Banach–Alaoglu theorem

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 $C([a,\; b])$ 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 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{{harvnb|Köthe|1983}}, Theorem (4) in §20.9.{{harvnb|Meise|Vogt|1997}}, 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.{{sfn|Narici|Beckenstein|2011|pp=235-240}}

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

If $X$ is a vector space over the field $\backslash mathbb\{K\}$ then $X^\{\backslash \#\}$ will denote the algebraic dual space of $X$ and these two spaces are henceforth associated with the bilinear {{em|evaluation map}} $\backslash left\backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash right\backslash rangle\; :\; X\; \backslash times\; X^\{\backslash \#\}\; \backslash to\; \backslash mathbb\{K\}$ defined by

$$\backslash left\backslash langle\; x,\; f\; \backslash right\backslash rangle\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; f(x)$$

where the triple $\backslash left\backslash langle\; X,\; X^\{\backslash \#\},\backslash left\backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash right\backslash rangle\; \backslash right\backslash rangle$ forms a dual system called the {{em|canonical dual system}}.

If $X$ is a topological vector space (TVS) then its continuous dual space will be denoted by $X^\{\backslash prime\},$ where $X^\{\backslash prime\}\; \backslash subseteq\; X^\{\backslash \#\}$ always holds.

Denote the weak-* topology on $X^\{\backslash \#\}$ by $\backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right)$ and denote the weak-* topology on $X^\{\backslash prime\}$ by $\backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right).$

The weak-* topology is also called the topology of pointwise convergence because given a map $f$ and a net of maps $f\_\{\backslash bull\}\; =\; \backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\},$ the net $f\_\{\backslash bull\}$ converges to $f$ in this topology if and only if for every point $x$ in the domain, the net of values $\backslash left(f\_i(x)\backslash right)\_\{i\; \backslash in\; I\}$ converges to the value $f(x).$

{{Math theorem|name=Alaoglu theorem{{sfn|Narici|Beckenstein|2011|pp=225-273}}|math_statement=

For any topological vector space (TVS) $X$ ({{em|not}} necessarily Hausdorff or locally convex) with continuous dual space $X^\{\backslash prime\},$ the polar

$$U^\{\backslash circ\}\; =\; \backslash left\backslash \{f\; \backslash in\; X^\{\backslash prime\}\; ~:~\; \backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1\backslash right\backslash \}$$

of any neighborhood $U$ of origin in $X$ is compact in the weak-* topologyExplicitly, a subset $B^\{\backslash prime\}\; \backslash subseteq\; X^\{\backslash prime\}$ is said to be "compact (resp. totally bounded, etc.) in the weak-* topology" if when $X^\{\backslash prime\}$ is given the weak-* topology and the subset $B^\{\backslash prime\}$ is given the subspace topology inherited from $\backslash left(X^\{\backslash prime\},\; \backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right)\backslash right),$ then $B^\{\backslash prime\}$ is a compact (resp. totally bounded, etc.) space. $\backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right)$ on $X^\{\backslash prime\}.$ Moreover, $U^\{\backslash circ\}$ is equal to the polar of $U$ with respect to the canonical system $\backslash left\backslash langle\; X,\; X^\{\backslash \#\}\; \backslash right\backslash rangle$ and it is also a compact subset of $\backslash left(X^\{\backslash \#\},\; \backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right)\backslash right).$

}}

{{math proof|drop=hidden|proof=

Denote by the underlying field of $X$ by $\backslash mathbb\{K\},$ which is either the real numbers $\backslash R$ or complex numbers $\backslash Complex.$

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 $X^\{\backslash \#\}$ is endowed with the weak-* topology $\backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right),$ then this Hausdorff locally convex topological vector space is denoted by $\backslash left(X^\{\backslash \#\},\; \backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right)\backslash right).$

The space $\backslash left(X^\{\backslash \#\},\; \backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right)\backslash right)$ is always a complete TVS; however, $\backslash left(X^\{\backslash prime\},\; \backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right)\backslash right)$ may fail to be a complete space, which is the reason why this proof involves the space $\backslash left(X^\{\backslash \#\},\; \backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right)\backslash right).$

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 $X^\{\backslash prime\}$ inherits from $\backslash left(X^\{\backslash \#\},\; \backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right)\backslash right)$ is equal to $\backslash sigma\backslash left(X^\{\backslash prime\},\; X\backslash right).$ This can be readily verified by showing that given any $f\; \backslash in\; X^\{\backslash prime\},$ a net in $X^\{\backslash prime\}$ converges to $f$ in one of these topologies if and only if it also converges to $f$ 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 $\backslash left\backslash langle\; X,\; X^\{\backslash prime\}\; \backslash right\backslash rangle$ is a dual pairing although unlike $\backslash left\backslash langle\; X,\; X^\{\backslash \#\}\; \backslash right\backslash rangle,$ 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 $\backslash left\backslash langle\; X,\; X^\{\backslash prime\}\; \backslash right\backslash rangle.$

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

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

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

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

}}

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 $X^\{\backslash prime\}$ of $X$ (with the usual dual norm).

Consequently, this theorem can be specialized to:

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

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

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 $\backslash mathbb\{K\}^X$ of all functions of the form $X\; \backslash to\; \backslash mathbb\{K\}$ is identified as the Cartesian product $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\},$ and the relationship between pointwise convergence, the product topology, and subspace topologies they induce on subsets such as the algebraic dual space $X^\{\backslash \#\}$ and products of subspaces such as $\backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\}.$

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 $r,$ $B\_r\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash \{c\; \backslash in\; \backslash mathbb\{K\}\; :\; |c|\; \backslash leq\; r\backslash \}$ will denote the closed ball of radius $r$ centered at $0$ and $r\; U\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash \{r\; u\; :\; u\; \backslash in\; U\backslash \}$ for any $U\; \backslash subseteq\; X,$

Identification of functions with tuples

The Cartesian product $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}$ is usually thought of as the set of all $X$-indexed tuples $s\_\{\backslash bull\}\; =\; \backslash left(s\_x\backslash right)\_\{x\; \backslash in\; X\}$ but, since tuples are technically just functions from an indexing set, it can also be identified with the space $\backslash mathbb\{K\}^X$ of all functions having prototype $X\; \backslash to\; \backslash mathbb\{K\},$ as is now described:

• {{em|Function $\backslash to$ Tuple}}: A function $s\; :\; X\; \backslash to\; \backslash mathbb\{K\}$ belonging to $\backslash mathbb\{K\}^X$ is identified with its ($X$-indexed) "{{em|tuple of values}}" $s\_\{\backslash bull\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; (s(x))\_\{x\; \backslash in\; X\}.$
• {{em|Tuple $\backslash to$ Function}}: A tuple $s\_\{\backslash bull\}\; =\; \backslash left(s\_x\backslash right)\_\{x\; \backslash in\; X\}$ in $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}$ is identified with the function $s\; :\; X\; \backslash to\; \backslash mathbb\{K\}$ defined by $s(x)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; s\_x$; this function's "tuple of values" is the original tuple $\backslash left(s\_x\backslash right)\_\{x\; \backslash in\; X\}.$

This is the reason why many authors write, often without comment, the equality

$$\backslash mathbb\{K\}^X\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}$$

and why the Cartesian product $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}$ is sometimes taken as the definition of the set of maps $\backslash mathbb\{K\}^X$ (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) {{em|projections}}.

The {{visible anchor|canonical projection of the Cartesian product}} at a given point $z\; \backslash in\; X$ is the function

$$\backslash Pr\{\}\_z\; :\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}\; \backslash to\; \backslash mathbb\{K\}\; \backslash quad\; \backslash text\{\; defined\; by\; \}\; \backslash quad\; s\_\{\backslash bull\}\; =\; \backslash left(s\_x\backslash right)\_\{x\; \backslash in\; X\}\; \backslash mapsto\; s\_z$$

where under the above identification, $\backslash Pr\{\}\_z$ sends a function $s\; :\; X\; \backslash to\; \backslash mathbb\{K\}$ to

$$\backslash Pr\{\}\_z(s)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; s(z).$$

Stated in words, for a point $z$ and function $s,$ "plugging $z$ into $s$" is the same as "plugging $s$ into $\backslash Pr\{\}\_z$".

In particular, suppose that $\backslash left(r\_x\backslash right)\_\{x\; \backslash in\; X\}$ are non-negative real numbers.

Then $\backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\}\; \backslash subseteq\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}\; =\; \backslash mathbb\{K\}^X,$ where under the above identification of tuples with functions, $\backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\}$ is the set of all functions $s\; \backslash in\; \backslash mathbb\{K\}^X$ such that $s(x)\; \backslash in\; B\_\{r\_x\}$ for every $x\; \backslash in\; X.$

If a subset $U\; \backslash subseteq\; X$ partitions $X$ into $X\; =\; U\; \backslash ,\; \backslash cup\; \backslash ,(X\; \backslash setminus\; U)$ then the linear bijection

$$\backslash begin\{alignat\}\{4\}$$

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}

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

$$\backslash begin\{alignat\}\{4\}$$

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}.

Notation for nets and function composition with nets

A net $x\_\{\backslash bull\}\; =\; \backslash left(x\_i\backslash right)\_\{i\; \backslash in\; I\}$ in $X$ is by definition a function $x\_\{\backslash bull\}\; :\; I\; \backslash to\; X$ from a non-empty directed set $(I,\; \backslash leq).$

Every sequence in $X,$ which by definition is just a function of the form $\backslash N\; \backslash to\; X,$ is also a net.

As with sequences, the value of a net $x\_\{\backslash bull\}$ at an index $i\; \backslash in\; I$ is denoted by $x\_i$; however, for this proof, this value $x\_i$ may also be denoted by the usual function parentheses notation $x\_\{\backslash bull\}(i).$

Similarly for function composition, if $F\; :\; X\; \backslash to\; Y$ is any function then the net (or sequence) that results from "plugging $x\_\{\backslash bull\}$ into $F$" is just the function $F\; \backslash circ\; x\_\{\backslash bull\}\; :\; I\; \backslash to\; Y,$ although this is typically denoted by $\backslash left(F\backslash left(x\_i\backslash right)\backslash right)\_\{i\; \backslash in\; I\}$ (or by $\backslash left(F\backslash left(x\_i\backslash right)\backslash right)\_\{i=1\}^\{\backslash infty\}$ if $x\_\{\backslash bull\}$ is a sequence).

In the proofs below, this resulting net may be denoted by any of the following notations

$$F\backslash left(x\_\{\backslash bull\}\backslash right)\; =\; \backslash left(F\backslash left(x\_i\backslash right)\backslash right)\_\{i\; \backslash in\; I\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; F\; \backslash circ\; x\_\{\backslash bull\},$$

depending on whichever notation is cleanest or most clearly communicates the intended information.

In particular, if $F\; :\; X\; \backslash to\; Y$ is continuous and $x\_\{\backslash bull\}\; \backslash to\; x$ in $X,$ then the conclusion commonly written as $\backslash left(F\backslash left(x\_i\backslash right)\backslash right)\_\{i\; \backslash in\; I\}\; \backslash to\; F(x)$ may instead be written as $F\backslash left(x\_\{\backslash bull\}\backslash right)\; \backslash to\; F(x)$ or $F\; \backslash circ\; x\_\{\backslash bull\}\; \backslash to\; F(x).$

Topology

The set $\backslash mathbb\{K\}^X\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}$ 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 $\backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\},$ where $f$ and every $f\_i$ is an element of $\backslash mathbb\{K\}^X\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\},$ then the net $\backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}\; \backslash to\; f$ converges in the product topology if and only if

:for every $z\; \backslash in\; X,$ the net $\backslash Pr\{\}\_z\backslash left(\backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}\backslash right)\; \backslash to\; \backslash Pr\{\}\_z(f)$ converges in $\backslash mathbb\{K\},$

where because $\backslash ;\backslash Pr\{\}\_z(f)\; =\; f(z)\backslash ;$ and

$\backslash Pr\{\}\_z\backslash left(\backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}\backslash right)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left(\backslash Pr\{\}\_z\backslash left(f\_i\backslash right)\backslash right)\_\{i\; \backslash in\; I\}\; =\; \backslash left(f\_i(z)\backslash right)\_\{i\; \backslash in\; I\},$

this happens if and only if

:for every $z\; \backslash in\; X,$ the net $\backslash left(f\_i(z)\backslash right)\_\{i\; \backslash in\; I\}\; \backslash to\; f(z)$ converges in $\backslash mathbb\{K\},$

Thus $\backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}$ 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 $x\; \backslash in\; X,$ $S\_x\; \backslash subseteq\; \backslash mathbb\{K\}$ is some (topological) subspace of $\backslash mathbb\{K\}$ then the topology of pointwise convergence (or equivalently, the product topology) on $\backslash prod\_\{x\; \backslash in\; X\}\; S\_x$ is equal to the subspace topology that the set $\backslash prod\_\{x\; \backslash in\; X\}\; S\_x$ inherits from $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}.$

And if $S\_x$ is closed in $\backslash mathbb\{K\}$ for every $x\; \backslash in\; X,$ then $\backslash prod\_\{x\; \backslash in\; X\}\; S\_x$ is a closed subset of $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}.$

Characterization of $\backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; r$

An important fact used by the proof is that for any real $r,$

$$\backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; r\; \backslash qquad\; \backslash text\{\; if\; and\; only\; if\; \}\; \backslash qquad\; f(U)\; \backslash subseteq\; B\_r$$

where $\backslash ,\backslash sup\backslash ,$ denotes the supremum and $f(U)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash \{f(u)\; :\; u\; \backslash in\; U\backslash \}.$

As a side note, this characterization does not hold if the closed ball $B\_r$ is replaced with the open ball (and replacing $\backslash ;\backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; r\backslash ;$ with the strict inequality will not change this; for counter-examples, consider $X\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash mathbb\{K\}$ and the identity map $f\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash operatorname\{Id\}$ on $X$).

{{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 $X$ to endowed with any topology.

{{Math theorem|name=Proposition{{sfn|Narici|Beckenstein|2011|pp=225-273}}|math_statement=

Let $U$ be a subset of a vector space $X$ over the field $\backslash mathbb\{K\}$ (where $\backslash mathbb\{K\}\; =\; \backslash R\; \backslash text\{\; or\; \}\; \backslash mathbb\{K\}\; =\; \backslash Complex$) and for every real number $r,$ endow the closed ball $B\_r\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash \{s\; \backslash in\; \backslash mathbb\{K\}\; :\; |s|\; \backslash leq\; r\backslash \}$ with its usual topology ($X$ need not be endowed with any topology, but $\backslash mathbb\{K\}$ has its usual Euclidean topology).

Define

$$U^\{\backslash \#\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash Big\backslash \{f\; \backslash in\; X^\{\backslash \#\}\; ~:~\; \backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1\backslash Big\backslash \}.$$

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

{{Math proof|drop=hidden|title=Proof that Banach–Alaoglu follows from the proposition above|proof=

Assume that $X$ is a topological vector space with continuous dual space $X^\{\backslash prime\}$ and that $U$ is a neighborhood of the origin.

Because $U$ is a neighborhood of the origin in $X,$ it is also an absorbing subset of $X,$ so for every $x\; \backslash in\; X,$ there exists a real number $r\_x\; >\; 0$ such that $x\; \backslash in\; r\_x\; U.$

Thus the hypotheses of the above proposition are satisfied, and so the set $U^\{\backslash \#\}$ is therefore compact in the weak-* topology.

The proof of the Banach–Alaoglu theorem will be complete once it is shown that $U^\{\backslash \#\}\; =\; U^\{\backslash circ\},$If $\backslash tau$ denotes the topology that $X$ is (originally) endowed with, then the equality $U^\{\backslash circ\}\; =\; U^\{\backslash \#\}$ shows that the polar $U^\{\backslash circ\}\; =\; \backslash Big\backslash \{f\; \backslash in\; X^\{\backslash prime\}\; ~:~\; \backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1\backslash Big\backslash \}$ of $U$ is dependent {{em|only}} on $U$ (and $X^\{\backslash \#\}$) and that the rest of the topology $\backslash tau$ can be ignored. To clarify what is meant, suppose $\backslash sigma$ is any TVS topology on $X$ such that the set $U$ is (also) a neighborhood of the origin in $(X,\; \backslash sigma).$ Denote the continuous dual space of $(X,\; \backslash sigma)$ by $(X,\; \backslash sigma)^\{\backslash prime\}$ and denote the polar of $U$ with respect to $(X,\; \backslash sigma)$ by

$$U^\{\backslash circ,\; \backslash sigma\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash Big\backslash \{f\; \backslash in\; (X,\; \backslash sigma)^\{\backslash prime\}\; ~:~\; \backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1\backslash Big\backslash \}$$

so that $U^\{\backslash circ,\; \backslash tau\}$ is just the set $U^\{\backslash circ\}$ from above.

Then $U^\{\backslash circ,\; \backslash tau\}\; =\; U^\{\backslash circ,\; \backslash sigma\}$ because both of these sets are equal to $U^\{\backslash \#\}.$ Said differently, the polar set $U^\{\backslash circ,\; \backslash sigma\}$'s defining "requirement" that $U^\{\backslash circ,\; \backslash sigma\}$ be a subset of the {{em|continuous}} dual space $(X,\; \backslash sigma)^\{\backslash prime\}$ is inconsequential and can be ignored because it does not have any effect on the resulting set of linear functionals. However, if $\backslash nu$ is a TVS topology on $X$ such that $U$ is {{em|not}} a neighborhood of the origin in $(X,\; \backslash nu)$ then the polar $U^\{\backslash circ,\; \backslash nu\}$ of $U$ with respect to $(X,\; \backslash nu)$ is not guaranteed to equal $U^\{\backslash \#\}$ and so the topology $\backslash nu$ can not be ignored.

where recall that $U^\{\backslash circ\}$ was defined as

$$U^\{\backslash circ\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash Big\backslash \{f\; \backslash in\; X^\{\backslash prime\}\; ~:~\; \backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1\backslash Big\backslash \}\; ~=~\; U^\{\backslash \#\}\; \backslash cap\; X^\{\backslash prime\}.$$

Proof that $U^\{\backslash circ\}\; =\; U^\{\backslash \#\}:$

Because $U^\{\backslash circ\}\; =\; U^\{\backslash \#\}\; \backslash cap\; X^\{\backslash prime\},$ the conclusion is equivalent to $U^\{\backslash \#\}\; \backslash subseteq\; X^\{\backslash prime\}.$

If $f\; \backslash in\; U^\{\backslash \#\}$ then $\backslash ;\backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1,\backslash ,$ which states exactly that the linear functional $f$ is bounded on the neighborhood $U;$ thus $f$ is a continuous linear functional (that is, $f\; \backslash in\; X^\{\backslash prime\}$), as desired.

$\backslash blacksquare$

}}

{{Math proof|drop=hidden|title=Proof of Proposition|proof=

The product space $\backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\}$ is compact by Tychonoff's theorem (since each closed ball $B\_\{r\_x\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash \{s\; \backslash in\; \backslash mathbb\{K\}\; :\; |s|\; \backslash leq\; r\_x\backslash \}$ is a HausdorffBecause every $B\_\{r\_x\}$ is also a Hausdorff space, the conclusion that $\backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\}$ 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. 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

$$U^\{\backslash \#\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash Big\backslash \{f\; \backslash in\; X^\{\backslash \#\}\; ~:~\; \backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1\backslash Big\backslash \}\; ~=~\; \backslash left\backslash \{f\; \backslash in\; X^\{\backslash \#\}\; ~:~\; f(U)\; \backslash subseteq\; B\_1\backslash right\backslash \}$$

is a closed subset of $\backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\}.$

The following statements guarantee this conclusion:

1. $U^\{\backslash \#\}\; \backslash subseteq\; \backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\}.$
2. $U^\{\backslash \#\}$ is a closed subset of the product space $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}\; =\; \backslash mathbb\{K\}^X.$

Proof of (1):

For any $z\; \backslash in\; X,$ let $\backslash Pr\{\}\_z\; :\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}\; \backslash to\; \backslash mathbb\{K\}$ denote the projection to the $z$^th coordinate (as defined above).

To prove that $U^\{\backslash \#\}\; \backslash subseteq\; \backslash prod\_\{x\; \backslash in\; X\}\; B\_\{r\_x\},$ it is sufficient (and necessary) to show that $\backslash Pr\{\}\_x\backslash left(U^\{\backslash \#\}\backslash right)\; \backslash subseteq\; B\_\{r\_x\}$ for every $x\; \backslash in\; X.$

So fix $x\; \backslash in\; X$ and let $f\; \backslash in\; U^\{\backslash \#\}.$

Because $\backslash Pr\{\}\_x(f)\; \backslash ,=\backslash ,\; f(x),$ it remains to show that $f(x)\; \backslash in\; B\_\{r\_x\}.$

Recall that $r\_x\; >\; 0$ was defined in the proposition's statement as being any positive real number that satisfies $x\; \backslash in\; r\_x\; U$ (so for example, $r\_u\; :=\; 1$ would be a valid choice for each $u\; \backslash in\; U$), which implies $\backslash ,u\_x\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash frac\{1\}\{r\_x\}\; \backslash ,\; x\; \backslash in\; U.\backslash ,$

Because $f$ is a positive homogeneous function that satisfies $\backslash ;\backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1,\backslash ,$

$$\backslash frac\{1\}\{r\_x\}|f(x)|$$

= \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 $|f(x)|\; \backslash leq\; r\_x,$ which shows that $f(x)\; \backslash in\; B\_\{r\_x\},$ as desired.

Proof of (2):

The algebraic dual space $X^\{\backslash \#\}$ is always a closed subset of $\backslash mathbb\{K\}^X\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}$ (this is proved in the lemma below for readers who are not familiar with this result).

The set

$$\backslash begin\{alignat\}\{9\}$$

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}

is closed in the product topology on $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}\; =\; \backslash mathbb\{K\}^X$ since it is a product of closed subsets of $\backslash mathbb\{K\}.$

Thus $U\_\{B\_1\}\; \backslash cap\; X^\{\backslash \#\}\; =\; U^\{\backslash \#\}$ is an intersection of two closed subsets of $\backslash mathbb\{K\}^X,$ which proves (2).The conclusion $U\_\{B\_1\}\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; C\_x$ can be written as $U\_\{B\_1\}\; ~=~\; \backslash Big(\backslash prod\_\{u\; \backslash in\; U\}\; B\_1\backslash Big)\; \backslash times\; \backslash prod\_\{x\; \backslash in\; X\; \backslash setminus\; U\}\; \backslash mathbb\{K\}.$ The set $U^\{\backslash \#\}$ may thus equivalently be defined by $U^\{\backslash \#\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; X^\{\backslash \#\}\; \backslash cap\; \backslash left[\backslash Big(\backslash prod\_\{u\; \backslash in\; U\}\; B\_1\backslash Big)\; \backslash times\; \backslash prod\_\{x\; \backslash in\; X\; \backslash setminus\; U\}\; \backslash mathbb\{K\}\backslash right].$ Rewriting the definition in this way helps make it apparent that the set $U^\{\backslash \#\}$ is closed in $\backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}$ because this is true of $X^\{\backslash \#\}.$

$\backslash blacksquare$

}}

The conclusion that the set $U\_\{B\_1\}\; =\; \backslash left\backslash \{f\; \backslash in\; \backslash mathbb\{K\}^X\; :\; f(U)\; \backslash subseteq\; B\_1\backslash right\backslash \}$ is closed can also be reached by applying the following more general result, this time proved using nets, to the special case $Y\; :=\; \backslash mathbb\{K\}$ and $B\; :=\; B\_1.$

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

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

{{math theorem|name=Lemma ($X^\{\backslash \#\}$ is closed in $\backslash mathbb\{K\}^X$){{anchor|Algebraic dual space is closed in the space of all functions}}|note=|style=|math_statement=

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

}}

{{collapse top|title=Proof of lemma|left=true}}

Let $f\; \backslash in\; \backslash mathbb\{K\}^X$ and suppose that $f\_\{\backslash bull\}\; =\; \backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}$ is a net in $X^\{\backslash \#\}$ the converges to $f$ in $\backslash mathbb\{K\}^X.$

To conclude that $f\; \backslash in\; X^\{\backslash \#\},$ it must be shown that $f$ is a linear functional. So let $s$ be a scalar and let $x,\; y\; \backslash in\; X.$

For any $z\; \backslash in\; X,$ let $f\_\{\backslash bull\}(z)\; :\; I\; \backslash to\; \backslash mathbb\{K\}$ denote {{em|$f\_\{\backslash bull\}$'s net of values at $z$}}

$$f\_\{\backslash bull\}(z)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left(f\_i(z)\backslash right)\_\{i\; \backslash in\; I\}.$$

Because $f\_\{\backslash bull\}\; \backslash to\; f$ in $\backslash mathbb\{K\}^X,$ which has the topology of pointwise convergence, $f\_\{\backslash bull\}(z)\; \backslash to\; f(z)$ in $\backslash mathbb\{K\}$ for every $z\; \backslash in\; X.$

By using $x,\; y,\; sx,\; \backslash text\{\; and\; \}\; x\; +\; y,$ in place of $z,$ it follows that each of the following nets of scalars converges in $\backslash mathbb\{K\}:$

$$f\_\{\backslash bull\}(x)\; \backslash to\; f(x),\; \backslash quad\; f\_\{\backslash bull\}(y)\; \backslash to\; f(y),\; \backslash quad\; f\_\{\backslash bull\}(x\; +\; y)\; \backslash to\; f(x\; +\; y),\; \backslash quad\; \backslash text\{\; and\; \}\; \backslash quad\; f\_\{\backslash bull\}(sx)\; \backslash to\; f(sx).$$

Proof that $f(s\; x)\; =\; s\; f(x):$

Let $M\; :\; \backslash mathbb\{K\}\; \backslash to\; \backslash mathbb\{K\}$ be the "multiplication by $s$" map defined by $M(c)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; s\; c.$

Because $M$ is continuous and $f\_\{\backslash bull\}(x)\; \backslash to\; f(x)$ in $\backslash mathbb\{K\},$ it follows that $M\backslash left(f\_\{\backslash bull\}(x)\backslash right)\; \backslash to\; M(f(x))$ where the right hand side is $M(f(x))\; =\; s\; f(x)$ and the left hand side is

$$\backslash begin\{alignat\}\{4\}$$

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}

which proves that $f\_\{\backslash bull\}(sx)\; \backslash to\; s\; f(x).$ Because also $f\_\{\backslash bull\}(sx)\; \backslash to\; f(sx)$ and limits in $\backslash mathbb\{K\}$ 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 $z\_\{\backslash bull\}\; =\; \backslash left(z\_i\backslash right)\_\{i\; \backslash in\; I\}\; :\; I\; \backslash to\; \backslash mathbb\{K\}\; \backslash times\; \backslash mathbb\{K\}$ by letting $z\_i\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left(f\_i(x),\; f\_i(y)\backslash right)$ for every $i\; \backslash in\; I.$

Because $f\_\{\backslash bull\}(x)\; =\; \backslash left(f\_i(x)\backslash right)\_\{i\; \backslash in\; I\}\; \backslash to\; f(x)$ and $f\_\{\backslash bull\}(y)\; =\; \backslash left(f\_i(y)\backslash right)\_\{i\; \backslash in\; I\}\; \backslash to\; f(y),$ it follows that $z\_\{\backslash bull\}\; \backslash to\; (\; f(x),\; f(y)\; )$ in $\backslash mathbb\{K\}\; \backslash times\; \backslash mathbb\{K\}.$

Let $A\; :\; \backslash mathbb\{K\}\; \backslash times\; \backslash mathbb\{K\}\; \backslash to\; \backslash mathbb\{K\}$ be the addition map defined by $A(x,\; y)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; x\; +\; y.$

The continuity of $A$ implies that $A\backslash left(z\_\{\backslash bull\}\backslash right)\; \backslash to\; A(f(x),\; f(y))$ in $\backslash mathbb\{K\}$ where the right hand side is $A(f(x),\; f(y))\; =\; f(x)\; +\; f(y)$ and the left hand side is

$$A\backslash left(z\_\{\backslash bull\}\backslash right)\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; A\; \backslash circ\; z\_\{\backslash bull\}$$

= \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)

which proves that $f\_\{\backslash bull\}(x\; +\; y)\; \backslash to\; f(x)\; +\; f(y).$ Because also $f\_\{\backslash bull\}(x\; +\; y)\; \backslash to\; f(x\; +\; y),$ it follows that $f(x\; +\; y)\; =\; f(x)\; +\; f(y),$ as desired.

$\backslash blacksquare$

{{collapse bottom}}

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

{{math theorem|name=Corollary to lemma ($X^\{\backslash \#\}$ is weak-* complete)|note=|style=|math_statement=

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

}}

{{collapse top|title=Proof of corollary to lemma|left=true}}

Because the underlying field $\backslash mathbb\{K\}$ is a complete Hausdorff locally convex topological vector space, the same is true of the product space $\backslash mathbb\{K\}^X\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash mathbb\{K\}.$

A closed subset of a complete space is complete, so by the lemma, the space $\backslash left(X^\{\backslash \#\},\; \backslash sigma\backslash left(X^\{\backslash \#\},\; X\backslash right)\backslash right)$ is complete.

$\backslash blacksquare$

{{collapse bottom}}

The above elementary proof of the Banach–Alaoglu theorem actually shows that if $U\; \backslash subseteq\; X$ is any subset that satisfies $X\; =\; (0,\; \backslash infty)\; U\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash \{r\; u\; :\; r\; >\; 0,\; u\; \backslash in\; U\backslash \}$ (such as any absorbing subset of $X$), then $U^\{\backslash \#\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left\backslash \{f\; \backslash in\; X^\{\backslash \#\}\; :\; f(U)\; \backslash subseteq\; B\_1\backslash right\backslash \}$ is a weak-* compact subset of $X^\{\backslash \#\}.$

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

For any non-empty subset $A\; \backslash subseteq\; [0,\; \backslash infty),$ the equality $\backslash cap\; \backslash left\backslash \{B\_a\; :\; a\; \backslash in\; A\backslash right\backslash \}\; =\; B\_\{\backslash inf\_\{\}\; A\}$ 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 $\backslash mathbb\{K\}$ and so must itself be closed). For every $x\; \backslash in\; X,$ let $m\_x\; =\; \backslash inf\_\{\}\; \backslash left\backslash \{\; R\_x\; :\; R\_\{\backslash bull\}\; \backslash in\; T\_P\; \backslash right\backslash \}$ so that the previous set equality implies $\backslash cap\; \backslash operatorname\{Box\}\_P\; =\; \backslash bigcap\_\{R\_\{\backslash bull\}\; \backslash in\; T\_P\}\; \backslash prod\_\{x\; \backslash in\; X\}\; B\_\{R\_x\}\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; \backslash bigcap\_\{R\_\{\backslash bull\}\; \backslash in\; T\_P\}\; B\_\{R\_x\}\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; B\_\{m\_x\}.$ From $P\; \backslash subseteq\; \backslash cap\; \backslash operatorname\{Box\}\_P$ it follows that $m\_\{\backslash bull\}\; \backslash in\; T\_P$ and $\backslash cap\; \backslash operatorname\{Box\}\_P\; \backslash in\; \backslash operatorname\{Box\}\_P,$ thereby making $\backslash cap\; \backslash operatorname\{Box\}\_P$ the least element of $\backslash operatorname\{Box\}\_P$ with respect to $\backslash ,\backslash subseteq.\backslash ,$ (In fact, the family $\backslash operatorname\{Box\}\_P$ is closed under (non-nullary) arbitrary intersections and also under finite unions of at least one set). The elementary proof showed that $T\_P$ and $\backslash operatorname\{Box\}\_P$ are not empty and moreover, it also even showed that $T\_P$ has an element $\backslash left(r\_x\backslash right)\_\{x\; \backslash in\; X\}$ that satisfies $r\_u\; =\; 1$ for every $u\; \backslash in\; U,$ which implies that $m\_u\; \backslash leq\; 1$ for every $u\; \backslash in\; U.$ The inclusion $P\; ~\backslash subseteq~\; \backslash left(\backslash cap\; \backslash operatorname\{Box\}\_P\backslash right)\; \backslash cap\; X^\{\backslash prime\}\; ~\backslash subseteq~\; \backslash left(\backslash cap\; \backslash operatorname\{Box\}\_P\backslash right)\; \backslash cap\; X^\{\backslash \#\}$ is immediate; to prove the reverse inclusion, let $f\; \backslash in\; \backslash left(\backslash cap\; \backslash operatorname\{Box\}\_P\backslash right)\; \backslash cap\; X^\{\backslash \#\}.$ By definition, $f\; \backslash in\; P\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; U^\{\backslash \#\}$ if and only if $\backslash sup\_\{u\; \backslash in\; U\}\; |f(u)|\; \backslash leq\; 1,$ so let $u\; \backslash in\; U$ and it remains to show that $|f(u)|\; \backslash leq\; 1.$ From $f\; \backslash in\; \backslash cap\; \backslash operatorname\{Box\}\_P\; =\; \backslash prod\; B\_\{m\_\backslash bull\},$ it follows that $f(u)\; =\; \backslash Pr\{\}\_u\; (f)\; \backslash in\; \backslash Pr\{\}\_u\; \backslash left(\backslash prod\_\{x\; \backslash in\; X\}\; B\_\{m\_x\}\backslash right)\; =\; B\_\{m\_u\},$ which implies that $|f(u)|\; \backslash leq\; m\_u\; \backslash leq\; 1,$ as desired.

$\backslash blacksquare$

that there exist $X$-indexed non-negative real numbers $m\_\{\backslash bull\}\; =\; \backslash left(m\_x\backslash right)\_\{x\; \backslash in\; X\}$ such that

$$\backslash begin\{alignat\}\{4\}$$

U^{\circ}

&= U^{\#} && \\

&= X^{\#} && \cap \prod_{x \in X} B_{m_x} \\

&= X^{\prime} && \cap \prod_{x \in X} B_{m_x} \\

\end{alignat}

where these real numbers $m\_\{\backslash bull\}$ can also be chosen to be "minimal" in the following sense:

using $P\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; U^\{\backslash circ\}$ (so $P\; =\; U^\{\backslash \#\}$ as in the proof) and defining the notation $\backslash prod\; B\_\{R\_\backslash bull\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash prod\_\{x\; \backslash in\; X\}\; B\_\{R\_x\}$ for any $R\_\{\backslash bull\}\; =\; \backslash left(R\_x\backslash right)\_\{x\; \backslash in\; X\}\; \backslash in\; \backslash R^X,$ if

$$T\_P\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left\backslash \{R\_\{\backslash bull\}\; \backslash in\; \backslash R^X\; ~:~\; P\; \backslash subseteq\; \backslash prod\; B\_\{R\_\backslash bull\}\backslash right\backslash \}$$

then $m\_\{\backslash bull\}\; \backslash in\; T\_P$ and for every $x\; \backslash in\; X,$ $m\_x\; =\; \backslash inf\; \backslash left\backslash \{\; R\_x\; :\; R\_\{\backslash bull\}\; \backslash in\; T\_P\; \backslash right\backslash \},$

which shows that these numbers $m\_\{\backslash bull\}$ are unique; indeed, this infimum formula can be used to define them.

In fact, if $\backslash operatorname\{Box\}\_P$ denotes the set of all such products of closed balls containing the polar set $P,$

$$\backslash operatorname\{Box\}\_P\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left\backslash \{\; \backslash prod\; B\_\{R\_\backslash bull\}\; ~:~\; R\_\{\backslash bull\}\; \backslash in\; T\_P\; \backslash right\backslash \}\; ~=~\; \backslash left\backslash \{\; \backslash prod\; B\_\{R\_\backslash bull\}\; ~:~\; P\; \backslash subseteq\; \backslash prod\; B\_\{R\_\backslash bull\}\; \backslash right\backslash \},$$

then

$\backslash prod\; B\_\{m\_\backslash bull\}\; =\; \backslash cap\; \backslash operatorname\{Box\}\_P\; \backslash in\; \backslash operatorname\{Box\}\_P$

where $\backslash bigcap\; \backslash operatorname\{Box\}\_P$ denotes the intersection of all sets belonging to $\backslash operatorname\{Box\}\_P.$

This implies (among other thingsThis tuple $m\_\{\backslash bull\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left(m\_x\backslash right)\_\{x\; \backslash in\; X\}$ is the least element of $T\_P$ with respect to natural induced pointwise partial order defined by $R\_\{\backslash bull\}\; \backslash leq\; S\_\{\backslash bull\}$ if and only if $R\_x\; \backslash leq\; S\_x$ for every $x\; \backslash in\; X.$ Thus, every neighborhood $U$ of the origin in $X$ can be associated with this unique (minimum) function $m\_\{\backslash bull\}\; :\; X\; \backslash to\; [0,\; \backslash infty).$ For any $x\; \backslash in\; X,$ if $r\; >\; 0$ is such that $x\; \backslash in\; r\; U$ then $m\_x\; \backslash leq\; r$ so that in particular, $m\_0\; =\; 0$ and $m\_u\; \backslash leq\; 1$ for every $u\; \backslash in\; U.$)

that $\backslash prod\; B\_\{m\_\backslash bull\}\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; B\_\{m\_x\}$ the unique least element of $\backslash operatorname\{Box\}\_P$ with respect to $\backslash ,\backslash subseteq;$ this may be used as an alternative definition of this (necessarily convex and balanced) set.

The function $m\_\{\backslash bull\}\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; \backslash left(m\_x\backslash right)\_\{x\; \backslash in\; X\}\; :\; X\; \backslash to\; [0,\; \backslash infty)$ is a seminorm and it is unchanged if $U$ is replaced by the convex balanced hull of $U$ (because $U^\{\backslash \#\}\; =\; [\backslash operatorname\{cobal\}\; U]^\{\backslash \#\}$).

Similarly, because $U^\{\backslash circ\}\; =\; \backslash left[\backslash operatorname\{cl\}\_X\; U\backslash right]^\{\backslash circ\},$ $m\_\{\backslash bull\}$ is also unchanged if $U$ is replaced by its closure in $X.$

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 $X^\{\backslash prime\}.$ Since $X$ is separable, let $x\_\{\backslash bull\}\; =\; \backslash left(x\_n\backslash right)\_\{n=1\}^\{\backslash infty\}$ be a countable dense subset.

Then the following defines a metric, where for any $x,\; y\; \backslash in\; B$

$$\backslash rho(x,y)\; =\; \backslash sum\_\{n=1\}^\backslash infty\; \backslash ,\; 2^\{-n\}\; \backslash ,\; \backslash frac\{\backslash left|\backslash langle\; x\; -\; y,\; x\_n\; \backslash rangle\backslash right|\}\{1\; +\; \backslash left|\backslash langle\; x\; -\; y,\; x\_n\; \backslash rangle\backslash right|\}$$

in which $\backslash langle\backslash cdot,\; \backslash cdot\backslash rangle$ denotes the duality pairing of $X^\{\backslash prime\}$ 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 $F\; :\; X^\{\backslash prime\}\; \backslash to\; \backslash R$ on the dual of a separable normed vector space $X,$ one common strategy is to first construct a minimizing sequence $x\_1,\; x\_2,\; \backslash ldots\; \backslash in\; X^\{\backslash prime\}$ 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 $X^\{\backslash prime\}$ is the space of finite Radon measures on the real line (so that $X\; =\; C\_0(\backslash R)$ 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.

{{math proof|drop=hidden|proof=

For every $x\; \backslash in\; X,$ let

$$D\_x\; =\; \backslash \{c\; \backslash in\; \backslash Complex\; :\; |c|\; \backslash leq\; \backslash |x\backslash |\backslash \}$$

and let

$$D\; =\; \backslash prod\_\{x\; \backslash in\; X\}\; D\_x$$

be endowed with the product topology.

Because every $D\_x$ is a compact subset of the complex plane, Tychonoff's theorem guarantees that their product $D$ is compact.

The closed unit ball in $X^\{\backslash prime\},$ denoted by $B\_1^\{\backslash ,\backslash prime\},$ can be identified as a subset of $D$ in a natural way:

$$\backslash begin\{alignat\}\{4\}$$

F :\;&& B_1^{\,\prime} &&\;\to \;& D \\[0.3ex]

&& f &&\;\mapsto\;& (f(x))_{x \in X}. \\

\end{alignat}

This map is injective and it is continuous when $B\_1^\{\backslash ,\backslash prime\}$ 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 $\backslash lambda\_\{\backslash bull\}\; =\; \backslash left(\backslash lambda\_x\backslash right)\_\{x\; \backslash in\; X\}\; \backslash in\; D$ and a net $\backslash left(f\_i(x)\backslash right)\_\{x\; \backslash in\; X\}$ in the image of $F$ indexed by $i\; \backslash in\; I$ such that

$$\backslash lim\_\{i\}\; \backslash left(f\_i(x)\backslash right)\_\{x\; \backslash in\; X\}\; \backslash to\; \backslash lambda\_\{\backslash bull\}\; \backslash quad\; \backslash text\{\; in\; \}\; D,$$

the functional $g\; :\; X\; \backslash to\; \backslash Complex$ defined by

$$g(x)\; =\; \backslash lambda\_x\; \backslash qquad\; \backslash text\{\; for\; every\; \}\; x\; \backslash in\; X,$$

lies in $B\_1^\{\backslash ,\backslash prime\}$ and $F(g)\; =\; \backslash lambda\_\{\backslash bull\}.$

$\backslash blacksquare$

}}

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

• The closed unit ball in $X^\{\backslash prime\}$ is weak-* compact.{{sfn|Narici|Beckenstein|2011|pp=225-273}} So if $X^\{\backslash prime\}$ 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).

• A Banach space is reflexive if and only if its closed unit ball is $\backslash sigma\backslash left(X,\; X^\{\backslash prime\}\backslash right)$-compact; this is known as James' theorem.{{sfn|Narici|Beckenstein|2011|pp=225-273}}
• 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 $L^p(\backslash mu)$ where and let $q$ satisfy $\backslash frac\{1\}\{p\}\; +\; \backslash frac\{1\}\{q\}\; =\; 1.$

  Let $f\_1,\; f\_2,\; \backslash ldots$ be a bounded sequence of functions in $X.$

  Then there exists a subsequence $\backslash left(f\_\{n\_k\}\backslash right)\_\{k=1\}^\{\backslash infty\}$ and an $f\; \backslash in\; X$ such that

  $$\backslash int\; f\_\{n\_k\}\; g\backslash ,d\backslash mu\; \backslash to\; \backslash int\; f\; g\backslash ,d\backslash mu\; \backslash qquad\; \backslash text\{\; for\; all\; \}\; g\; \backslash in\; L^q(\backslash mu)\; =\; X^\{\backslash prime\}.$$

  The corresponding result for $p\; =\; 1$ is not true, as $L^1(\backslash mu)$ is not reflexive.

• 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.

{{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 case (see 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 spaces (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{{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}} 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}}|math_statement=

Let $X$ be a normed space and let $B$ denote the closed unit ball of its continuous dual space $X^\{\backslash prime\}.$ Then $B$ has the following property, which is called (weak-*) {{em|{{visible anchor|quasicompactness}}}} or {{em|{{visible anchor|convex compactness}}}}: whenever $\backslash mathcal\{C\}$ is a cover of $B$ by {{em|convex}} weak-* closed subsets of $X^\{\backslash prime\}$ such that $\backslash \{B\; \backslash cap\; C\; :\; C\; \backslash in\; \backslash mathcal\{C\}\backslash \}$ has the finite intersection property, then $B\; \backslash cap\; \backslash bigcap\_\{C\; \backslash in\; \backslash mathcal\{C\}\}\; C$ is not empty.

}}

Compactness implies convex compactness because a topological space is compact if and only if every family of closed subsets having the finite intersection property (FIP) has non-empty intersection.

The definition of convex compactness is similar to this characterization of compact spaces in terms of the FIP, except that it only involves those closed subsets that are also convex (rather than all closed subsets).

• {{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}}

{{reflist|group=note}}

Proofs

{{reflist|group=proof}}

{{reflist}}

{{refbegin}}

• {{Köthe Topological Vector Spaces I}}
• {{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}}
• {{Rudin Walter Functional Analysis|edition=2}} See Theorem 3.15, p. 68.
• {{Schaefer Wolff Topological Vector Spaces|edition=2}}
• {{Schechter Handbook of Analysis and Its Foundations}}
• {{Trèves François Topological vector spaces, distributions and kernels}}

{{refend}}

• {{Conway A Course in Functional Analysis|edition=2}}
• {{Lax Functional Analysis}}

{{Duality and spaces of linear maps}}

{{Functional analysis}}

{{Topological vector spaces}}

{{DEFAULTSORT:Banach-Alaoglu theorem}}

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