Triangle inequality in Lp spaces
Inmathematical analysis , theMinkowski inequality establishes that theL p {\displaystyle L^{p}} triangle inequality in the definition ofnormed vector spaces . The inequality is named after the German mathematicianHermann Minkowski .
LetS {\textstyle S} measure space , let1 ≤ p ≤ ∞ {\textstyle 1\leq p\leq \infty } f {\textstyle f} g {\textstyle g} L p ( S ) . {\textstyle L^{p}(S).} f + g {\textstyle f+g} L p ( S ) , {\textstyle L^{p}(S),}
‖ f + g ‖ p ≤ ‖ f ‖ p + ‖ g ‖ p {\displaystyle \|f+g\|_{p}\leq \|f\|_{p}+\|g\|_{p}}
with equality for1 < p < ∞ {\textstyle 1<p<\infty } f {\textstyle f} g {\textstyle g} linearly dependent ; that is,f = λ g {\textstyle f=\lambda g} λ ≥ 0 {\textstyle \lambda \geq 0} g = 0. {\textstyle g=0.}
‖ f ‖ p = ( ∫ | f | p d μ ) 1 p {\displaystyle \|f\|_{p}=\left(\int |f|^{p}d\mu \right)^{\frac {1}{p}}}
ifp < ∞ , {\textstyle p<\infty ,} p = ∞ {\textstyle p=\infty } 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 inL p ( S ) . {\textstyle L^{p}(S).}
‖ 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.
LikeHölder's inequality , the Minkowski inequality can be specialized to sequences and vectors by using thecounting 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 allreal (orcomplex ) numbersx 1 , … , x n , y 1 , … , y n {\textstyle x_{1},\dots ,x_{n},y_{1},\dots ,y_{n}} n {\textstyle n} cardinality ofS {\textstyle S} S {\textstyle S}
In probabilistic terms, given theprobability space ( Ω , F , P ) , {\displaystyle (\Omega ,{\mathcal {F}},\mathbb {P} ),} E {\displaystyle \mathbb {E} } expectation operator for every real- or complex-valuedrandom variables X {\displaystyle X} Y {\displaystyle Y} Ω , {\displaystyle \Omega ,}
( 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}}.} [ edit ] First, we prove thatf + g {\textstyle f+g} p {\textstyle p} f {\textstyle f} g {\textstyle g}
| 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 thath ( x ) = | x | p {\textstyle h(x)=|x|^{p}} convex overR + {\textstyle \mathbb {R} ^{+}} p > 1 {\textstyle p>1}
| 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}} ‖ f + g ‖ p {\textstyle \|f+g\|_{p}} 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 [ edit ] Givent ∈ ( 0 , 1 ) {\displaystyle t\in (0,1)} Jensen's inequality ), for everyx ∈ 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.
[ edit ] Suppose that( S 1 , μ 1 ) {\textstyle (S_{1},\mu _{1})} ( S 2 , μ 2 ) {\textstyle (S_{2},\mu _{2})} F : S 1 × S 2 → R {\textstyle F:S_{1}\times S_{2}\to \mathbb {R} } [ 1] [ 2]
[ ∫ 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 casep = ∞ . {\textstyle p=\infty .} p > 1 , {\textstyle p>1,} | F ( x , y ) | = φ ( x ) ψ ( y ) {\textstyle |F(x,y)|=\varphi (x)\,\psi (y)} φ {\textstyle \varphi } ψ {\textstyle \psi }
Ifμ 1 {\textstyle \mu _{1}} S 1 = { 1 , 2 } , {\textstyle S_{1}=\{1,2\},} f i ( y ) = F ( i , y ) {\textstyle f_{i}(y)=F(i,y)} i = 1 , 2 , {\textstyle i=1,2,}
‖ 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 functionF : S 1 × S 2 → R {\textstyle F:S_{1}\times S_{2}\to \mathbb {R} } 1 ≤ p ≤ q ≤ ∞ , {\textstyle 1\leq p\leq q\leq \infty ,} [ 3]
‖ ‖ F ( ⋅ , s 2 ) ‖ L p ( S 1 , μ 1 ) ‖ L q ( S 2 , μ 2 ) ≤ ‖ ‖ F ( s 1 , ⋅ ) ‖ L q ( S 2 , μ 2 ) ‖ L p ( S 1 , μ 1 ) . {\displaystyle \left\|\left\|F(\,\cdot ,s_{2})\right\|_{L^{p}(S_{1},\mu _{1})}\right\|_{L^{q}(S_{2},\mu _{2})}~\leq ~\left\|\left\|F(s_{1},\cdot )\right\|_{L^{q}(S_{2},\mu _{2})}\right\|_{L^{p}(S_{1},\mu _{1})}\ .}
This notation has been generalized to
‖ f ‖ p , q = ( ∫ R m [ ∫ R n | f ( x , y ) | q d y ] p q d x ) 1 p {\displaystyle \|f\|_{p,q}=\left(\int _{\mathbb {R} ^{m}}\left[\int _{\mathbb {R} ^{n}}|f(x,y)|^{q}\mathrm {d} y\right]^{\frac {p}{q}}\mathrm {d} x\right)^{\frac {1}{p}}}
forf : R m + n → E , {\textstyle f:\mathbb {R} ^{m+n}\to E,} L p , q ( R m + n , E ) = { f ∈ E R m + n : ‖ f ‖ p , q < ∞ } . {\textstyle {\mathcal {L}}_{p,q}(\mathbb {R} ^{m+n},E)=\{f\in E^{\mathbb {R} ^{m+n}}:\|f\|_{p,q}<\infty \}.} p < q , {\textstyle p<q,} ‖ f ‖ q , p ≤ ‖ f ‖ p , q {\textstyle \|f\|_{q,p}\leq \|f\|_{p,q}}
Whenp < 1 {\textstyle p<1}
‖ f + g ‖ p ≥ ‖ f ‖ p + ‖ g ‖ p . {\displaystyle \|f+g\|_{p}\geq \|f\|_{p}+\|g\|_{p}.}
We further need the restriction that bothf {\textstyle f} g {\textstyle g} f = − 1 , g = 1 {\textstyle f=-1,g=1} p = 1 {\textstyle p=1}
‖ f + g ‖ 1 = 0 < 2 = ‖ f ‖ 1 + ‖ g ‖ 1 . {\textstyle \|f+g\|_{1}=0<2=\|f\|_{1}+\|g\|_{1}.} The reverse inequality follows from the same argument as the standard Minkowski, but uses that Holder's inequality is also reversed in this range.
Using the Reverse Minkowski, we may prove that power means withp ≤ 1 , {\textstyle p\leq 1,} harmonic mean and thegeometric mean are concave.
Generalizations to other functions [ edit ] The Minkowski inequality can be generalized to other functionsϕ ( x ) {\textstyle \phi (x)} x p {\textstyle x^{p}}
ϕ − 1 ( ∑ i = 1 n ϕ ( x i + y i ) ) ≤ ϕ − 1 ( ∑ i = 1 n ϕ ( x i ) ) + ϕ − 1 ( ∑ i = 1 n ϕ ( y i ) ) . {\displaystyle \phi ^{-1}\left(\textstyle \sum \limits _{i=1}^{n}\phi (x_{i}+y_{i})\right)\leq \phi ^{-1}\left(\textstyle \sum \limits _{i=1}^{n}\phi (x_{i})\right)+\phi ^{-1}\left(\textstyle \sum \limits _{i=1}^{n}\phi (y_{i})\right).}
Various sufficient conditions onϕ {\textstyle \phi } [ 4] x ≥ 0 {\textstyle x\geq 0}
ϕ ( x ) {\textstyle \phi (x)} ϕ ( 0 ) = 0. {\textstyle \phi (0)=0.} ϕ ( x ) {\textstyle \phi (x)} x . {\textstyle x.} log ϕ ( x ) {\textstyle \log \phi (x)} log ( x ) . {\textstyle \log(x).} Bahouri, Hajer ;Chemin, Jean-Yves ;Danchin, Raphaël (2011).Fourier Analysis and Nonlinear Partial Differential Equations ISBN 978-3-642-16830-7 OCLC 704397128 .Hardy, G. H. ;Littlewood, J. E. ;Pólya, G. (1988).Inequalities ISBN 0-521-35880-9 Minkowski, H. (1953).Geometrie der Zahlen . Chelsea.Stein, Elias (1970).Singular integrals and differentiability properties of functions . Princeton University Press.M.I. Voitsekhovskii (2001) [1994],"Minkowski inequality" ,Encyclopedia of Mathematics EMS Press Lohwater, Arthur J. (1982)."Introduction to Inequalities" .
