Hahn–Banach theorem#Hahn–Banach separation theorem

The theorem is named for the mathematicians Hans Hahn and Stefan Banach, who proved it independently in the late 1920s.

The special case of the theorem for the space $C[a,\; b]$ of continuous functions on an interval was proved earlier (in 1912) by Eduard Helly,{{MacTutor Biography|id=Helly}} and a more general extension theorem, the M. Riesz extension theorem, from which the Hahn–Banach theorem can be derived, was proved in 1923 by Marcel Riesz.See M. Riesz extension theorem.

According to {{cite journal|mr=0256837|last=Gårding|first=L.|author-link=Lars Gårding|title=Marcel Riesz in memoriam|journal=Acta Math.|volume=124|year=1970|issue=1|pages=I–XI |doi=10.1007/bf02394565|doi-access=free}}, the argument was known to Riesz already in 1918.

The first Hahn–Banach theorem was proved by Eduard Helly in 1912 who showed that certain linear functionals defined on a subspace of a certain type of normed space ($\backslash Complex^\{\backslash N\}$) had an extension of the same norm. Helly did this through the technique of first proving that a one-dimensional extension exists (where the linear functional has its domain extended by one dimension) and then using induction. In 1927, Hahn defined general Banach spaces and used Helly's technique to prove a norm-preserving version of Hahn–Banach theorem for Banach spaces (where a bounded linear functional on a subspace has a bounded linear extension of the same norm to the whole space). In 1929, Banach, who was unaware of Hahn's result, generalized it by replacing the norm-preserving version with the dominated extension version that uses sublinear functions. Whereas Helly's proof used mathematical induction, Hahn and Banach both used transfinite induction.{{sfn|Narici|Beckenstein|2011|pp=177-220}}

The Hahn–Banach theorem arose from attempts to solve infinite systems of linear equations. This is needed to solve problems such as the moment problem, whereby given all the potential moments of a function one must determine if a function having these moments exists, and, if so, find it in terms of those moments. Another such problem is the Fourier cosine series problem, whereby given all the potential Fourier cosine coefficients one must determine if a function having those coefficients exists, and, again, find it if so.

Riesz and Helly solved the problem for certain classes of spaces (such as $L^p([0,\; 1])$ and $C([a,\; b])$) where they discovered that the existence of a solution was equivalent to the existence and continuity of certain linear functionals. In effect, they needed to solve the following problem:{{sfn|Narici|Beckenstein|2011|pp=177-220}}

:({{visible anchor|The vector problem}}) Given a collection $\backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}$ of bounded linear functionals on a normed space $X$ and a collection of scalars $\backslash left(c\_i\backslash right)\_\{i\; \backslash in\; I\},$ determine if there is an $x\; \backslash in\; X$ such that $f\_i(x)\; =\; c\_i$ for all $i\; \backslash in\; I.$

If $X$ happens to be a reflexive space then to solve the vector problem, it suffices to solve the following dual problem:{{sfn|Narici|Beckenstein|2011|pp=177-220}}

:(The functional problem) Given a collection $\backslash left(x\_i\backslash right)\_\{i\; \backslash in\; I\}$ of vectors in a normed space $X$ and a collection of scalars $\backslash left(c\_i\backslash right)\_\{i\; \backslash in\; I\},$ determine if there is a bounded linear functional $f$ on $X$ such that $f\backslash left(x\_i\backslash right)\; =\; c\_i$ for all $i\; \backslash in\; I.$

Riesz went on to define $L^p([0,\; 1])$ space () in 1910 and the $\backslash ell^p$ spaces in 1913. While investigating these spaces he proved a special case of the Hahn–Banach theorem. Helly also proved a special case of the Hahn–Banach theorem in 1912. In 1910, Riesz solved the functional problem for some specific spaces and in 1912, Helly solved it for a more general class of spaces. It wasn't until 1932 that Banach, in one of the first important applications of the Hahn–Banach theorem, solved the general functional problem. The following theorem states the general functional problem and characterizes its solution.{{sfn|Narici|Beckenstein|2011|pp=177-220}}

{{Math theorem|name=Theorem{{sfn|Narici|Beckenstein|2011|pp=177-220}}|note=The functional problem|math_statement=

Let $\backslash left(x\_i\backslash right)\_\{i\; \backslash in\; I\}$ be vectors in a real or complex normed space $X$ and let $\backslash left(c\_i\backslash right)\_\{i\; \backslash in\; I\}$ be scalars also indexed by $I\; \backslash neq\; \backslash varnothing.$

There exists a continuous linear functional $f$ on $X$ such that $f\backslash left(x\_i\backslash right)\; =\; c\_i$ for all $i\; \backslash in\; I$ if and only if there exists a $K\; >\; 0$ such that for any choice of scalars $\backslash left(s\_i\backslash right)\_\{i\; \backslash in\; I\}$ where all but finitely many $s\_i$ are $0,$ the following holds:

$$\backslash left|\backslash sum\_\{i\; \backslash in\; I\}\; s\_i\; c\_i\backslash right|\; \backslash leq\; K\; \backslash left\backslash |\backslash sum\_\{i\; \backslash in\; I\}\; s\_i\; x\_i\backslash right\backslash |.$$

}}

The Hahn–Banach theorem can be deduced from the above theorem.{{sfn|Narici|Beckenstein|2011|pp=177-220}} If $X$ is reflexive then this theorem solves the vector problem.

A real-valued function $f\; :\; M\; \backslash to\; \backslash R$ defined on a subset $M$ of $X$ is said to be {{em|{{visible anchor|dominated real functional|text=dominated (above) by}}}} a function $p\; :\; X\; \backslash to\; \backslash R$ if $f(m)\; \backslash leq\; p(m)$ for every $m\; \backslash in\; M.$

For this reason, the following version of the Hahn–Banach theorem is called {{em|the dominated extension theorem}}.

{{Math theorem

| name = {{visible anchor|Hahn–Banach dominated extension theorem}} (for real linear functionals){{sfn|Rudin|1991|pp=56-62}}{{harvnb|Rudin|1991}}, Th. 3.2{{sfn|Narici|Beckenstein|2011|pp=177-183}}

| math_statement =

If $p\; :\; X\; \backslash to\; \backslash R$ is a sublinear function (such as a norm or seminorm for example) defined on a real vector space $X$ then any linear functional defined on a vector subspace of $X$ that is dominated above by $p$ has at least one linear extension to all of $X$ that is also dominated above by $p.$

Explicitly, if $p\; :\; X\; \backslash to\; \backslash R$ is a sublinear function, which by definition means that it satisfies

$$p(x\; +\; y)\; \backslash leq\; p(x)\; +\; p(y)\; \backslash quad\; \backslash text\{\; and\; \}\; \backslash quad\; p(t\; x)\; =\; t\; p(x)\; \backslash qquad\; \backslash text\{\; for\; all\; \}\; \backslash ;\; x,\; y\; \backslash in\; X\; \backslash ;\; \backslash text\{\; and\; all\; real\; \}\; \backslash ;\; t\; \backslash geq\; 0,$$

and if $f\; :\; M\; \backslash to\; \backslash R$ is a linear functional defined on a vector subspace $M$ of $X$ such that

$$f(m)\; \backslash leq\; p(m)\; \backslash quad\; \backslash text\{\; for\; all\; \}\; m\; \backslash in\; M$$

then there exists a linear functional $F\; :\; X\; \backslash to\; \backslash R$ such that

$$F(m)\; =\; f(m)\; \backslash quad\; \backslash text\{\; for\; all\; \}\; m\; \backslash in\; M,$$

$$F(x)\; \backslash leq\; p(x)\; \backslash quad\; ~\backslash ;\backslash ,\; \backslash text\{\; for\; all\; \}\; x\; \backslash in\; X.$$

Moreover, if $p$ is a seminorm then $|F(x)|\; \backslash leq\; p(x)$ necessarily holds for all $x\; \backslash in\; X.$

}}

The theorem remains true if the requirements on $p$ are relaxed to require only that $p$ be a convex function:{{Sfn|Schechter|1996|pp=318-319}}{{Sfn|Reed|Simon|1980|p=}}

A function $p\; :\; X\; \backslash to\; \backslash R$ is convex and satisfies $p(0)\; \backslash leq\; 0$ if and only if $p(a\; x\; +\; b\; y)\; \backslash leq\; a\; p(x)\; +\; b\; p(y)$ for all vectors $x,\; y\; \backslash in\; X$ and all non-negative real $a,\; b\; \backslash geq\; 0$ such that $a\; +\; b\; \backslash leq\; 1.$ Every sublinear function is a convex function.

On the other hand, if $p\; :\; X\; \backslash to\; \backslash R$ is convex with $p(0)\; \backslash geq\; 0,$ then the function defined by $p\_0(x)\; \backslash ;\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}\backslash ;\; \backslash inf\_\{t\; >\; 0\}\; \backslash frac\{p(tx)\}\{t\}$ is positively homogeneous

(because for all $x$ and $r>0$ one has $p\_0(rx)=\backslash inf\_\{t\; >\; 0\}\; \backslash frac\{p(trx)\}\{t\}\; =r\backslash inf\_\{t\; >\; 0\}\; \backslash frac\{p(trx)\}\{tr\}\; =\; r\backslash inf\_\{\backslash tau\; >\; 0\}\; \backslash frac\{p(\backslash tau\; x)\}\{\backslash tau\}=rp\_0(x)$), hence, being convex, it is sublinear. It is also bounded above by $p\_0\; \backslash leq\; p,$ and satisfies $F\; \backslash leq\; p\_0$ for every linear functional $F\; \backslash leq\; p.$ So the extension of the Hahn–Banach theorem to convex functionals does not have a much larger content than the classical one stated for sublinear functionals.

If $F\; :\; X\; \backslash to\; \backslash R$ is linear then $F\; \backslash leq\; p$ if and only if{{sfn|Rudin|1991|pp=56-62}} $$-p(-x)\; \backslash leq\; F(x)\; \backslash leq\; p(x)\; \backslash quad\; \backslash text\{\; for\; all\; \}\; x\; \backslash in\; X,$$

which is the (equivalent) conclusion that some authors{{sfn|Rudin|1991|pp=56-62}} write instead of $F\; \backslash leq\; p.$

It follows that if $p\; :\; X\; \backslash to\; \backslash R$ is also {{em|symmetric}}, meaning that $p(-x)\; =\; p(x)$ holds for all $x\; \backslash in\; X,$ then $F\; \backslash leq\; p$ if and only $|F|\; \backslash leq\; p.$

Every norm is a seminorm and both are symmetric balanced sublinear functions. A sublinear function is a seminorm if and only if it is a balanced function. On a real vector space (although not on a complex vector space), a sublinear function is a seminorm if and only if it is symmetric. The identity function $\backslash R\; \backslash to\; \backslash R$ on $X\; :=\; \backslash R$ is an example of a sublinear function that is not a seminorm.

The dominated extension theorem for real linear functionals implies the following alternative statement of the Hahn–Banach theorem that can be applied to linear functionals on real or complex vector spaces.

{{Math theorem

| name = {{visible anchor|Hahn–Banach theorem for real or complex vector spaces|text=Hahn–Banach theorem}}{{sfn|Narici|Beckenstein|2011|pp=177-220}}{{harvnb|Rudin|1991}}, Th. 3.2

| math_statement = Suppose $p\; :\; X\; \backslash to\; \backslash R$ a seminorm on a vector space $X$ over the field $\backslash mathbf\{K\},$ which is either $\backslash R$ or $\backslash Complex.$

If $f\; :\; M\; \backslash to\; \backslash mathbf\{K\}$ is a linear functional on a vector subspace $M$ such that

$$|f(m)|\; \backslash leq\; p(m)\; \backslash quad\; \backslash text\{\; for\; all\; \}\; m\; \backslash in\; M,$$

then there exists a linear functional $F\; :\; X\; \backslash to\; \backslash mathbf\{K\}$ such that

$$F(m)\; =\; f(m)\; \backslash quad\; \backslash ;\; \backslash text\{\; for\; all\; \}\; m\; \backslash in\; M,$$

$$|F(x)|\; \backslash leq\; p(x)\; \backslash quad\; \backslash ;\backslash ,\; \backslash text\{\; for\; all\; \}\; x\; \backslash in\; X.$$

}}

The theorem remains true if the requirements on $p$ are relaxed to require only that for all $x,\; y\; \backslash in\; X$ and all scalars $a$ and $b$ satisfying $|a|\; +\; |b|\; \backslash leq\; 1,${{Sfn|Reed|Simon|1980|p=}}

$$p(a\; x\; +\; b\; y)\; \backslash leq\; |a|\; p(x)\; +\; |b|\; p(y).$$

This condition holds if and only if $p$ is a convex and balanced function satisfying $p(0)\; \backslash leq\; 0,$ or equivalently, if and only if it is convex, satisfies $p(0)\; \backslash leq\; 0,$ and $p(u\; x)\; \backslash leq\; p(x)$ for all $x\; \backslash in\; X$ and all unit length scalars $u.$

A complex-valued functional $F$ is said to be {{em|{{visible anchor|dominated complex functional|text=dominated by $p$}}}} if $|F(x)|\; \backslash leq\; p(x)$ for all $x$ in the domain of $F.$

With this terminology, the above statements of the Hahn–Banach theorem can be restated more succinctly:

:Hahn–Banach dominated extension theorem: If $p\; :\; X\; \backslash to\; \backslash R$ is a seminorm defined on a real or complex vector space $X,$ then every dominated linear functional defined on a vector subspace of $X$ has a dominated linear extension to all of $X.$ In the case where $X$ is a real vector space and $p\; :\; X\; \backslash to\; \backslash R$ is merely a convex or sublinear function, this conclusion will remain true if both instances of "dominated" (meaning $|F|\; \backslash leq\; p$) are weakened to instead mean "#dominated real functional" (meaning $F\; \backslash leq\; p$).{{Sfn|Schechter|1996|pp=318-319}}{{Sfn|Reed|Simon|1980|p=}}

Proof

The following observations allow the Hahn–Banach theorem for real vector spaces to be applied to (complex-valued) linear functionals on complex vector spaces.

Every linear functional $F\; :\; X\; \backslash to\; \backslash Complex$ on a complex vector space is completely determined by its real part $\backslash ;\; \backslash operatorname\{Re\}\; F\; :\; X\; \backslash to\; \backslash R\; \backslash ;$ through the formula{{sfn|Narici|Beckenstein|2011|pp=177-183}}If $z\; =\; a\; +\; i\; b\; \backslash in\; \backslash Complex$ has real part $\backslash operatorname\{Re\}\; z\; =\; a$ then $-\; \backslash operatorname\{Re\}\; (i\; z)\; =\; b,$ which proves that $z\; =\; \backslash operatorname\{Re\}\; z\; -\; i\; \backslash operatorname\{Re\}\; (i\; z).$ Substituting $F(x)$ in for $z$ and using $i\; F(x)\; =\; F(i\; x)$ gives $F(x)\; =\; \backslash operatorname\{Re\}\; F(x)\; -\; i\; \backslash operatorname\{Re\}\; F(i\; x).$ $\backslash blacksquare$

$$F(x)\; \backslash ;=\backslash ;\; \backslash operatorname\{Re\}\; F(x)\; -\; i\; \backslash operatorname\{Re\}\; F(i\; x)\; \backslash qquad\; \backslash text\{\; for\; all\; \}\; x\; \backslash in\; X$$

and moreover, if $\backslash |\backslash cdot\backslash |$ is a norm on $X$ then their dual norms are equal: $\backslash |F\backslash |\; =\; \backslash |\backslash operatorname\{Re\}\; F\backslash |.${{sfn|Narici|Beckenstein|2011|pp=126-128}}

In particular, a linear functional on $X$ extends another one defined on $M\; \backslash subseteq\; X$ if and only if their real parts are equal on $M$ (in other words, a linear functional $F$ extends $f$ if and only if $\backslash operatorname\{Re\}\; F$ extends $\backslash operatorname\{Re\}\; f$).

The real part of a linear functional on $X$ is always a {{visible anchor|real-linear functional}} (meaning that it is linear when $X$ is considered as a real vector space) and if $R\; :\; X\; \backslash to\; \backslash R$ is a real-linear functional on a complex vector space then $x\; \backslash mapsto\; R(x)\; -\; i\; R(i\; x)$ defines the unique linear functional on $X$ whose real part is $R.$

If $F$ is a linear functional on a (complex or real) vector space $X$ and if $p\; :\; X\; \backslash to\; \backslash R$ is a seminorm then{{sfn|Narici|Beckenstein|2011|pp=177-183}}Let $F$ be any homogeneous scalar-valued map on $X$ (such as a linear functional) and let $p\; :\; X\; \backslash to\; \backslash R$ be any map that satisfies $p(u\; x)\; =\; p(x)$ for all $x$ and unit length scalars $u$ (such as a seminorm). If $|F|\; \backslash leq\; p$ then $\backslash operatorname\{Re\}\; F\; \backslash leq\; |\backslash operatorname\{Re\}\; F|\; \backslash leq\; |F|\; \backslash leq\; p.$ For the converse, assume $\backslash operatorname\{Re\}\; F\; \backslash leq\; p$ and fix $x\; \backslash in\; X.$ Let $r\; =\; |F(x)|$ and pick any $\backslash theta\; \backslash in\; \backslash R$ such that $F(x)\; =\; r\; e^\{i\; \backslash theta\};$ it remains to show $r\; \backslash leq\; p(x).$ Homogeneity of $F$ implies $F\backslash left(e^\{-i\; \backslash theta\}\; x\backslash right)\; =\; r$ is real so that $\backslash operatorname\{Re\}\; F\backslash left(e^\{-i\; \backslash theta\}\; x\backslash right)\; =\; F\backslash left(e^\{-i\; \backslash theta\}\; x\backslash right).$ By assumption, $\backslash operatorname\{Re\}\; F\; \backslash leq\; p$ and $p\backslash left(e^\{-i\; \backslash theta\}\; x\backslash right)\; =\; p(x),$ so that $r\; =\; \backslash operatorname\{Re\}\; F\backslash left(e^\{-i\; \backslash theta\}\; x\backslash right)\; \backslash leq\; p\backslash left(e^\{-i\; \backslash theta\}\; x\backslash right)\; =\; p(x),$ as desired. $\backslash blacksquare$

$$|F|\; \backslash ,\backslash leq\backslash ,\; p\; \backslash quad\; \backslash text\{\; if\; and\; only\; if\; \}\; \backslash quad\; \backslash operatorname\{Re\}\; F\; \backslash ,\backslash leq\backslash ,\; p.$$

Stated in simpler language, a linear functional is dominated by a seminorm $p$ if and only if its real part is dominated above by $p.$

{{Math proof|title=Proof of Hahn–Banach for complex vector spaces by reduction to real vector spaces{{sfn|Narici|Beckenstein|2011|pp=177-220}}|drop=hidden|proof=

Suppose $p\; :\; X\; \backslash to\; \backslash R$ is a seminorm on a complex vector space $X$ and let $f\; :\; M\; \backslash to\; \backslash Complex$ be a linear functional defined on a vector subspace $M$ of $X$ that satisfies $|f|\; \backslash leq\; p$ on $M.$

Consider $X$ as a real vector space and apply the Hahn–Banach theorem for real vector spaces to the real-linear functional $\backslash ;\; \backslash operatorname\{Re\}\; f\; :\; M\; \backslash to\; \backslash R\; \backslash ;$ to obtain a real-linear extension $R\; :\; X\; \backslash to\; \backslash R$ that is also dominated above by $p,$ so that it satisfies $R\; \backslash leq\; p$ on $X$ and $R\; =\; \backslash operatorname\{Re\}\; f$ on $M.$

The map $F\; :\; X\; \backslash to\; \backslash Complex$ defined by $F(x)\; \backslash ;=\backslash ;\; R(x)\; -\; i\; R(i\; x)$ is a linear functional on $X$ that extends $f$ (because their real parts agree on $M$) and satisfies $|F|\; \backslash leq\; p$ on $X$ (because $\backslash operatorname\{Re\}\; F\; \backslash leq\; p$ and $p$ is a seminorm).

$\backslash blacksquare$

}}

The proof above shows that when $p$ is a seminorm then there is a one-to-one correspondence between dominated linear extensions of $f\; :\; M\; \backslash to\; \backslash Complex$ and dominated real-linear extensions of $\backslash operatorname\{Re\}\; f\; :\; M\; \backslash to\; \backslash R;$ the proof even gives a formula for explicitly constructing a linear extension of $f$ from any given real-linear extension of its real part.

Continuity

A linear functional $F$ on a topological vector space is continuous if and only if this is true of its real part $\backslash operatorname\{Re\}\; F;$ if the domain is a normed space then $\backslash |F\backslash |\; =\; \backslash |\backslash operatorname\{Re\}\; F\backslash |$ (where one side is infinite if and only if the other side is infinite).{{sfn|Narici|Beckenstein|2011|pp=126-128}}

Assume $X$ is a topological vector space and $p\; :\; X\; \backslash to\; \backslash R$ is sublinear function.

If $p$ is a continuous sublinear function that dominates a linear functional $F$ then $F$ is necessarily continuous.{{sfn|Narici|Beckenstein|2011|pp=177-183}} Moreover, a linear functional $F$ is continuous if and only if its absolute value $|F|$ (which is a seminorm that dominates $F$) is continuous.{{sfn|Narici|Beckenstein|2011|pp=177-183}} In particular, a linear functional is continuous if and only if it is dominated by some continuous sublinear function.

The Hahn–Banach theorem for real vector spaces ultimately follows from Helly's initial result for the special case where the linear functional is extended from $M$ to a larger vector space in which $M$ has codimension $1.${{sfn|Narici|Beckenstein|2011|pp=177-220}}

{{Math theorem

| name = Lemma{{sfn|Narici|Beckenstein|2011|pp=177-183}}

| note = {{visible anchor|One–dimensional dominated extension theorem}}

| math_statement = Let $p\; :\; X\; \backslash to\; \backslash R$ be a sublinear function on a real vector space $X,$ let $f\; :\; M\; \backslash to\; \backslash R$ a linear functional on a proper vector subspace $M\; \backslash subsetneq\; X$ such that $f\; \backslash leq\; p$ on $M$ (meaning $f(m)\; \backslash leq\; p(m)$ for all $m\; \backslash in\; M$), and let $x\; \backslash in\; X$ be a vector {{em|not}} in $M$ (so $M\; \backslash oplus\; \backslash R\; x\; =\; \backslash operatorname\{span\}\; \backslash \{M,\; x\backslash \}$).

There exists a linear extension $F\; :\; M\; \backslash oplus\; \backslash R\; x\; \backslash to\; \backslash R$ of $f$ such that $F\; \backslash leq\; p$ on $M\; \backslash oplus\; \backslash R\; x.$

}}

{{Math proof|title=Proof{{sfn|Narici|Beckenstein|2011|pp=177-183}}|drop=hidden|proof=

Given any real number $b,$ the map $F\_b\; :\; M\; \backslash oplus\; \backslash R\; x\; \backslash to\; \backslash R$ defined by $F\_b(m\; +\; r\; x)\; =\; f(m)\; +\; r\; b$ is always a linear extension of $f$ to $M\; \backslash oplus\; \backslash R\; x$This definition means, for instance, that $F\_b(x)\; =\; F\_b(0\; +\; 1\; x)\; =\; f(0)\; +\; 1\; b\; =\; b$ and if $m\; \backslash in\; M$ then $F\_b(m)\; =\; F\_b(m\; +\; 0\; x)\; =\; f(m)\; +\; 0\; b\; =\; f(m).$ In fact, if $G\; :\; M\; \backslash oplus\; \backslash R\; x\; \backslash to\; \backslash R$ is any linear extension of $f$ to $M\; \backslash oplus\; \backslash R\; x$ then $G\; =\; F\_b$ for $b\; :=\; G(x).$ In other words, every linear extension of $f$ to $M\; \backslash oplus\; \backslash R\; x$ is of the form $F\_b$ for some (unique) $b.$ but it might not satisfy $F\_b\; \backslash leq\; p.$

It will be shown that $b$ can always be chosen so as to guarantee that $F\_b\; \backslash leq\; p,$ which will complete the proof.

If $m,\; n\; \backslash in\; M$ then

$$f(m)\; -\; f(n)\; =\; f(m\; -\; n)\; \backslash leq\; p(m\; -\; n)\; =\; p(m\; +\; x\; -\; x\; -\; n)\; \backslash leq\; p(m\; +\; x)\; +\; p(-\; x\; -\; n)$$

which implies

$$-p(-n\; -\; x)\; -\; f(n)\; ~\backslash leq~\; p(m\; +\; x)\; -\; f(m).$$

So define

$$a\; =\; \backslash sup\_\{n\; \backslash in\; M\}[-p(-n\; -\; x)\; -\; f(n)]\; \backslash qquad\; \backslash text\{\; and\; \}\; \backslash qquad\; c\; =\; \backslash inf\_\{m\; \backslash in\; M\}\; [p(m\; +\; x)\; -\; f(m)]$$

where $a\; \backslash leq\; c$ are real numbers.

To guarantee $F\_b\; \backslash leq\; p,$ it suffices that $a\; \backslash leq\; b\; \backslash leq\; c$ (in fact, this is also necessaryExplicitly, for any real number $b\; \backslash in\; \backslash R,$ $F\_b\; \backslash leq\; p$ on $M\; \backslash oplus\; \backslash R\; x$ if and only if $a\; \backslash leq\; b\; \backslash leq\; c.$ Combined with the fact that $F\_b(x)\; =\; b,$ it follows that the dominated linear extension of $f$ to $M\; \backslash oplus\; \backslash R\; x$ is unique if and only if $a\; =\; c,$ in which case this scalar will be the extension's values at $x.$ Since every linear extension of $f$ to $M\; \backslash oplus\; \backslash R\; x$ is of the form $F\_b$ for some $b,$ the bounds $a\; \backslash leq\; b\; =\; F\_b(x)\; \backslash leq\; c$ thus also limit the range of possible values (at $x$) that can be taken by any of $f$'s dominated linear extensions. Specifically, if $F\; :\; X\; \backslash to\; \backslash R$ is any linear extension of $f$ satisfying $F\; \backslash leq\; p$ then for every $x\; \backslash in\; X\; \backslash setminus\; M,$ $\backslash sup\_\{m\; \backslash in\; M\}[-p(-m\; -\; x)\; -\; f(m)]\; ~\backslash leq~\; F(x)\; ~\backslash leq~\; \backslash inf\_\{m\; \backslash in\; M\}\; [p(m\; +\; x)\; -\; f(m)].$) because then $b$ satisfies "the decisive inequality"{{sfn|Narici|Beckenstein|2011|pp=177-183}}

$$-p(-n\; -\; x)\; -\; f(n)\; ~\backslash leq~\; b\; ~\backslash leq~\; p(m\; +\; x)\; -\; f(m)\; \backslash qquad\; \backslash text\{\; for\; all\; \}\backslash ;\; m,\; n\; \backslash in\; M.$$

To see that $f(m)\; +\; r\; b\; \backslash leq\; p(m\; +\; r\; x)$ follows, assume $r\; \backslash neq\; 0$ and substitute $\backslash tfrac\{1\}\{r\}\; m$ in for both $m$ and $n$ to obtain

$$-p\backslash left(-\; \backslash tfrac\{1\}\{r\}\; m\; -\; x\backslash right)\; -\; \backslash tfrac\{1\}\{r\}\; f\backslash left(m\backslash right)\; ~\backslash leq~\; b\; ~\backslash leq~\; p\backslash left(\backslash tfrac\{1\}\{r\}\; m\; +\; x\backslash right)\; -\; \backslash tfrac\{1\}\{r\}\; f\backslash left(m\backslash right).$$

If $r\; >\; 0$ (respectively, if ) then the right (respectively, the left) hand side equals $\backslash tfrac\{1\}\{r\}\; \backslash left[p(m\; +\; r\; x)\; -\; f(m)\backslash right]$ so that multiplying by $r$ gives $r\; b\; \backslash leq\; p(m\; +\; r\; x)\; -\; f(m).$

$\backslash blacksquare$

}}

This lemma remains true if $p\; :\; X\; \backslash to\; \backslash R$ is merely a convex function instead of a sublinear function.{{Sfn|Schechter|1996|pp=318-319}}{{Sfn|Reed|Simon|1980|p=}}

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

Assume that $p$ is convex, which means that $p(t\; y\; +\; (1\; -\; t)\; z)\; \backslash leq\; t\; p(y)\; +\; (1\; -\; t)\; p(z)$ for all $0\; \backslash leq\; t\; \backslash leq\; 1$ and $y,\; z\; \backslash in\; X.$ Let $M,$ $f\; :\; M\; \backslash to\; \backslash R,$ and $x\; \backslash in\; X\; \backslash setminus\; M$ be as in the lemma's statement. Given any $m,\; n\; \backslash in\; M$ and any positive real $r,\; s\; >\; 0,$ the positive real numbers $t\; :=\; \backslash tfrac\{s\}\{r\; +\; s\}$ and $\backslash tfrac\{r\}\{r\; +\; s\}\; =\; 1\; -\; t$ sum to $1$ so that the convexity of $p$ on $X$ guarantees

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

p\left(\tfrac{s}{r + s} m + \tfrac{r}{r + s} n\right)

~&=~ p\big(\tfrac{s}{r + s} (m - r x) &&+ \tfrac{r}{r + s} (n + s x)\big) && \\

&\leq~ \tfrac{s}{r + s} \; p(m - r x) &&+ \tfrac{r}{r + s} \; p(n + s x) && \\

\end{alignat}

and hence

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

s f(m) + r f(n)

~&=~ (r + s) \; f\left(\tfrac{s}{r + s} m + \tfrac{r}{r + s} n\right) && \qquad \text{ by linearity of } f \\

&\leq~ (r + s) \; p\left(\tfrac{s}{r + s} m + \tfrac{r}{r + s} n\right) && \qquad f \leq p \text{ on } M \\

&\leq~ s p(m - r x) + r p(n + s x) \\

\end{alignat}

thus proving that $-\; s\; p(m\; -\; r\; x)\; +\; s\; f(m)\; ~\backslash leq~\; r\; p(n\; +\; s\; x)\; -\; r\; f(n),$ which after multiplying both sides by $\backslash tfrac\{1\}\{rs\}$ becomes

$$\backslash tfrac\{1\}\{r\}\; [-\; p(m\; -\; r\; x)\; +\; f(m)]\; ~\backslash leq~\; \backslash tfrac\{1\}\{s\}\; [p(n\; +\; s\; x)\; -\; f(n)].$$

This implies that the values defined by

$$a\; =\; \backslash sup\_\{\backslash stackrel\{m\; \backslash in\; M\}\{r\; >\; 0\}\}\; \backslash tfrac\{1\}\{r\}\; [-\; p(m\; -\; r\; x)\; +\; f(m)]\; \backslash qquad\; \backslash text\{\; and\; \}\; \backslash qquad\; c\; =\; \backslash inf\_\{\backslash stackrel\{n\; \backslash in\; M\}\{s\; >\; 0\}\}\; \backslash tfrac\{1\}\{s\}\; [p(n\; +\; s\; x)\; -\; f(n)]$$

are real numbers that satisfy $a\; \backslash leq\; c.$ As in the above proof of the one–dimensional dominated extension theorem above, for any real $b\; \backslash in\; \backslash R$ define $F\_b\; :\; M\; \backslash oplus\; \backslash R\; x\; \backslash to\; \backslash R$ by $F\_b(m\; +\; r\; x)\; =\; f(m)\; +\; r\; b.$

It can be verified that if $a\; \backslash leq\; b\; \backslash leq\; c$ then $F\_b\; \backslash leq\; p$ where $r\; b\; \backslash leq\; p(m\; +\; r\; x)\; -\; f(m)$ follows from $b\; \backslash leq\; c$ when $r\; >\; 0$ (respectively, follows from $a\; \backslash leq\; b$ when ).

$\backslash blacksquare$

{{collapse bottom}}

The lemma above is the key step in deducing the dominated extension theorem from Zorn's lemma.

{{Math proof|title=Proof of dominated extension theorem using Zorn's lemma|drop=hidden|proof=

The set of all possible dominated linear extensions of $f$ are partially ordered by extension of each other, so there is a maximal extension $F.$ By the codimension-1 result, if $F$ is not defined on all of $X,$ then it can be further extended. Thus $F$ must be defined everywhere, as claimed.

$\backslash blacksquare$

}}

When $M$ has countable codimension, then using induction and the lemma completes the proof of the Hahn–Banach theorem. The standard proof of the general case uses Zorn's lemma although the strictly weaker ultrafilter lemma{{sfn|Luxemburg|1962|p=}} (which is equivalent to the compactness theorem and to the Boolean prime ideal theorem) may be used instead. Hahn–Banach can also be proved using Tychonoff's theorem for compact Hausdorff spaces{{sfn|Łoś|Ryll-Nardzewski|1951|pp=233–237}} (which is also equivalent to the ultrafilter lemma)

The Mizar project has completely formalized and automatically checked the proof of the Hahn–Banach theorem in the HAHNBAN file.[http://mizar.uwb.edu.pl/JFM/Vol5/hahnban.html HAHNBAN file]

The Hahn–Banach theorem can be used to guarantee the existence of continuous linear extensions of continuous linear functionals.

{{Math theorem

|name={{visible anchor|Hahn–Banach continuous extension theorem}}{{sfn|Narici|Beckenstein|2011|pp=182,498}}

|math_statement=

Every continuous linear functional $f$ defined on a vector subspace $M$ of a (real or complex) locally convex topological vector space $X$ has a continuous linear extension $F$ to all of $X.$ If in addition $X$ is a normed space, then this extension can be chosen so that its dual norm is equal to that of $f.$

}}

In category-theoretic terms, the underlying field of the vector space is an injective object in the category of locally convex vector spaces.

On a normed (or seminormed) space, a linear extension $F$ of a bounded linear functional $f$ is said to be {{em|{{visible anchor|norm-preserving linear extension|text=norm-preserving}}}} if it has the same dual norm as the original functional: $\backslash |F\backslash |\; =\; \backslash |f\backslash |.$

Because of this terminology, the second part of the above theorem is sometimes referred to as the "norm-preserving" version of the Hahn–Banach theorem.{{sfn|Narici|Beckenstein|2011|p=184}} Explicitly:

{{Math theorem

|name={{visible anchor|Norm-preserving Hahn–Banach continuous extension theorem}}{{sfn|Narici|Beckenstein|2011|p=184}}

|math_statement=Every continuous linear functional $f$ defined on a vector subspace $M$ of a (real or complex) normed space $X$ has a continuous linear extension $F$ to all of $X$ that satisfies $\backslash |f\backslash |\; =\; \backslash |F\backslash |.$

}}

The following observations allow the continuous extension theorem to be deduced from the Hahn–Banach theorem.{{sfn|Narici|Beckenstein|2011|p=182}}

The absolute value of a linear functional is always a seminorm. A linear functional $F$ on a topological vector space $X$ is continuous if and only if its absolute value $|F|$ is continuous, which happens if and only if there exists a continuous seminorm $p$ on $X$ such that $|F|\; \backslash leq\; p$ on the domain of $F.${{sfn|Narici|Beckenstein|2011|p=126}}

If $X$ is a locally convex space then this statement remains true when the linear functional $F$ is defined on a {{em|proper}} vector subspace of $X.$

{{Math proof|title=Proof of the continuous extension theorem for locally convex spaces{{sfn|Narici|Beckenstein|2011|p=182}}{{anchor|Proof of the continuous extension theorem for locally convex topological vector spaces}}|drop=hidden|proof=

Let $f$ be a continuous linear functional defined on a vector subspace $M$ of a locally convex topological vector space $X.$

Because $X$ is locally convex, there exists a continuous seminorm $p\; :\; X\; \backslash to\; \backslash Reals$ on $X$ that dominates $f$ (meaning that $|f(m)|\; \backslash leq\; p(m)$ for all $m\; \backslash in\; M$).

By the Hahn–Banach theorem, there exists a linear extension of $f$ to $X,$ call it $F,$ that satisfies $|F|\; \backslash leq\; p$ on $X.$

This linear functional $F$ is continuous since $|F|\; \backslash leq\; p$ and $p$ is a continuous seminorm.

}}

Proof for normed spaces

A linear functional $f$ on a normed space is continuous if and only if it is bounded, which means that its dual norm

$$\backslash |f\backslash |\; =\; \backslash sup\; \backslash \{|f(m)|\; :\; \backslash |m\backslash |\; \backslash leq\; 1,\; m\; \backslash in\; \backslash operatorname\{domain\}\; f\backslash \}$$

is finite, in which case $|f(m)|\; \backslash leq\; \backslash |f\backslash |\; \backslash |m\backslash |$ holds for every point $m$ in its domain.

Moreover, if $c\; \backslash geq\; 0$ is such that $|f(m)|\; \backslash leq\; c\; \backslash |m\backslash |$ for all $m$ in the functional's domain, then necessarily $\backslash |f\backslash |\; \backslash leq\; c.$

If $F$ is a linear extension of a linear functional $f$ then their dual norms always satisfy $\backslash |f\backslash |\; \backslash leq\; \backslash |F\backslash |$

so that equality $\backslash |f\backslash |\; =\; \backslash |F\backslash |$ is equivalent to $\backslash |F\backslash |\; \backslash leq\; \backslash |f\backslash |,$ which holds if and only if $|F(x)|\; \backslash leq\; \backslash |f\backslash |\; \backslash |x\backslash |$ for every point $x$ in the extension's domain.

This can be restated in terms of the function $\backslash |f\backslash |\; \backslash ,\; \backslash |\backslash cdot\backslash |\; :\; X\; \backslash to\; \backslash Reals$ defined by $x\; \backslash mapsto\; \backslash |f\backslash |\; \backslash ,\; \backslash |x\backslash |,$ which is always a seminorm:Like every non-negative scalar multiple of a norm, this seminorm $\backslash |f\backslash |\; \backslash ,\; \backslash |\backslash cdot\backslash |$ (the product of the non-negative real number $\backslash |f\backslash |$ with the norm $\backslash |\backslash cdot\backslash |$) is a norm when $\backslash |f\backslash |$ is positive, although this fact is not needed for the proof.

:A linear extension of a bounded linear functional $f$ is norm-preserving if and only if the extension is dominated by the seminorm $\backslash |f\backslash |\; \backslash ,\; \backslash |\backslash cdot\backslash |.$

Applying the Hahn–Banach theorem to $f$ with this seminorm $\backslash |f\backslash |\; \backslash ,\; \backslash |\backslash cdot\backslash |$ thus produces a dominated linear extension whose norm is (necessarily) equal to that of $f,$ which proves the theorem:

{{Math proof|title=Proof of the norm-preserving Hahn–Banach continuous extension theorem{{sfn|Narici|Beckenstein|2011|p=184}}|drop=hidden|proof=

Let $f$ be a continuous linear functional defined on a vector subspace $M$ of a normed space $X.$

Then the function $p\; :\; X\; \backslash to\; \backslash Reals$ defined by $p(x)\; =\; \backslash |f\backslash |\; \backslash ,\; \backslash |x\backslash |$ is a seminorm on $X$ that dominates $f,$ meaning that $|f(m)|\; \backslash leq\; p(m)$ holds for every $m\; \backslash in\; M.$

By the Hahn–Banach theorem, there exists a linear functional $F$ on $X$ that extends $f$ (which guarantees $\backslash |f\backslash |\; \backslash leq\; \backslash |F\backslash |$) and that is also dominated by $p,$ meaning that $|F(x)|\; \backslash leq\; p(x)$ for every $x\; \backslash in\; X.$

The fact that $\backslash |f\backslash |$ is a real number such that $|F(x)|\; \backslash leq\; \backslash |f\backslash |\; \backslash |x\backslash |$ for every $x\; \backslash in\; X,$ guarantees $\backslash |F\backslash |\; \backslash leq\; \backslash |f\backslash |.$

Since $\backslash |F\backslash |\; =\; \backslash |f\backslash |$ is finite, the linear functional $F$ is bounded and thus continuous.

}}

The continuous extension theorem might fail if the topological vector space (TVS) $X$ is not locally convex. For example, for the Lebesgue space $L^p([0,\; 1])$ is a complete metrizable TVS (an F-space) that is {{em|not}} locally convex (in fact, its only convex open subsets are itself $L^p([0,\; 1])$ and the empty set) and the only continuous linear functional on $L^p([0,\; 1])$ is the constant $0$ function {{harv|Rudin|1991|loc=§1.47}}. Since $L^p([0,\; 1])$ is Hausdorff, every finite-dimensional vector subspace $M\; \backslash subseteq\; L^p([0,\; 1])$ is linearly homeomorphic to Euclidean space $\backslash Reals^\{\backslash dim\; M\}$ or $\backslash Complex^\{\backslash dim\; M\}$ (by F. Riesz's theorem) and so every non-zero linear functional $f$ on $M$ is continuous but none has a continuous linear extension to all of $L^p([0,\; 1]).$

However, it is possible for a TVS $X$ to not be locally convex but nevertheless have enough continuous linear functionals that its continuous dual space $X^*$ separates points; for such a TVS, a continuous linear functional defined on a vector subspace {{em|might}} have a continuous linear extension to the whole space.

If the TVS $X$ is not locally convex then there might not exist any continuous seminorm $p\; :\; X\; \backslash to\; \backslash R$ {{em|defined on $X$}} (not just on $M$) that dominates $f,$ in which case the Hahn–Banach theorem can not be applied as it was in the above proof of the continuous extension theorem.

However, the proof's argument can be generalized to give a characterization of when a continuous linear functional has a continuous linear extension: If $X$ is any TVS (not necessarily locally convex), then a continuous linear functional $f$ defined on a vector subspace $M$ has a continuous linear extension $F$ to all of $X$ if and only if there exists some continuous seminorm $p$ on $X$ that dominates $f.$ Specifically, if given a continuous linear extension $F$ then $p\; :=\; |F|$ is a continuous seminorm on $X$ that dominates $f;$ and conversely, if given a continuous seminorm $p\; :\; X\; \backslash to\; \backslash Reals$ on $X$ that dominates $f$ then any dominated linear extension of $f$ to $X$ (the existence of which is guaranteed by the Hahn–Banach theorem) will be a continuous linear extension.

{{See also|Hyperplane separation theorem}}

The key element of the Hahn–Banach theorem is fundamentally a result about the separation of two convex sets: $\backslash \{-p(-\; x\; -\; n)\; -\; f(n)\; :\; n\; \backslash in\; M\backslash \},$ and $\backslash \{p(m\; +\; x)\; -\; f(m)\; :\; m\; \backslash in\; M\backslash \}.$ This sort of argument appears widely in convex geometry,{{cite journal|last1=Harvey|first1=R.|last2=Lawson|first2=H. B.|year=1983|title=An intrinsic characterisation of Kähler manifolds|journal=Invent. Math.|volume=74|issue=2|pages=169–198|doi=10.1007/BF01394312|bibcode=1983InMat..74..169H|s2cid=124399104}} optimization theory, and economics. Lemmas to this end derived from the original Hahn–Banach theorem are known as the Hahn–Banach separation theorems.{{cite book|last=Zălinescu|first=C.|title=Convex analysis in general vector spaces|publisher=World Scientific Publishing Co., Inc|location= River Edge, NJ |date= 2002|pages=5–7|isbn=981-238-067-1|mr=1921556}}Gabriel Nagy, [http://www.math.ksu.edu/~nagy/real-an/ap-e-h-b.pdf Real Analysis] [http://www.math.ksu.edu/~nagy/real-an/ lecture notes]

They are generalizations of the hyperplane separation theorem, which states that two disjoint nonempty convex subsets of a finite-dimensional space $\backslash R^n$ can be separated by some {{em|affine hyperplane}}, which is a fiber (level set) of the form $f^\{-1\}(s)\; =\; \backslash \{x\; :\; f(x)\; =\; s\backslash \}$ where $f\; \backslash neq\; 0$ is a non-zero linear functional and $s$ is a scalar.

{{Math theorem

| name = Theorem

| math_statement = Let $A$ and $B$ be non-empty convex subsets of a real locally convex topological vector space $X.$

If $\backslash operatorname\{Int\}\; A\; \backslash neq\; \backslash varnothing$ and $B\; \backslash cap\; \backslash operatorname\{Int\}\; A\; =\; \backslash varnothing$ then there exists a continuous linear functional $f$ on $X$ such that $\backslash sup\; f(A)\; \backslash leq\; \backslash inf\; f(B)$ and for all $a\; \backslash in\; \backslash operatorname\{Int\}\; A$ (such an $f$ is necessarily non-zero).

}}

When the convex sets have additional properties, such as being open or compact for example, then the conclusion can be substantially strengthened:

{{Math theorem

| name = Theorem{{sfn|Narici|Beckenstein|2011|pp=177-220}}{{cite book|last=Brezis|first=Haim|publisher=Springer|location=New York|date=2011|pages=6–7|title=Functional Analysis, Sobolev Spaces, and Partial Differential Equations}}

| math_statement = Let $A$ and $B$ be convex non-empty disjoint subsets of a real topological vector space $X.$

• If $A$ is open then $A$ and $B$ are separated by a closed hyperplane. Explicitly, this means that there exists a continuous linear map $f\; :\; X\; \backslash to\; \backslash mathbf\{K\}$ and $s\; \backslash in\; \backslash R$ such that for all $a\; \backslash in\; A,\; b\; \backslash in\; B.$ If both $A$ and $B$ are open then the right-hand side may be taken strict as well.
• If $X$ is locally convex, $A$ is compact, and $B$ closed, then $A$ and $B$ are strictly separated: there exists a continuous linear map $f\; :\; X\; \backslash to\; \backslash mathbf\{K\}$ and $s,\; t\; \backslash in\; \backslash R$ such that for all $a\; \backslash in\; A,\; b\; \backslash in\; B.$

If $X$ is complex (rather than real) then the same claims hold, but for the real part of $f.$

}}

Then following important corollary is known as the Geometric Hahn–Banach theorem or Mazur's theorem (also known as Ascoli–Mazur theorem{{cite book|first1=Semen|last1=Kutateladze|date=1996|pages=40|title=Fundamentals of Functional Analysis|series=Kluwer Texts in the Mathematical Sciences |volume=12|isbn=978-90-481-4661-1|url=https://www.researchgate.net/publication/240011075|doi=10.1007/978-94-015-8755-6}}). It follows from the first bullet above and the convexity of $M.$

{{Math theorem

| name = Theorem (Mazur){{sfn|Trèves|2006|p=184}}

| math_statement = Let $M$ be a vector subspace of the topological vector space $X$ and suppose $K$ is a non-empty convex open subset of $X$ with $K\; \backslash cap\; M\; =\; \backslash varnothing.$

Then there is a closed hyperplane (codimension-1 vector subspace) $N\; \backslash subseteq\; X$ that contains $M,$ but remains disjoint from $K.$

}}

Mazur's theorem clarifies that vector subspaces (even those that are not closed) can be characterized by linear functionals.

{{Math theorem

| name = Corollary{{sfn|Narici|Beckenstein|2011|pp=195}}

| note = Separation of a subspace and an open convex set

| math_statement = Let $M$ be a vector subspace of a locally convex topological vector space $X,$ and $U$ be a non-empty open convex subset disjoint from $M.$ Then there exists a continuous linear functional $f$ on $X$ such that $f(m)\; =\; 0$ for all $m\; \backslash in\; M$ and $\backslash operatorname\{Re\}\; f\; >\; 0$ on $U.$

}}

Since points are trivially convex, geometric Hahn–Banach implies that functionals can detect the boundary of a set. In particular, let $X$ be a real topological vector space and $A\; \backslash subseteq\; X$ be convex with $\backslash operatorname\{Int\}\; A\; \backslash neq\; \backslash varnothing.$ If $a\_0\; \backslash in\; A\; \backslash setminus\; \backslash operatorname\{Int\}\; A$ then there is a functional that is vanishing at $a\_0,$ but supported on the interior of $A.$

Call a normed space $X$ smooth if at each point $x$ in its unit ball there exists a unique closed hyperplane to the unit ball at $x.$ Köthe showed in 1983 that a normed space is smooth at a point $x$ if and only if the norm is Gateaux differentiable at that point.{{sfn|Narici|Beckenstein|2011|pp=177-220}}

Let $U$ be a convex balanced neighborhood of the origin in a locally convex topological vector space $X$ and suppose $x\; \backslash in\; X$ is not an element of $U.$ Then there exists a continuous linear functional $f$ on $X$ such that{{sfn|Narici|Beckenstein|2011|pp=177-220}}

$\backslash sup\; |f(U)|\; \backslash leq\; |f(x)|.$

The Hahn–Banach theorem is the first sign of an important philosophy in functional analysis: to understand a space, one should understand its continuous functionals.

For example, linear subspaces are characterized by functionals: if {{mvar|X}} is a normed vector space with linear subspace {{mvar|M}} (not necessarily closed) and if $z$ is an element of {{mvar|X}} not in the closure of {{mvar|M}}, then there exists a continuous linear map $f\; :\; X\; \backslash to\; \backslash mathbf\{K\}$ with $f(m)\; =\; 0$ for all $m\; \backslash in\; M,$ $f(z)\; =\; 1,$ and $\backslash |f\backslash |\; =\; \backslash operatorname\{dist\}(z,\; M)^\{-1\}.$ (To see this, note that $\backslash operatorname\{dist\}(\backslash cdot,\; M)$ is a sublinear function.) Moreover, if $z$ is an element of {{mvar|X}}, then there exists a continuous linear map $f\; :\; X\; \backslash to\; \backslash mathbf\{K\}$ such that $f(z)\; =\; \backslash |z\backslash |$ and $\backslash |f\backslash |\; \backslash leq\; 1.$ This implies that the natural injection $J$ from a normed space {{mvar|X}} into its double dual $V^\{**\}$ is isometric.

That last result also suggests that the Hahn–Banach theorem can often be used to locate a "nicer" topology in which to work. For example, many results in functional analysis assume that a space is Hausdorff or locally convex. However, suppose {{mvar|X}} is a topological vector space, not necessarily Hausdorff or locally convex, but with a nonempty, proper, convex, open set {{mvar|M}}. Then geometric Hahn–Banach implies that there is a hyperplane separating {{mvar|M}} from any other point. In particular, there must exist a nonzero functional on {{mvar|X}} — that is, the continuous dual space $X^*$ is non-trivial.{{sfn|Narici|Beckenstein|2011|pp=177-220}}{{sfn|Schaefer|Wolff|1999|p=47}} Considering {{mvar|X}} with the weak topology induced by $X^*,$ then {{mvar|X}} becomes locally convex; by the second bullet of geometric Hahn–Banach, the weak topology on this new space separates points.

Thus {{mvar|X}} with this weak topology becomes Hausdorff. This sometimes allows some results from locally convex topological vector spaces to be applied to non-Hausdorff and non-locally convex spaces.

The Hahn–Banach theorem is often useful when one wishes to apply the method of a priori estimates. Suppose that we wish to solve the linear differential equation $P\; u\; =\; f$ for $u,$ with $f$ given in some Banach space {{mvar|X}}. If we have control on the size of $u$ in terms of $\backslash |f\backslash |\_X$ and we can think of $u$ as a bounded linear functional on some suitable space of test functions $g,$ then we can view $f$ as a linear functional by adjunction: $(f,\; g)\; =\; (u,\; P^*g).$ At first, this functional is only defined on the image of $P,$ but using the Hahn–Banach theorem, we can try to extend it to the entire codomain {{mvar|X}}. The resulting functional is often defined to be a weak solution to the equation.

{{Math theorem

| name = Theorem{{sfn|Narici|Beckenstein|2011|p=212}}

| math_statement = A real Banach space is reflexive if and only if every pair of non-empty disjoint closed convex subsets, one of which is bounded, can be strictly separated by a hyperplane.

}}

To illustrate an actual application of the Hahn–Banach theorem, we will now prove a result that follows almost entirely from the Hahn–Banach theorem.

{{Math theorem

| name = Proposition

| math_statement = Suppose $X$ is a Hausdorff locally convex TVS over the field $\backslash mathbf\{K\}$ and $Y$ is a vector subspace of $X$ that is TVS–isomorphic to $\backslash mathbf\{K\}^I$ for some set $I.$

Then $Y$ is a closed and complemented vector subspace of $X.$

}}

{{Math proof|drop=hidden|proof=

Since $\backslash mathbf\{K\}^I$ is a complete TVS so is $Y,$ and since any complete subset of a Hausdorff TVS is closed, $Y$ is a closed subset of $X.$

Let $f\; =\; \backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}\; :\; Y\; \backslash to\; \backslash mathbf\{K\}^I$ be a TVS isomorphism, so that each $f\_i\; :\; Y\; \backslash to\; \backslash mathbf\{K\}$ is a continuous surjective linear functional.

By the Hahn–Banach theorem, we may extend each $f\_i$ to a continuous linear functional $F\_i\; :\; X\; \backslash to\; \backslash mathbf\{K\}$ on $X.$

Let $F\; :=\; \backslash left(F\_i\backslash right)\_\{i\; \backslash in\; I\}\; :\; X\; \backslash to\; \backslash mathbf\{K\}^I$ so $F$ is a continuous linear surjection such that its restriction to $Y$ is $F\backslash big\backslash vert\_Y\; =\; \backslash left(F\_i\backslash big\backslash vert\_Y\backslash right)\_\{i\; \backslash in\; I\}\; =\; \backslash left(f\_i\backslash right)\_\{i\; \backslash in\; I\}\; =\; f.$

Let $P\; :=\; f^\{-1\}\; \backslash circ\; F\; :\; X\; \backslash to\; Y,$ which is a continuous linear map whose restriction to $Y$ is $P\backslash big\backslash vert\_Y\; =\; f^\{-1\}\; \backslash circ\; F\backslash big\backslash vert\_Y\; =\; f^\{-1\}\; \backslash circ\; f\; =\; \backslash mathbf\{1\}\_Y,$ where $\backslash mathbb\{1\}\_Y$ denotes the identity map on $Y.$

This shows that $P$ is a continuous linear projection onto $Y$ (that is, $P\; \backslash circ\; P\; =\; P$).

Thus $Y$ is complemented in $X$ and $X\; =\; Y\; \backslash oplus\; \backslash ker\; P$ in the category of TVSs. $\backslash blacksquare$

}}

The above result may be used to show that every closed vector subspace of $\backslash R^\{\backslash N\}$ is complemented because any such space is either finite dimensional or else TVS–isomorphic to $\backslash R^\{\backslash N\}.$

General template

There are now many other versions of the Hahn–Banach theorem. The general template for the various versions of the Hahn–Banach theorem presented in this article is as follows:

:$p\; :\; X\; \backslash to\; \backslash R$ is a sublinear function (possibly a seminorm) on a vector space $X,$ $M$ is a vector subspace of $X$ (possibly closed), and $f$ is a linear functional on $M$ satisfying $|f|\; \backslash leq\; p$ on $M$ (and possibly some other conditions). One then concludes that there exists a linear extension $F$ of $f$ to $X$ such that $|F|\; \backslash leq\; p$ on $X$ (possibly with additional properties).

{{Math theorem

| name = Theorem{{sfn|Narici|Beckenstein|2011|pp=177-220}}

| math_statement = If $D$ is an absorbing disk in a real or complex vector space $X$ and if $f$ be a linear functional defined on a vector subspace $M$ of $X$ such that $|f|\; \backslash leq\; 1$ on $M\; \backslash cap\; D,$ then there exists a linear functional $F$ on $X$ extending $f$ such that $|F|\; \backslash leq\; 1$ on $D.$

}}

{{Math theorem

| name = {{visible anchor|Hahn–Banach theorem for seminorms}}{{sfn|Wilansky|2013|pp=18-21}}{{sfn|Narici|Beckenstein|2011|pp=150}}

| math_statement = If $p\; :\; M\; \backslash to\; \backslash Reals$ is a seminorm defined on a vector subspace $M$ of $X,$ and if $q\; :\; X\; \backslash to\; \backslash Reals$ is a seminorm on $X$ such that $p\; \backslash leq\; q\backslash big\backslash vert\_M,$ then there exists a seminorm $P\; :\; X\; \backslash to\; \backslash Reals$ on $X$ such that $P\backslash big\backslash vert\_M\; =\; p$ on $M$ and $P\; \backslash leq\; q$ on $X.$

}}

{{Math proof|title=Proof of the Hahn–Banach theorem for seminorms|drop=hidden|proof=

Let $S$ be the convex hull of $\backslash \{m\; \backslash in\; M\; :\; p(m)\; \backslash leq\; 1\backslash \}\; \backslash cup\; \backslash \{x\; \backslash in\; X\; :\; q(x)\; \backslash leq\; 1\backslash \}.$ Because $S$ is an absorbing disk in $X,$ its Minkowski functional $P$ is a seminorm. Then $p\; =\; P$ on $M$ and $P\; \backslash leq\; q$ on $X.$

}}

So for example, suppose that $f$ is a bounded linear functional defined on a vector subspace $M$ of a normed space $X,$ so its the operator norm $\backslash |f\backslash |$ is a non-negative real number.

Then the linear functional's absolute value $p\; :=\; |f|$ is a seminorm on $M$ and the map $q\; :\; X\; \backslash to\; \backslash Reals$ defined by $q(x)\; =\; \backslash |f\backslash |\; \backslash ,\; \backslash |x\backslash |$ is a seminorm on $X$ that satisfies $p\; \backslash leq\; q\backslash big\backslash vert\_M$ on $M.$

The Hahn–Banach theorem for seminorms guarantees the existence of a seminorm $P\; :\; X\; \backslash to\; \backslash Reals$ that is equal to $|f|$ on $M$ (since $P\backslash big\backslash vert\_M\; =\; p\; =\; |f|$) and is bounded above by $P(x)\; \backslash leq\; \backslash |f\backslash |\; \backslash ,\; \backslash |x\backslash |$ everywhere on $X$ (since $P\; \backslash leq\; q$).

{{Math theorem

| name = {{visible anchor|Hahn–Banach sandwich theorem}}{{sfn|Narici|Beckenstein|2011|pp=177-220}}

| math_statement = Let $p\; :\; X\; \backslash to\; \backslash R$ be a sublinear function on a real vector space $X,$ let $S\; \backslash subseteq\; X$ be any subset of $X,$ and let $f\; :\; S\; \backslash to\; \backslash R$ be {{em|any}} map.

If there exist positive real numbers $a$ and $b$ such that

$$0\; \backslash geq\; \backslash inf\_\{s\; \backslash in\; S\}\; [p(s\; -\; a\; x\; -\; b\; y)\; -\; f(s)\; -\; a\; f(x)\; -\; b\; f(y)]\; \backslash qquad\; \backslash text\{\; for\; all\; \}\; x,\; y\; \backslash in\; S,$$

then there exists a linear functional $F\; :\; X\; \backslash to\; \backslash R$ on $X$ such that $F\; \backslash leq\; p$ on $X$ and $f\; \backslash leq\; F\; \backslash leq\; p$ on $S.$

}}

{{Math theorem

| name = Theorem{{sfn|Narici|Beckenstein|2011|pp=177-220}}

| note = Andenaes, 1970

| math_statement = Let $p\; :\; X\; \backslash to\; \backslash R$ be a sublinear function on a real vector space $X,$ let $f\; :\; M\; \backslash to\; \backslash R$ be a linear functional on a vector subspace $M$ of $X$ such that $f\; \backslash leq\; p$ on $M,$ and let $S\; \backslash subseteq\; X$ be any subset of $X.$

Then there exists a linear functional $F\; :\; X\; \backslash to\; \backslash R$ on $X$ that extends $f,$ satisfies $F\; \backslash leq\; p$ on $X,$ and is (pointwise) maximal on $S$ in the following sense: if $\backslash widehat\{F\}\; :\; X\; \backslash to\; \backslash R$ is a linear functional on $X$ that extends $f$ and satisfies $\backslash widehat\{F\}\; \backslash leq\; p$ on $X,$ then $F\; \backslash leq\; \backslash widehat\{F\}$ on $S$ implies $F\; =\; \backslash widehat\{F\}$ on $S.$

}}

If $S\; =\; \backslash \{s\backslash \}$ is a singleton set (where $s\; \backslash in\; X$ is some vector) and if $F\; :\; X\; \backslash to\; \backslash R$ is such a maximal dominated linear extension of $f\; :\; M\; \backslash to\; \backslash R,$ then $F(s)\; =\; \backslash inf\_\{m\; \backslash in\; M\}\; [f(s)\; +\; p(s\; -\; m)].${{sfn|Narici|Beckenstein|2011|pp=177-220}}

{{See also|Vector-valued Hahn–Banach theorems}}

{{Math theorem

| name = {{visible anchor|Vector–valued Hahn–Banach theorem}}{{sfn|Narici|Beckenstein|2011|pp=177-220}}

| math_statement = If $X$ and $Y$ are vector spaces over the same field and if $f\; :\; M\; \backslash to\; Y$ is a linear map defined on a vector subspace $M$ of $X,$ then there exists a linear map $F\; :\; X\; \backslash to\; Y$ that extends $f.$

}}

{{See also|Vector-valued Hahn–Banach theorems}}

A set $\backslash Gamma$ of maps $X\; \backslash to\; X$ is {{em|{{visible anchor|commutative set of maps|text=commutative}}}} (with respect to function composition $\backslash ,\backslash circ\backslash ,$) if $F\; \backslash circ\; G\; =\; G\; \backslash circ\; F$ for all $F,\; G\; \backslash in\; \backslash Gamma.$

Say that a function $f$ defined on a subset $M$ of $X$ is {{em|{{visible anchor|invariant map|text=$\backslash Gamma$-invariant}}}} if $L(M)\; \backslash subseteq\; M$ and $f\; \backslash circ\; L\; =\; f$ on $M$ for every $L\; \backslash in\; \backslash Gamma.$

{{Math theorem

| name = {{visible anchor|An invariant Hahn–Banach theorem}}{{sfn|Rudin|1991|p=141}}

| math_statement =

Suppose $\backslash Gamma$ is a commutative set of continuous linear maps from a normed space $X$ into itself and let $f$ be a continuous linear functional defined some vector subspace $M$ of $X$ that is $\backslash Gamma$-invariant, which means that $L(M)\; \backslash subseteq\; M$ and $f\; \backslash circ\; L\; =\; f$ on $M$ for every $L\; \backslash in\; \backslash Gamma.$

Then $f$ has a continuous linear extension $F$ to all of $X$ that has the same operator norm $\backslash |f\backslash |\; =\; \backslash |F\backslash |$ and is also $\backslash Gamma$-invariant, meaning that $F\; \backslash circ\; L\; =\; F$ on $X$ for every $L\; \backslash in\; \backslash Gamma.$

}}

This theorem may be summarized:

:Every $\backslash Gamma$-invariant continuous linear functional defined on a vector subspace of a normed space $X$ has a $\backslash Gamma$-invariant Hahn–Banach extension to all of $X.${{sfn|Rudin|1991|p=141}}

The following theorem of Mazur–Orlicz (1953) is equivalent to the Hahn–Banach theorem.

{{Math theorem

| name = {{visible anchor|Mazur–Orlicz theorem}}{{sfn|Narici|Beckenstein|2011|pp=177–220}}

| math_statement = Let $p\; :\; X\; \backslash to\; \backslash R$ be a sublinear function on a real or complex vector space $X,$ let $T$ be any set, and let $R\; :\; T\; \backslash to\; \backslash R$ and $v\; :\; T\; \backslash to\; X$ be any maps. The following statements are equivalent:

1. there exists a real-valued linear functional $F$ on $X$ such that $F\; \backslash leq\; p$ on $X$ and $R\; \backslash leq\; F\; \backslash circ\; v$ on $T$;
2. for any finite sequence $s\_1,\; \backslash ldots,\; s\_n$ of $n\; >\; 0$ non-negative real numbers, and any sequence $t\_1,\; \backslash ldots,\; t\_n\; \backslash in\; T$ of elements of $T,$ $$\backslash sum\_\{i=1\}^n\; s\_i\; R\backslash left(t\_i\backslash right)\; \backslash leq\; p\backslash left(\backslash sum\_\{i=1\}^n\; s\_i\; v\backslash left(t\_i\backslash right)\backslash right).$$

}}

The following theorem characterizes when {{em|any}} scalar function on $X$ (not necessarily linear) has a continuous linear extension to all of $X.$

{{Math theorem

| name = Theorem

| note = {{visible anchor|The extension principle}}{{sfn|Edwards|1995|pp=124-125}}

| math_statement = Let $f$ a scalar-valued function on a subset $S$ of a topological vector space $X.$

Then there exists a continuous linear functional $F$ on $X$ extending $f$ if and only if there exists a continuous seminorm $p$ on $X$ such that

$$\backslash left|\backslash sum\_\{i=1\}^n\; a\_i\; f(s\_i)\backslash right|\; \backslash leq\; p\backslash left(\backslash sum\_\{i=1\}^n\; a\_is\_i\backslash right)$$

for all positive integers $n$ and all finite sequences $a\_1,\; \backslash ldots,\; a\_n$ of scalars and elements $s\_1,\; \backslash ldots,\; s\_n$ of $S.$

}}

Let {{mvar|X}} be a topological vector space. A vector subspace {{mvar|M}} of {{mvar|X}} has the extension property if any continuous linear functional on {{mvar|M}} can be extended to a continuous linear functional on {{mvar|X}}, and we say that {{mvar|X}} has the Hahn–Banach extension property (HBEP) if every vector subspace of {{mvar|X}} has the extension property.{{sfn|Narici|Beckenstein|2011|pp=225-273}}

The Hahn–Banach theorem guarantees that every Hausdorff locally convex space has the HBEP. For complete metrizable topological vector spaces there is a converse, due to Kalton: every complete metrizable TVS with the Hahn–Banach extension property is locally convex.{{sfn|Narici|Beckenstein|2011|pp=225-273}} On the other hand, a vector space {{mvar|X}} of uncountable dimension, endowed with the finest vector topology, then this is a topological vector spaces with the Hahn–Banach extension property that is neither locally convex nor metrizable.{{sfn|Narici|Beckenstein|2011|pp=225-273}}

A vector subspace {{mvar|M}} of a TVS {{mvar|X}} has the separation property if for every element of {{mvar|X}} such that $x\; \backslash not\backslash in\; M,$ there exists a continuous linear functional $f$ on {{mvar|X}} such that $f(x)\; \backslash neq\; 0$ and $f(m)\; =\; 0$ for all $m\; \backslash in\; M.$ Clearly, the continuous dual space of a TVS {{mvar|X}} separates points on {{mvar|X}} if and only if $\backslash \{0\backslash \},$ has the separation property. In 1992, Kakol proved that any infinite dimensional vector space {{mvar|X}}, there exist TVS-topologies on {{mvar|X}} that do not have the HBEP despite having enough continuous linear functionals for the continuous dual space to separate points on {{mvar|X}}. However, if {{mvar|X}} is a TVS then {{em|every}} vector subspace of {{mvar|X}} has the extension property if and only if {{em|every}} vector subspace of {{mvar|X}} has the separation property.{{sfn|Narici|Beckenstein|2011|pp=225-273}}

{{See also|Krein–Milman theorem#Relation to other statements}}

{{anchor|Relation to axiom of choice}}The proof of the Hahn–Banach theorem for real vector spaces (HB) commonly uses Zorn's lemma, which in the axiomatic framework of Zermelo–Fraenkel set theory (ZF) is equivalent to the axiom of choice (AC). It was discovered by Łoś and Ryll-Nardzewski{{sfn|Łoś|Ryll-Nardzewski|1951|pp=233–237}} and independently by Luxemburg{{sfn|Luxemburg|1962|p=}} that HB can be proved using the ultrafilter lemma (UL), which is equivalent (under ZF) to the Boolean prime ideal theorem (BPI). BPI is strictly weaker than the axiom of choice and it was later shown that HB is strictly weaker than BPI.{{sfn|Pincus|1974|pp=203–205}}

The ultrafilter lemma is equivalent (under ZF) to the Banach–Alaoglu theorem,{{sfn|Schechter|1996|pp=766–767}} which is another foundational theorem in functional analysis. Although the Banach–Alaoglu theorem implies HB,{{cite book|last=Muger|first= Michael|title=Topology for the Working Mathematician|year=2020}} it is not equivalent to it (said differently, the Banach–Alaoglu theorem is strictly stronger than HB).

However, HB is equivalent to a certain weakened version of the Banach–Alaoglu theorem for normed spaces.{{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}}

The Hahn–Banach theorem is also equivalent to the following statement:{{cite book|last=Schechter|first=Eric|title=Handbook of Analysis and its Foundations|page=620|author-link=Eric Schechter}}

:(∗): On every Boolean algebra {{mvar|B}} there exists a "probability charge", that is: a non-constant finitely additive map from $B$ into $[0,\; 1].$

(BPI is equivalent to the statement that there are always non-constant probability charges which take only the values 0 and 1.)

In ZF, the Hahn–Banach theorem suffices to derive the existence of a non-Lebesgue measurable set.{{cite journal|last1=Foreman|first1=M.|last2=Wehrung|first2=F.|year=1991|title=The Hahn–Banach theorem implies the existence of a non-Lebesgue measurable set|url=http://matwbn.icm.edu.pl/ksiazki/fm/fm138/fm13812.pdf|journal=Fundamenta Mathematicae|volume=138|pages=13–19|doi=10.4064/fm-138-1-13-19|doi-access=free}} Moreover, the Hahn–Banach theorem implies the Banach–Tarski paradox.{{cite journal|last=Pawlikowski|first=Janusz|year=1991|title=The Hahn–Banach theorem implies the Banach–Tarski paradox|journal=Fundamenta Mathematicae|volume=138|pages=21–22|doi=10.4064/fm-138-1-21-22|doi-access=free}}

For separable Banach spaces, D. K. Brown and S. G. Simpson proved that the Hahn–Banach theorem follows from WKL₀, a weak subsystem of second-order arithmetic that takes a form of Kőnig's lemma restricted to binary trees as an axiom. In fact, they prove that under a weak set of assumptions, the two are equivalent, an example of reverse mathematics.{{cite journal|last1=Brown|first1=D. K.|last2=Simpson|first2=S. G.|year=1986|title=Which set existence axioms are needed to prove the separable Hahn–Banach theorem?|journal=Annals of Pure and Applied Logic|volume=31|pages=123–144|doi=10.1016/0168-0072(86)90066-7 |doi-access=}} [http://www.math.psu.edu/simpson/papers/hilbert/node7.html#3 Source of citation].Simpson, Stephen G. (2009), Subsystems of second order arithmetic, Perspectives in Logic (2nd ed.), Cambridge University Press, {{ISBN|978-0-521-88439-6}}, {{MR|2517689}}

• {{annotated link|Farkas' lemma}}
• {{annotated link|Fichera's existence principle}}
• {{annotated link|M. Riesz extension theorem}}
• {{annotated link|Separating axis theorem}}
• {{annotated link|Vector-valued Hahn–Banach theorems}}

{{reflist|group=note|refs=

• Geometric illustration:

The geometric idea of the above proof can be fully presented in the case of $X\; =\; \backslash R^2,\; M\; =\; \backslash \{(x,\; 0)\; :\; x\; \backslash in\; \backslash R\backslash \}.$

First, define the simple-minded extension $f\_0(x,\; y)\; =\; f(x),$ It doesn't work, since maybe $f\_0\; \backslash leq\; p$. But it is a step in the right direction. $p-f\_0$ is still convex, and $p-f\_0\; \backslash geq\; f-f\_0.$ Further, $f-f\_0$ is identically zero on the x-axis. Thus we have reduced to the case of $f\; =\; 0,\; p\; \backslash geq\; 0$ on the x-axis.

If $p\; \backslash geq\; 0$ on $\backslash R^2,$ then we are done. Otherwise, pick some $v\; \backslash in\; \backslash R^2,$ such that

The idea now is to perform a simultaneous bounding of $p$ on $v\; +\; M$ and $-v+M$ such that $p\; \backslash geq\; b$ on $v+M$ and $p\; \backslash geq\; -b$ on $-v+M,$ then defining $\backslash tilde\; f(w\; +\; rv)\; =\; rb$ would give the desired extension.

Since $-v+M,\; v+M$ are on opposite sides of $M,$ and at some point on $v+M,$ by convexity of $p,$ we must have $p\; \backslash geq\; 0$ on all points on $-v+M.$ Thus $\backslash inf\_\{u\backslash in\; -v\; +\; M\}\; p(u)$ is finite.

Geometrically, this works because is a convex set that is disjoint from $M,$ and thus must lie entirely on one side of $M.$

Define $b\; =\; -\backslash inf\_\{u\backslash in\; -v\; +\; M\}\; p(u).$ This satisfies $p\backslash geq\; -b$ on $-v+M.$ It remains to check the other side.

For all $v+w\; \backslash in\; v+M,$ convexity implies that for all $-v+w\text{'}\; \backslash in\; -v+M,\; p(v+w)\; +\; p(-v\; +w\text{'})\; \backslash geq\; 2p((w+w\text{'})/2)\; =\; 0,$ thus

$p(v\; +\; w)\; \backslash geq\; \backslash sup\_\{u\backslash in\; -v\; +\; M\}\; -p(u)\; =\; b.$

Since during the proof, we only used convexity of $p$, we see that the lemma remains true for merely convex $p.$

}}

Proofs

{{reflist|group=proof|refs=

• The map $F$ being an extension of $f$ means that $\backslash operatorname\{domain\}\; f\; \backslash subseteq\; \backslash operatorname\{domain\}\; F$ and $F(m)\; =\; f(m)$ for every $m\; \backslash in\; \backslash operatorname\{domain\}\; f.$ Consequently,

$$\backslash \{|f(m)|\; :\; \backslash |m\backslash |\; \backslash leq\; 1,\; m\; \backslash in\; \backslash operatorname\{domain\}\; f\backslash \}\; =\; \backslash \{|F(m)|:\; \backslash |m\backslash |\; \backslash leq\; 1,\; m\; \backslash in\; \backslash operatorname\{domain\}\; f\backslash \}\; \backslash subseteq\; \backslash \{|F(x)\backslash ,|\; :\; \backslash |x\backslash |\; \backslash leq\; 1,\; x\; \backslash in\; \backslash operatorname\{domain\}\; F\backslash \}$$ and so the supremum of the set on the left hand side, which is $\backslash |f\backslash |,$ does not exceed the supremum of the right hand side, which is $\backslash |F\backslash |.$ In other words, $\backslash |f\backslash |\; \backslash leq\; \backslash |F\backslash |.$

}}

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