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Section 3.2. Riesz Representation of functionals Page 11
3.2 Riesz Representation of functionals
Theorem 3.2.1 ([Red53], 5·4 Theorem (Riesz representation theorem)) Every bounded linear
functional ϕ on a Hilbert space H can be represented in terms of the inner product
ϕ(x) = 〈x, u〉, (3.6)
where u depends on ϕ and is uniquely determined by ϕ and has norm ϕ = u.
We want to prove that
(a) ϕ has a representation ϕ(x) = 〈x, u〉,
(b) u is unique,
(c) ϕ = u holds.
Proof.
If ϕ = 0 then (a), (b) and (c) hold, if we take u = 0. So suppose that ϕ = 0.
(a) Now N(ϕ) is a closed subspace of H. Furthermore since ϕ = 0, we have that N(ϕ) = H,
which implies that N(ϕ) ⊥ = 0 by Theorem 2.3.7. Hence there must be at least one nonzero
element, say u0 in N(ϕ) ⊥ .
Now set z = ϕ(x)u0 − ϕ(u0)x, then apply ϕ to give ϕ(z) = ϕ(x)ϕ(u0) − ϕ(u0)ϕ(x) =
0, and so z ∈ N(ϕ). Also, since u0 ∈ N(ϕ) ⊥ we have,
where
which implies that
If we set
then ϕ(x) = 〈x, u〉. Hence (a) is proved.
0 = 〈z, u0〉 = 〈ϕ(x)u0 − ϕ(u0)x, u0〉
= ϕ(x)〈u0, u0〉 − ϕ(u0)〈x, u0〉
〈u0, u0〉 = u0 2 = 0,
ϕ(x) = ϕ(u0)〈x, u0〉
u02 .
u = ϕ(u0)u0
,
u02 (b) To prove that u is unique, suppose that for all x ∈ H we have,
Then
ϕ(x) = 〈x, u1〉 = 〈x, u2〉.
〈x, u1 − u2〉 = 0 for all x.