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Spring 2012 - Problem 5

Hilbert spaces, Riesz representation

State and prove the Riesz representation theorem for linear functionals on a (separable) Hilbert space.
Solution.

Let H\mathcal{H} be a Hilbert space and let f∈H∗f \in \mathcal{H}^* be a bounded linear functional. Then there exists a unique y∈Hy \in \mathcal{H} such that f(x)=⟨x,y⟩f\p{x} = \inner{x, y} for all x∈Hx \in \mathcal{H}.

We first show uniqueness: suppose y,y′∈Hy, y' \in \mathcal{H} are such that ⟨x,y⟩=⟨x,y′⟩\inner{x, y} = \inner{x, y'} for all x∈Hx \in \mathcal{H}. In particular, if we set x=y−y′x = y - y', then

0=⟨y−y′,y−y′⟩=∥y−y′∥2  ⟹  y=y′.0 = \inner{y - y', y - y'} = \norm{y - y'}^2 \implies y = y'.

Next, we need to show existence. If f=0f = 0, then f(x)=⟨x,0⟩f\p{x} = \inner{x, 0}. Otherwise, ker⁡f\ker{f} is a closed, proper subspace of H\mathcal{H}. In particular, there exists z∈(ker⁡f)⊥z \in \p{\ker{f}}^\perp, and by normalizing, we may assume that ∥z∥=1\norm{z} = 1.

Let x∈Hx \in \mathcal{H}. By linearity, if we set u=xf(z)−zf(x)u = xf\p{z} - zf\p{x}, then

f(u)=f(x)f(z)−f(z)f(x)=0,f\p{u} = f\p{x}f\p{z} - f\p{z}f\p{x} = 0,

so u∈ker⁡fu \in \ker{f}. Thus, since z∈(ker⁡f)⊥z \in \p{\ker{f}}^\perp,

0=⟨u,z⟩=f(z)⟨x,z⟩−f(x)⟨z,z⟩  ⟹  f(x)=⟨x,f(z)‾z⟩.\begin{aligned} 0 = \inner{u, z} &= f\p{z}\inner{x, z} - f\p{x}\inner{z, z} \\ \implies f\p{x} &= \inner{x, \conj{f\p{z}}z}. \end{aligned}

Thus, if we set y=f(z)‾zy = \conj{f\p{z}}z, we have f(x)=⟨x,y⟩f\p{x} = \inner{x, y} for any x∈Hx \in \mathcal{H}, which completes the proof.