关于Minkowski与Holder不等式的疑问：等式成立条件探究

Great question! Since you already have a solid grasp of Hölder's inequality (and its equality condition), we can use that as a foundation to figure out when Minkowski's inequality hits equality. Let's walk through this step by step.

First, let's recap the standard proof of Minkowski's inequality using Hölder's, since that's where the equality condition comes from. For non-negative $a_i, b_i$, $p > 1$, and $\frac{1}{p} + \frac{1}{q} = 1$, we start by rewriting the left-hand side:

$$\sum_i (a_i + b_i)^p = \sum_i (a_i + b_i)^{p-1}a_i + \sum_i (a_i + b_i)^{p-1}b_i$$

Notice that $p-1 = \frac{p}{q}$ (since $q = \frac{p}{p-1}$), so we can apply Hölder's inequality to each of the two sums on the right:

1. For the first sum:
   $$\sum_i (a_i + b_i)^{p-1}a_i \leq \left( \sum_i a_i^p \right)^{\frac{1}{p}} \cdot \left( \sum_i (a_i + b_i)^{(p-1)q} \right)^{\frac{1}{q}}$$
   Since $(p-1)q = p$, this simplifies to:
   $$\sum_i (a_i + b_i)^{p-1}a_i \leq \left( \sum_i a_i^p \right)^{\frac{1}{p}} \cdot \left( \sum_i (a_i + b_i)^p \right)^{\frac{1}{q}}$$

2. For the second sum, similarly:
   $$\sum_i (a_i + b_i)^{p-1}b_i \leq \left( \sum_i b_i^p \right)^{\frac{1}{p}} \cdot \left( \sum_i (a_i + b_i)^p \right)^{\frac{1}{q}}$$

Adding these two inequalities together and factoring out the common term gives:
$$\sum_i (a_i + b_i)^p \leq \left[ \left( \sum_i a_i^p \right)^{\frac{1}{p}} + \left( \sum_i b_i^p \right)^{\frac{1}{p}} \right] \cdot \left( \sum_i (a_i + b_i)^p \right)^{\frac{1}{q}}$$

Dividing both sides by $\left( \sum_i (a_i + b_i)^p \right)^{\frac{1}{q}}$ (assuming this term is non-zero; we'll cover the zero case separately) gives us Minkowski's inequality.

When Does Equality Hold? #

Equality in Minkowski's inequality requires that equality holds in both applications of Hölder's inequality, plus consistency between the two conditions.

From Hölder's equality condition (you noted this: $a_i^p = c b_i^q$ for some constant $c \geq 0$), let's translate that to our two sums:

1. For the first sum, equality means there exists a constant $\lambda \geq 0$ such that:
   $$a_i^p = \lambda \cdot (a_i + b_i)^p \quad \text{for all } i$$
   Since all terms are non-negative, we can rewrite this as:
   $$a_i = \lambda^{\frac{1}{p}} (a_i + b_i) \implies a_i(1 - \lambda^{\frac{1}{p}}) = \lambda^{\frac{1}{p}} b_i$$

2. For the second sum, equality means there exists a constant $\mu \geq 0$ such that:
   $$b_i^p = \mu \cdot (a_i + b_i)^p \quad \text{for all } i$$
   Again, non-negativity lets us rewrite this as:
   $$b_i = \mu^{\frac{1}{p}} (a_i + b_i) \implies b_i(1 - \mu^{\frac{1}{p}}) = \mu^{\frac{1}{p}} a_i$$

Now let's break down the cases:

• Case 1: $\sum_i (a_i + b_i)^p = 0$
  This only happens if $a_i + b_i = 0$ for all $i$, which (since $a_i, b_i \geq 0$) means $a_i = b_i = 0$ for all $i$. Equality obviously holds here.

• Case 2: $\sum_i (a_i + b_i)^p > 0$
  From the two proportionality conditions above, we can see that $a_i$ and $b_i$ must be scalar multiples of each other. Specifically:

  • Either all $a_i = 0$ (then equality holds trivially, since both sides become $\left( \sum_i b_i^p \right)^{\frac{1}{p}}$)
  • Or all $b_i = 0$ (same trivial equality)
  • Or there exists a non-negative constant $k$ such that $b_i = k a_i$ for every $i$ (or equivalently, $a_i = k b_i$ for some non-negative $k$)

To confirm this, substitute $b_i = k a_i$ into Minkowski's inequality:

• Left-hand side: $\left( \sum_i (a_i + k a_i)^p \right)^{\frac{1}{p}} = (1 + k) \left( \sum_i a_i^p \right)^{\frac{1}{p}}$
• Right-hand side: $\left( \sum_i a_i^p \right)^{\frac{1}{p}} + \left( \sum_i (k a_i)^p \right)^{\frac{1}{p}} = (1 + k) \left( \sum_i a_i^p \right)^{\frac{1}{p}}$

Both sides are equal, so this checks out.

Final Summary of Equality Conditions #

Minkowski's inequality $$\left( \sum_i (a_i + b_i)^p \right)^{\frac{1}{p}} \leq \left( \sum_i a_i^p \right)^{\frac{1}{p}} + \left( \sum_i b_i^p \right)^{\frac{1}{p}}$$ holds with equality if and only if:

• All $a_i = b_i = 0$, or
• One of the sequences is identically zero, or
• The two sequences are non-negative scalar multiples of each other (i.e., there exists $k \geq 0$ such that $b_i = k a_i$ for all $i$)

内容的提问来源于stack exchange，提问作者Oria Gruber