The dual and the Riesz theorem
A linear functional on X is a linear map f: X -> scalars; it is bounded when |f(x)| ≤ C ||x||. The collection of all bounded linear functionals, with the operator norm, is the dual space X*. Because the scalars are complete, X* is always a Banach space. The dual is where we put the test instruments that probe X one number at a time.
In a Hilbert space the dual is completely understood. The Riesz representation theorem says every bounded functional f on H is f(x) = ⟨x, y⟩ for a unique y in H, and ||f|| = ||y||. So a Hilbert space is isometrically identified with its own dual: there is no new space to discover. The proof is pure projection geometry.
Riesz representation (sketch with the geometry shown).
Let f ∈ H* be nonzero. Let M = ker f = { x : f(x) = 0 }.
M is a closed subspace (f is continuous), and M ≠ H.
Step 1. By the projection theorem M⊥ is nonzero; pick
z ∈ M⊥ with f(z) = 1 (rescale).
Step 2. For any x, the vector x - f(x) z lies in M because
f(x - f(x) z) = f(x) - f(x)·f(z) = f(x) - f(x) = 0.
Since z ⊥ M:
⟨x - f(x) z, z⟩ = 0
⇒ ⟨x, z⟩ = f(x) ⟨z, z⟩ = f(x) ||z||^2
⇒ f(x) = ⟨x, z / ||z||^2⟩.
So y = z / ||z||^2 represents f: f(x) = ⟨x, y⟩.
Step 3 (norm). By Cauchy–Schwarz |f(x)| ≤ ||y|| ||x||, and
f(y) = ⟨y, y⟩ = ||y||^2 = ||y|| · ||y||,
so the bound is attained: ||f|| = ||y||. ∎Hahn–Banach: enough functionals always exist
Outside Hilbert spaces we have no inner product to manufacture functionals, so it is not obvious the dual is rich. The Hahn–Banach theorem settles this: a bounded functional defined on a subspace M of a normed space X can be extended to all of X without increasing its norm. The extension is rarely unique, but it always exists, and that existence is the engine of the whole theory.
- Norming functionals. For any nonzero x in X there is f in X* with ||f|| = 1 and f(x) = ||x||. (Define f on the line through x, then extend by Hahn–Banach.)
- Separation of points. If x ≠ y, the functional above on x − y distinguishes them, so X* separates points of X.
- Norm recovery. ||x|| = sup{ |f(x)| : f in X*, ||f|| ≤ 1 }. The dual sees the norm exactly.