Which Lᵖ space is the only one that is also a Hilbert space, and why?
AL¹, because it contains the most integrable functions and therefore has the richest structure
BL², because the exponent p = 2 enables a genuine inner product ⟨f, g⟩ = ∫fg dμ satisfying all inner product axioms
CL∞, because bounded functions have the most regularity and smoothness
DAll Lᵖ spaces are Hilbert spaces for p ≥ 1
A Hilbert space requires an inner product, not just a norm. An inner product must satisfy the parallelogram law: ‖f + g‖² + ‖f − g‖² = 2‖f‖² + 2‖g‖². This identity holds in L² — where ‖f‖₂² = ∫|f|² dμ corresponds to a genuine dot-product structure — but fails for all Lᵖ with p ≠ 2. L² has the inner product ⟨f,g⟩ = ∫fḡ dμ, making it the natural home for Fourier analysis, quantum mechanics, and spectral theory, all of which rely on orthogonality and projections.
Question 2 Multiple Choice
A function f: [0,1] → ℝ is defined as f(x) = 0 for all x except f(1/2) = 10⁶. In L²([0,1]) with Lebesgue measure, this function is:
AA non-trivial L² element with ‖f‖₂ = 10⁶, since it takes a large value at x = 1/2
BIdentified with the zero function, because {1/2} has Lebesgue measure zero so ‖f‖₂ = 0
CNot in L² because its pointwise value exceeds 1
DIn L² with norm equal to 1, because Lebesgue measure normalizes point masses
‖f‖₂² = ∫₀¹ |f|² dμ = 0, because the single point {1/2} has Lebesgue measure zero — it contributes nothing to a Lebesgue integral. Since ‖f‖₂ = 0 while f is not identically zero, the norm would be degenerate if we treated such functions as distinct elements. The solution is that Lᵖ elements are equivalence classes: f ~ g if they agree almost everywhere. This function is identified with the zero function because they differ only on a null set.
Question 3 True / False
Lᵖ spaces are Banach spaces — complete normed vector spaces — meaning every Cauchy sequence in Lᵖ converges to an element that is also in Lᵖ.
TTrue
FFalse
Answer: True
Completeness is the key structural property that makes Lᵖ spaces analytically useful. Without it, limits of sequences of Lᵖ functions might escape the space, making analysis intractable. The Riesz-Fischer theorem establishes that Lᵖ is complete for all 1 ≤ p ≤ ∞. This property underpins all convergence theorems in functional analysis and makes Lᵖ spaces suitable for solving differential equations, optimization problems, and approximation questions.
Question 4 True / False
On a probability space (total measure 1), most function in L¹ is also in L², since a finite integral automatically implies a finite squared integral.
TTrue
FFalse
Answer: False
The inclusion goes the other direction: on a probability space, L∞ ⊆ L² ⊆ L¹. A function can have finite integral (L¹) without having finite squared integral (L²). For example, f(x) = x^(−3/4) on (0,1]: ∫₀¹ x^(−3/4) dx = 4 (finite, so f ∈ L¹), but ∫₀¹ x^(−3/2) dx diverges (f ∉ L²). Higher Lᵖ membership is the more restrictive condition — being in L² is a stronger requirement than being in L¹ on probability spaces.
Question 5 Short Answer
Explain why elements of Lᵖ spaces are defined as equivalence classes of functions rather than individual functions, and what property of the norm makes this identification necessary.
Think about your answer, then reveal below.
Model answer: A norm must satisfy positive-definiteness: ‖f‖ₚ = 0 should imply f is 'zero.' But ‖f‖ₚ = (∫|f|ᵖ dμ)^(1/p) = 0 whenever f = 0 almost everywhere — including functions that are nonzero only on a set of measure zero. If these were treated as distinct elements, the norm would not be positive-definite. The fix is to identify any two functions that agree almost everywhere as the same Lᵖ element. On equivalence classes, ‖[f]‖ₚ = 0 if and only if f = 0 a.e., which is the same as [f] = [0], restoring positive-definiteness.
This identification reflects the fundamental measure-theoretic principle that integration cannot detect what a function does on a null set. Modifying a function on a null set produces an indistinguishable function from the measure's perspective. The equivalence class construction is not a technicality to be ignored — it is essential to the mathematical integrity of Lᵖ as a normed space, and understanding it clarifies why 'a.e.' appears throughout functional analysis.