In functional analysis, a recurring task is to construct bounded linear functionals with prescribed behaviour. You may need a functional that attains a certain value on a specific vector. You may need one that vanishes on a given subspace but not on a particular element outside it. You may need to certify that two distinct points in a [Banach space](/page/Banach%20Space) are genuinely different by exhibiting a functional that separates them. In finite dimensions, such constructions are routine: extend a basis, define the functional by its values on each basis vector, and check linearity. In infinite dimensions, this strategy collapses. There is no countable algebraic basis to work with in general, and even when one exists (a Hamel basis), the resulting functional may fail to be bounded, rendering it useless for analysis.

The Hahn-Banach theorem resolves this difficulty. It guarantees that a bounded linear functional defined on a subspace of a [normed vector space](/page/Normed%20Vector%20Space) can always be extended to the entire space without increasing its norm. The theorem comes in several forms --- an algebraic version for arbitrary real vector spaces, an analytic version for normed spaces, and a geometric version that separates convex sets by hyperplanes --- and each form serves as a foundational tool for different parts of the theory.

The consequences are profound. The theorem ensures that the [dual space](/page/Dual%20Space) $X^*$ of any normed space $X$ is rich enough to separate points, to recover the norm of every element, and to support a robust duality theory. Without it, functionals on subspaces would be trapped there, dual spaces might be trivial, and the entire program of weak topologies, reflexivity, and variational methods would be untenable.

[example: The Extension Problem in Finite Dimensions] Consider $\mathbb{R}^3$ with the Euclidean norm and the subspace $W = \{(x_1, x_2, 0) : x_1, x_2 \in \mathbb{R}\}$. Define the functional $f \in W^*$ by $f(x_1, x_2, 0) = 3x_1 - x_2$. Its operator norm on $W$ is: \begin{align*} \|f\|_{W^*} = \sup_{\substack{(x_1, x_2, 0) \neq 0}} \frac{|3x_1 - x_2|}{(x_1^2 + x_2^2)^{1/2}} = (3^2 + (-1)^2)^{1/2} = \sqrt{10}. \end{align*} Any extension $\tilde{f}: \mathbb{R}^3 \to \mathbb{R}$ must take the form $\tilde{f}(x_1, x_2, x_3) = 3x_1 - x_2 + cx_3$ for some $c \in \mathbb{R}$. Its norm on $\mathbb{R}^3$ is $\|\tilde{f}\|_{(\mathbb{R}^3)^*} = (9 + 1 + c^2)^{1/2} = (10 + c^2)^{1/2}$, which is minimised when $c = 0$. The unique norm-preserving extension is $\tilde{f}(x_1, x_2, x_3) = 3x_1 - x_2$. In finite dimensions, the extension is straightforward and the norm-preserving choice is often unique (here, by orthogonality). In infinite dimensions, neither the construction nor the uniqueness is guaranteed without the Hahn-Banach theorem. Extensions always exist algebraically (via a Hamel basis), but they may increase the norm or fail to be continuous altogether. [/example]

## Definition

The central objects in the Hahn-Banach theorem are sublinear functionals, which serve as the dominating functions that control the extension. A sublinear functional is not a norm --- it need not be symmetric --- but it captures just enough structure (positive homogeneity and the triangle inequality) to constrain the values of a linear functional from above.

[definition: Sublinear Functional] Let $V$ be a real vector space. A function $p: V \to \mathbb{R}$ is called a **sublinear functional** if it satisfies: 1. **Positive homogeneity:** $p(\lambda v) = \lambda \, p(v)$ for all $\lambda > 0$ and $v \in V$. 2. **Subadditivity:** $p(v_1 + v_2) \le p(v_1) + p(v_2)$ for all $v_1, v_2 \in V$. [/definition]

Every norm is a sublinear functional (with the additional property that $p(-v) = p(v)$), but the class is strictly larger. For instance, if $f: V \to \mathbb{R}$ is any linear functional, then $p(v) = f(v)$ is sublinear (it satisfies both conditions by linearity), even though $p(-v) = -p(v) \neq p(v)$ in general. The Minkowski functional (gauge) of a convex set containing the origin is another important example, and it plays a key role in the geometric form of the theorem.

[example: The Minkowski Functional] Let $X$ be a real [normed vector space](/page/Normed%20Vector%20Space), and let $C \subset X$ be a convex, open set with $0 \in C$. The **Minkowski functional** (or gauge) of $C$ is \begin{align*} \mu_C: X &\to \mathbb{R} \\ x &\mapsto \inf\{t > 0 : x \in tC\}. \end{align*} We verify that $\mu_C$ is sublinear. For positive homogeneity: given $\lambda > 0$, \begin{align*} \mu_C(\lambda x) = \inf\{t > 0 : \lambda x \in tC\} = \inf\{t > 0 : x \in (t/\lambda) C\} = \lambda \inf\{s > 0 : x \in sC\} = \lambda\, \mu_C(x), \end{align*} where we substituted $s = t/\lambda$. For subadditivity: given $x_1, x_2 \in X$ and $\varepsilon > 0$, choose $t_1, t_2 > 0$ with $x_1 \in t_1 C$, $x_2 \in t_2 C$, $t_1 < \mu_C(x_1) + \varepsilon$, and $t_2 < \mu_C(x_2) + \varepsilon$. Write $x_1 = t_1 c_1$ and $x_2 = t_2 c_2$ with $c_1, c_2 \in C$. Then \begin{align*} x_1 + x_2 = (t_1 + t_2)\left(\frac{t_1}{t_1 + t_2} c_1 + \frac{t_2}{t_1 + t_2} c_2\right). \end{align*} The term in parentheses is a convex combination of $c_1$ and $c_2$, so it lies in $C$ by convexity. Therefore $x_1 + x_2 \in (t_1 + t_2)C$, giving $\mu_C(x_1 + x_2) \le t_1 + t_2 < \mu_C(x_1) + \mu_C(x_2) + 2\varepsilon$. Since $\varepsilon > 0$ was arbitrary, $\mu_C(x_1 + x_2) \le \mu_C(x_1) + \mu_C(x_2)$. The Minkowski functional also characterises membership in $C$: since $C$ is open, $x \in C$ if and only if $\mu_C(x) < 1$. This equivalence is what makes $\mu_C$ the bridge between the analytic and geometric forms of the Hahn-Banach theorem. [/example]

## The Algebraic Extension Theorem

The core of the Hahn-Banach theory is a purely algebraic result that requires no topology. It states that a linear functional dominated by a sublinear functional on a subspace can be extended to the full vector space while preserving the domination. The proof relies on Zorn's lemma (equivalently, the Axiom of Choice), and this non-constructive character is an essential feature: in general, the extension cannot be made explicit, and it is not unique.

[quotetheorem:879]

Several aspects of this statement deserve emphasis. First, the domination condition $f(w) \le p(w)$ is a *one-sided* inequality --- it does not assert $|f(w)| \le p(w)$. This asymmetry is deliberate: a sublinear functional $p$ satisfies $p(-v) \neq p(v)$ in general, so the two-sided bound is not available from $p$ alone. However, when $p$ happens to be a seminorm (so that $p(-v) = p(v)$), the one-sided bound $\tilde{f}(v) \le p(v)$ applied to both $v$ and $-v$ gives the absolute bound $|\tilde{f}(v)| \le p(v)$.

Second, the theorem asserts existence but not uniqueness. The extension depends on the choice of $\tilde{f}(v_0)$ at each step of the inductive process (or, in the Zorn's lemma formulation, on the choice of maximal element). Different choices of $\tilde{f}(v_0)$ at the codimension-one step yield different extensions. Uniqueness holds only when $W$ is dense in $V$ and $p$ is continuous (forcing the extension to be determined by continuity).

Third, the theorem is purely algebraic: no topology is assumed on $V$, no continuity is required of $f$ or $p$, and the extension $\tilde{f}$ is not claimed to be continuous. Continuity enters only in the normed-space version, where the choice of a specific $p$ (namely, $p(v) = \|f\|_{W^*} \cdot \|v\|_V$) ensures it.

## The Norm-Preserving Extension

The algebraic theorem becomes a powerful analytic tool once we specialise the sublinear functional $p$ to the norm. The resulting statement answers the motivating question directly: bounded linear functionals on subspaces extend to the whole space with no increase in norm.

To pass from the algebraic to the analytic version, the key observation is that if $f: W \to \mathbb{R}$ is a bounded linear functional on a subspace $W$ of a normed space $V$, then the function

\begin{align*} p: V &\to \mathbb{R} \\ v &\mapsto \|f\|_{W^*} \cdot \|v\|_V \end{align*}

is a sublinear functional (positive homogeneity comes from the homogeneity of the norm, and subadditivity from the triangle inequality). Moreover, $f(w) \le |f(w)| \le \|f\|_{W^*} \cdot \|w\|_V = p(w)$ for all $w \in W$, so the domination condition of the algebraic theorem is satisfied. The algebraic extension $\tilde{f}$ then satisfies $\tilde{f}(v) \le \|f\|_{W^*} \cdot \|v\|_V$ for all $v \in V$. Applying this to $-v$ as well gives $|\tilde{f}(v)| \le \|f\|_{W^*} \cdot \|v\|_V$, which is precisely the statement that $\tilde{f}$ is bounded with $\|\tilde{f}\|_{V^*} \le \|f\|_{W^*}$. The reverse inequality $\|\tilde{f}\|_{V^*} \ge \|f\|_{W^*}$ is immediate since $W \subset V$.