Hahn-Banach Theorem

Extending a Linear Functional

The Closed Graph Theorem used completeness and the Bounded Inverse Theorem to obtain continuity from a structural condition on an operator. The Hahn-Banach Theorem addresses a different problem: given a linear functional on a subspace, can it be extended to the whole space without losing a prescribed bound? Its answer is yes when the bound is expressed through a sublinear function. The theorem is algebraic in its main form: it does not require a norm, a metric, or completeness.

We work with real vector spaces and real-valued linear functionals. The central idea is to extend the functional one vector at a time. At each step, its value on the new vector must be chosen from an interval that preserves domination. Zorn’s lemma then supplies an extension whose domain cannot be enlarged, and the one-vector step shows that this domain must already be the whole space.

Sublinearity is weaker than linearity: the inequality in the first condition need not be an equality, and homogeneity is required only for nonnegative scalars. Domination is a one-sided inequality. In a normed space, taking \(p(x)=C\|x\|\) will turn this one-sided condition into an absolute-value bound, because it can also be applied to \(-x\).

The One-Dimensional Extension Step

Suppose a functional \(f\) is already defined on a subspace \(M\), and choose a vector \(x_0\notin M\). Every vector in the enlarged subspace \(M+\mathbb{R}x_0\) has a unique expression \(m+t x_0\), where \(m\in M\) and \(t\in\mathbb{R}\). To extend \(f\), it is enough to choose a real number \(c\) as the proposed value at \(x_0\), and set the value at \(m+t x_0\) to \(f(m)+tc\).

The choice of \(c\) must work for positive and negative \(t\). The following lemma identifies the required interval and proves that it is nonempty.

Proof. To ensure domination for vectors \(m+t x_0\) with \(t>0\), a candidate value \(c=g(x_0)\) should satisfy \(c\leq p(u+x_0)-f(u)\) for every \(u\in M\). To ensure it for vectors with \(t<0\), it should also satisfy \(c\geq f(u)-p(u-x_0)\) for every \(u\in M\). We show that every proposed lower bound is at most every proposed upper bound. For \(u,v\in M\), subadditivity and domination give \[ f(u)+f(v)=f(u+v)\leq p(u+v)\leq p(u-x_0)+p(v+x_0). \] Rearranging yields \[ f(u)-p(u-x_0)\leq p(v+x_0)-f(v). \]

The sets of lower and upper bounds are nonempty because \(0\in M\). Fixing any \(v\in M\), the displayed inequality shows that all lower bounds are bounded above by the finite number \(p(v+x_0)-f(v)\). Thus their supremum is finite. Fixing any \(u\in M\) shows that all upper bounds are bounded below by the finite number \(f(u)-p(u-x_0)\), so their infimum is finite. The inequality between every lower and upper bound implies that the supremum of the lower bounds is at most the infimum of the upper bounds. Choose \(c\) between them.

Define \(g(m+t x_0)=f(m)+tc\). The representation is unique: if \(m+t x_0=m'+t'x_0\) and \(t\ne t'\), then \(x_0\in M\), contrary to the choice of \(x_0\); if \(t=t'\), then \(m=m'\). The definition is therefore well-defined and linear. For \(t=0\), domination is the original inequality for \(f\). If \(t>0\), write \(m=t u\) with \(u\in M\). Then \[ g(m+t x_0)=t(f(u)+c)\leq t p(u+x_0)=p(m+t x_0). \] If \(t<0\), write \(s=-t>0\) and \(m=s u\) with \(u\in M\). The lower-bound condition on \(c\) gives \[ g(m+t x_0)=s(f(u)-c)\leq s p(u-x_0)=p(m+t x_0). \] These cases prove domination on the enlarged subspace. \(\square\)

The Hahn-Banach Theorem

Proof. Consider the collection \(\mathcal{P}\) of all pairs \((N,g)\) such that \(N\) is a subspace of \(V\) containing \(M\), \(g:N\to\mathbb{R}\) is linear, \(g\) extends \(f\), and \(g(n)\leq p(n)\) for every \(n\in N\). This collection is nonempty because \((M,f)\in\mathcal{P}\). Order it by extension: \((N_1,g_1)\leq(N_2,g_2)\) if \(N_1\subseteq N_2\) and \(g_2\) agrees with \(g_1\) on \(N_1\).

Every chain in \(\mathcal{P}\) has an upper bound. Take the union \(N\) of the subspaces in the chain. Since the subspaces are nested along the chain, \(N\) is a subspace. Define \(g(n)=g_0(n)\) using any pair \((N_0,g_0)\) in the chain with \(n\in N_0\). This value is independent of the chosen pair: any two such pairs are comparable, and the larger functional extends the smaller. The same comparison shows that \(g\) is linear. Each value satisfies \(g(n)\leq p(n)\), because \(n\) belongs to one of the subspaces in the chain. Thus \((N,g)\) is an upper bound in \(\mathcal{P}\).

By Zorn’s lemma, \(\mathcal{P}\) has a maximal element, say \((N,g)\). If \(N\ne V\), choose \(x_0\in V\setminus N\). The one-dimensional extension lemma, applied to \(N\), \(g\), and \(x_0\), produces a dominated linear extension to \(N+\mathbb{R}x_0\). This gives an element of \(\mathcal{P}\) strictly larger than \((N,g)\), contradicting maximality. Therefore \(N=V\), and \(F=g\) is the required extension. \(\square\)

Neither completeness nor boundedness appears in this theorem’s hypotheses. Those concepts enter an important consequence when the sublinear function is built from a norm. The general theorem also explains why the one-sided formulation is useful: the one-dimensional step needs only subadditivity and nonnegative homogeneity, not symmetry such as \(p(-x)=p(x)\).

The Norm-Preserving Form

Proof. Set \(C=\|f\|\), so \(|f(m)|\leq C\|m\|\) for every \(m\in M\). In particular, \(f(m)\leq p(m)\) for the sublinear function \(p(x)=C\|x\|\). The Hahn-Banach Theorem gives an extension \(F\) satisfying \(F(x)\leq C\|x\|\) for every \(x\in X\). Applying the same bound to \(-x\) and using linearity gives \(-F(x)=F(-x)\leq C\|x\|\). Hence \(|F(x)|\leq C\|x\|\), so \(\|F\|\leq C\). Since \(F\) extends \(f\), the supremum defining \(\|F\|\) is at least the supremum over unit vectors in \(M\), and therefore \(\|F\|\geq\|f\|=C\). Thus \(\|F\|=C\). \(\square\)

If \(M=\{0\}\), its only functional has norm zero; the same argument applies with \(C=0\), and the resulting extension is the zero functional. The norm-preserving statement does not require \(X\) or \(M\) to be complete.

Worked Examples

Why the Theorem Matters

The norm-preserving corollary gives a useful supply of bounded linear functionals. For any nonzero \(x_0\) in a real normed space, define \(f(t x_0)=t\|x_0\|\) on the span of \(x_0\). The calculation \(|f(t x_0)|=\|t x_0\|\) shows that \(\|f\|=1\). Extending \(f\) without increasing its norm produces a functional \(F\) with \(\|F\|=1\) and \(F(x_0)=\|x_0\|\). Thus each nonzero vector can be tested by a bounded linear functional that attains its norm at that vector.

A common pitfall is to replace the one-sided domination hypothesis with a claim that \(f(m)\leq p(m)\) automatically means \(|f(m)|\leq p(m)\). It does not in general. The absolute-value estimate follows in the normed-space application because \(p(x)=C\|x\|\) has the symmetry \(p(-x)=p(x)\), and the domination inequality can be applied to both \(x\) and \(-x\). Another important distinction is that the general theorem uses Zorn’s lemma to obtain an extension on the whole vector space; the one-dimensional lemma alone only extends to one additional direction at a time.