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Hölder 與 Minkowski:兩條不等式

使 Lᵖ 成為賦範空間的一切都依賴兩條不等式。Hölder 控制共軛指數間的乘積;Minkowski 是偽裝的三角不等式。我們從 Young 不等式出發證明二者。

共軛指數與 Young 不等式

(1, ∞) 中的兩個指數 p、q 當 1/p + 1/q = 1 時稱為 共軛。於是 q = p/(p−1);自共軛情形是 p = q = 2。Lᵖ 中乘積的全部理論都由這一配對支配,其種子是兩個非負數之間的逐點不等式。

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. ∎
Young 不等式不過是 log 的凹性,披上指數的外衣。

Hölder 不等式

Hölder 不等式 說:若 f ∈ Lᵖ、g ∈ Lq 且 p、q 共軛,則 fg ∈ L¹ 且 ‖fg‖₁ ≤ ‖f‖ₚ · ‖g‖q。當 p = q = 2 時這正是 Cauchy–Schwarz 不等式。證明是在把兩函數歸一化為單位範數後對 Young 不等式積分。

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. ∎
歸一化為單位範數,對 Young 積分,再乘回去。

Minkowski 不等式即三角不等式

Minkowski 不等式 斷言 ‖f + g‖ₚ ≤ ‖f‖ₚ + ‖g‖ₚ。這恰是 ‖·‖ₚ 的 三角不等式,是稱 Lᵖ 為 賦範向量空間 所需的最後一條公理。技巧是把 |f+g|ᵖ 拆成 |f+g|·|f+g|^(p−1) 並對每塊用 Hölder。

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. ∎
拆開冪次,用兩次 Hölder,再約掉公因子。