Proof Architecture for Hahn-Banach

From a Local Step to a Global Extension

The Hahn-Banach Theorem, Real Dominated Extension Form, established earlier in this course, turns a linear functional on a subspace into one on the whole vector space while preserving domination by a sublinear function. Its proof architecture separates into two tasks. First, the One-Dimensional Extension Lemma shows how to enlarge a functional’s domain by one direction. Second, a maximality argument shows that a domain which cannot be enlarged must already be the whole space.

The second task is not achieved by adding one direction repeatedly in a sequence. A vector space can have too many directions for any countable list of such steps to exhaust it. Instead, we collect all admissible partial extensions into a partially ordered set. Zorn’s lemma supplies a maximal one, provided every chain in that order has an upper bound. The union of a chain gives precisely that upper bound.

This tutorial develops that architecture rather than introducing a new form of the Hahn-Banach conclusion. The central bookkeeping issue is compatibility: extensions in a chain agree wherever their domains overlap, so their union is a single well-defined linear functional.

The Ordered Set of Partial Extensions

Let \(V\) be a real vector space, let \(p:V\to\mathbb{R}\) be sublinear, and suppose \(f:M\to\mathbb{R}\) is linear on a subspace \(M\subseteq V\), with \(f(m)\leq p(m)\) for all \(m\in M\). Consider every pair \((N,g)\) such that \(N\) is a subspace of \(V\), \(M\subseteq N\), and \(g:N\to\mathbb{R}\) is a linear extension of \(f\) satisfying \(g(n)\leq p(n)\) for all \(n\in N\).

Definition (Order by Extension): For two admissible pairs \((N_1,g_1)\) and \((N_2,g_2)\), write \((N_1,g_1)\preceq(N_2,g_2)\) if \(N_1\subseteq N_2\) and \(g_2|_{N_1}=g_1\). Thus the second pair extends the first both in its domain and in its values.

This relation is a partial order. It is reflexive because every pair extends itself. It is transitive because domain inclusion and agreement of restrictions are both transitive. Finally, if each pair extends the other, then their domains are equal and their functionals agree there, so the pairs are equal. The ordered set is nonempty: \((M,f)\) is one of its elements.

A chain is a subcollection in which every two elements are comparable. In this setting, comparability means that one partial extension extends the other. The chain condition matters: arbitrary admissible extensions need not agree, so they cannot necessarily be combined into one functional.

Why a Chain Has an Upper Bound

Proposition (Union of a Chain of Dominated Extensions): Every nonempty chain of admissible pairs \((N,g)\), ordered by extension, has an admissible upper bound. The upper bound is obtained by taking the union of the domains and joining the functionals on those domains.

Proof. Let \(\mathcal{C}\) be a nonempty chain of admissible pairs. Define \(N_*=\bigcup_{(N,g)\in\mathcal{C}}N\). First, \(N_*\) is a subspace. It contains \(0\), since each domain in the chain is a subspace. If \(x,y\in N_*\), choose \((N_1,g_1),(N_2,g_2)\in\mathcal{C}\) with \(x\in N_1\) and \(y\in N_2\). Because \(\mathcal{C}\) is a chain, one of these pairs extends the other. In particular, one of \(N_1,N_2\) contains both \(x\) and \(y\). That domain contains \(ax+by\) for every \(a,b\in\mathbb{R}\), so \(ax+by\in N_*\).

For \(x\in N_*\), choose any \((N,g)\in\mathcal{C}\) with \(x\in N\), and define \(g_*(x)=g(x)\). This value does not depend on the choice. Indeed, if \(x\) belongs to domains \(N_1\) and \(N_2\), comparability makes one pair an extension of the other, and the extending functional agrees with the earlier one on its domain. Thus both choices give the same value.

To verify linearity, take \(x,y\in N_*\) and \(a,b\in\mathbb{R}\). As above, a single domain \(N\) in the chain contains both vectors. On that domain, \(g_*\) agrees with its functional \(g\); hence \(g_*(ax+by)=g(ax+by)=ag(x)+bg(y)=ag_*(x)+bg_*(y)\). For any \(x\in N_*\), choose a pair whose domain contains \(x\). Domination on that domain gives \(g_*(x)=g(x)\leq p(x)\). The same argument shows that \(g_*\) extends \(f\), because every domain in the ordered set contains \(M\) and every functional there extends \(f\). Therefore \((N_*,g_*)\) is admissible and extends every member of \(\mathcal{C}\). It is an upper bound, as claimed. \(\square\)

The argument depends on the chain condition in two distinct places. It puts two chosen vectors into one domain, allowing linearity to be checked there. It also ensures that two functionals assigned to the same vector agree. Without comparability, neither conclusion is automatic.

The Maximality Principle

Theorem (Maximality Turns Local Extension into Global Extension): Let \(\mathcal{P}\) be a nonempty partially ordered set in which every chain has an upper bound. Suppose that whenever \(q\in\mathcal{P}\) is not maximal, there is a strictly larger element of \(\mathcal{P}\) that extends \(q\). Then Zorn’s lemma gives a maximal element; if every element that is not already global admits a strict extension, every maximal element is global.

Proof. By Zorn’s lemma, \(\mathcal{P}\) has a maximal element \(q_*\). If \(q_*\) were not global, the stated extension property would give an element \(q\in\mathcal{P}\) strictly larger than \(q_*\). That contradicts maximality, which means that no strictly larger element exists. Therefore \(q_*\) is global. \(\square\)

For dominated linear extensions, “global” means that the domain is \(V\). The union proposition verifies the chain upper-bound hypothesis. The remaining step is to show that every admissible pair with a proper domain has a strict admissible extension. That is exactly where the One-Dimensional Extension Lemma from the earlier Hahn-Banach tutorial enters.

Application to the Hahn-Banach proof architecture. Apply Zorn’s lemma to the ordered set of admissible pairs. Let \((N,g)\) be a maximal pair. If \(N\neq V\), choose \(x\in V\setminus N\). The One-Dimensional Extension Lemma applies to the subspace \(N\), the dominated functional \(g\), and the new direction \(x\). It provides a dominated linear extension to \(N+\mathbb{R}x\). Since \(x\notin N\), this enlarged subspace strictly contains \(N\), contradicting maximality. Therefore \(N=V\). The maximal pair is a global dominated extension. This is the assembly of the local lemma and Zorn’s lemma that underlies the already established Hahn-Banach theorem. \(\square\)

Worked Examples

Worked Example: A Chain of Compatible Extensions

Let \(V=\mathbb{R}^3\), let \(p(x,y,z)=|x|+2|y|+3|z|\), and let \(M\) be the span of \(e_1=(1,0,0)\). Define \(f(te_1)=\tfrac12t\). This is dominated on \(M\), since \(\tfrac12t\leq\tfrac12|t|\leq |t|=p(te_1)\). Consider the subspaces \(N_0=M\), \(N_1=\operatorname{span}\{e_1,e_2\}\), and \(N_2=V\), with functionals \(g_0(x,0,0)=\tfrac12x\), \(g_1(x,y,0)=\tfrac12x+y\), and \(g_2(x,y,z)=\tfrac12x+y+2z\).

They form a chain: \(g_1\) restricts to \(g_0\) on \(N_0\), and \(g_2\) restricts to \(g_1\) on \(N_1\). Each is dominated on its domain. In fact, for every \((x,y,z)\in V\), \(g_2(x,y,z)\leq \tfrac12|x|+|y|+2|z|\leq |x|+2|y|+3|z|=p(x,y,z)\). The union construction assigns the value prescribed by whichever member of the chain contains the vector; compatibility ensures this is one functional. Here the largest domain already equals \(V\), so the union is global.

Worked Example: Why Competing Extensions Cannot Simply Be United

Let \(V=\mathbb{R}^2\), \(M=\operatorname{span}\{(1,0)\}\), \(p(x,y)=|x|+|y|\), and let \(f\) be zero on \(M\). Two dominated extensions to all of \(V\) are \(F_+(x,y)=y\) and \(F_-(x,y)=-y\). They extend \(f\), and for every \((x,y)\), \(F_+(x,y)=y\leq |y|\leq p(x,y)\) and \(F_-(x,y)=-y\leq |y|\leq p(x,y)\).

Neither extension extends the other: their common domain is all of \(V\), but \(F_+(0,1)=1\) while \(F_-(0,1)=-1\). They are incomparable in the order by extension. Attempting to “take their union” would assign two different values to \((0,1)\), so it would not define a function. This illustrates why the chain upper-bound argument cannot be replaced by a union over all admissible extensions.

Worked Example: The Available Values in a One-Direction Extension

Again let \(V=\mathbb{R}^2\), \(p(x,y)=|x|+2|y|\), and \(M=\operatorname{span}\{(1,0)\}\), but now define \(f(t,0)=\tfrac12t\). Any linear extension \(F\) to \(V\) has the form \(F(x,y)=\tfrac12x+dy\) for some \(d\in\mathbb{R}\). If \(|d|\leq2\), then \(F(x,y)\leq\tfrac12|x|+|d||y|\leq |x|+2|y|=p(x,y)\). Thus each such \(d\) gives a dominated extension.

The bound \(|d|\leq2\) is also necessary. Substituting \((0,1)\) into domination gives \(d\leq2\), while substituting \((0,-1)\) gives \(-d\leq2\). Together these inequalities give \(-2\leq d\leq2\). In particular, the direction \((0,1)\) can be added with \(d=1\), producing \(F(x,y)=\tfrac12x+y\). The calculation illustrates the local choice made by a one-dimensional extension step; the maximality argument does not need to choose a unique value.

Where the Argument Can Go Wrong

A common mistake is to assume that repeatedly applying the one-dimensional lemma must eventually produce a functional on all of \(V\). That reasoning has no valid stopping point in a general vector space. The maximality argument avoids any countability assumption: it considers the entire ordered set of admissible partial extensions at once. Zorn’s lemma supplies a maximal element because every chain has an upper bound, not because a sequence of domains has been shown to converge to \(V\).

Another mistake is to treat every family of extensions as compatible. The ordered set records both domain inclusion and agreement of values. When a chain is united, these conditions make the definition unambiguous and preserve linearity. If two extensions are incomparable, their values can conflict, as the second worked example shows.

Finally, the algebraic extension argument and the continuity conclusion are separate parts of the overall Hahn-Banach picture. The maximality construction gives a linear functional dominated by \(p\). In the normed-space specialization, taking \(p(x)=\|x\|\) and using domination on both \(x\) and \(-x\) yields \(|F(x)|\leq\|x\|\), which gives boundedness and hence continuity. The chain argument itself does not establish continuity; it preserves the domination condition that later implies it.

Check Your Understanding

Use the order by extension and the maximality argument to answer the following questions.

What does it mean for one admissible pair \((N_1,g_1)\) to be below another pair \((N_2,g_2)\)?
Why does the union of a chain of domains form a subspace?
Where is the chain condition used to show that the union functional is well-defined?
How does the One-Dimensional Extension Lemma contradict maximality when a domain is a proper subspace of \(V\)?
Why can two dominated extensions that are not comparable fail to have a functional as their union?
In the normed-space specialization, what additional step turns domination into continuity?