# 柯西-施瓦茨不等式

## 叙述

$\big| \langle x,y\rangle \big|^2 \leq \langle x,x\rangle \cdot \langle y,y\rangle$

$|\langle x,y\rangle| \leq \|x\| \cdot \|y\|.\,$

$x_1,\ldots, x_n\in\mathbb C$$y_1,\ldots, y_n\in\mathbb C$有虚部，内积即为标准内积，用拔标记共轭复数那么这个不等式可以更明确的表述为

$|x_1 \bar{y}_1 + \cdots + x_n \bar{y}_n|^2 \leq (|x_1|^2 + \cdots + |x_n|^2) (|y_1|^2 + \cdots + |y_n|^2).$

## 特例

$\left(\sum_{i=1}^n x_i y_i\right)^2 \leq \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right)$

$\frac {x_1}{y_1} = \frac {x_2}{y_2} = \cdots = \frac {x_n}{y_n}.$

$(x_1^2 +x_2^2 + \cdots + x_n^2)(y_1^2 +y_2^2 + \cdots + y_n^2) \ge (x_1 y_1 + x_2 y_2 + \cdots + x_n y_n)^2$

$(x_1 t + y_1)^2 + \cdots + (x_n t + y_n)^2 = 0$

$(x_1^2 +x_2^2 + \cdots + x_n^2) t^2 + 2 (x_1 y_1 + x_2 y_2 + \cdots + x_n y_n) t + (y_1^2 +y_2^2 + \cdots + y_n^2) = 0$

$D = 4 (x_1 y_1 + x_2 y_2 + \cdots + x_n y_n)^2 - 4 (x_1^2 +x_2^2 + \cdots + x_n^2) (y_1^2 +y_2^2 + \cdots + y_n^2) \leq 0$

$(x_1^2 +x_2^2 + \cdots + x_n^2)(y_1^2 +y_2^2 + \cdots + y_n^2) \ge (x_1 y_1 + x_2 y_2 + \cdots + x_n y_n)^2$

$(x_1 t + y_1)^2 + \cdots + (x_n t + y_n)^2 = 0$

$(x_1^2 +x_2^2 + \cdots + x_n^2)(y_1^2 +y_2^2 + \cdots + y_n^2) \ge (x_1 y_1 + x_2 y_2 + \cdots + x_n y_n)^2$

$\frac {x_1}{y_1} = \frac {x_2}{y_2} = \cdots = \frac {x_n}{y_n}.$

• 對平方可積的複值函數，有
$\left|\int f^*(x)g(x)\,dx\right|^2\leq\int \left|f(x)\right|^2\,dx \cdot \int\left|g(x)\right|^2\,dx$

$\langle x,x\rangle \cdot \langle y,y\rangle = |\langle x,y\rangle|^2 + |x \times y|^2$

$\left(\sum_{i=1}^n x_i y_i\right)^2 = \left(\sum_{i=1}^n x_i^2\right) \left(\sum_{i=1}^n y_i^2\right)-\left(\sum_{1\le i < j\le n}(x_i y_j - x_j y_i)^2\right)$
n=3 时的特殊情况。

## 矩阵不等式

$x,y$列向量，则$|x^*y|^2\le x^*x\cdot y^*y$[註 1]

x=0時不等式成立，设x非零，$z=y-\cfrac{x^*y}{\|x\|}x$，则$x^*z=0$
$0\le \|z\|^2=z^*y=\|y\|^2-\cfrac{x^*y}{\|x\|^2}x^*y=\|y\|^2-\cfrac{|x^*y|^2}{\|x\|^2}$
$|x^*y|^2\le\|x\|^2\|y\|^2$

$A$$n\times n$Hermite阵，且$A\ge 0$，则$|x^*Ay|^2\le x^*Ax\cdot y^*Ay$

$|u^*v|^2\le u^*u\cdot v^*v$
$|x^*A^{1/2}A^{1/2}y|^2\le x^*A^{1/2}A^{1/2}x\cdot y^*A^{1/2}A^{1/2}y$
$|x^*Ay|^2\le x^*Ax\cdot y^*Ay$

$A$$n\times n$Hermite阵，且$A>0$，则$|x^*y|^2\le x^*Ax\cdot y^*A^{-1}y$

$|u^*v|^2\le u^*u\cdot v^*v$
$|x^*A^{1/2}A^{-1/2}y|^2\le x^*A^{1/2}A^{1/2}x\cdot y^*A^{-1/2}A^{-1/2}y$
$|x^*y|^2\le x^*Ax\cdot y^*A^{-1}y$

$\displaystyle q_i\ge 0,\sum_i q_i=1$，则$\displaystyle (x^*A^{\sum_i a_i q_i}x)\le \prod_i (x^*A^{a_i}x)^{q_i}$[2]

## 其它推广

$\sqrt{\sum_{i=1}^n (\sum_{j=1}^m a_{ij})^2} \le \sum_{j=1}^m \sqrt{\sum_{i=1}^n a_{ij}^2}$[3]

$m\ge \alpha >0,(\sum_{i=1}^n \prod_{j=1}^m a_{ij})^{\alpha} \le \prod_{j=1}^m \sum_{i=1}^n a_{ij}^{\alpha}$[4]

## 注释

1. ^ $x^*$表示x的共轭转置

## 参考资料

1. ^ 王松桂. 矩阵不等式-(第二版).
2. ^
3. ^
4. ^