \left\,$
then
$\prod B_ = \cap \operatorname_P \in \operatorname_P$
where $\bigcap \operatorname_P$ denotes the intersection of all sets belonging to $\operatorname_P.$

This implies (among other thingsThis tuple $m_ ~\stackrel~ \left(m_x\right)_$ is the least element of $T_P$ with respect to natural induced pointwise partial order defined by $R_ \leq S_$ if and only if $R_x \leq S_x$ for every $x \in X.$ Thus, every neighborhood $U$ of the origin in $X$ can be associated with this unique (minimum) function $m_ : X \to$least element of $\operatorname_P$ with respect to $\,\subseteq;$ this may be used as an alternative definition of this (necessarily convex