Hahn–Banach Theorem — Mathematics Knowledge System

Overview

The Hahn–Banach theorem is one of the three pillars of functional analysis. In its most basic form it asserts that a bounded linear functional defined on a subspace of a normed space can be extended to the entire space without increasing its norm. The theorem has two flavours: the analytic extension form and the geometric separation form. The latter—that two disjoint convex sets can be separated by a hyperplane—is the cornerstone of convex analysis, optimization, and mathematical economics. The theorem requires no completeness (it holds for all normed spaces) and is proved via Zorn's lemma.

Intuition

Suppose you can measure a quantity along a subspace—e.g., a linear price functional on a subset of commodities. The Hahn–Banach theorem guarantees you can extend that price to all goods without distorting its size. The geometric version says: if two convex sets do not overlap, there is a 'price vector' (a hyperplane) that separates them—one set pays more than the other. This is the mathematical backbone of duality in linear programming and the supporting hyperplane theorem in convex geometry.

Formal Definition

Definition

Let $X$ be a real vector space, $p: X \to \mathbb{R}$ a sublinear functional ($p(x+y) \le p(x)+p(y)$, $p(tx)=tp(x)$ for $t \ge 0$), $M \subset X$ a subspace, and $f: M \to \mathbb{R}$ a linear functional dominated by $p$ (i.e., $f(x) \le p(x)$ for all $x \in M$). Then there exists a linear extension $F: X \to \mathbb{R}$ with $F|_M = f$ and $F(x) \le p(x)$ for all $x \in X$. In the normed-space setting with $p(x)=\|f\|\cdot\|x\|$, this gives $\|F\| = \|f\|_{M^*}$.

F\big|_M = f, \quad F(x) \le p(x) \quad \forall x \in X

Extension with domination

\|F\|_{X^*} = \|f\|_{M^*}

Norm-preserving extension

Proofs

Hahn–Banach Extension (one-dimensional step)

$\text{Let } M \subset X \text{ and pick } z \in X \setminus M. \text{ Set } M_1 = M + \mathbb{R}z.$ (We extend $f$ from $M$ to $M_1$ by choosing $f(z) = c$ freely.)
$\text{Need: } f(m) + tc \le p(m + tz) \text{ for all } m \in M, t \in \mathbb{R}.$ (This is the domination condition on $M_1$.)
$\sup_{m \in M}[f(m) - p(m - z)] \le c \le \inf_{m \in M}[p(m + z) - f(m)].$ (Rearranging for $t = 1$ and $t = -1$; the sup $\le$ inf by sublinearity of $p$, so a valid $c$ exists.)
$\text{Apply Zorn's lemma to the set of dominated extensions ordered by inclusion of domain.}$ (Every chain has an upper bound; maximal element must have domain $X$.)

Theorems

Theorem 3.1: Hahn–Banach Extension Theorem

Let M be a subspace of a normed space X and f \in M^*. There exists F \in X^* with F|_M = f and \|F\|_{X^*} = \|f\|_{M^*}.

Theorem 3.2: Hahn–Banach Separation Theorem (First Geometric Form)

Let A and B be nonempty disjoint convex subsets of a normed space X with A open. Then there exists f \in X^* and c \in \mathbb{R} such that f(a) < c \le f(b) for all a \in A, b \in B.

Theorem 3.3: Hahn–Banach Separation Theorem (Second Geometric Form)

Let A and B be nonempty disjoint closed convex subsets of a Banach space X with A compact. Then there exists f \in X^* and c_1 < c_2 such that f(a) \le c_1 < c_2 \le f(b) for all a \in A, b \in B.

Theorem 3.4: Corollary: Dual separates points

If x \neq 0 in a normed space X, there exists f \in X^* with \|f\| = 1 and f(x) = \|x\|.

Worked Examples

Define $M = \text{span}\{x_0\}$ and $f_0(tx_0) = t\|x_0\|$ for $t \in \mathbb{R}$.
Check: $|f_0(tx_0)| = |t|\|x_0\| = \|tx_0\|$, so $f_0$ is bounded with $\|f_0\|_{M^*} = 1$.
$|f_0(tx_0)| = \|tx_0\|$
By Hahn–Banach, extend $f_0$ to $F \in X^*$ with $\|F\| = \|f_0\|_{M^*} = 1$.
Then $F(x_0) = f_0(x_0) = \|x_0\|$, as required.
$F(x_0) = \|x_0\|, \quad \|F\| = 1$

Answer: Such an $F \in X^*$ is produced directly by Hahn–Banach applied to the one-dimensional subspace.

Practice Problems

Difficulty 8/10

Prove that the canonical embedding $J: X \to X^{**}$, $Jx(f) = f(x)$, is an isometric injection.

Difficulty 9/10

A subspace $M$ of a normed space $X$ is dense in $X$ if and only if every $f \in X^*$ that vanishes on $M$ must be identically zero.

Difficulty 8/10

Use the geometric Hahn–Banach theorem to prove that in a normed space, a closed convex set $C$ equals the intersection of all closed half-spaces containing it.

Common Mistakes

Common Mistake

The extension is unique

Extensions are generally not unique. The Hahn–Banach theorem guarantees existence but not uniqueness. Uniqueness holds only in special cases, e.g., when $M$ is dense in $X$.

Common Mistake

Hahn–Banach requires the Axiom of Choice only in an avoidable way

For separable spaces, Hahn–Banach can be proved without full AC (only dependent choice is needed). For general spaces, Zorn's lemma (equivalent to AC) is essential.

Quiz

Summary

Hahn–Banach (analytic form): a bounded linear functional on a subspace extends to the whole space preserving its norm.
The proof uses Zorn's lemma to extend one dimension at a time while maintaining domination by the sublinear functional.
Key corollary: the dual space $X^*$ separates points—for each $x \neq 0$ there is $f$ with $f(x) = \|x\|$.
Geometric form: disjoint convex sets (with a compactness/openness condition) can be separated by a closed hyperplane.
Applications include the canonical embedding $X \hookrightarrow X^{**}$ being isometric, the characterisation of reflexivity, and the supporting hyperplane theorem in convex analysis.

References

BookRudin, W. Functional Analysis. 2nd ed., McGraw-Hill, 1991. Chapter 3.
BookBrezis, H. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer, 2011. Chapter 1.
WebsiteWikipedia — Hahn–Banach theorem