Questions: Riesz Representation Theorem for Hilbert Spaces

Question 1

A bounded linear functional f: H → ℝ is defined on a Hilbert space H. According to the Riesz Representation Theorem, which statement is correct?

• A. f(x) = ⟨x, y⟩ for some y ∈ H, but y may not be unique
• B. f(x) = ⟨x, y⟩ for a unique y ∈ H, and ‖f‖ = ‖y‖
• C. This representation holds only when H is finite-dimensional
• D. f(x) = ‖x‖ · ‖y‖ for some y ∈ H by the Cauchy-Schwarz inequality

Answer: (B)

The theorem guarantees both existence and uniqueness of the representing vector y. Uniqueness follows because if f(x) = ⟨x, y₁⟩ = ⟨x, y₂⟩ for all x, then ⟨x, y₁ − y₂⟩ = 0 for all x, which forces y₁ − y₂ = 0. The norm equality ‖f‖ = ‖y‖ makes the map y ↦ f_y an isometry. Option C is a common mistake: the theorem holds for all Hilbert spaces, finite or infinite-dimensional. Option D confuses the Cauchy-Schwarz bound |⟨x,y⟩| ≤ ‖x‖‖y‖ with the functional itself.

Question 2

What makes Hilbert spaces special compared to general Banach spaces regarding their dual?

• A. Hilbert spaces are always finite-dimensional, making their dual trivially equal to themselves
• B. The norm of a Hilbert space is always defined by an inner product, which forces H ≅ H* isometrically
• C. Bounded linear functionals only exist on Hilbert spaces, not on general Banach spaces
• D. Every Banach space is isometrically isomorphic to its dual, just as every Hilbert space is

Answer: (B)

The inner product is the key. The map y ↦ f_y where f_y(x) = ⟨x, y⟩ is an isometric isomorphism from H onto H*. For general Banach spaces X, the dual X* can be a completely different space — for example, the dual of L¹ is L^∞, not L¹. Self-duality is a special feature of Hilbert space geometry, arising because the inner product provides a canonical way to identify vectors with functionals.

Question 3

The proof of the Riesz Representation Theorem uses the fact that the kernel of a bounded linear functional is a closed subspace of H.

• T. True
• F. False

Answer: True

True, and this is the pivotal step. Continuity of f (which follows from boundedness) ensures that ker(f) = f⁻¹({0}) is closed. The orthogonal complement ker(f)^⊥ is then non-trivial (since f ≠ 0), and the representing vector y is constructed from a unit vector in ker(f)^⊥. Without closedness, the orthogonal decomposition H = ker(f) ⊕ ker(f)^⊥ would not be valid.

Question 4

For any Banach space X, there is typically an isometric isomorphism between X and its dual X*.

• T. True
• F. False

Answer: False

False. Hilbert spaces are self-dual (H ≅ H* isometrically) by the Riesz theorem, but this is special. For Banach spaces the dual can be entirely different. The dual of L^p is L^q where 1/p + 1/q = 1 (for p ≠ 2), so (L^p)* ≅ L^q ≇ L^p when p ≠ 2. The dual of L^1 is L^∞, not L^1. Only for p = 2, where L² is a Hilbert space, does self-duality hold.

Question 5

Explain in your own words why the Riesz Representation Theorem implies that a Hilbert space is 'self-dual,' and why this is a special feature not shared by all Banach spaces.

Model answer: The Riesz theorem says every bounded linear functional on H has the form f(x) = ⟨x, y⟩ for a unique y ∈ H. This means the map φ: H → H* defined by φ(y) = f_y is a bijection — every element of H gives a functional and every functional comes from an element of H. Since the map is also an isometry (‖f_y‖ = ‖y‖) and conjugate-linear, it is an isometric isomorphism H ≅ H*. This self-duality is special because it requires the rich structure of an inner product. In a general Banach space, the norm does not provide a canonical way to pair vectors with functionals, so X and X* can have completely different structures (e.g., the dual of L¹ is L^∞).

Intuition: the inner product ⟨·, y⟩ is already a bounded linear functional for every y, and every bounded functional arises this way. So H and H* are in perfect one-to-one correspondence. In Banach spaces without inner products, there is no such built-in pairing, and the dual can look nothing like the original space.