Minkowski inequality

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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

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:

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

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

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:

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


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

Indeed, here we use the fact that is convex over R+ (for p > 1) and so, by the definition of convexity,

This means that

Now, we can legitimately talk about . If it is zero, then Minkowski's inequality holds. We now assume that is not zero. Using the triangle inequality and then Hölder's inequality, we find that

We obtain Minkowski's inequality by multiplying both sides by

Minkowski's integral inequality[edit]

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):

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

See also[edit]