Basic Analysis – Gently Done Topological Vector Spaces

48 Basic Analysis
NextweconsiderthefirstofseveralversionsoftheHahn-Banachtheorem. This
one is for real vector spaces and uses the order structure in R. Some terminology
is needed.
Definition 5.19 We say that a map p : X → R on a real vector space X is
subadditive if
and that p is positively homogeneous if
p(x+y) ≤ p(x)+p(y), for x,y ∈ X
p(tx) = tp(x), for any x ∈ X and t > 0.
Note that if p is positively homogeneous, then, with x = 0 and t = 2, we see
that p(0) = 2p(0) so that p(0) = 0. Hence p(tx) = tp(x) for all t ≥ 0. If p is
also subadditive, then, setting y = −x, we get 0 = p(0) ≤ p(x)+p(−x), that is,
−p(−x) ≤ p(x), for x ∈ X.
Theorem 5.20 (Hahn-Banach) Let X be a real vector space and suppose that
f : M → R is a linear mapping defined on a linear subspace M of X such that
f(y) ≤ p(y), for all y ∈ M, for some subadditive and positively homogeneous map
p : X → R. Then there is a linear functional Λ : X → R such that Λ(x) = f(x)
for x ∈ M and
−p(−x) ≤ Λ(x) ≤ p(x) for x ∈ X.
(In other words, f can be extended to X whilst retaining the same bound.)
Proof Wefirstshow howwecanextendthedefinitionoff byoneextradimension.
If M �= X, let x 1 ∈ X with x 1 /∈ M. Define M 1 = {x + tx 1 : x ∈ M, t ∈ R}.
Then M 1 is a linear subspace of X—the subspace of X spanned by {x 1 }∪M. Any
z ∈ M 1 can be written uniquely as z = x+tx 1 for x ∈ M and t ∈ R. Indeed, if
x+tx 1 = x ′ +t ′ x 1 for x,x ′ ∈ M and t,t ′ ∈ R, then 0 = (x−x ′ )+(t−t ′ )x 1 . Since
x 1 /∈ M, this implies that t = t ′ and therefore x = x ′ .
The implications of the existence of a suitable extension to f will provide the
idea of how to actually construct such an extension. So suppose for the moment
that f 1 is an extension of f to M 1 , satisfying the stated bound in terms of p. Let
x+tx 1 ∈ M 1 , x ∈ M, t ∈ R. Then f 1 (x+tx 1 ) = f(x)+tf 1 (x 1 ) = f(x)+tµ, where
we have set µ = f 1 (x 1 ). The given bound demands that f(x)+tµ ≤ p(x+tx 1 )
for all t ∈ R and x ∈ M. For t = 0, this is nothing other than the hypothesis on
f. Replacing x by tx, we get f(tx)+tµ ≤ p(tx+tx 1 ). Setting t = 1 and t = −1,
we obtain
f(x)+µ ≤ p(x+x 1 ) for x ∈ M (taking t = 1)
Department of Mathematics King’s College, London