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柯西-施瓦茨不等式

在許多不同的設置中遇到的有用的不等式,例如線性代數,分析,概率論,向量代數和其他領域。 它被認為是所有數學中最重要的不等式之一 来自维基百科,自由的百科全书

柯西不等式
叙述特例Rn-n维欧几里得空间L2矩阵不等式复变函数中的柯西不等式其它推广参见注释参考资料

柯西-施瓦茨不等式(英语:Cauchy–Schwarz inequality),又称施瓦茨不等式或柯西-布尼亚科夫斯基-施瓦茨不等式或柯西不等式,在多个数学领域中均有应用的不等式;例如线性代数的矢量,数学分析的无穷级数和乘积的积分,和概率论的方差和协方差。它被认为是最重要的数学不等式之一。它有一些推广,如赫尔德不等式。

不等式以奥古斯丁·路易·柯西(Augustin Louis Cauchy),赫尔曼·阿曼杜斯·施瓦茨(Hermann Amandus Schwarz),和维克托·雅科夫列维奇·布尼亚科夫斯基(英语:Viktor_Bunyakovsky)(Виктор Яковлевич Буняковский)命名。

V {\displaystyle V} {\displaystyle V} 是个复内积空间,则对所有的 v , w ∈ V {\displaystyle v,\,w\in V} {\displaystyle v,\,w\in V} 有:

(a) ‖ v ‖ ‖ w ‖ ≥ | ⟨ v , w ⟩ | {\displaystyle \|v\|\|w\|\geq |\langle v,\,w\rangle |} {\displaystyle \|v\|\|w\|\geq |\langle v,\,w\rangle |}
(b) ‖ v ‖ ‖ w ‖ = | ⟨ v , w ⟩ | {\displaystyle \|v\|\|w\|=|\langle v,\,w\rangle |} {\displaystyle \|v\|\|w\|=|\langle v,\,w\rangle |} ⇔ {\displaystyle \Leftrightarrow } {\displaystyle \Leftrightarrow } 存在 λ ∈ C {\displaystyle \lambda \in \mathbb {C} } {\displaystyle \lambda \in \mathbb {C} } 使 v = λ ⋅ w {\displaystyle v=\lambda \cdot w} {\displaystyle v=\lambda \cdot w}

证明请见内积空间#范数。

Rn-n维欧几里得空间

对欧几里得空间Rn,有

( ∑ i = 1 n x i y i ) 2 ≤ ( ∑ i = 1 n x i 2 ) ( ∑ i = 1 n y i 2 ) {\displaystyle \left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}\leq \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)} {\displaystyle \left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}\leq \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)}。

等式成立时:

x 1 y 1 = x 2 y 2 = ⋯ = x n y n . {\displaystyle {\frac {x_{1}}{y_{1}}}={\frac {x_{2}}{y_{2}}}=\cdots ={\frac {x_{n}}{y_{n}}}.} {\displaystyle {\frac {x_{1}}{y_{1}}}={\frac {x_{2}}{y_{2}}}=\cdots ={\frac {x_{n}}{y_{n}}}.}

也可以表示成

( x 1 2 + x 2 2 + ⋯ + x n 2 ) ( y 1 2 + y 2 2 + ⋯ + y n 2 ) ≥ ( x 1 y 1 + x 2 y 2 + ⋯ + x n y n ) 2 {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq (x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}} {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq (x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}}

证明则须考虑一个关于 t {\displaystyle t} {\displaystyle t}的一个一元二次方程式 ( x 1 t + y 1 ) 2 + ⋯ + ( x n t + y n ) 2 = 0 {\displaystyle (x_{1}t+y_{1})^{2}+\cdots +(x_{n}t+y_{n})^{2}=0} {\displaystyle (x_{1}t+y_{1})^{2}+\cdots +(x_{n}t+y_{n})^{2}=0}

很明显的,此方程式无实数解或有重根,故其判别式 D ≤ 0 {\displaystyle D\leq 0} {\displaystyle D\leq 0}

注意到

( x 1 t + y 1 ) 2 + ⋯ + ( x n t + y n ) 2 ≥ 0 {\displaystyle (x_{1}t+y_{1})^{2}+\cdots +(x_{n}t+y_{n})^{2}\geq 0} {\displaystyle (x_{1}t+y_{1})^{2}+\cdots +(x_{n}t+y_{n})^{2}\geq 0}

⇒ ( x 1 2 + x 2 2 + ⋯ + x n 2 ) t 2 + 2 ( x 1 y 1 + x 2 y 2 + ⋯ + x n y n ) t + ( y 1 2 + y 2 2 + ⋯ + y n 2 ) ≥ 0 {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})t^{2}+2(x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})t+(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq 0} {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})t^{2}+2(x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})t+(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq 0}

则

D = 4 ( x 1 y 1 + x 2 y 2 + ⋯ + x n y n ) 2 − 4 ( x 1 2 + x 2 2 + ⋯ + x n 2 ) ( y 1 2 + y 2 2 + ⋯ + y n 2 ) ≤ 0 {\displaystyle D=4(x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}-4(x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\leq 0} {\displaystyle D=4(x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}-4(x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\leq 0}

即

( x 1 2 + x 2 2 + ⋯ + x n 2 ) ( y 1 2 + y 2 2 + ⋯ + y n 2 ) ≥ ( x 1 y 1 + x 2 y 2 + ⋯ + x n y n ) 2 {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq (x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}} {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq (x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}}

( x 1 t + y 1 ) 2 + ⋯ + ( x n t + y n ) 2 = 0 {\displaystyle (x_{1}t+y_{1})^{2}+\cdots +(x_{n}t+y_{n})^{2}=0} {\displaystyle (x_{1}t+y_{1})^{2}+\cdots +(x_{n}t+y_{n})^{2}=0}

( x 1 2 + x 2 2 + ⋯ + x n 2 ) ( y 1 2 + y 2 2 + ⋯ + y n 2 ) ≥ ( x 1 y 1 + x 2 y 2 + ⋯ + x n y n ) 2 {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq (x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}} {\displaystyle (x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots +y_{n}^{2})\geq (x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})^{2}}

而等号成立于判别式 D = 0 {\displaystyle D=0} {\displaystyle D=0}时

也就是此时方程式有重根,故

x 1 y 1 = x 2 y 2 = ⋯ = x n y n . {\displaystyle {\frac {x_{1}}{y_{1}}}={\frac {x_{2}}{y_{2}}}=\cdots ={\frac {x_{n}}{y_{n}}}.} {\displaystyle {\frac {x_{1}}{y_{1}}}={\frac {x_{2}}{y_{2}}}=\cdots ={\frac {x_{n}}{y_{n}}}.}

  • 对平方可积的复值函数,有
| ∫ f ( x ) g ( x ) d x | 2 ≤ ∫ | f ( x ) | 2 d x ⋅ ∫ | g ( x ) | 2 d x {\displaystyle \left|\int f(x)g(x)\,dx\right|^{2}\leq \int \left|f(x)\right|^{2}\,dx\cdot \int \left|g(x)\right|^{2}\,dx} {\displaystyle \left|\int f(x)g(x)\,dx\right|^{2}\leq \int \left|f(x)\right|^{2}\,dx\cdot \int \left|g(x)\right|^{2}\,dx}。

这两例可更一般化为赫尔德不等式。

  • 在3维空间,有一个较强结果值得注意:原不等式可以增强至拉格朗日恒等式
⟨ x , x ⟩ ⋅ ⟨ y , y ⟩ = | ⟨ x , y ⟩ | 2 + | x × y | 2 {\displaystyle \langle x,x\rangle \cdot \langle y,y\rangle =|\langle x,y\rangle |^{2}+|x\times y|^{2}} {\displaystyle \langle x,x\rangle \cdot \langle y,y\rangle =|\langle x,y\rangle |^{2}+|x\times y|^{2}}。
这是
( ∑ i = 1 n x i y i ) 2 = ( ∑ i = 1 n x i 2 ) ( ∑ i = 1 n y i 2 ) − ( ∑ 1 ≤ i < j ≤ n ( x i y j − x j y i ) 2 ) {\displaystyle \left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}=\left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)-\left(\sum _{1\leq i<j\leq n}(x_{i}y_{j}-x_{j}y_{i})^{2}\right)} {\displaystyle \left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}=\left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)-\left(\sum _{1\leq i<j\leq n}(x_{i}y_{j}-x_{j}y_{i})^{2}\right)}
在n=3 时的特殊情况。

L2

对于平方可积复值函数的内积空间,有如下不等式:

| ∫ R n f ( x ) g ( x ) ¯ d x | 2 ≤ ∫ R n | f ( x ) | 2 d x ∫ R n | g ( x ) | 2 d x . {\displaystyle \left|\int _{\mathbb {R} ^{n}}f(x){\overline {g(x)}}\,dx\right|^{2}\leq \int _{\mathbb {R} ^{n}}|f(x)|^{2}\,dx\int _{\mathbb {R} ^{n}}|g(x)|^{2}\,dx.} {\displaystyle \left|\int _{\mathbb {R} ^{n}}f(x){\overline {g(x)}}\,dx\right|^{2}\leq \int _{\mathbb {R} ^{n}}|f(x)|^{2}\,dx\int _{\mathbb {R} ^{n}}|g(x)|^{2}\,dx.}

赫尔德不等式是该式的推广。

设 x , y {\displaystyle x,y} {\displaystyle x,y}为列向量,则 | x ∗ y | 2 ≤ x ∗ x ⋅ y ∗ y {\displaystyle |x^{*}y|^{2}\leq x^{*}x\cdot y^{*}y} {\displaystyle |x^{*}y|^{2}\leq x^{*}x\cdot y^{*}y}[a]

x = 0 {\displaystyle x=0} {\displaystyle x=0} 时不等式成立,设 x {\displaystyle x} {\displaystyle x}非零, z = y − y ∗ x ‖ x ‖ 2 x {\displaystyle z=y-{\cfrac {y^{*}x}{\|x\|^{2}}}x} {\displaystyle z=y-{\cfrac {y^{*}x}{\|x\|^{2}}}x},则 x ∗ z = 0 {\displaystyle x^{*}z=0} {\displaystyle x^{*}z=0}
0 ≤ ‖ z ‖ 2 = z ∗ y = ‖ y ‖ 2 − x ∗ y ‖ x ‖ 2 x ∗ y = ‖ y ‖ 2 − | x ∗ y | 2 ‖ x ‖ 2 {\displaystyle 0\leq \|z\|^{2}=z^{*}y=\|y\|^{2}-{\cfrac {x^{*}y}{\|x\|^{2}}}x^{*}y=\|y\|^{2}-{\cfrac {|x^{*}y|^{2}}{\|x\|^{2}}}} {\displaystyle 0\leq \|z\|^{2}=z^{*}y=\|y\|^{2}-{\cfrac {x^{*}y}{\|x\|^{2}}}x^{*}y=\|y\|^{2}-{\cfrac {|x^{*}y|^{2}}{\|x\|^{2}}}}
| x ∗ y | 2 ≤ ‖ x ‖ 2 ‖ y ‖ 2 {\displaystyle |x^{*}y|^{2}\leq \|x\|^{2}\|y\|^{2}} {\displaystyle |x^{*}y|^{2}\leq \|x\|^{2}\|y\|^{2}}
等号成立 ⇔ z = 0 ⇔ y {\displaystyle \Leftrightarrow z=0\Leftrightarrow y} {\displaystyle \Leftrightarrow z=0\Leftrightarrow y}与 x {\displaystyle x} {\displaystyle x}线性相关

设 A {\displaystyle A} {\displaystyle A}为 n × n {\displaystyle n\times n} {\displaystyle n\times n}Hermite阵,且 A ≥ 0 {\displaystyle A\geq 0} {\displaystyle A\geq 0},则 | x ∗ A y | 2 ≤ x ∗ A x ⋅ y ∗ A y {\displaystyle |x^{*}Ay|^{2}\leq x^{*}Ax\cdot y^{*}Ay} {\displaystyle |x^{*}Ay|^{2}\leq x^{*}Ax\cdot y^{*}Ay}

存在 A 1 / 2 {\displaystyle A^{1/2}} {\displaystyle A^{1/2}},设 u = A 1 / 2 x , v = A 1 / 2 y {\displaystyle u=A^{1/2}x,v=A^{1/2}y} {\displaystyle u=A^{1/2}x,v=A^{1/2}y}
| u ∗ v | 2 ≤ u ∗ u ⋅ v ∗ v {\displaystyle |u^{*}v|^{2}\leq u^{*}u\cdot v^{*}v} {\displaystyle |u^{*}v|^{2}\leq u^{*}u\cdot v^{*}v}
| x ∗ A 1 / 2 A 1 / 2 y | 2 ≤ x ∗ A 1 / 2 A 1 / 2 x ⋅ y ∗ A 1 / 2 A 1 / 2 y {\displaystyle |x^{*}A^{1/2}A^{1/2}y|^{2}\leq x^{*}A^{1/2}A^{1/2}x\cdot y^{*}A^{1/2}A^{1/2}y} {\displaystyle |x^{*}A^{1/2}A^{1/2}y|^{2}\leq x^{*}A^{1/2}A^{1/2}x\cdot y^{*}A^{1/2}A^{1/2}y}
| x ∗ A y | 2 ≤ x ∗ A x ⋅ y ∗ A y {\displaystyle |x^{*}Ay|^{2}\leq x^{*}Ax\cdot y^{*}Ay} {\displaystyle |x^{*}Ay|^{2}\leq x^{*}Ax\cdot y^{*}Ay}
等号成立 ⇔ y {\displaystyle \Leftrightarrow y} {\displaystyle \Leftrightarrow y}与 x {\displaystyle x} {\displaystyle x}线性相关

设 A {\displaystyle A} {\displaystyle A}为 n × n {\displaystyle n\times n} {\displaystyle n\times n}Hermite阵,且 A > 0 {\displaystyle A>0} {\displaystyle A>0},则 | x ∗ y | 2 ≤ x ∗ A x ⋅ y ∗ A − 1 y {\displaystyle |x^{*}y|^{2}\leq x^{*}Ax\cdot y^{*}A^{-1}y} {\displaystyle |x^{*}y|^{2}\leq x^{*}Ax\cdot y^{*}A^{-1}y}

存在 A 1 / 2 , A − 1 / 2 {\displaystyle A^{1/2},A^{-1/2}} {\displaystyle A^{1/2},A^{-1/2}},设 u = A 1 / 2 x , v = A − 1 / 2 y {\displaystyle u=A^{1/2}x,v=A^{-1/2}y} {\displaystyle u=A^{1/2}x,v=A^{-1/2}y}
| u ∗ v | 2 ≤ u ∗ u ⋅ v ∗ v {\displaystyle |u^{*}v|^{2}\leq u^{*}u\cdot v^{*}v} {\displaystyle |u^{*}v|^{2}\leq u^{*}u\cdot v^{*}v}
| x ∗ A 1 / 2 A − 1 / 2 y | 2 ≤ x ∗ A 1 / 2 A 1 / 2 x ⋅ y ∗ A − 1 / 2 A − 1 / 2 y {\displaystyle |x^{*}A^{1/2}A^{-1/2}y|^{2}\leq x^{*}A^{1/2}A^{1/2}x\cdot y^{*}A^{-1/2}A^{-1/2}y} {\displaystyle |x^{*}A^{1/2}A^{-1/2}y|^{2}\leq x^{*}A^{1/2}A^{1/2}x\cdot y^{*}A^{-1/2}A^{-1/2}y}
| x ∗ y | 2 ≤ x ∗ A x ⋅ y ∗ A − 1 y {\displaystyle |x^{*}y|^{2}\leq x^{*}Ax\cdot y^{*}A^{-1}y} {\displaystyle |x^{*}y|^{2}\leq x^{*}Ax\cdot y^{*}A^{-1}y}
等号成立 ⇔ x {\displaystyle \Leftrightarrow x} {\displaystyle \Leftrightarrow x}与 A − 1 y {\displaystyle A^{-1}y} {\displaystyle A^{-1}y}线性相关[1]

若 q i ≥ 0 , ∑ i q i = 1 {\displaystyle \displaystyle q_{i}\geq 0,\sum _{i}q_{i}=1} {\displaystyle \displaystyle q_{i}\geq 0,\sum _{i}q_{i}=1},则 ( x ∗ A ∑ i a i q i x ) ≤ ∏ i ( x ∗ A a i x ) q i {\displaystyle \displaystyle (x^{*}A^{\sum _{i}a_{i}q_{i}}x)\leq \prod _{i}(x^{*}A^{a_{i}}x)^{q_{i}}} {\displaystyle \displaystyle (x^{*}A^{\sum _{i}a_{i}q_{i}}x)\leq \prod _{i}(x^{*}A^{a_{i}}x)^{q_{i}}}[2]

设 f ( z ) {\displaystyle f(z)} {\displaystyle f(z)}在区域 D {\displaystyle D} {\displaystyle D}及其边界上解析, a {\displaystyle a} {\displaystyle a} 为 D {\displaystyle D} {\displaystyle D}内一点,以 a {\displaystyle a} {\displaystyle a}为圆心做圆周 C R : | z − a | = R {\displaystyle C_{R}:|z-a|=R} {\displaystyle C_{R}:|z-a|=R},只要 C R {\displaystyle C_{R}} {\displaystyle C_{R}}及其内部 G {\displaystyle G} {\displaystyle G}均被 D {\displaystyle D} {\displaystyle D}包含,则有:

| f ( n ) ( z 0 ) | ≤ n ! M R n ( n = 1 , 2 , 3 , . . . ) {\displaystyle \left|f^{(n)}(z_{0})\right|\leq {\frac {n!M}{R^{n}}}\qquad (n=1,2,3,...)} {\displaystyle \left|f^{(n)}(z_{0})\right|\leq {\frac {n!M}{R^{n}}}\qquad (n=1,2,3,...)}

其中,M是 | f ( z ) | {\displaystyle |f(z)|} {\displaystyle |f(z)|}的最大值, M = max | x − a | ∈ R | f ( x ) | {\displaystyle M=\max \limits _{|x-a|\in R}|f(x)|} {\displaystyle M=\max \limits _{|x-a|\in R}|f(x)|} 。

∑ i = 1 n ( ∑ j = 1 m a i j ) 2 ≤ ∑ j = 1 m ∑ i = 1 n a i j 2 {\displaystyle {\sqrt {\sum _{i=1}^{n}(\sum _{j=1}^{m}a_{ij})^{2}}}\leq \sum _{j=1}^{m}{\sqrt {\sum _{i=1}^{n}a_{ij}^{2}}}} {\displaystyle {\sqrt {\sum _{i=1}^{n}(\sum _{j=1}^{m}a_{ij})^{2}}}\leq \sum _{j=1}^{m}{\sqrt {\sum _{i=1}^{n}a_{ij}^{2}}}}[3]

m ≥ α > 0 , ( ∑ i = 1 n ∏ j = 1 m a i j ) α ≤ ∏ j = 1 m ∑ i = 1 n a i j α {\displaystyle m\geq \alpha >0,(\sum _{i=1}^{n}\prod _{j=1}^{m}a_{ij})^{\alpha }\leq \prod _{j=1}^{m}\sum _{i=1}^{n}a_{ij}^{\alpha }} {\displaystyle m\geq \alpha >0,(\sum _{i=1}^{n}\prod _{j=1}^{m}a_{ij})^{\alpha }\leq \prod _{j=1}^{m}\sum _{i=1}^{n}a_{ij}^{\alpha }}[4]

  • 三角不等式
  • 内积空间
  1. [a]
    x ∗ {\displaystyle x^{*}} {\displaystyle x^{*}}表示x的共轭转置。
  1. [1]
    王松桂. 矩阵不等式-(第二版).
  2. [2]
    程伟丽 齐静. Cauchy不等式矩阵形式的推广. 郑州轻工业学院学报(自然科学版). 2008, (4) [2015-03-24]. (原始内容存档于2019-06-08).
  3. [3]
    赵明方. Cauchy不等式的推广. 四川师范大学学报(自然科学版). 1981, (2) [2015-03-24]. (原始内容存档于2019-06-03).
  4. [4]
    洪勇. 推广的Cauchy不等式的再推广. 曲靖师范学院学报. 1993, (S1) [2015-03-24]. (原始内容存档于2019-06-03).
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叙述

特例

矩阵不等式

复变函数中的柯西不等式

其它推广

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参考资料

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