Conjugate exponents and Young's inequality
Two exponents p, q in (1, ∞) are conjugate when 1/p + 1/q = 1. So q = p/(p−1); the self-conjugate case is p = q = 2. The whole theory of products in Lᵖ is governed by this pairing, and the seed is a pointwise inequality between two non-negative numbers.
Young's inequality. For a, b ≥ 0 and conjugate p, q:
a·b ≤ aᵖ/p + b^q/q.
Proof (via concavity of log). If a = 0 or b = 0 it is clear.
For a, b > 0, the logarithm is concave, so for weights 1/p + 1/q = 1:
log( (1/p)·aᵖ + (1/q)·b^q )
≥ (1/p)·log(aᵖ) + (1/q)·log(b^q)
= log(a) + log(b)
= log(a·b).
Applying the increasing function exp to both sides:
(1/p)·aᵖ + (1/q)·b^q ≥ a·b. ∎Hölder's inequality
Hölder's inequality says: if f ∈ Lᵖ and g ∈ Lq with conjugate p, q, then fg ∈ L¹ and ‖fg‖₁ ≤ ‖f‖ₚ · ‖g‖q. When p = q = 2 this is exactly the Cauchy–Schwarz inequality. The proof integrates Young's inequality after normalizing both functions to unit norm.
Hölder. ∫ |fg| dμ ≤ ‖f‖ₚ · ‖g‖q.
If ‖f‖ₚ = 0 or ‖g‖q = 0 then fg = 0 a.e. and both sides are 0.
Otherwise set F = |f|/‖f‖ₚ , G = |g|/‖g‖q , so ‖F‖ₚ = ‖G‖q = 1.
Apply Young pointwise with a = F(x), b = G(x):
F·G ≤ Fᵖ/p + G^q/q.
Integrate over X:
∫ F·G dμ ≤ (1/p)∫Fᵖ dμ + (1/q)∫G^q dμ
= (1/p)·1 + (1/q)·1 = 1.
Multiply back through by ‖f‖ₚ·‖g‖q:
∫ |f g| dμ ≤ ‖f‖ₚ · ‖g‖q. ∎Minkowski's inequality is the triangle inequality
Minkowski's inequality states ‖f + g‖ₚ ≤ ‖f‖ₚ + ‖g‖ₚ. This is precisely the triangle inequality for ‖·‖ₚ, the last axiom needed to call Lᵖ a normed vector space. The trick is to split |f+g|ᵖ as |f+g|·|f+g|^(p−1) and apply Hölder to each piece.
Minkowski. ‖f+g‖ₚ ≤ ‖f‖ₚ + ‖g‖ₚ (1 < p < ∞). Write |f+g|ᵖ = |f+g| · |f+g|^(p−1) ≤ (|f|+|g|)·|f+g|^(p−1). Integrate and split: ‖f+g‖ₚᵖ ≤ ∫ |f|·|f+g|^(p−1) + ∫ |g|·|f+g|^(p−1). Let q = p/(p−1) be the conjugate. Note |f+g|^(p−1) ∈ Lq because ∫ (|f+g|^(p−1))^q = ∫ |f+g|ᵖ = ‖f+g‖ₚᵖ < ∞. Apply Hölder to each term: ∫ |f|·|f+g|^(p−1) ≤ ‖f‖ₚ · ‖ |f+g|^(p−1) ‖q = ‖f‖ₚ · ‖f+g‖ₚ^(p−1). Same for g. Therefore ‖f+g‖ₚᵖ ≤ (‖f‖ₚ + ‖g‖ₚ) · ‖f+g‖ₚ^(p−1). If ‖f+g‖ₚ > 0, divide by ‖f+g‖ₚ^(p−1): ‖f+g‖ₚ ≤ ‖f‖ₚ + ‖g‖ₚ. ∎