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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— echos throughout functional analysis.” In 1912, Helly proved that the unit ball of the continuous dual space of ${\displaystyle C()}$ 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 , 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 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 ${\displaystyle X}$ is a vector space over the field ${\displaystyle \mathbb {K} }$ then ${\displaystyle X^{\#}}$ will denote the algebraic dual space of ${\displaystyle X}$ and these two spaces are henceforth associated with the bilinear evaluation map ${\displaystyle \left\langle \cdot ,\cdot \right\rangle :X\times X^{\#}\to \mathbb {K} }$ defined by $${\displaystyle \left\langle x,f\right\rangle ~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~f(x)}$$ where the triple ${\displaystyle \left\langle X,X^{\#},\left\langle \cdot ,\cdot \right\rangle \right\rangle }$ forms a dual system called the canonical dual system.

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

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

Proof involving duality theory

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

Let ${\displaystyle U}$ be a neighborhood of the origin in ${\displaystyle X}$ and let:

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

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

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

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

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

When the continuous dual space ${\displaystyle X^{\prime }}$ of ${\displaystyle X}$ is an infinite dimensional normed space then it is impossible for the closed unit ball in ${\displaystyle X^{\prime }}$ to be a compact subset when ${\displaystyle X^{\prime }}$ 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 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 ${\displaystyle \mathbb {K} ^{X}}$ of all functions of the form ${\displaystyle X\to \mathbb {K} }$ is identified as the Cartesian product ${\textstyle \prod _{x\in X}\mathbb {K} ,}$ and the relationship between pointwise convergence, the product topology, and subspace topologies they induce on subsets such as the algebraic dual space ${\displaystyle X^{\#}}$ and products of subspaces such as ${\textstyle \prod _{x\in X}B_{r_{x}}.}$ An explanation of these details is now given for readers who are interested.

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 not require the vector space ${\displaystyle X}$ to endowed with any topology.

Proposition — Let ${\displaystyle U}$ be a subset of a vector space ${\displaystyle X}$ over the field ${\displaystyle \mathbb {K} }$ (where ${\displaystyle \mathbb {K} =\mathbb {R} {\text{ or }}\mathbb {K} =\mathbb {C} }$) and for every real number ${\displaystyle r,}$ endow the closed ball ${\textstyle B_{r}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{s\in \mathbb {K} :|s|\leq r\}}$ with its usual topology (${\displaystyle X}$ need not be endowed with any topology, but ${\displaystyle \mathbb {K} }$ has its usual Euclidean topology). Define $${\displaystyle U^{\#}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\Big \{}f\in X^{\#}~:~\sup _{u\in U}|f(u)|\leq 1{\Big \}}.}$$

If for every ${\displaystyle x\in X,}$ ${\displaystyle r_{x}>0}$ is a real number such that ${\displaystyle x\in r_{x}U,}$ then ${\displaystyle U^{\#}}$ is a closed and compact subspace of the product space ${\displaystyle \prod _{x\in X}B_{r_{x}}}$ (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 ${\displaystyle U^{\#}}$ 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 ${\displaystyle X}$ is a topological vector space (TVS) and that ${\displaystyle U}$ is a neighborhood of the origin).

Assume that ${\displaystyle X}$ is a topological vector space with continuous dual space ${\displaystyle X^{\prime }}$ and that ${\displaystyle U}$ is a neighborhood of the origin. Because ${\displaystyle U}$ is a neighborhood of the origin in ${\displaystyle X,}$ it is also an absorbing subset of ${\displaystyle X,}$ so for every ${\displaystyle x\in X,}$ there exists a real number ${\displaystyle r_{x}>0}$ such that ${\displaystyle x\in r_{x}U.}$ Thus the hypotheses of the above proposition are satisfied, and so the set ${\displaystyle U^{\#}}$ is therefore compact in the weak-* topology. The proof of the Banach–Alaoglu theorem will be complete once it is shown that ${\displaystyle U^{\#}=U^{\circ },}$ where recall that ${\displaystyle U^{\circ }}$ was defined as $${\displaystyle U^{\circ }~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\Big \{}f\in X^{\prime }~:~\sup _{u\in U}|f(u)|\leq 1{\Big \}}~=~U^{\#}\cap X^{\prime }.}$$

Proof that ${\displaystyle U^{\circ }=U^{\#}:}$ Because ${\displaystyle U^{\circ }=U^{\#}\cap X^{\prime },}$ the conclusion is equivalent to ${\displaystyle U^{\#}\subseteq X^{\prime }.}$ If ${\displaystyle f\in U^{\#}}$ then ${\displaystyle \;\sup _{u\in U}|f(u)|\leq 1,\,}$ which states exactly that the linear functional ${\displaystyle f}$ is bounded on the neighborhood ${\displaystyle U;}$ thus ${\displaystyle f}$ is a continuous linear functional (that is, ${\displaystyle f\in X^{\prime }}$), as desired. ${\displaystyle \blacksquare }$

The product space ${\textstyle \prod _{x\in X}B_{r_{x}}}$ is compact by Tychonoff's theorem (since each closed ball ${\displaystyle B_{r_{x}}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{s\in \mathbb {K} :|s|\leq r_{x}\}}$ is a Hausdorff 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 $${\displaystyle U^{\#}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\Big \{}f\in X^{\#}~:~\sup _{u\in U}|f(u)|\leq 1{\Big \}}~=~\left\{f\in X^{\#}~:~f(U)\subseteq B_{1}\right\}}$$ is a closed subset of ${\textstyle \prod _{x\in X}B_{r_{x}}.}$ The following statements guarantee this conclusion:

1. ${\displaystyle U^{\#}\subseteq \prod _{x\in X}B_{r_{x}}.}$
2. ${\displaystyle U^{\#}}$ is a closed subset of the product space ${\displaystyle \prod _{x\in X}\mathbb {K} =\mathbb {K} ^{X}.}$

Proof of (1):

For any ${\displaystyle z\in X,}$ let ${\textstyle \Pr {}_{z}:\prod _{x\in X}\mathbb {K} \to \mathbb {K} }$ denote the projection to the ${\displaystyle z}$ coordinate (as defined above). To prove that ${\textstyle U^{\#}\subseteq \prod _{x\in X}B_{r_{x}},}$ it is sufficient (and necessary) to show that ${\displaystyle \Pr {}_{x}\left(U^{\#}\right)\subseteq B_{r_{x}}}$ for every ${\displaystyle x\in X.}$ So fix ${\displaystyle x\in X}$ and let ${\displaystyle f\in U^{\#}.}$ Because ${\displaystyle \Pr {}_{x}(f)\,=\,f(x),}$ it remains to show that ${\displaystyle f(x)\in B_{r_{x}}.}$ Recall that ${\displaystyle r_{x}>0}$ was defined in the proposition's statement as being any positive real number that satisfies ${\displaystyle x\in r_{x}U}$ (so for example, ${\displaystyle r_{u}:=1}$ would be a valid choice for each ${\displaystyle u\in U}$), which implies ${\displaystyle \,u_{x}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\frac {1}{r_{x}}}\,x\in U.\,}$ Because ${\displaystyle f}$ is a positive homogeneous function that satisfies ${\displaystyle \;\sup _{u\in U}|f(u)|\leq 1,\,}$ $${\displaystyle {\frac {1}{r_{x}}}|f(x)|=\left|{\frac {1}{r_{x}}}f(x)\right|=\left|f\left({\frac {1}{r_{x}}}x\right)\right|=\left|f\left(u_{x}\right)\right|\leq \sup _{u\in U}|f(u)|\leq 1.}$$

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

Proof of (2):

The algebraic dual space ${\displaystyle X^{\#}}$ 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). The set $${\displaystyle {\begin{alignedat}{9}U_{B_{1}}&\,{\stackrel {\scriptscriptstyle {\text{def}}}{=}}\,{\Big \{}~~\;~~\;~~\;~~f\ \in \mathbb {K} ^{X}~~\;~~:\sup _{u\in U}|f(u)|\leq 1{\Big \}}\\&={\big \{}~~\;~~\;~~\;~~f\,\in \mathbb {K} ^{X}~~\;~~:f(u)\in B_{1}{\text{ for all }}u\in U{\big \}}\\&={\Big \{}\left(f_{x}\right)_{x\in X}\in \prod _{x\in X}\mathbb {K} \,~:~\;~f_{u}~\in B_{1}{\text{ for all }}u\in U{\Big \}}\\&=\prod _{x\in X}C_{x}\quad {\text{ where }}\quad C_{x}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\begin{cases}B_{1}&{\text{ if }}x\in U\\\mathbb {K} &{\text{ if }}x\not \in U\\\end{cases}}\\\end{alignedat}}}$$ is closed in the product topology on ${\displaystyle \prod _{x\in X}\mathbb {K} =\mathbb {K} ^{X}}$ since it is a product of closed subsets of ${\displaystyle \mathbb {K} .}$ Thus ${\displaystyle U_{B_{1}}\cap X^{\#}=U^{\#}}$ is an intersection of two closed subsets of ${\displaystyle \mathbb {K} ^{X},}$ which proves (2). ${\displaystyle \blacksquare }$

The conclusion that the set ${\displaystyle U_{B_{1}}=\left\{f\in \mathbb {K} ^{X}:f(U)\subseteq B_{1}\right\}}$ is closed can also be reached by applying the following more general result, this time proved using nets, to the special case ${\displaystyle Y:=\mathbb {K} }$ and ${\displaystyle B:=B_{1}.}$

Observation: If ${\displaystyle U\subseteq X}$ is any set and if ${\displaystyle B\subseteq Y}$ is a closed subset of a topological space ${\displaystyle Y,}$ then ${\displaystyle U_{B}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{f\in Y^{X}:f(U)\subseteq B\right\}}$ is a closed subset of ${\displaystyle Y^{X}}$ in the topology of pointwise convergence.

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

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

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

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

The above elementary proof of the Banach–Alaoglu theorem actually shows that if ${\displaystyle U\subseteq X}$ is any subset that satisfies ${\displaystyle X=(0,\infty )U~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\{ru:r>0,u\in U\}}$ (such as any absorbing subset of ${\displaystyle X}$), then ${\displaystyle U^{\#}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{f\in X^{\#}:f(U)\subseteq B_{1}\right\}}$ is a weak-* compact subset of ${\displaystyle X^{\#}.}$

As a side note, with the help of the above elementary proof, it may be shown (see this footnote) that there exist ${\displaystyle X}$-indexed non-negative real numbers ${\displaystyle m_{\bullet }=\left(m_{x}\right)_{x\in X}}$ such that $${\displaystyle {\begin{alignedat}{4}U^{\circ }&=U^{\#}&&\\&=X^{\#}&&\cap \prod _{x\in X}B_{m_{x}}\\&=X^{\prime }&&\cap \prod _{x\in X}B_{m_{x}}\\\end{alignedat}}}$$ where these real numbers ${\displaystyle m_{\bullet }}$ can also be chosen to be "minimal" in the following sense: using ${\displaystyle P~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~U^{\circ }}$ (so ${\displaystyle P=U^{\#}}$ as in the proof) and defining the notation ${\displaystyle \prod B_{R_{\bullet }}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\prod _{x\in X}B_{R_{x}}}$ for any ${\displaystyle R_{\bullet }=\left(R_{x}\right)_{x\in X}\in \mathbb {R} ^{X},}$ if $${\displaystyle T_{P}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{R_{\bullet }\in \mathbb {R} ^{X}~:~P\subseteq \prod B_{R_{\bullet }}\right\}}$$ then ${\displaystyle m_{\bullet }\in T_{P}}$ and for every ${\displaystyle x\in X,}$ ${\displaystyle m_{x}=\inf \left\{R_{x}:R_{\bullet }\in T_{P}\right\},}$ which shows that these numbers ${\displaystyle m_{\bullet }}$ are unique; indeed, this infimum formula can be used to define them.