# Minkowski inequality

(Redirected from Minkowski's inequality)

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

${\displaystyle \|f+g\|_{p}\leq \|f\|_{p}+\|g\|_{p}}$

with equality for 1 < p < ∞ if and only if f and g are positively linearly dependent, i.e., f = λg for some λ ≥ 0 or g = 0. Here, the norm is given by:

${\displaystyle \|f\|_{p}=\left(\int |f|^{p}d\mu \right)^{\frac {1}{p}}}$

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

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

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

${\displaystyle \|f\|_{p}=\sup _{\|g\|_{q}=1}\int |fg|d\mu ,\qquad {\tfrac {1}{p}}+{\tfrac {1}{q}}=1}$

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:

${\displaystyle \left(\sum _{k=1}^{n}|x_{k}+y_{k}|^{p}\right)^{\frac {1}{p}}\leq \left(\sum _{k=1}^{n}|x_{k}|^{p}\right)^{\frac {1}{p}}+\left(\sum _{k=1}^{n}|y_{k}|^{p}\right)^{\frac {1}{p}}}$

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

## Proof

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

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

Indeed, here we use the fact that ${\displaystyle h(x)=x^{p}}$ is convex over R+ (for p > 1) and so, by the definition of convexity,

${\displaystyle \left|{\tfrac {1}{2}}f+{\tfrac {1}{2}}g\right|^{p}\leq \left|{\tfrac {1}{2}}|f|+{\tfrac {1}{2}}|g|\right|^{p}\leq {\tfrac {1}{2}}|f|^{p}+{\tfrac {1}{2}}|g|^{p}.}$

This means that

${\displaystyle |f+g|^{p}\leq {\tfrac {1}{2}}|2f|^{p}+{\tfrac {1}{2}}|2g|^{p}=2^{p-1}|f|^{p}+2^{p-1}|g|^{p}.}$

Now, we can legitimately talk about ${\displaystyle (\|f+g\|_{p})}$. If it is zero, then Minkowski's inequality holds. We now assume that ${\displaystyle (\|f+g\|_{p})}$ is not zero. Using the triangle inequality and then Hölder's inequality, we find that

{\displaystyle {\begin{aligned}\|f+g\|_{p}^{p}&=\int |f+g|^{p}\,\mathrm {d} \mu \\&=\int |f+g|\cdot |f+g|^{p-1}\,\mathrm {d} \mu \\&\leq \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 \\&\leq \left(\left(\int |f|^{p}\,\mathrm {d} \mu \right)^{\frac {1}{p}}+\left(\int |g|^{p}\,\mathrm {d} \mu \right)^{\frac {1}{p}}\right)\left(\int |f+g|^{(p-1)\left({\frac {p}{p-1}}\right)}\,\mathrm {d} \mu \right)^{1-{\frac {1}{p}}}&&{\text{ Hölder's inequality}}\\&=\left(\|f\|_{p}+\|g\|_{p}\right){\frac {\|f+g\|_{p}^{p}}{\|f+g\|_{p}}}\end{aligned}}}

We obtain Minkowski's inequality by multiplying both sides by

${\displaystyle {\frac {\|f+g\|_{p}}{\|f+g\|_{p}^{p}}}.}$

## Minkowski's integral inequality

Suppose that (S1, μ1) and (S2, μ2) are two σ-finite measure spaces and F: S1 × S2R is measurable. Then Minkowski's integral inequality is (Stein 1970, §A.1), (Hardy, Littlewood & Pólya 1988, Theorem 202):

${\displaystyle \left[\int _{S_{2}}\left|\int _{S_{1}}F(x,y)\,\mu _{1}(\mathrm {d} x)\right|^{p}\mu _{2}(\mathrm {d} y)\right]^{\frac {1}{p}}\leq \int _{S_{1}}\left(\int _{S_{2}}|F(x,y)|^{p}\,\mu _{2}(\mathrm {d} y)\right)^{\frac {1}{p}}\mu _{1}(\mathrm {d} x),}$

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 μ1 is the counting measure on a two-point set S1 = {1,2}, then Minkowski's integral inequality gives the usual Minkowski inequality as a special case: for putting fi(y) = F(i,y) for i = 1, 2, the integral inequality gives

${\displaystyle \|f_{1}+f_{2}\|_{p}=\left(\int _{S_{2}}\left|\int _{S_{1}}F(x,y)\,\mu _{1}(\mathrm {d} x)\right|^{p}\mu _{2}(\mathrm {d} y)\right)^{\frac {1}{p}}\leq \int _{S_{1}}\left(\int _{S_{2}}|F(x,y)|^{p}\,\mu _{2}(\mathrm {d} y)\right)^{\frac {1}{p}}\mu _{1}(\mathrm {d} x)=\|f_{1}\|_{p}+\|f_{2}\|_{p}.}$