Minkowski's Inequality

Proof
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Prerequisites

For the p\ell^p norm to deserve the name norm, it must satisfy the triangle inequality: the length of a sum is at most the sum of the lengths. That statement is Minkowski’s inequality, and it is the property that makes p\ell^p a genuine normed space. The proof is a clever splitting followed by a single application of Hölder’s inequality.

The theorem

For p1p \geq 1, write the p\ell^p norm of a finite sequence as ap(iaip)1/p\|a\|_p \coloneqq \big(\sum_i |a_i|^p\big)^{1/p}.

Theorem (Minkowski’s Inequality). Let p1p \geq 1 and let a1,,ana_1, \ldots, a_n and b1,,bnb_1, \ldots, b_n be real (or complex) numbers. Then

(i=1nai+bip)1/p    (i=1naip)1/p+(i=1nbip)1/p,(1)\left(\sum_{i=1}^{n} |a_i + b_i|^{p}\right)^{1/p} \;\leq\; \left(\sum_{i=1}^{n} |a_i|^{p}\right)^{1/p} + \left(\sum_{i=1}^{n} |b_i|^{p}\right)^{1/p}, \tag{1}

that is, a+bpap+bp\|a + b\|_p \leq \|a\|_p + \|b\|_p.

The two easy exponents

Case p=1p = 1. Inequality (1)(1) is just the ordinary triangle inequality summed termwise: ai+biai+bi|a_i + b_i| \leq |a_i| + |b_i|, so iai+biiai+ibi\sum_i |a_i + b_i| \leq \sum_i |a_i| + \sum_i |b_i|.

So assume p>1p > 1 for the rest, and let q=pp1q = \dfrac{p}{p-1} be its conjugate exponent, so that (p1)q=p(p-1)q = p and 1p+1q=1\dfrac{1}{p} + \dfrac{1}{q} = 1.

Proof for p>1p > 1: split, then apply Hölder

Let Si=1nai+bipS \coloneqq \sum_{i=1}^{n} |a_i + b_i|^{p}. If S=0S = 0 both sides of (1)(1) are zero, so assume S>0S > 0.

Split the power. Write one factor of ai+bi|a_i + b_i| off the front and bound it with the triangle inequality ai+biai+bi|a_i + b_i| \leq |a_i| + |b_i|:

ai+bip=ai+biai+bip1    aiai+bip1+biai+bip1.|a_i + b_i|^{p} = |a_i + b_i|\,\cdot\,|a_i + b_i|^{p-1} \;\leq\; |a_i|\,|a_i + b_i|^{p-1} + |b_i|\,|a_i + b_i|^{p-1}.

Summing over ii,

S    i=1naiai+bip1(I)  +  i=1nbiai+bip1(II).(2)S \;\leq\; \underbrace{\sum_{i=1}^{n} |a_i|\,|a_i + b_i|^{p-1}}_{(\mathrm{I})} \;+\; \underbrace{\sum_{i=1}^{n} |b_i|\,|a_i + b_i|^{p-1}}_{(\mathrm{II})}. \tag{2}

Apply Hölder to each piece. Treat (I)(\mathrm{I}) as the pairing of the sequence (ai)(|a_i|) with (ai+bip1)(|a_i + b_i|^{p-1}) and apply Hölder’s inequality with exponents pp and qq:

(I)    (i=1naip)1/p(i=1nai+bi(p1)q)1/q.(\mathrm{I}) \;\leq\; \left(\sum_{i=1}^{n} |a_i|^{p}\right)^{1/p} \left(\sum_{i=1}^{n} |a_i + b_i|^{(p-1)q}\right)^{1/q}.

Because (p1)q=p(p-1)q = p, the second factor is (iai+bip)1/q=S1/q\big(\sum_i |a_i + b_i|^{p}\big)^{1/q} = S^{1/q}. Hence

(I)    apS1/q,and likewise(II)    bpS1/q.(\mathrm{I}) \;\leq\; \|a\|_p \, S^{1/q}, \qquad\text{and likewise}\qquad (\mathrm{II}) \;\leq\; \|b\|_p \, S^{1/q}.

Combine. Substituting both bounds into (2)(2),

S    (ap+bp)S1/q.S \;\leq\; \big(\|a\|_p + \|b\|_p\big)\, S^{1/q}.

Since S>0S > 0, divide both sides by S1/qS^{1/q}. Using 11q=1p1 - \dfrac{1}{q} = \dfrac{1}{p},

S1/p=S11/q    ap+bp,S^{1/p} = S^{\,1 - 1/q} \;\leq\; \|a\|_p + \|b\|_p,

which is exactly (1)(1). \square

Why this matters

Minkowski’s inequality is the triangle inequality for the p\ell^p norm. Together with the easy facts ap0\|a\|_p \geq 0 (with equality only for the zero sequence) and λap=λap\|\lambda a\|_p = |\lambda|\,\|a\|_p, it certifies that p\|\cdot\|_p is a genuine norm — and therefore that p\ell^p is a normed vector space for every p1p \geq 1. The same argument, with sums replaced by integrals, gives Minkowski’s inequality for LpL^p function spaces.

Summary

  • Minkowski’s inequality: a+bpap+bp\|a + b\|_p \leq \|a\|_p + \|b\|_p for p1p \geq 1 — the triangle inequality for the p\ell^p norm.
  • Case p=1p = 1 is the termwise triangle inequality; the work is in p>1p > 1.
  • Proof for p>1p > 1: split ai+bip=ai+biai+bip1|a_i + b_i|^p = |a_i + b_i|\cdot|a_i + b_i|^{p-1}, bound the first factor by ai+bi|a_i| + |b_i|, and apply Hölder’s inequality to each resulting sum; the exponent identity (p1)q=p(p-1)q = p makes the leftover factor collapse to S1/qS^{1/q}.
  • Consequence: p\|\cdot\|_p is a norm, so p\ell^p (and LpL^p) are normed spaces.