Definition 13.3.3Let X be a real vector space ρ : X → ℝ is called a gauge functionif
ρ(x +y) ≤ ρ(x )+ ρ(y),
ρ(ax) = aρ(x)ifa ≥ 0. (13.17)
(13.17)
Suppose M is a subspace of X and z
∈∕
M. Suppose also that f is a linear real-valued function having the
property that f(x) ≤ ρ(x) for all x ∈ M. Consider the problem of extending f to M ⊕ ℝz such that if F is
the extended function, F(y) ≤ ρ(y) for all y ∈ M ⊕ ℝz and F is linear. Since F is to be linear, it suffices to
determine how to define F(z). Letting a > 0, it is required to define F
(z)
such that the following hold for
all x,y ∈ M.
f(x)
◜◞◟-◝
F (x)+ aF (z) = F(x+ az) ≤ ρ(x + az),
◜f(y◞◟)◝
F (y)− aF (z) = F(y− az) ≤ ρ(y − az). (13.18)
(13.18)
Now if these inequalities hold for all y∕a, they hold for all y because M is given to be a subspace.
Therefore, multiplying by a^{−1}13.17 implies that what is needed is to choose F
(z)
such that for all
x,y ∈ M,
f (x) +F (z) ≤ ρ(x+ z),f(y) − ρ(y − z) ≤ F(z)
and that if F
(z)
can be chosen in this way, this will satisfy 13.18 for all x,y and the problem of extending
f will be solved. Hence it is necessary to choose F(z) such that for all x,y ∈ M
f(y) − ρ(y − z) ≤ F(z) ≤ ρ(x+ z) − f (x). (13.19)
(13.19)
Is there any such number between f(y) − ρ(y − z) and ρ(x + z) − f(x) for every pair x,y ∈ M? This is
where f(x) ≤ ρ(x) on M and that f is linear is used. Forx,y ∈ M,
ρ(x + z)− f(x)− [f(y)− ρ(y− z)]
= ρ(x + z) + ρ(y − z)− (f (x) +f (y))
≥ ρ(x+ y)− f(x +y) ≥ 0.
Therefore there exists a number between
sup {f(y)− ρ(y − z) : y ∈ M }
and
inf{ρ(x+ z) − f (x) : x ∈ M }
Choose F(z) to satisfy 13.19. This has proved the following lemma.
Lemma 13.3.4Let M be a subspace of X, a real linear space, and let ρ be a gauge function onX. Suppose f : M → ℝ is linear, z
∕∈
M, and f
(x)
≤ ρ
(x)
for all x ∈ M. Then f can be extended toM ⊕ ℝz such that, if F is the extended function, F is linear and F(x) ≤ ρ(x) for all x ∈ M ⊕ ℝz.
With this lemma, the Hahn Banach theorem can be proved.
Theorem 13.3.5(Hahn Banach theorem) Let X be a real vector space, let M be a subspace of X,let f : M → ℝ be linear, let ρ be a gauge function on X, and suppose f(x) ≤ ρ(x) for all x ∈ M.Then there exists a linear function, F : X → ℝ, such that
a.)F(x) = f(x) for all x ∈ M
b.) F(x) ≤ ρ(x) for all x ∈ X.
Proof: Let ℱ = {(V,g) : V ⊇ M,V is a subspace of X,g : V → ℝ is linear, g(x) = f(x) for all x ∈ M,
and g(x) ≤ ρ(x) for x ∈ V }. Then (M,f) ∈ℱ so ℱ≠∅. Define a partial order by the following
rule.
(V,g) ≤ (W,h)
means
V ⊆ W and h(x) = g(x) if x ∈ V.
By Theorem 13.3.2, there exists a maximal chain, C⊆ℱ. Let Y = ∪{V : (V,g) ∈C} and let h : Y → ℝ be
defined by h(x) = g(x) where x ∈ V and
(V,g)
∈C. This is well defined because if x ∈ V_{1} and V_{2} where
(V ,g )
1 1
and
(V ,g )
2 2
are both in the chain, then since C is a chain, the two element related.
Therefore, g_{1}
(x)
= g_{2}
(x)
. Also h is linear because if ax + by ∈ Y , then x ∈ V_{1} and y ∈ V_{2} where
(V ,g )
1 1
and
(V ,g )
2 2
are elements of C. Therefore, letting V denote the larger of the two V_{i},
and g be the function that goes with V , it follows ax + by ∈ V where
≤ ρ(x) for any x ∈ Y because for such x, x ∈ V where
(V,g)
∈C.
Is Y = X? If not, there exists z ∈ X ∖ Y and there exists an extension of h to Y ⊕ ℝz using Lemma
13.3.4. Letting h denote this extended function, contradicts the maximality of C. Indeed, C∪{
( -)
Y ⊕ ℝz,h
}
would be a longer chain. ■
This is the original version of the theorem. There is also a version of this theorem for complex vector
spaces which is based on a trick.