advanced31 min studyLesson 782 in Mathematics

Hahn–Banach extension and separation

Extend scalar functionals without increasing their norm and state real, complex and geometric versions with their proper hypotheses.

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What you’ll understand

  • Separate the real sublinear domination form from complex modulus domination.
  • Construct and compare norm-preserving scalar extensions.
  • Use a bounded functional to certify distance from a closed subspace.

Hahn–Banach turns local scalar information into global tests without enlarging its norm. It does not provide a canonical extension, a projection onto every subspace, or an extension of an arbitrary vector-valued operator. Its strength is exactly that a bounded scalar functional can be continued from a subspace while preserving its bound. The same principle yields supporting and separating hyperplanes, with real parts used over complex spaces. Distinguishing these versions prevents a convenient real inequality from being applied to complex numbers, where an order relation is unavailable.

Real domination and complex norm preservation

In the real algebraic version, a sublinear function p is subadditive and positively homogeneous. If a real-linear functional on a linear subspace satisfies f≤p there, it extends to a real-linear functional with the same domination on the whole vector space. Completeness is not required. Choosing p(x)=C∥x∥ gives the familiar bounded extension theorem, and applying domination to x and −x produces the absolute bound. A maximal-extension argument supplies existence, but generally does not identify a unique extension.

For a complex vector space, the useful domination form uses a seminorm and an absolute-value inequality, not an order comparison of complex values. In particular, a bounded complex-linear functional extends complex-linearly with unchanged norm. One proof extends its real part on the underlying real space, obtaining g, and reconstructs the complex functional as g(x)−ig(ix). Multiplying x by a scalar of modulus one then verifies the full modulus bound. This reconstruction is essential: an arbitrary real-linear extension need not respect multiplication by i.

f≤p ⟹ ∃F:F∣M=f, F≤pover R,F(x)=g(x)−i g(ix)in the complex reconstructionf\leq p\ \Longrightarrow\ \exists F: F|_M=f,\ F\leq p\quad\text{over }\mathbb R,\qquad F(x)=g(x)-i\,g(ix)\quad\text{in the complex reconstruction}
The real statement uses a sublinear majorant. The complex reconstruction begins with an extension g of the real part and recovers complex linearity.↔ Scroll wide equations horizontally.

The dual detects the norm and quotient distance

Given a nonzero vector x, define a functional on its one-dimensional span that takes value ∥x∥ at x and has norm one. Hahn–Banach extends it to the ambient space. Hence the norm of a vector is the supremum of its evaluations by dual unit-ball functionals, and that supremum is attained for each fixed vector. This does not say that every fixed functional attains its own norm on the primal unit ball; the order of the quantifiers is different. Such norm attainment can fail in nonreflexive spaces.

For a closed subspace M and a point outside it, apply the same argument in X/M and pull the functional back through the quotient map. The resulting norm-one functional vanishes on M and evaluates the chosen point at its distance from M, with a suitable phase in the complex case. This gives a lower-bound certificate for every candidate approximation from M. It still does not produce a nearest point in M. A dual certificate can identify an infimum even when no primal representative achieves that infimum.

∥x∥=sup⁡∥f∥≤1∣f(x)∣,dist⁡(x,M)=sup⁡{∣f(x)∣:f∈X∗, ∥f∥≤1, f∣M=0}(M closed)\|x\|=\sup_{\|f\|\leq1}|f(x)|,\qquad\operatorname{dist}(x,M)=\sup\{|f(x)|:f\in X^*,\ \|f\|\leq1,\ f|_M=0\}\quad(M\text{ closed})
Hahn–Banach gives exact dual tests for norm and quotient distance. These are scalar-functional results, without a general projection theorem.↔ Scroll wide equations horizontally.

Separation needs topological hypotheses

In a real normed space, a point outside a nonempty closed convex set can be strictly separated from that set by a continuous linear functional with a positive gap. Over a complex space, use the real part of a complex-linear functional. One route separates the convex set from a sufficiently small open ball around the point. More general separation statements have different strengths: two disjoint convex sets with one open admit a separating functional, but two arbitrary disjoint closed convex sets need not admit a uniform positive separating gap.

Convexity and closedness play different roles. Convexity makes supporting linear inequalities describe the set, while closedness keeps an exterior point a positive distance from a small surrounding ball disjoint from the set. As a consequence, a norm-closed convex subset is also weakly closed, a fact used later in weak compactness arguments. None of these statements makes every closed nonconvex set weakly closed. Nor does scalar extension imply that every closed subspace has a bounded complementary projection; that stronger property distinguishes special Banach-space geometry.

Many extensions can preserve the same norm

In real R² with the supremum norm, define f(t,t)=t on the diagonal. Find every linear norm-preserving extension F(a,b)=αa+βb.

  1. On the diagonal ∥(t,t)∥∞=|t|, so the original functional has norm one. Agreement requires α+β=1.

  2. The dual norm of F is |α|+|β|, since independent signs of a and b attain that value on the square unit ball.

  3. Combining α+β=1 with |α|+|β|=1 forces α,β≥0. Conversely, every such pair has the required norm and agrees on the diagonal, including α=1/4,β=3/4.

The result

All norm-preserving extensions are F(a,b)=αa+(1−α)b with 0≤α≤1. Existence is guaranteed, while uniqueness fails even in this simple finite-dimensional setting.

Explore the model

Interactive lab

Interactive lab · vectors

A norm-one functional need not equal the vector norm

Pause and Reset, then set angle=0°,magnitude=2. For the functional F(x,y)=x, predict evaluation 2, equal to the vector norm. For a fresh comparison Reset and set angle=60°,magnitude=2; predict evaluation 1 and y-component √3, while the norm stays 2. The horizontal projection is a scalar functional bounded by the Euclidean norm. This plane illustrates a dual norm bound and its attainment, not a general Hahn–Banach extension algorithm.

A norm-one functional need not equal the vector normPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA norming functional can attain the norm at a chosen vector without giving equality for every other vector.

Check your understanding

Which statement is guaranteed by the scalar normed-space Hahn–Banach theorem?

Try it yourself

In Euclidean R² let M={(t,2t):t∈R} and x=(3,1). Find a norm-one functional annihilating M and use it to compute the distance to M.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A normal vector to the line is (2,−1). Normalize it, then find a residual parallel to that normal.

Compare with the solution

Take F(a,b)=(2a−b)/√5, whose Euclidean dual norm is one and which vanishes on M. Its value at x is √5, so every distance ∥x−m∥ is at least √5. Orthogonal projection, available in this particular Euclidean example, gives the point (1,2) on M and residual (2,−1) of norm √5. Thus the distance is exactly √5. The functional certifies the lower bound, while the explicit point proves attainment here.

Take it with you

  • Real sublinear domination and complex modulus domination are different formulations.
  • Scalar functionals detect vector norms and closed-subspace distances.
  • Existence of an extension does not imply uniqueness or a complementary projection.

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