In mathematical analysis, the Minkowski inequality establishes that the L p {\displaystyle L^{p}} spaces satisfy the triangle inequality in the definition of normed vector spaces. The inequality is named after the German mathematician Hermann Minkowski. Let S {\textstyle S} be a measure space, let 1 ≤ p ≤ ∞ {\textstyle 1\leq p\leq \infty } and let f {\textstyle f} and g {\textstyle g} be elements of L p ( S ) . {\textstyle L^{p}(S).} Then f + g {\textstyle f+g} is in L p ( S ) , {\textstyle L^{p}(S),} and we have the triangle inequality
‖ f + g ‖ p ≤ ‖ f ‖ p + ‖ g ‖ p {\displaystyle \|f+g\|_{p}\leq \|f\|_{p}+\|g\|_{p}}
with equality for 1 < p < ∞ {\textstyle 1<p<\infty } if and only if f {\textstyle f} and g {\textstyle g} are positively linearly dependent; that is, f = λ g {\textstyle f=\lambda g} for some λ ≥ 0 {\textstyle \lambda \geq 0} or g = 0. {\textstyle g=0.} Here, the norm is given by:
‖ f ‖ p = ( ∫ | f | p d μ ) 1 p {\displaystyle \|f\|_{p}=\left(\int |f|^{p}d\mu \right)^{\frac {1}{p}}}
if p < ∞ , {\textstyle p<\infty ,} or in the case p = ∞ {\textstyle p=\infty } by the essential supremum
‖ f ‖ ∞ = e s s s u p x ∈ S | f ( x ) | . {\displaystyle \|f\|_{\infty }=\operatorname {ess\ sup} _{x\in S}|f(x)|.}
The Minkowski inequality is the triangle inequality in L p ( S ) . {\textstyle L^{p}(S).} In fact, it is a special case of the more general fact
‖ f ‖ p = sup ‖ g ‖ q = 1 ∫ | f g | d μ , 1 p + 1 q = 1 {\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:
( ∑ k = 1 n | x k + y k | p ) 1 / p ≤ ( ∑ k = 1 n | x k | p ) 1 / p + ( ∑ k = 1 n | y k | p ) 1 / p {\displaystyle \left(\sum _{k=1}^{n}|x_{k}+y_{k}|^{p}\right)^{1/p}\leq \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 x 1 , … , x n , y 1 , … , y n {\textstyle x_{1},\dots ,x_{n},y_{1},\dots ,y_{n}} and where n {\textstyle n} is the cardinality of S {\textstyle S} (the number of elements in S {\textstyle S} ). In probabilistic terms, given the probability space ( Ω , F , P ) , {\displaystyle (\Omega ,{\mathcal {F}},\mathbb {P} ),} and E {\displaystyle \mathbb {E} } denote the expectation operator for every real- or complex-valued random variables X {\displaystyle X} and Y {\displaystyle Y} on Ω , {\displaystyle \Omega ,} Minkowski's inequality reads
( E [ | X + Y | p ] ) 1 p ⩽ ( E [ | X | p ] ) 1 p + ( E [ | Y | p ] ) 1 p . {\displaystyle \left(\mathbb {E} [|X+Y|^{p}]\right)^{\frac {1}{p}}\leqslant \left(\mathbb {E} [|X|^{p}]\right)^{\frac {1}{p}}+\left(\mathbb {E} [|Y|^{p}]\right)^{\frac {1}{p}}.}
Proof
Proof by Hölder's inequality First, we prove that f + g {\textstyle f+g} has finite p {\textstyle p} -norm if f {\textstyle f} and g {\textstyle g} both do, which follows by
| f + g | p ≤ 2 p − 1 ( | f | p + | g | p ) . {\displaystyle |f+g|^{p}\leq 2^{p-1}(|f|^{p}+|g|^{p}).}
Indeed, here we use the fact that h ( x ) = | x | p {\textstyle h(x)=|x|^{p}} is convex over R + {\textstyle \mathbb {R} ^{+}} (for p > 1 {\textstyle p>1} ) and so, by the definition of convexity,
| 1 2 f + 1 2 g | p ≤ | 1 2 | f | + 1 2 | g | | p ≤ 1 2 | f | p + 1 2 | g | p . {\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
| f + g | p ≤ 1 2 | 2 f | p + 1 2 | 2 g | p = 2 p − 1 | f | p + 2 p − 1 | g | p . {\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 ‖ f + g ‖ p {\textstyle \|f+g\|_{p}} . If it is zero, then Minkowski's inequality holds. We now assume that ‖ f + g ‖ p {\textstyle \|f+g\|_{p}} is not zero. Using the triangle inequality and then Hölder's inequality, we find that
‖ f + g ‖ p p = ∫ | f + g | p d μ = ∫ | f + g | ⋅ | f + g | p − 1 d μ ≤ ∫ ( | f | + | g | ) | f + g | p − 1 d μ = ∫ | f | | f + g | p − 1 d μ + ∫ | g | | f + g | p − 1 d μ ≤ ( ( ∫ | f | p d μ ) 1 p + ( ∫ | g | p d μ ) 1 p ) ( ∫ | f + g | ( p − 1 ) ( p p − 1 ) d μ ) 1 − 1 p Hölder's inequality = ( ‖ f ‖ p + ‖ g ‖ p ) ‖ f + g ‖ p p ‖ f + g ‖ p {\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
‖ f + g ‖ p ‖ f + g ‖ p p . {\displaystyle {\frac {\|f+g\|_{p}}{\|f+g\|_{p}^{p}}}.}
Proof by a direct convexity argument Given t ∈ ( 0 , 1 ) {\displaystyle t\in (0,1)} , one has, by convexity (Jensen's inequality), for every x ∈ S {\displaystyle x\in S}
| f ( x ) + g ( x ) | p = | ( 1 − t ) f ( x ) 1 − t + t g ( x ) t | p ≤ ( 1 − t ) | f ( x ) 1 − t | p + t | g ( x ) t | p = | f ( x ) | p ( 1 − t ) p − 1 + | g ( x ) | p t p − 1 . {\displaystyle |f(x)+g(x)|^{p}={\Bigl |}(1-t){\frac {f(x)}{1-t}}+t{\frac {g(x)}{t}}{\Bigr |}^{p}\leq (1-t){\Bigl |}{\frac {f(x)}{1-t}}{\Bigr |}^{p}+t{\Bigl |}{\frac {g(x)}{t}}{\Bigr |}^{p}={\frac {|f(x)|^{p}}{(1-t)^{p-1}}}+{\frac {|g(x)|^{p}}{t^{p-1}}}.}
By integration this leads to
∫ S | f + g | p d μ ≤ 1 ( 1 − t ) p − 1 ∫ S | f | p d μ + 1 t p − 1 ∫ S | g | p d μ . {\displaystyle \int _{S}|f+g|^{p}\,\mathrm {d} \mu \leq {\frac {1}{(1-t)^{p-1}}}\int _{S}|f|^{p}\,\mathrm {d} \mu +{\frac {1}{t^{p-1}}}\int _{S}|g|^{p}\,\mathrm {d} \mu .}
One takes then
t = ‖ g ‖ p ‖ f ‖ p + ‖ g ‖ p {\displaystyle t={\frac {\Vert g\Vert _{p}}{\Vert f\Vert _{p}+\Vert g\Vert _{p}}}}
to reach the conclusion.
Minkowski's integral inequality Suppose that ( S 1 , μ 1 ) {\textstyle (S_{1},\mu _{1})} and ( S 2 , μ 2 ) {\textstyle (S_{2},\mu _{2})} are two 𝜎-finite measure spaces and F : S 1 × S 2 → R {\textstyle F:S_{1}\times S_{2}\to \mathbb {R} } is measurable. Then Minkowski's integral inequality is:
[ ∫ S 2 | ∫ S 1 F ( x , y ) μ 1 ( d x ) | p μ 2 ( d y ) ] 1 p ≤ ∫ S 1 ( ∫ S 2 | F ( x , y ) | p μ 2 ( d y ) ) 1 p μ 1 ( d x ) , p ∈ [ 1 , ∞ ) {\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),\quad p\in [1,\infty )}
with obvious modifications in the case p = ∞ . {\textstyle p=\infty .} If p > 1 , {\textstyle p>1,} and both sides are finite, then equality holds only if | F ( x , y ) | = φ ( x ) ψ ( y ) {\textstyle |F(x,y)|=\varphi (x)\,\psi (y)} a.e. for some non-negative measurable functions φ {\textstyle \varphi } and ψ {\textstyle \psi } . If μ 1 {\textstyle \mu _{1}} is the counting measure on a two-point set S 1 = { 1 , 2 } , {\textstyle S_{1}=\{1,2\},} then Minkowski's integral inequality gives the usual Minkowski inequality as a special case: for putting f i ( y ) = F ( i , y ) {\textstyle f_{i}(y)=F(i,y)} for i = 1 , 2 , {\textstyle i=1,2,} the integral inequality gives
‖ f 1 + f 2 ‖ p = ( ∫ S 2 | ∫ S 1 F ( x , y ) μ 1 ( d x ) | p μ 2 ( d y ) ) 1 p ≤ ∫ S 1 ( ∫ S 2 | F ( x , y ) | p μ 2 ( d y ) ) 1 p μ 1 ( d x ) = ‖ f 1 ‖ p + ‖ f 2 ‖ p . {\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}.}
If the measurable function F : S 1 × S 2 → R {\textstyle F:S_{1}\times S_{2}\to \mathbb {R} } is non-negative then for all 1 ≤ p