Questions: Hölder's and Minkowski's Inequalities

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

Why is Hölder's inequality needed to prove that ‖f + g‖ₚ ≤ ‖f‖ₚ + ‖g‖ₚ (Minkowski's inequality), rather than just applying the ordinary triangle inequality for integrals?

AThe ordinary triangle inequality applies to real numbers, not integrals, so a completely new approach is required
BThe ordinary triangle inequality gives ∫|f+g|ᵖ ≤ (∫|f|ᵖ + ∫|g|ᵖ), but the pth root then creates a term (∫|f|ᵖ + ∫|g|ᵖ)^(1/p) that does not split into ‖f‖ₚ + ‖g‖ₚ; Hölder's inequality is needed to handle this
CHölder's inequality is needed only when p = 2; for other values of p, the result is trivial
DMinkowski's inequality is actually equivalent to the triangle inequality for real numbers and needs no further proof
Question 2 Multiple Choice

Hölder's inequality ∫|fg| dμ ≤ ‖f‖ₚ‖g‖_q is an equality (the bound is sharp) when:

Af and g are both square-integrable (both in L²)
B|f|ᵖ and |g|^q are proportional almost everywhere
Cf = g almost everywhere
DThe measure μ is a probability measure
Question 3 True / False

Minkowski's inequality ‖f+g‖ₚ ≤ ‖f‖ₚ + ‖g‖ₚ would hold trivially from basic properties of integrals, so Hölder's inequality is not actually needed in its proof.

TTrue
FFalse
Question 4 True / False

Without Minkowski's inequality, ‖f‖ₚ = (∫|f|ᵖ dμ)^(1/p) would still define a valid norm on Lᵖ, since the positivity and homogeneity axioms are easily verified.

TTrue
FFalse
Question 5 Short Answer

Explain why Hölder's inequality is called a 'duality' result and how it serves as the engine for proving Minkowski's inequality as a 'convexity' result.

Think about your answer, then reveal below.