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The Hahn–Banach Theorem

We saw that X∗X^* could be trivial (LpL^p for 0<p<10 < p < 1), and asked whether the dual is always a complete measurement system for Banach spaces. The following theorem answers yes: every bounded linear functional on a subspace extends to the whole space with the same norm. This guarantees that X∗X^* is rich enough to separate points, recover the norm, and support the weak topology.

Theorem 1 (Hahn–Banach theorem)

Let XX be a normed space and M⊂XM \subset X a linear subspace. Let f∈M∗f \in M^* with ∣f(x)∣≤k∥x∥|f(x)| \leq k\|x\| for all x∈Mx \in M. Then there exists an extension F∈X∗F \in X^* of ff to all of XX with ∣F(x)∣≤k∥x∥|F(x)| \leq k\|x\| for all x∈Xx \in X.

Note that FF is not necessarily unique.

Proof 1

The proof has three ingredients. The first two are structural; the third is where the analysis lives.

Ingredient 1: Zorn’s lemma setup.

Define the poset:

E={(g,G):G⊇M is a subspace,  g extends f on G,  ∣g(x)∣≤k∥x∥  ∀x∈G}E = \{(g, G) : G \supseteq M \text{ is a subspace}, \; g \text{ extends } f \text{ on } G, \; |g(x)| \leq k\|x\| \;\forall x \in G\}
(1)

Order by extension: (h,H)≤(g,G)(h, H) \leq (g, G) if H⊆GH \subseteq G and g∣H=hg|_H = h. This is nonempty ((f,M)∈E(f, M) \in E), chains have upper bounds (take the union), so Zorn gives a maximal element (F,W)(F, W).

Ingredient 2: Maximal implies total.

Suppose W≠XW \neq X. Pick z∈X∖Wz \in X \setminus W and set Z=W+span(z)Z = W + \text{span}(z). By the codimension-1 decomposition, every element of ZZ can be written uniquely as w+αzw + \alpha z with w∈Ww \in W, α∈R\alpha \in \mathbb{R}. We want to define:

F~(w+αz)=F(w)+αc\tilde{F}(w + \alpha z) = F(w) + \alpha c
(2)

for some constant cc. If we can find any valid cc, then (F~,Z)(\tilde{F}, Z) strictly extends (F,W)(F, W), contradicting maximality.

Ingredient 3: The constant cc exists.

The requirement ∣F~(w+αz)∣≤k∥w+αz∥|\tilde{F}(w + \alpha z)| \leq k\|w + \alpha z\| constrains cc to lie in an interval. The existing bound ∣F(⋅)∣≤k∥⋅∥|F(\cdot)| \leq k\|\cdot\| on WW, combined with the triangle inequality, guarantees that the lower bound on cc is ≤\leq the upper bound. The interval is nonempty.

Since cc exists, we can extend, contradicting maximality. Therefore W=XW = X.

What to take away

  • Ingredients 1 and 2 follow the same Zorn’s lemma template as the Hamel basis proof, combined with the codimension-1 decomposition. No new ideas are needed.

  • Ingredient 3 is the only piece that requires actual work, and the only tool it uses is the triangle inequality. This is where the assumption that we’re in a normed space matters — and it’s exactly the property that fails in LpL^p for 0<p<10 < p < 1, which is why the dual collapses there.

Geometric Consequences

The corollaries of Hahn–Banach are all separation results — they say you can always find a hyperplane that separates things.

Corollary 1 (The distinguishing property)

Let x,y∈Xx, y \in X. If f(x)=f(y)f(x) = f(y) for all f∈X∗f \in X^* then x=yx = y.

Proof 2

If x≠yx \neq y, define φ(α(x−y))=α∥x−y∥\varphi(\alpha(x - y)) = \alpha\|x - y\| on span(x−y)\text{span}(x - y). Then ∥φ∥=1\|\varphi\| = 1 and Hahn–Banach extends φ\varphi to F∈X∗F \in X^* with F(x)−F(y)=F(x−y)=∥x−y∥≠0F(x) - F(y) = F(x-y) = \|x-y\| \neq 0, contradicting the assumption.

The contrapositive gives the intuition: if two points are on the same level set for every possible foliation, no matter how you orient the hyperplanes, then they must be the same point.

Source
import numpy as np
import matplotlib.pyplot as plt

fig, axes = plt.subplots(1, 2, figsize=(10, 4))

x_pt = np.array([1.0, 0.8])
y_pt = np.array([-0.5, 0.8])

# --- Left panel: f(u,v) = v (horizontal level sets) ---
ax = axes[0]

# Foliation by f(u,v) = v (horizontal lines)
for level in np.arange(-2, 2.5, 0.4):
    alpha = 0.4 if abs(level - round(level)) < 0.01 else 0.12
    lw = 0.8 if abs(level - round(level)) < 0.01 else 0.4
    ax.axhline(level, color='C1', alpha=alpha, lw=lw)

# Level sets through x and y (same height!)
ax.axhline(x_pt[1], color='C1', lw=2, alpha=0.7, ls='--')

ax.plot(*x_pt, 'C3o', ms=9, zorder=5)
ax.plot(*y_pt, 'C2o', ms=9, zorder=5)
ax.text(x_pt[0] + 0.12, x_pt[1] + 0.15, r'$x$', fontsize=12, color='C3')
ax.text(y_pt[0] - 0.25, y_pt[1] + 0.15, r'$y$', fontsize=12, color='C2')

ax.text(0.25, 0.3, r'$f(x) = f(y) = 0.8$', fontsize=10, color='C1',
        ha='center')

ax.set_xlim(-2.2, 2.2)
ax.set_ylim(-2.4, 2.2)
ax.set_aspect('equal')
ax.set_title(r'$f(u,v) = v$: same reading, no separation', fontsize=11)
ax.set_xlabel(r'$u$', fontsize=10)
ax.set_ylabel(r'$v$', fontsize=10)
ax.tick_params(labelsize=9)

# --- Right panel: g(u,v) = u (vertical level sets) ---
ax = axes[1]

# Foliation by g(u,v) = u (vertical lines)
for level in np.arange(-2, 2.5, 0.4):
    alpha = 0.4 if abs(level - round(level)) < 0.01 else 0.12
    lw = 0.8 if abs(level - round(level)) < 0.01 else 0.4
    ax.axvline(level, color='C4', alpha=alpha, lw=lw)

# Level sets through x and y (different!)
ax.axvline(x_pt[0], color='C3', lw=2, alpha=0.6, ls='--')
ax.axvline(y_pt[0], color='C2', lw=2, alpha=0.6, ls='--')

ax.plot(*x_pt, 'C3o', ms=9, zorder=5)
ax.plot(*y_pt, 'C2o', ms=9, zorder=5)
ax.text(x_pt[0] + 0.12, x_pt[1] + 0.15, r'$x$', fontsize=12, color='C3')
ax.text(y_pt[0] - 0.25, y_pt[1] + 0.15, r'$y$', fontsize=12, color='C2')

ax.text(x_pt[0] + 0.08, -1.7, rf'$g(x) = {x_pt[0]}$', fontsize=10,
        color='C3', ha='left')
ax.text(y_pt[0] - 0.08, -1.7, rf'$g(y) = {y_pt[0]}$', fontsize=10,
        color='C2', ha='right')

ax.set_xlim(-2.2, 2.2)
ax.set_ylim(-2.4, 2.2)
ax.set_aspect('equal')
ax.set_title(r'$g(u,v) = u$: different readings, separation!', fontsize=11)
ax.set_xlabel(r'$u$', fontsize=10)
ax.set_ylabel(r'$v$', fontsize=10)
ax.tick_params(labelsize=9)

plt.tight_layout()
plt.show()
<Figure size 1000x400 with 2 Axes>

Left: the functional f(u,v)=vf(u,v) = v gives the same reading for xx and yy (they lie on the same horizontal level set), so ff alone cannot distinguish them. Right: the functional g(u,v)=ug(u,v) = u gives different readings (different vertical level sets) and separates the two points. No single functional suffices; we need the entire dual X∗X^* to guarantee that distinct points are always distinguishable.

Corollary 2 (Norming property)

For each x∈Xx \in X, there exists f∈X∗f \in X^* with f(x)=∥x∥f(x) = \|x\| and ∥f∥op=1\|f\|_{\text{op}} = 1.

Proof 3

Define f(αx)=α∥x∥f(\alpha x) = \alpha\|x\| on span(x)\text{span}(x). Then ∣f(αx)∣=∣α∣∥x∥=∥αx∥|f(\alpha x)| = |\alpha|\|x\| = \|\alpha x\|, so ∥f∥=1\|f\| = 1 on span(x)\text{span}(x). Hahn–Banach extends to F∈X∗F \in X^* with ∥F∥=1\|F\| = 1 and F(x)=∥x∥F(x) = \|x\|.

Source
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Polygon

fig, ax = plt.subplots(figsize=(5, 4))

# Unit ball (ell^infty)
square_verts = np.array([[-1, -1], [1, -1], [1, 1], [-1, 1]])
sq = Polygon(square_verts, fill=True, facecolor='C0', alpha=0.12,
             edgecolor='C0', lw=2)
ax.add_patch(sq)
ax.text(0, 0, r'$B_X$', fontsize=12, ha='center', va='center', color='C0')

# The strip |f| <= 1 (same as unit ball width in f-direction)
ax.axvspan(-1, 1, alpha=0.04, color='C1')
ax.axvline(-1, color='C1', lw=1.5, ls=':', alpha=0.5)
ax.axvline(1, color='C1', lw=1.5, ls=':', alpha=0.5)

# The point x0
x0 = np.array([1.8, 0.5])
ax.plot(*x0, 'C3o', ms=9, zorder=5)
ax.text(x0[0] + 0.1, x0[1] + 0.15, r'$x_0$', fontsize=12, color='C3')

# Level set through x0
ax.axvline(x0[0], color='C3', lw=2, alpha=0.6, ls='--')
ax.text(x0[0] - 0.08, -1.6, rf'$f(x_0) = {x0[0]} = \|x_0\|_\infty$',
        fontsize=10, color='C3', ha='right')

ax.set_xlim(-2.4, 2.8)
ax.set_ylim(-2.2, 2.2)
ax.set_aspect('equal')
ax.set_title(r'Norming functional: $f(u,v) = u$, $\|f\| = 1$', fontsize=11)
ax.set_xlabel(r'$u$', fontsize=10)
ax.set_ylabel(r'$v$', fontsize=10)
ax.tick_params(labelsize=9)

plt.tight_layout()
plt.show()
<Figure size 500x400 with 1 Axes>

The functional f(u,v)=uf(u,v) = u has ∥f∥=1\|f\| = 1 (the unit ball fits inside the strip ∣f∣≤1|f| \leq 1) and achieves f(x0)=∥x0∥∞=1.8f(x_0) = \|x_0\|_\infty = 1.8. The reading equals the norm exactly.

Corollary 3 (The sup formula)

Let XX be a normed space and BX∗={f∈X∗:∥f∥≤1}B_{X^*} = \{f \in X^* : \|f\| \leq 1\} the closed unit ball of the dual. Then for all x∈Xx \in X:

∥x∥=sup⁡{∣f(x)∣:f∈BX∗}=max⁡{∣f(x)∣:f∈BX∗}\|x\| = \sup\{|f(x)| : f \in B_{X^*}\} = \max\{|f(x)| : f \in B_{X^*}\}
(3)

In particular, the supremum is attained: there exists f∈BX∗f \in B_{X^*} with ∣f(x)∣=∥x∥|f(x)| = \|x\|. The dual X∗X^* is a complete measurement system with no information loss.

Proof 4

The inequality ≤\leq: for any ff with ∥f∥≤1\|f\| \leq 1, we have ∣f(x)∣≤∥f∥∥x∥≤∥x∥|f(x)| \leq \|f\|\|x\| \leq \|x\|, so sup⁡≤∥x∥\sup \leq \|x\|.

The inequality ≥\geq (and that it is a max): by the norming property, there exists f∈X∗f \in X^* with ∥f∥=1\|f\| = 1 and f(x)=∥x∥f(x) = \|x\|. This ff attains the supremum.

Source
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Polygon

fig, ax = plt.subplots(figsize=(5.5, 5))

# Unit ball
square_verts = np.array([[-1, -1], [1, -1], [1, 1], [-1, 1]])
sq = Polygon(square_verts, fill=True, facecolor='C0', alpha=0.1,
             edgecolor='C0', lw=2)
ax.add_patch(sq)
ax.text(0, 0, r'$B_X$', fontsize=11, ha='center', va='center', color='C0')

# The point
x0 = np.array([1.5, 1.0])
ax.plot(*x0, 'C3o', ms=9, zorder=10)
ax.text(x0[0] + 0.12, x0[1] + 0.12, r'$x_0$', fontsize=12, color='C3')

# Three functionals: draw only the level set through x0 (full extent)
functionals = [
    ((1, 0), 'C1', r'$f_1 = u$'),
    ((0, 1), 'C2', r'$f_2 = v$'),
    ((1/np.sqrt(2), 1/np.sqrt(2)), 'C4', r'$f_3 = \frac{u+v}{\sqrt{2}}$'),
]

t = np.linspace(-3, 3, 300)
for (a, b), color, label in functionals:
    reading = a * x0[0] + b * x0[1]
    if abs(b) > 0.01:
        # Kernel (f = 0)
        y_ker = (-a * t) / b
        ax.plot(t, y_ker, color=color, alpha=0.35, lw=1.2, ls='--')
        # Level set through x0 (f = reading)
        y_line = (reading - a * t) / b
        ax.plot(t, y_line, color=color, alpha=0.5, lw=1.5, label=label)
    else:
        ax.axvline(0, color=color, alpha=0.35, lw=1.2, ls='--')
        ax.axvline(reading, color=color, alpha=0.5, lw=1.5, label=label)

ax.set_xlim(-2.5, 2.5)
ax.set_ylim(-2.5, 2.5)
ax.set_aspect('equal')
ax.legend(fontsize=9, loc='lower left')
ax.set_title(r'The sup formula: $\|x_0\| = \max_{\|f\| \leq 1} |f(x_0)|$',
             fontsize=11)
ax.set_xlabel(r'$u$', fontsize=10)
ax.set_ylabel(r'$v$', fontsize=10)
ax.tick_params(labelsize=9)

plt.tight_layout()
plt.show()
<Figure size 550x500 with 1 Axes>

Three unit-norm functionals give different readings of x0=(1.5,1.0)x_0 = (1.5, 1.0): f1(u,v)=uf_1(u,v) = u reads 1.5, f2(u,v)=vf_2(u,v) = v reads 1.0, and f3=(u+v)/2f_3 = (u+v)/\sqrt{2} reads ≈1.25\approx 1.25. The best reading equals ∥x0∥∞=1.5\|x_0\|_\infty = 1.5. The norming property guarantees such an optimal functional always exists.

Corollary 4 (Classification of closures)

Let MM be a linear subspace of a normed space XX and let x0∈Xx_0 \in X. Then x0∈M‾x_0 \in \overline{M} if and only if there exists no bounded linear functional ff such that f(x)=0f(x) = 0 for all x∈Mx \in M but f(x0)≠0f(x_0) \neq 0.

Proof 5

(⇒)(\Rightarrow) If x0∈M‾x_0 \in \overline{M}, pick xn∈Mx_n \in M with xn→x0x_n \to x_0. For any f∈X∗f \in X^* vanishing on MM, continuity gives f(x0)=lim⁡f(xn)=0f(x_0) = \lim f(x_n) = 0.

(⇐)(\Leftarrow) Suppose x0∉M‾x_0 \notin \overline{M}. Then M‾\overline{M} is a proper closed subspace and d=dist(x0,M‾)>0d = \text{dist}(x_0, \overline{M}) > 0. Define gg on M‾+span(x0)\overline{M} + \text{span}(x_0) by g(m+αx0)=αg(m + \alpha x_0) = \alpha. Then ∣g(m+αx0)∣=∣α∣≤∥m+αx0∥/d|g(m + \alpha x_0)| = |\alpha| \leq \|m + \alpha x_0\|/d (since ∥m+αx0∥≥∣α∣d\|m + \alpha x_0\| \geq |\alpha| d). So ∥g∥≤1/d\|g\| \leq 1/d. Hahn-Banach extends gg to f∈X∗f \in X^*. By construction f=0f = 0 on MM but f(x0)=1≠0f(x_0) = 1 \neq 0.

Geometric Hahn–Banach: Separation of Convex Sets

The corollaries above separate points from points and points from the unit ball. The geometric form of Hahn–Banach separates a point from any closed convex set.

Theorem 2 (Separation of point from closed convex set)

Let C⊂XC \subset X be a closed convex set and x0∉Cx_0 \notin C. Then there exists f∈X∗f \in X^* and a constant γ\gamma such that

f(x0)>γ≥f(c)for all c∈Cf(x_0) > \gamma \geq f(c) \quad \text{for all } c \in C
(4)

That is, there is a closed hyperplane f−1(γ)f^{-1}(\gamma) that strictly separates x0x_0 from CC.

Source
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Polygon, Ellipse

fig, axes = plt.subplots(1, 2, figsize=(12, 5))

# --- Panel 1: Separation from a convex set ---
ax = axes[0]

# Draw a convex "blob" (ellipse)
ellipse = Ellipse((-0.5, 0), 1.8, 1.2, angle=20, facecolor='C0', alpha=0.2,
                  edgecolor='C0', lw=2)
ax.add_patch(ellipse)
ax.text(-0.5, 0, r'$C$', fontsize=14, ha='center', va='center', color='C0')

# Point outside
x0 = np.array([1.5, 0.8])
ax.plot(*x0, 'C3o', ms=8, zorder=5)
ax.text(x0[0]+0.1, x0[1]+0.1, r'$x_0$', fontsize=13, color='C3')

# Separating hyperplane (a line)
# Normal direction pointing from C toward x0
normal = np.array([0.8, 0.3])
normal = normal / np.linalg.norm(normal)
gamma_pt = np.array([0.6, 0.4])  # point on the hyperplane

perp = np.array([-normal[1], normal[0]])
t = np.linspace(-3, 3, 100)
line = gamma_pt[:, None] + perp[:, None] * t[None, :]
ax.plot(line[0], line[1], 'C1-', lw=2, label=r'$f^{-1}(\gamma)$')

# Shade the side containing C
# Draw parallel level sets
for offset in [-0.5, -1.0, 0.5, 1.0]:
    shifted = gamma_pt + offset * normal
    line_s = shifted[:, None] + perp[:, None] * t[None, :]
    ax.plot(line_s[0], line_s[1], 'gray', alpha=0.2, lw=0.8)

ax.set_xlim(-2.5, 2.5)
ax.set_ylim(-2, 2)
ax.set_aspect('equal')
ax.set_title('Geometric Hahn–Banach:\nseparation from a closed convex set', fontsize=11)
ax.legend(fontsize=10, loc='lower right')
ax.set_xlabel(r'$x$'); ax.set_ylabel(r'$y$')

# --- Panel 2: Why convexity and closedness matter ---
ax = axes[1]

# Non-convex set: annulus
theta = np.linspace(0, 2*np.pi, 200)
r_out = 1.5
r_in = 0.5
ax.fill_between(r_out*np.cos(theta), r_out*np.sin(theta),
                alpha=0.2, color='C0')
ax.fill_between(r_in*np.cos(theta), r_in*np.sin(theta),
                alpha=1.0, color='white')
ax.plot(r_out*np.cos(theta), r_out*np.sin(theta), 'C0', lw=2)
ax.plot(r_in*np.cos(theta), r_in*np.sin(theta), 'C0', lw=2, ls='--')

ax.plot(0, 0, 'C3o', ms=8, zorder=5)
ax.text(0.1, 0.15, r'$x_0$', fontsize=13, color='C3')
ax.text(0, -1.0, r'$C$ (non-convex)', fontsize=11, ha='center', color='C0')

# Try to draw a line — it can't separate
ax.plot([-2, 2], [0, 0], 'C1--', lw=1.5, alpha=0.5, label='No line separates!')
ax.text(1.5, 0.15, '?', fontsize=16, color='C1', fontweight='bold')

ax.set_xlim(-2.2, 2.2)
ax.set_ylim(-2.2, 2.2)
ax.set_aspect('equal')
ax.set_title('Why convexity matters:\nnon-convex sets can surround a point', fontsize=11)
ax.legend(fontsize=10, loc='lower right')
ax.set_xlabel(r'$x$'); ax.set_ylabel(r'$y$')

plt.tight_layout()
plt.show()
<Figure size 1200x500 with 2 Axes>

Why closedness and convexity matter

Applied Functional Analysis
The Geometry of Linear Functionals
Applied Functional Analysis
Application: The Universal Approximation Theorem