The Riesz Representation Theorem guarantees that for every bounded linear functional f: H → ℝ on a Hilbert space H, there exists:
AA sequence of vectors whose inner products with x converge to f(x) for every x
BA unique y ∈ H such that f(x) = ⟨x, y⟩ for all x ∈ H
CAn orthonormal basis {eₙ} such that f is determined by the values f(eₙ)
DA closed subspace K ⊂ H such that f(x) = ‖proj_K x‖
The theorem says every bounded linear functional is exactly inner product with a unique fixed vector y — no approximation, no basis dependence. Option C is true (knowing f on a basis determines f), but it is not what the Riesz theorem says; the theorem's content is that this determining data has the specific form of an inner product with a single vector. Option D confuses the representation with projection, which computes distance, not a linear functional.
Question 2 Multiple Choice
The Banach space L^p (with 1 < p < ∞, p ≠ 2) has dual (L^p)* ≅ L^q where 1/p + 1/q = 1. Why does the Hilbert-space Riesz Representation Theorem not apply to L^p?
AL^p is infinite-dimensional, and the theorem requires finite-dimensional spaces
BL^p has no inner product, so the self-duality H ≅ H* fails — the dual L^q is a genuinely different space unless p = 2
CBounded linear functionals on L^p are not continuous, so the theorem's hypotheses fail
DThe theorem only applies to real Hilbert spaces, not function spaces
The self-duality H ≅ H* is a special consequence of the inner product, not just the norm. In L^p with p ≠ 2, there is no inner product, and the dual is the different space L^q (conjugate exponent). When p = 2, L^p = L² is a Hilbert space, q = 2 as well, and indeed L^2 ≅ (L^2)* — the theorem applies. The inner product structure is essential: it is what allows 'measurements' (functionals) to be identified with 'states' (vectors) in the same space.
Question 3 True / False
The Riesz Representation Theorem shows that nearly every Banach space is self-dual (isomorphic to its dual space), since bounded linear functionals on any normed space can be represented as inner products.
TTrue
FFalse
Answer: False
This is the most important misconception to correct. Self-duality is SPECIFIC to Hilbert spaces — it depends on the existence of an inner product. General Banach spaces are not self-dual: the dual of L^p is L^q (conjugate exponent), which is a different space unless p = 2. The Riesz theorem characterizes a special feature of Hilbert geometry, not a universal property of Banach spaces. The inner product is the structure that collapses the distinction between a space and its dual.
Question 4 True / False
The self-duality H ≅ H* of a Hilbert space depends on the inner product structure, not merely on the completeness or the norm.
TTrue
FFalse
Answer: True
A Hilbert space is a Banach space (complete normed space) with the extra structure of an inner product. The Riesz theorem uses the inner product crucially: the proof constructs y via orthogonal decomposition relative to ker(f), which requires orthogonality — an inner product concept. Two Banach spaces can be isometric (same norm structure) while having different duals. The inner product is the additional ingredient that identifies vectors with functionals.
Question 5 Short Answer
Explain geometrically why every bounded linear functional f on a Hilbert space has the form f(x) = ⟨x, y⟩ for some unique y. What role does orthogonality play in the proof?
Think about your answer, then reveal below.
Model answer: A nonzero bounded functional f has a kernel ker(f) — the closed subspace of all vectors it maps to zero. The key geometric fact is that any closed subspace of a Hilbert space has an orthogonal complement. So H = ker(f) ⊕ ker(f)^⊥, and ker(f)^⊥ is one-dimensional (since f is scalar-valued). Pick a unit vector z in ker(f)^⊥. Any vector x decomposes as x = (x − f(x)/f(z)·z) + f(x)/f(z)·z — the first part is in ker(f), the second is a scalar multiple of z. Setting y = f(z)·z̄ (conjugated in the complex case), we get f(x) = ⟨x, y⟩. Uniqueness holds because if ⟨x, y⟩ = ⟨x, y'⟩ for all x, then y = y'. The whole proof hinges on orthogonal decomposition — which requires an inner product, not just a norm.
The geometric picture is: 'evaluating f at x' is the same as 'projecting x onto the direction y and scaling.' Orthogonality provides the decomposition that makes this possible. In a general Banach space, you have no such decomposition, which is why the identification of functionals with vectors fails.