Minkowski inequality

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In mathematical analysis, the Minkowski inequality establishes that the L^p spaces are normed vector spaces. Let S be a measure space, let 1 ≤ p ≤ ∞ and let f and g be elements of L^p(S). Then f + g is in L^p(S), and we have

$\|f+g\|_p \le \|f\|_p + \|g\|_p$

with equality for 1 < p < ∞ if and only if f and g are positively linearly dependent (which means f = $λ$ g or g = $λ$ f for some $λ$ ≥ 0). Here, the norm is given by:

$\|f\|_p = \left( \int |f|^p d\mu \right)^{1/p}$

if p < ∞, or in the case p = ∞ by the essential supremum

$\|f\|_\infty = \operatorname{ess\ sup}_{x\in S}|f(x)|.$

The Minkowski inequality is the triangle inequality in L^p(S). In fact, it is a special case of the more general fact

$\|f\|_p = \sup_{\|g\|_q = 1} \int |fg| d\mu, \qquad 1/p + 1/q = 1$

where it is easy to see that the right-hand side satisfies the triangular inequality.

Like Hölder's inequality, the Minkowski inequality can be specialized to sequences and vectors by using the counting measure:

$\left( \sum_{k=1}^n |x_k + y_k|^p \right)^{1/p} \le \left( \sum_{k=1}^n |x_k|^p \right)^{1/p} + \left( \sum_{k=1}^n |y_k|^p \right)^{1/p}$

for all real (or complex) numbers x₁, ..., x_n, y₁, ..., y_n and where n is the cardinality of S (the number of elements in S).

1 Proof
2 Minkowski's integral inequality
3 See also
4 References

[ Proof

First, we prove that f+g has finite p-norm if f and g both do, which follows by

$|f + g|^p \le 2^{p-1}(|f|^p + |g|^p).$

Indeed, here we use the fact that $h (x) = x p$ is convex over $\mathbb{R}^+$ (for $p$ greater than one) and so, if a and b are both positive then

$\left(\frac{1}{2} a + \frac{1}{2} b\right)^p \le \frac{1}{2}a^p + \frac{1}{2} b^p.$

This means that

$(a+b)^p \le 2^{p-1}a^p + 2^{p-1}b^p.$

Now, we can legitimately talk about $(\|f + g\|_p)$ . If it is zero, then Minkowski's inequality holds. We now assume that $(\|f + g\|_p)$ is not zero. Using Hölder's inequality

$\|f + g\|_p^p = \int |f + g|^p \, \mathrm{d}\mu$

$\le \int (|f| + |g|)|f + g|^{p-1} \, \mathrm{d}\mu$

$=\int |f||f + g|^{p-1} \, \mathrm{d}\mu+\int |g||f + g|^{p-1} \, \mathrm{d}\mu$

$\stackrel{\text{H}\ddot{\text{o}}\text{lder}}{\le} \left( \left(\int |f|^p \, \mathrm{d}\mu\right)^{1/p} + \left (\int |g|^p \,\mathrm{d}\mu\right)^{1/p} \right) \left(\int |f + g|^{(p-1)\left(\frac{p}{p-1}\right)} \, \mathrm{d}\mu \right)^{1-\frac{1}{p}}$

$= (\|f\|_p + \|g\|_p)\frac{\|f + g\|_p^p}{\|f + g\|_p}.$

We obtain Minkowski's inequality by multiplying both sides by $\frac{\|f + g\|_p}{\|f + g\|_p^p}.$

[ Minkowski's integral inequality

Suppose that (S₁,μ₁) and (S₂,μ₂) are two measure spaces and F : S₁×S₂ → R is measurable. Then Minkowski's integral inequality is (Stein 1970, §A.1), (Hardy, Littlewood & Pólya 1988, Theorem 202):

$\left[\int_{S_2}\left(\int_{S_1}|F(x,y)|\,d\mu_1(x)\right)^pd\mu_2(y)\right]^{1/p} \le \int_{S_1}\left(\int_{S_2}|F(x,y)|^p\,d\mu_2(y)\right)^{1/p}d\mu_1(x),$

with obvious modifications in the case p = ∞. If p > 1, and both sides are finite, then equality holds only if |F(x,y)| = φ(x)ψ(y) a.e. for some non-negative measurable functions φ and ψ.

If μ₂ is the counting measure on a two-point set S₂ = {1,2}, then Minkowski's integral inequality gives the usual Minkowski inequality as a special case: for putting ƒ_i(x) = F(x,i) for i = 1,2, the integral inequality gives

$\begin{align} \|f_1 + f_2\|_p &= \left[\int_{S_2}\left|\int_{S_1}F(x,y)\,d\mu_1(x)\right|^pd\mu_2(y)\right]^{1/p} \\ &\le\int_{S_1}\left(\int_{S_2}|F(x,y)|^p\,d\mu_2(y)\right)^{1/p}d\mu_1(x)\\ &=\|f_1\|_p + \|f_2\|_p. \end{align}$