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 the most important fact about the weak-* topology - [that] echos throughout functional analysis." In 1912, Helly proved that the unit ball of the continuous dual space of $C([a,b])$ 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 [2007], there are at least 12 mathematicians who can lay claim to this theorem or an important predecessor to it.

The Bourbaki–Alaoglu theorem is a generalization 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

See also: Topological vector space § Dual space, Dual system, and Polar set

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

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

Alaoglu theorem — For any topological vector space (TVS) $X$ (not necessarily Hausdorff or locally convex) with continuous dual space $X^{\prime },$ the polar $$U^{\circ }=\left\{x^{\prime }\in X^{\prime }~:~\sup _{x\in U}\left|x^{\prime }(x)\right|\leq 1\right\}$$ of any neighborhood $U$ of origin in $X$ is compact in the weak-* topology $\sigma \left(X^{\prime },X\right)$ on $X^{\prime }.$ Moreover, $U^{\circ }$ is equal to the polar of $U$ with respect to the canonical system $\left\langle X,X^{\#}\right\rangle$ and it is also a compact subset of $\left(X^{\#},\sigma \left(X^{\#},X\right)\right).$

Proof involving duality theory

Proof

Denote by the underlying field of $X$ by $\mathbb {K} ,$ which is either the real numbers $\mathbb {R}$ or complex numbers $\mathbb {C} .$ 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 $X^{\#}$ is endowed with the weak-* topology $\sigma \left(X^{\#},X\right),$ then this Hausdorff locally convex topological vector space is denoted by $\left(X^{\#},\sigma \left(X^{\#},X\right)\right).$ The space $\left(X^{\#},\sigma \left(X^{\#},X\right)\right)$ is always a complete TVS; however, $\left(X^{\prime },\sigma \left(X^{\prime },X\right)\right)$ may fail to be a complete space, which is the reason why this proof involves the space $\left(X^{\#},\sigma \left(X^{\#},X\right)\right).$ 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. Importantly, the subspace topology that $X^{\prime }$ inherits from $\left(X^{\#},\sigma \left(X^{\#},X\right)\right)$ is equal to $\sigma \left(X^{\prime },X\right).$ This can be readily verified by showing that given any $x^{\prime }\in X^{\prime },$ a net in $X^{\prime }$ converges to $x^{\prime }$ in one of these topologies if and only if it also converges to $x^{\prime }$ 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 $\left\langle X,X^{\prime }\right\rangle$ is a dual pairing although unlike $\left\langle X,X^{\#}\right\rangle ,$ 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 pairing $\left\langle X,X^{\prime }\right\rangle .$

Let $U$ be a neighborhood of the origin in $X$ and let:

• $U^{\circ }=\left\{f\in X^{\prime }~:~\sup _{u\in U}|f(u)|\leq 1\right\}$ be the polar of $U$ with respect to the canonical pairing $\left\langle X,X^{\prime }\right\rangle$;
• $U^{\circ \circ }=\left\{x\in X~:~\sup _{f\in U^{\circ }}|f(x)|\leq 1\right\}$ be the bipolar of $U$ with respect to $\left\langle X,X^{\prime }\right\rangle$;
• $U^{\#}=\left\{f\in X^{\#}~:~\sup _{u\in U}|f(u)|\leq 1\right\}$ be the polar of $U$ with respect to the canonical dual system $\left\langle X,X^{\#}\right\rangle .$

A well known fact about polars of sets is that $U^{\circ \circ \circ }\subseteq U^{\circ }.$

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

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 $X^{\prime }$ of $X$ (with the usual dual norm). Consequently, this theorem can be specialized to:

Banach–Alaoglu theorem — If $X$ is a normed space then the closed unit ball in the continuous dual space $X^{\prime }$ (endowed with its usual operator norm) is compact with respect to the weak-* topology.

When the continuous dual space $X^{\prime }$ of $X$ is an infinite dimensional normed space then it is impossible for the closed unit ball in $X^{\prime }$ to be a compact subset when $X^{\prime }$ 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 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 Weil that all locally compact Hausdorff topological vector spaces must be finite-dimensional.

Elementary proof

The following proof involves only elementary concepts from set theory, topology, and functional analysis. In particular, what is need from topology is a working knowledge of nets in topological spaces, the product topology, and their relationship to pointwise convergence (some details of this relationship are given in the proof). Familiarity with the fact that a linear functional is continuous if and only if it is bounded on a neighborhood of the origin is also needed (this is described in the article on sublinear functionals).

Proof

Denote by the underlying field of $X$ by $\mathbb {K} ,$ which is either the real numbers $\mathbb {R}$ or complex numbers $\mathbb {C} .$ For any real $r,$ let $$B_{r}:=\{s\in \mathbb {K} :|s|\leq r\}$$ denote the closed ball of radius $r$ at the origin in $\mathbb {K} ,$ which is a compact and closed subset of $\mathbb {K} .$

Because $U$ is a neighborhood of the origin in $X,$ it is also an absorbing subset of $X,$ so for every $x\in X,$ there exists a real number $r_{x}>0$ such that $x\in r_{x}U:=\left\{r_{x}u:u\in U\right\}.$ Let $$U^{\#}:=\left\{f\in X^{\#}~:~\sup _{u\in U}|f(u)|\leq 1\right\}~=~\left\{f\in X^{\#}~:~f(U)\subseteq B_{1}\right\}$$ denote the polar of $U$ with respect to the canonical dual system $\left\langle X,X^{\#}\right\rangle .$ As is now shown, this polar set $U^{\#}$ is the same as the polar $U^{\circ }$ of $U$ with respect to $\left\langle X,X^{\prime }\right\rangle .$

Proof that $U^{\circ }=U^{\#}:$ The inclusion $U^{\circ }\subseteq U^{\#}$ holds because every continuous linear functional is (in particular) a linear functional. For the reverse inclusion $\,U^{\#}\subseteq U^{\circ },\,$ let $f\in U^{\#}$ so that $\;\sup _{u\in U}|f(u)|\leq 1,\,$ which states exactly that the linear functional $f$ is bounded on the neighborhood $U$; thus $f$ is a continuous linear functional (that is, $f\in X^{\prime }$) and so $f\in U^{\circ },$ as desired. Q.E.D.

The rest of this proof requires a proper understanding how the Cartesian product ${\textstyle \prod _{x\in X}\mathbb {K} }$ is identified as the space $\mathbb {K} ^{X}$ of all functions of the form $X\to \mathbb {K} .$ An explanation is now given for readers who are interested.

Premiere on identification of functions with tuples

The Cartesian product ${\textstyle \prod _{x\in X}\mathbb {K} }$ is usually thought of as the set of all $X$-indexed tuples $s_{\bullet }=\left(s_{x}\right)_{x\in X}$ but, as is now described, it can also be identified with the space $\mathbb {K} ^{X}$ of all functions having prototype $X\to \mathbb {K} .$ Function $\to$ Tuple: A function $s:X\to \mathbb {K}$ belonging to $\mathbb {K} ^{X}$ is identified with its ($X$-indexed) "tuple of values" $s_{\bullet }:=(s(x))_{x\in X}.$ Tuple $\to$ Function: A tuple $s_{\bullet }=\left(s_{x}\right)_{x\in X}$ in ${\textstyle \prod _{x\in X}\mathbb {K} }$ is identified with the function $s:X\to \mathbb {K}$ defined by $s(x):=s_{x}$; this function's "tuple of values" is the original tuple $\left(s_{x}\right)_{x\in X}.$ This is the reason why many authors write, often without comment, the equality $$\mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K}$$ and why the Cartesian product ${\textstyle \prod _{x\in X}\mathbb {K} }$ is sometimes taken as the definition of the set of maps $\mathbb {K} ^{X}.$ 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) projections. The Cartesian product's canonical projection at any given $z\in X$ is the function $$\Pr {}_{z}:\prod _{x\in X}\mathbb {K} \to \mathbb {K} \quad {\text{ defined by }}\quad s_{\bullet }=\left(s_{x}\right)_{x\in X}\mapsto s_{z}$$ where under the above identification, $\Pr {}_{z}$ sends a function $s:X\to \mathbb {K}$ to $$\Pr {}_{z}(s):=s(z).$$ In words, for a point $z$ and function $s,$ "plugging $z$ into $s$" is the same as "plugging $s$ into $\Pr {}_{z}$". Topology The set ${\textstyle \mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K} }$ 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 $\left(f_{i}\right)_{i\in I},$ where $f$ and every $f_{i}$ is an element of ${\textstyle \mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K} ,}$ then the net $\left(f_{i}\right)_{i\in I}\to f$ converges in the product topology if and only if for every $z\in X,$ the net $\Pr {}_{z}\left(\left(f_{i}\right)_{i\in I}\right)\to \Pr {}_{z}(f)$ converges in $\mathbb {K} ,$ where $\;\Pr {}_{z}(f)=f(z)\;$ and $$\Pr {}_{z}\left(\left(f_{i}\right)_{i\in I}\right):=\left(\Pr {}_{z}\left(f_{i}\right)\right)_{i\in I}=\left(f_{i}(z)\right)_{i\in I}.$$ Thus $\left(f_{i}\right)_{i\in I}$ converges to $f$ in the product topology if and only if it converges to $f$ pointwise on $X.$ Also used in this proof will be the fact that the topology of pointwise convergence is preserved when passing to topological subspaces. This means, for example, that if for every $x\in X,$ $S_{x}\subseteq \mathbb {K}$ is some (topological) subspace of $\mathbb {K}$ then the topology of pointwise convergence (or equivalently, the product topology) on ${\textstyle \prod _{x\in X}S_{x}}$ is equal to the subspace topology that the set ${\textstyle \prod _{x\in X}S_{x}}$ inherits from ${\textstyle \prod _{x\in X}\mathbb {K} .}$

Having established that $U^{\circ }=U^{\#},$ to reduce symbol clutter, this $P$olar set will be denoted by $$P:=U^{\circ }$$ unless an attempt is being made to draw attention to the definition of $U^{\circ }$ or $U^{\#}.$

The proof of the theorem will be complete once the following statements are verified:

1. $P$ is a closed subset of $\mathbb {K} ^{X}.$
   • Here $\mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K}$ is endowed with the topology of pointwise convergence, which is identical to the product topology.
2. $P\subseteq \prod _{x\in X}B_{r_{x}}.$
   • $B_{r_{x}}\subseteq \mathbb {K}$ denotes the closed ball of radius $r_{x}$ centered at $0.$ For each $x\in X,$ $r_{x}$ was defined at the start of this proof as any real $r_{x}>0$ that satisfies $x\in r_{x}U$ (so in particular, $r_{u}:=1$ is a valid choice for each $u\in U$).

These statements imply that $P$ is a closed subset of ${\textstyle \prod _{x\in X}B_{r_{x}},}$ where this product space is compact by Tychonoff's theorem (because every closed ball $B_{r_{x}}$ is a compact space). Because a closed subset of a compact space is compact, it follows that $U^{\circ }=P$ is compact, which is the main conclusion of the Banach–Alaoglu theorem.

Proof of (1):

The algebraic dual space $X^{\#}$ is always a closed subset of ${\textstyle \mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K} }$ (this is proved in the lemma below for readers who are not familiar with this result). To prove that $P$ is closed in $\mathbb {K} ^{X},$ it suffices to show that the set $P_{U}$ defined by $$P_{U}~:=~\left\{f\in \mathbb {K} ^{X}~:~\sup _{u\in U}|f(u)|\leq 1\right\}$$ is a closed subset of $\mathbb {K} ^{X}$ because then $P_{U}\cap X^{\#}=U^{\#}=P$ is an intersection of two closed subsets of $\mathbb {K} ^{X}.$ Let $f\in \mathbb {K} ^{X}$ and suppose that $f_{\bullet }=\left(f_{i}\right)_{i\in I}$ is a net in $P_{U}$ that converges to $f$ in $\mathbb {K} ^{X}.$ To conclude that $f\in P_{U},$ it is sufficient (and necessary) to show that for every $u\in U,$ $|f(u)|\leq 1$ (or equivalently, that $f(u)\in B_{1}$). Because $\left(f_{i}(u)\right)_{i\in I}\to f(u)$ in the scalar field $\mathbb {K}$ and every value $f_{i}(u)$ belongs to the closed (in $\mathbb {K}$) subset $B_{1}=\left\{s\in \mathbb {K} :|s|\leq 1\right\},$ so too must this net's limit $f(u)$ belong to this closed set. Thus $f(u)\in B_{1},$ which completes the proof of (1). Q.E.D.

As a side note, this proof can be generalized to prove the following more general result, from which the above conclusion follows as the special case $Y:=\mathbb {K}$ and $B:=B_{1}.$

Proposition: If $U\subseteq X$ is any set and if $B\subseteq Y$ is a closed subset of a topological space $Y,$ then $P_{U}:=\left\{f\in Y^{X}~:~f(U)\subseteq B\right\}$ is a closed subset of $Y^{X}$ with respect to the topology of pointwise convergence.

Proof of (2):

For any $z\in X,$ let ${\textstyle \Pr {}_{z}:\prod _{x\in X}\mathbb {K} \to \mathbb {K} }$ denote the projection to the $z$^th coordinate (as defined above). To prove that ${\textstyle P\subseteq \prod _{x\in X}B_{r_{x}},}$ it is sufficient (and necessary) to show that $\Pr {}_{x}(P)\subseteq B_{r_{x}}$ for every $x\in X.$ So fix $x\in X$ and let $f\in P$; it remains to show that $\Pr {}_{x}(f):=f(x)\in B_{r_{x}}.$ The defining condition on $r_{x}>0$ was that $x\in r_{x}U,\,$ which implies that ${\textstyle \,u_{x}:=\left({\frac {1}{r_{x}}}\right)x\in U.\,}$ Because $f\in P=U^{\#},$ the linear functional $f$ satisfies ${\textstyle \;\sup _{u\in U}|f(u)|\leq 1\;}$ and so $u_{x}\in U$ implies

${\frac {1}{r_{x}}}|f(x)|=\left|{\frac {1}{r_{x}}}f(x)\right|=\left|f\left({\frac {1}{r_{x}}}x\right)\right|=\left|f\left(u_{x}\right)\right|\leq \sup _{u\in U}|f(u)|\leq 1.$

Thus $|f(x)|\leq r_{x},$ which shows that $f(x)\in B_{r_{x}},$ as desired. Q.E.D.

The elementary proof above actually shows that if $U\subseteq X$ is any subset that satisfies $X=(0,\infty )U:=\{ru:r>0,u\in U\}$ (such as any absorbing subset of $X$), then $U^{\#}:=\left\{f\in X^{\#}~:~f(U)\subseteq B_{1}\right\}$ 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) that $${\begin{alignedat}{4}U^{\circ }&=U^{\#}&&\\&=X^{\#}&&\cap \prod _{x\in X}B_{m_{x}}\\&=X^{\prime }&&\cap \prod _{x\in X}B_{m_{x}}\\\end{alignedat}}$$ where the real numbers $m_{x}\geq 0$ are "minimal" in the following sense: every $m_{x}$ is defined by $m_{x}:=\inf \left\{R_{x}:R_{\bullet }\in T_{P}\right\}$ for every $x\in X,$ with $P:=U^{\circ }$ (as in the proof) and $$T_{P}~:=~\left\{R_{\bullet }=\left(R_{x}\right)_{x\in X}\in \prod _{x\in X}[0,\infty )~:~P\subseteq \prod _{x\in X}B_{R_{x}}\right\}.$$

In fact, $$\left(m_{x}\right)_{x\in X}~\in ~T_{P}\quad {\text{ and }}\quad \prod _{x\in X}B_{m_{x}}~=~\cap {\mathcal {B}}_{P}~\in ~{\mathcal {B}}_{P}$$ where ${\textstyle \bigcap {\mathcal {B}}_{P}}$ denotes the intersection of all sets belonging to $${\mathcal {B}}_{P}:=\left\{\prod _{x\in X}B_{R_{x}}~:~R_{\bullet }\in T_{P}\right\}~=~\left\{\prod _{x\in X}B_{R_{x}}~:~P\subseteq \prod _{x\in X}B_{R_{x}}{\text{ and }}R_{x}\geq 0{\text{ for all }}x\in X\right\}.$$

This implies (among other things) that ${\textstyle \prod _{x\in X}B_{m_{x}}}$ the unique least element of ${\mathcal {B}}_{P}$ with respect to $\,\subseteq$; this may be used as an alternative definition of this (necessarily convex and balanced) set. The function $m_{\bullet }:=\left(m_{x}\right)_{x\in X}:X\to [0,\infty )$ is a seminorm and it is unchanged if $U$ is replaced by the convex balanced hull of $U$ (because $U^{\#}=[\operatorname {cobal} U]^{\#}$). Similarly, because $U^{\circ }=\left[\operatorname {cl} _{X}U\right]^{\circ },$ $m_{\bullet }$ is also unchanged if $U$ is replaced by its closure in $X.$

Lemma — The algebraic dual space $X^{\#}$ of any vector space $X$ over a field $\mathbb {K}$ (where $\mathbb {K}$ is $\mathbb {R}$ or $\mathbb {C}$) is a closed subset of ${\textstyle \mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K} }$ in the topology of pointwise convergence. (The vector space $X$ need not be endowed with any topology).

Proof of lemma

Notation for nets and function composition with nets A net $x_{\bullet }=\left(x_{i}\right)_{i\in I}$ in $X$ is by definition a function $x_{\bullet }:I\to X$ from a non-empty directed set $(I,\leq ).$ Every sequence in $X,$ which by definition is just a function of the form $\mathbb {N} \to X,$ is also a net. As with sequences, the value of a net $x_{\bullet }$ 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 $x_{\bullet }(i).$ Similarly for function composition, if $F:X\to Y$ is any function then the net (or sequence) that results from "plugging $x_{\bullet }$ into $F$" is just the function $F\circ x_{\bullet }:I\to Y,$ although this is typically denoted by $\left(F\left(x_{i}\right)\right)_{i\in I}$ (or by $\left(F\left(x_{i}\right)\right)_{i=1}^{\infty }$ if $x_{\bullet }$ is a sequence). In this proof, this resulting net may be denoted by any of the following notations $$F\left(x_{\bullet }\right)=\left(F\left(x_{i}\right)\right)_{i\in I}:=F\circ x_{\bullet },$$ depending on whichever notation is cleanest or most clearly communicates the intended information. In particular, if $F:X\to Y$ is continuous and $x_{\bullet }\to x$ in $X,$ then the conclusion commonly written as $\left(F\left(x_{i}\right)\right)_{i\in I}\to F(x)$ may instead be written as $F\left(x_{\bullet }\right)\to F(x)$ or $F\circ x_{\bullet }\to F(x).$ Start of proof: Let $f\in \mathbb {K} ^{X}$ and suppose that $f_{\bullet }=\left(f_{i}\right)_{i\in I}$ is a net in $X^{\#}$ the converges to $f$ in $\mathbb {K} ^{X}.$ If $z\in X$ then $f_{\bullet }(z):I\to \mathbb {K}$ will denote $f_{\bullet }$'s net of values at $z$ $$f_{\bullet }(z):=\left(f_{i}(z)\right)_{i\in I}.$$ To conclude that $f\in X^{\#},$ it must be shown that $f$ is a linear functional so let $s$ be a scalar and let $x,y\in X.$ The topology on $\mathbb {K} ^{X}$ is the topology of pointwise convergence so by considering the points $x,y,sx,$ and $x+y,$ the convergence of $f_{\bullet }\to f$ in $\mathbb {K} ^{X}$ implies that each of the following nets of scalars converges in $\mathbb {K} :$ $$f_{\bullet }(x)\to f(x),\quad f_{\bullet }(y)\to f(y),\quad f_{\bullet }(x+y)\to f(x+y),\quad {\text{ and }}\quad f_{\bullet }(sx)\to f(sx).$$ Proof that $f(sx)=sf(x):$ Let $M:\mathbb {K} \to \mathbb {K}$ be the "multiplication by $s$" map defined by $M(c):=sc.$ Because $M$ is continuous and $f_{\bullet }(x)\to f(x)$ in $\mathbb {K} ,$ it follows that $M\left(f_{\bullet }(x)\right)\to M(f(x))$ where the right hand side is $M(f(x))=sf(x)$ and the left hand side is $${\begin{alignedat}{4}M\left(f_{\bullet }(x)\right)=&~\left(M\left(f_{i}(x)\right)\right)_{i\in I}~~~&&{\text{ because }}M\left(f_{\bullet }(x)\right):=M\circ f_{\bullet }(x){\text{ where }}f_{\bullet }(x)=\left(f_{i}(x)\right)_{i\in I}:I\to \mathbb {K} \\=&~\left(sf_{i}(x)\right)_{i\in I}&&M\left(f_{i}(x)\right):=sf_{i}(x)\\=&~\left(f_{i}(sx)\right)_{i\in I}&&{\text{ by linearity of }}f_{i}\\=&~f_{\bullet }(sx)&&{\text{ notation }}\end{alignedat}}$$ which proves that $f_{\bullet }(sx)\to sf(x).$ Because also $f_{\bullet }(sx)\to f(sx)$ and limits in $\mathbb {K}$ are unique, it follows that $sf(x)=f(sx),$ as desired. Proof that $f(x+y)=f(x)+f(y):$ Define a net $z_{\bullet }=\left(z_{i}\right)_{i\in I}:I\to \mathbb {K} \times \mathbb {K}$ by letting $z_{i}:=\left(f_{i}(x),f_{i}(y)\right)$ for every $i\in I.$ Because $f_{\bullet }(x)=\left(f_{i}(x)\right)_{i\in I}\to f(x)$ and $f_{\bullet }(y)=\left(f_{i}(y)\right)_{i\in I}\to f(y),$ it follows that $z_{\bullet }\to (f(x),f(y))$ in $\mathbb {K} \times \mathbb {K} .$ Let $A:\mathbb {K} \times \mathbb {K} \to \mathbb {K}$ be the addition map defined by $A(x,y):=x+y.$ The continuity of $A$ implies that $A\left(z_{\bullet }\right)\to A(f(x),f(y))$ in $\mathbb {K}$ where the right hand side is $A(f(x),f(y))=f(x)+f(y)$ and the left hand side is $$A\left(z_{\bullet }\right):=A\circ z_{\bullet }=\left(A\left(z_{i}\right)\right)_{i\in I}=\left(A\left(f_{i}(x),f_{i}(y)\right)\right)_{i\in I}=\left(f_{i}(x)+f_{i}(y)\right)_{i\in I}=\left(f_{i}(x+y)\right)_{i\in I}=f_{\bullet }(x+y)$$ which proves that $f_{\bullet }(x+y)\to f(x)+f(y).$ Because also $f_{\bullet }(x+y)\to f(x+y),$ it follows that $f(x+y)=f(x)+f(y),$ as desired. Q.E.D.

Corollary to lemma — When the algebraic dual space $X^{\#}$ of a vector space $X$ is equipped with the topology $\sigma \left(X^{\#},X\right)$ of pointwise convergence (also known as the weak-* topology) then the resulting topological vector space (TVS) $\left(X^{\#},\sigma \left(X^{\#},X\right)\right)$ is a complete Hausdorff locally convex TVS.

Proof of corollary

Because the underlying field $\mathbb {K}$ is a complete Hausdorff locally convex TVS, the same is true of the Cartesian product ${\textstyle \mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K} .}$ 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.

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 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 $X$ be a separable normed space and $B$ the closed unit ball in $X^{\prime }.$ Since $X$ is separable, let $x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty }$ be a countable dense subset. Then the following defines a metric, where for any $x,y\in B$ $$\rho (x,y)=\sum _{n=1}^{\infty }\,2^{-n}\,{\frac {\left|\langle x-y,x_{n}\rangle \right|}{1+\left|\langle x-y,x_{n}\rangle \right|}}$$ in which $\langle \cdot ,\cdot \rangle$ denotes the duality pairing of $X^{\prime }$ with $X.$ Sequential compactness of $B$ 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 variational problems. For instance, if one wants to minimize a functional $F:X^{\prime }\to \mathbb {R}$ on the dual of a separable normed vector space $X,$ one common strategy is to first construct a minimizing sequence $x_{1},x_{2},\ldots \in X^{\prime }$ which approaches the infimum of $F,$ use the sequential Banach–Alaoglu theorem to extract a subsequence that converges in the weak* topology to a limit $x,$ and then establish that $x$ is a minimizer of $F.$ The last step often requires $F$ to obey a (sequential) lower semi-continuity property in the weak* topology.

When $X^{\prime }$ is the space of finite Radon measures on the real line (so that $X=C_{0}(\mathbb {R} )$ is the space of continuous functions vanishing at infinity, by the Riesz representation theorem), the sequential Banach–Alaoglu theorem is equivalent to the Helly selection theorem.

Proof

For every $x\in X,$ let $$D_{x}=\{z\in \mathbb {C} :|z|\leq \|x\|\},$$ and $$D=\prod _{x\in X}D_{x}.$$

Because each $D_{x}$ is a compact subset of the complex plane, $D$ is also compact in the product topology by Tychonoff's theorem.

The closed unit ball in $X^{\prime },$ $B_{1}\left(X^{\prime }\right)$ can be identified as a subset of $D$ in a natural way: $$f\in B_{1}\left(X^{\prime }\right)\mapsto (f(x))_{x\in X}\in D.$$

This map is injective and continuous, with $B_{1}\left(X^{\prime }\right)$ having the weak-* topology and $D$ the product topology. This map's inverse, defined on its range, is also continuous.

To finish proving this theorem, it will now be shown that the range of the above map is closed. Given a net $$\left(f_{\alpha }(x)\right)_{x\in X}\to \left(\lambda _{x}\right)_{x\in X}$$ in $D,$ the functional defined by $$g(x)=\lambda _{x}$$ lies in $B_{1}(X^{\prime }).$

Consequences

Consequences for normed spaces

Assume that $X$ is a normed space and endow its continuous dual space $X^{\prime }$ with the usual dual norm.

• The closed unit ball in $X^{\prime }$ is weak-* compact. So if $X^{\prime }$ is infinite dimensional then its closed unit ball is necessarily not compact in the norm topology by the F. Riesz theorem (despite it being weak-* compact).
• A Banach space is reflexive if and only if its closed unit ball is $\sigma \left(X^{\prime },X\right)$-compact.
• If $X$ is a reflexive Banach space, then every bounded sequence in $X$ has a weakly convergent subsequence. (This follows by applying the Banach–Alaoglu theorem to a weakly metrizable subspace of $X$; or, more succinctly, by applying the Eberlein–Šmulian theorem.) For example, suppose that $X$ is the space Lp space $L^{p}(\mu ),$ $1<p<\infty .$ Let $f_{n}$ be a bounded sequence of functions in $X.$ Then there exists a subsequence $f_{n_{k}}$ and an $f\in X$ such that $$\int f_{n_{k}}g\,d\mu \to \int fg\,d\mu$$ for all $g\in L^{q}(\mu )=X^{\prime }$ where ${\frac {1}{p}}+{\frac {1}{q}}=1$). The corresponding result for $p=1$ is not true, as $L^{1}(\mu )$ is not reflexive.

Consequences for Hilbert spaces

• 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).
• 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.
• Closed and bounded sets in $B(H)$ 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 $B(H),$ the trace class operators). Hence bounded sequences of operators have a weak accumulation point. As a consequence, $B(H)$ has the Heine–Borel property, if equipped with either the weak operator or the ultraweak topology.

Relation to the axiom of choice

Since the Banach–Alaoglu theorem is usually proven via Tychonoff's theorem, it relies on the ZFC axiomatic framework, and in particular the axiom of choice. Most mainstream functional analysis also relies on ZFC. However, the theorem does not rely upon the axiom of choice in the separable case (see above): in this case one actually has a constructive proof. In the non-separable case, the ultrafilter Lemma, which is strictly weaker than the axiom of choice, suffices for the proof of the Banach-Alaoglu theorem, and is in fact equivalent to it.

See also

• Bishop–Phelps theorem
• Banach–Mazur theorem
• Delta-compactness theorem
• Eberlein–Šmulian theorem – Relates three different kinds of weak compactness in a Banach space
• Goldstine theorem
• James' theorem – theorem in mathematicsPages displaying wikidata descriptions as a fallback
• Krein-Milman theorem – On when a space equals the closed convex hull of its extreme pointsPages displaying short descriptions of redirect targets
• Mazur's lemma – On strongly convergent combinations of a weakly convergent sequence in a Banach space
• Topological vector space – Vector space with a notion of nearness

Notes

Proofs

References

• Köthe, Gottfried (1969). Topological Vector Spaces I. New York: Springer-Verlag. See §20.9.
• Meise, Reinhold; Vogt, Dietmar (1997). Introduction to Functional Analysis. Oxford: Clarendon Press. ISBN 0-19-851485-9. See Theorem 23.5, p. 264.
• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
• Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277. See Theorem 3.15, p. 68.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
• Schechter, Eric (1997). Handbook of Analysis and its Foundations. San Diego: Academic Press.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.

Further reading

• John B. Conway (1994). A course in functional analysis (2nd ed.). Berlin: Springer-Verlag. ISBN 0-387-97245-5. See Chapter 5, section 3.
• Peter B. Lax (2002). Functional Analysis. Wiley-Interscience. pp. 120–121. ISBN 0-471-55604-1.