Theorem of Hahn-Banach

This theorem , to which was given the name of the two mathematicians Hans Hahn and Stefan Banach , guarantees the existence of a linear Forme checking certain conditions (specified values on part of space, but limited everywhere).

While making it possible abstractedly to prove the existence of many continuous functions, it is a fundamental tool of the functional Analyze.

By its geometrical interpretation in term of hyperplane S avoiding a convex fixed, it also plays a central role in the study of the geometry of convex, and beyond in convex Analyze.

Analytical form and geometrical form

The statements called “theorem of Hahn-Banach” in the scientific literature are numerous, differing from/to each other sometimes by simple details and sometimes to a significant degree. They are divided nevertheless clearly into two classes: some guarantee to be able to prolong a linear Forme, under certain requirements of increase (“analytical” forms of the theorem); others ensure that one can separate two units convex S by a Hyperplan refines (“geometrical” forms of the theorem).

Let us give to begin an example of statement for each one of these two categories.

A statement of the analytical form of the theorem

A statement of the geometrical form of the theorem

The analytical form of the theorem is due to Banach (1932) generalizing a result of Hahn which is interested as of 1920 in the normalized vector spaces. There exists a generalization of the theorem of Hahn-Banach to the vector spaces on the body of the complexes due to Bohnenblust and Sobczyk (1938). The difficulties of the generalization of the theorem of Hahn-Banach appear even for vector spaces of finished size.

Relations between the two statements and proof of the “geometrical” form

The geometrical form of the theorem - of where one can then deduce a succession from various alternatives relating to the Séparation of convex the - is the retranscription of the analytical form for the particular case where the convex function which intervenes there is the gauge of open convex of a normalized space. It is besides the case in the simplest and fundamental uses of the theorem in functional Analyze which one can according to his tastes lira since a version or the other (one will see a lower example of it).

Let us see more closely how the geometrical form results from the analytical form: Even if it means to make a translation beforehand, it will be supposed that the origin is in $C$. Consequently, since $L$ does not meet $C$, it is thus a subspace closely connected avoiding the origin.

Let us note $p$ the gauge convex $C$. It is under-linear and thus convex like any gauge; by definition even of a gauge it is obvious that for all $x$ in $C$, $p\; (X)\; \backslash \; Leq\; 1$. As one supposed $C$ opened, one can go a little further: on the one hand $C$ is a vicinity of $0$ and any open half-line resulting from $0$ thus contains points of $C$, that which one deduces that $p$ does not take the value $+\; \backslash \; infty$; in addition one can improve the broad inequality $p\; (X)\; \backslash \; Leq\; 1$ and specify without sorrow which the points of $C$ are

Let us note $G$ the vectorial Sous-espace generated by $L$. Since $0\; \backslash \; not\; \backslash \; in\; L$, the subvariety refines $L$ is of Codimension 1 in $G$ and there exists one (and only one) linear form $f$ on $G$ such as $L$ is the part of $G$ of equation $f\; (X)\; =1$. Here is for the linear form to prolong.

Lastly, for $x$ in $L$, $1\; \backslash \; Leq\; p\; (X)$ (since $x\; \backslash \; not\; \backslash \; in\; C$) while $f\; (X)\; =1$. The condition $f\; (X)\; \backslash \; Leq\; p\; (X)$ is thus checked on $L$. While exploiting the positive homogeneity of $f$ and $p$, one extends his field of validity to a strict half space of $G$; on the other half space $f$ takes negative or null values while, like everywhere, $p$ is with positive or null values. The inequality $f\; (X)\; \backslash \; Leq\; p\; (X)$ is thus true everywhere in $G$.

All the assumptions of the version known as “analytical” of the theorem are in place. Thus let us apply it. It offers a new still noted linear form to us $f$, this time defined on entire $E$. Let us note $H$ the hyperplane closely connected of equation $f\; (X)\; =1$: by construction, it is well a hyperplane containing $L$.

That is to say maintaining a point $x$ $C$: for this point, $f\; (X)\; \backslash \; Leq\; p\; (X)$ (since $f$ was produced by the analytical form of Hahn-Banach) and (since one is in open convex the $C$). Thus $f\; (X)\; \backslash \; not=1$, and $x$ is not in $H$. It was well checked that $C$ and $H$ do not meet.

Finally the hyperplanes of a topological vector Space are necessarily closed or dense. However $H$ is not dense since it does not meet the vicinity $C$ $0$. It is thus that it is closed.

One can be astonished that the geometrical form utilizes a topology while the analytical form relates to a vector space without additional structure. In fact, it is completely possible to state a geometrical form in an unspecified vector space: it will have then to be supposed that very relocated convex $C$ containing the origin is absorbent, in the absence of being able to give a direction to “open”; there is not of course any more the complement on the closed character of the hyperplane obtained fall. The demonstration is the same one.

The proof of the “analytical” form

Two types of quite distinct ideas are to be put end to end to lead to a proof within the framework of general information where the theorem was stated. Initially, some rather simple calculations make it possible to justify the extension of the linear form $f$ in the particular case where $G$ is of Codimension 1 in $V$. Once this reached stage, one has already the theorem in finished dimension (it is enough to enlarge step by step the subspace where one succeeded in extending $f$, of a dimension to each step, and until reaching the dimension of $V$). On the other hand, for the uses in infinite dimension, it is necessary to adapt this extremely simple method of methodical projection and to call some standardized enough techniques of Set theory: one carries out a transfinite Récurrence thus, generally written in the form of a call to the Lemme of Zorn.

First part: to gain a dimension

Initially, one will prolong the linear form $f$ with a space larger than $G$ by gaining a dimension. Let us take an element $v\_0$ $V$ apart from $G$ (if there is not $G=V$ of it and one finished before to have even started).

The prolongation of $f$ to the vectorial subspace $G\; \backslash \; oplus\; \{\backslash \; Bbb\; R\}\; v\_0$ takes place by defining it by the formula:

$f\; (X\; +\; \backslash \; lambda\; v\_0):\; =\; F\; (X)\; +\; \backslash \; lambda\; \backslash \; alpha,\; X\; \backslash \; in\; G,\; \backslash \; lambda\; \backslash \; in\; \{\backslash \; Bbb\; R\}$

in which $\backslash \; alpha$ is a reality which it will be necessary to choose judiciously so that the condition of increase of $f$ remains checked in $G\; \backslash \; oplus\; \{\backslash \; Bbb\; R\}\; v\_0$.

It is clear that this method of extension provides a linear form, whatever the choice of $\backslash \; alpha$.

The condition of increase will be checked if and only if, for each $x$ of $G$ and each $\backslash \; lambda$ real, the following inequality is checked:

$f\; (X)\; +\; \backslash \; lambda\; \backslash \; alpha\; \backslash \; Leq\; p\; (X\; +\; \backslash \; lambda\; v\_0)$.

After having noted that for $\backslash \; lambda=0$ it is justified by assumption, one can be interested only in the constraints corresponding to $\backslash \; lambda\; \backslash \; not=0$. It is judicious then to separate the conditions corresponding to $\backslash \; lambda\; >0$ and those where ; by noting $\backslash \; mu=\; \backslash \; lambda$ in the first case, and $\backslash \; mu=-\; \backslash \; lambda$ in the second, one has thus to check two families of inequalities:

$f\; (X)\; +\; \backslash \; driven\; \backslash \; alpha\; \backslash \; Leq\; p\; (X\; +\; \backslash \; driven\; v\_0)$ (where $x$ traverses $G$ and $\backslash \; mu$ traverses $\backslash \; R^\; \{+*\}$).

$f\; (X)\; -\; \backslash \; driven\; \backslash \; alpha\; \backslash \; Leq\; p\; (X\; \backslash \; driven\; v\_0)$ (where $x$ traverses $G$ and $\backslash \; mu$ traverses $\backslash \; R^\; \{+*\}$)

Elementary handling makes it possible to gather them in the form:

$\{1\; \backslash \; over\; \backslash \; driven\}\; \backslash \; left\; v\_0)\; \backslash \; right\; \backslash \; Leq\; \backslash \; alpha\; \backslash \; Leq\; \{1\; \backslash \; over\; \backslash \; driven\}\; \backslash \; left\; v\_0)\; -\; F\; (X)\; \backslash \; right.$

Let us note $a\_\; \{X,\; \backslash \; driven\}\; =\; \{1\; \backslash \; over\; \backslash \; driven\}\; \backslash \; left\; v\_0)\; \backslash \; right$ and $b\_\; \{X,\; \backslash \; driven\}\; =\; \{1\; \backslash \; over\; \backslash \; driven\}\; \backslash \; left\; v\_0)\; -\; F\; (X)\; \backslash \; right$ the ends of the preceding inequality. The requirement and sufficient so that one can define a valid prolongation of $f$ is as well as the intervals $;\; b\_\; \{X,\; \backslash \; driven\}$ (where $x$ traverses $G$ and $\backslash \; mu$ traverses $\backslash \; R^\; \{+*\}$) has a nonempty intersection. However this is equivalent to:

for all $x,\; there\; \backslash \; in\; G$, and all $\backslash \; driven,\; \backslash \; naked\; >\; 0,\; a\_\; \{X,\; \backslash \; driven\}\; \backslash \; Leq\; b\_\; \{there,\; \backslash \; naked\}$.

But this condition is carried out, by a heavy but easy checking which exploit the convexity of $p$, the linearity of $f$ and presumedly true increase on $G$: indeed, for $x$, $y$ in $G$ and all $\backslash \; driven,\; \backslash \; naked\; >\; 0$:

$b\_\; \{there,\; \backslash \; naked\}\; -\; a\_\; \{X,\; \backslash \; driven\}\; =\; \{1\; \backslash \; over\; \backslash \; naked\}\; \backslash \; left\; v\_0)\; -\; F\; (there)\; \backslash \; right\; -\; \{1\; \backslash \; over\; \backslash \; driven\}\; \backslash \; left\; v\_0)\; \backslash \; right$

$=\; \backslash \; left\; v\_0)\; +\; \{\backslash \; driven\; \backslash \; over\; \{\backslash \; mu+\; \backslash \; naked\}\}\; p\; (y+\; \backslash \; naked\; v\_0)\; -\; \{\backslash \; naked\; \backslash \; over\; \{\backslash \; mu+\; \backslash \; naked\}\}\; F\; (X)\; -\; \{\backslash \; driven\; \backslash \; over\; \{\backslash \; mu+\; \backslash \; naked\}\}\; F\; (there)\; \backslash \; right$

$\backslash \; geq\; \backslash \; left\; v\_0)\; +\; \{\backslash \; driven\; \backslash \; over\; \{\backslash \; mu+\; \backslash \; naked\}\}\; (y+\; \backslash \; naked\; v\_0)\; \backslash \; right)\; there\; -\; F\; \backslash \; left\; there\; (\{\{\backslash \; naked\; X\}\; \backslash \; over\; \{\backslash \; mu+\; \backslash \; naked\}\}\; +\; \{\{\backslash \; driven\}\; \backslash \; over\; \{\backslash \; mu+\; \backslash \; naked\}\}\; \backslash \; right)\; \backslash \; right$

$=\; \backslash \; left\; x+\; \backslash \; driven\}\; \backslash \; over\; \{\backslash \; driven\; +\; \backslash \; naked\}\}\; \backslash \; right)\; -\; F\; \backslash \; left\; there\; (\{\{\backslash \; naked\; x+\; \backslash \; driven\}\; \backslash \; over\; \{\backslash \; driven\; +\; \backslash \; naked\}\}\; \backslash \; right)\; \backslash \; right\; \backslash \; geq\; 0$.

Second part: execution of a transfinite Recurrence

While reasoning gradually, one sees that one can prolong $f$ with increasingly large spaces. If $G$ is of codimension finished in $V$, then the process thus determined stops. If not, one uses the Axiome of the choice.

For that, one considers the whole of the couples $(M,\; G)$ in which $M$ is a vectorial subspace of $V$ container $G$ and $g$ is a linear form on M$$ prolonging $f$ (by respecting the constraint of increase by $p$), and one partially orders it by:

$(M\_1,\; g\_1)\; \backslash \; Leq\; (M\_2,\; g\_2)\; \backslash \; Longleftrightarrow\; M\_1\; \backslash \; subset\; M\_2$ and $\backslash \; forall\; X\; \backslash \; in\; M\_1,\; g\_1\; (X)\; =\; g\_2\; (X)$.

The whole of the couples is inductive. Indeed if $(M\_i,\; g\_i)\; \_\; \{I\; \backslash \; in\; I\}$ is a completely ordered chain, then one poses:

$M\; =\; \backslash \; bigcup\_\; \{I\; \backslash \; in\; I\}\; M\_i$

M is a vectorial subspace. (In general, a union of vector spaces is not a vector space, on the other hand, here, it is the case because the family of the $(M\_i)\; \_\; \{I\; \backslash \; in\; I\}$ is completely ordered).

One defines the linear form G on space M by:

$g\; (X)\; =\; g\_i\; (X)$ if $x\; \backslash \; in\; M\_i$.

It is checked easily that this definition of G is correct. ( M , G ) is then one raising of the chain $(M\_i,\; g\_i)\; \_\; \{I\; \backslash \; in\; I\}$. The Lemme of Zorn applies, and one can then find a subspace NR maximum on which $f$ is prolonged.

Now if NR is not equal to V , then the first part of the demonstration shows that one can prolong $f$ (definite on NR ) with a space strictly larger than NR , which in contradiction with the maximality of NR .

An example of application in functional analysis

The following corollary illustrates how the theorem of Hahn-Banach can very easily produce significant results of functional Analyze.

Is noted $c=\; \backslash |F\; \backslash |$ and one applies the theorem to the convex function $p\; (X)\; =c\; \backslash |X\; \backslash |$.

It is unnecessarily long but instructive to solve the question by using the geometrical form of the theorem of Hahn-Banach: instead of thinking of the convex function $p$, one can also think of convex open the $C$ of which it is the gauge, namely the open ball of center $0$ and $1/c$. If one wants to launch out in this way, the subspace should be introduced then refines $L$, together points of $G$ where $f\; (X)\; =1$. One extends it in a hyperplane closed by applying Hahn-Banach; the linear form continues $g$ for which this hyperplane is the whole of equation $g\; (X)\; =1$ answers the schedule of conditions then.

Some other versions of the theorem

One will find below two alternatives of the “analytical form” which result easily from that highlighted. The first provides an alternative of the result for the complex vector spaces; second specifies that under a good assumption of symmetry of $p$, in particular checked when $p$ is a Semi-norme, one can obtain an increase of the absolute value (or module in the case complexes) prolonged linear form.

One will find alternatives of the geometrical form to the article Séparation of convex the.

The role of the Axiom of the choice

As one saw, the Lemme of Zorn (equivalent with the axiom of the choice) involves the theorem of Hahn-Banach. Actually, the Lemme of the ultrafilters, which is a proposal weaker than the axiom of the choice, is sufficient to show the theorem of Hahn-Banach. But conversely, one knows since work of D. Pinas of 1972 that the theorem of Hahn-Banach is not sufficient to show the lemma of the ultrafilters. Thus, the theorem of Hahn-Banach is not equivalent to the axiom of the choice in the system of axioms of Zermelo-Fraenkel. One must add to that the only system of Zermelo-Fraenkel is not with him sufficient to only show Hahn-Banach, of which any proof must thus rests inevitably on one or another alternative of the axiom of the choice.

References

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