Minkowski inequality

Minkowski 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

\|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 Lp(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, ..., xn, y1, ..., yn and where n is the cardinality of S (the number of elements in S).

Contents

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) = xp is convex over \mathbb{R}^+ (for p greater than one) and so, if a and b are both positive then, by Jensen's inequality,

\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 (S11) and (S22) are two measure spaces and F : S1×S2R 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 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 ƒi(y) = F(i,y) 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}

See also

References


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