柯西不等式的加强

hbghlyj

标题“柯西不等式的加强”不太合适. 因为这就是$\Bbb C^5$上的柯西不等式, 没有加强.

力工

hbghlyj 发表于 2022-9-14 08:26
补上链接
https://kuing.cjhb.site/forum.php?mod=viewthread&tid=416

谢谢大佬!

hbghlyj

A complex inner product space is a complex vector space $V$ and a function $V \times V \rightarrow \mathbb{C}$ satisfying:
(CI1) $\langle u, u\rangle \geqslant 0$ for any $u \in V$, and $\langle u, u\rangle=0$ iff $u=0$.
(CI2) $\langle u, v\rangle=\overline{\langle v, u\rangle}$ for all $u, v \in V$.
(CI3) $\left\langle\alpha u_1+\beta u_2, v\right\rangle=\alpha\left\langle u_1, v\right\rangle+\beta\left\langle u_2, v\right\rangle$ for all $u_1, u_2, v \in V$ and all $\alpha, \beta \in \mathbb{C}$.
在上面的定义中,
$V=\Bbb C^5$,
$⟨·,·⟩:V\times V\to\Bbb R^{\ge0}$定义为$⟨\left(z_{1}, \ldots, z_5\right),\left(w_{1}, \ldots, w_5\right)⟩=\sum_{i=1}^5 z_{i} \overline{w_{i}}$
就组成一个复内积空间.

hbghlyj

S.5.pdf
Proposition S.5.4 Let $(V,\langle\rangle)$ be a real inner product space, and for any $u \in V$ define $\|u\|$ by $\|u\|=\sqrt{\langle u, u\rangle}$. Then for any $u, v \in V$,
\[
|\langle u, v\rangle| \leqslant\|u\|\|v\| .
\]
Proof For any real number $x$, the inner product $\langle x u+v, x u+v\rangle=\|x u+v\|^2 \geqslant 0$. This says that $x^2\|u\|^2+2 x\langle u, v\rangle+\|v\|^2 \geqslant 0$ for all $x \in \mathbb{R}$. Think of this as a quadratic expression in $x$. The fact that is it always non-negative means that the quadratic equation $x^2\|u\|^2+2 x\langle u, v\rangle+\|v\|^2=0$ has either no real roots, or else one repeated real root; for otherwise the graph of the quadratic function would cross the real axis twice and would be negative for some values of $x$. Hence the discriminant " $b^2-4 a c$ " of the quadratic does not exceed zero. This says $\langle u, v\rangle^2 \leqslant\|u\|^2\|v\|^2$, from which the result follows.

hbghlyj

The version of Cauchy's inequality in Proposition S.5.4 holds also in the complex case, where again we define $\|u\|=\sqrt{\langle u, u\rangle}$; the proof has an added complication, which we now explain.
We first note that any non-zero complex number $w$ may be written in polar form, $w=|w| \mathrm{e}^{i \theta}$ for some (real) $\theta \in[0,2 \pi)$. We also note that Cauchy's inequality is trivially true if $\langle u, v\rangle=0$, so we may assume that $\langle u, v\rangle \neq 0$, and as above we may write it as $|\langle u, v\rangle| \mathrm{e}^{i \theta}$ for some $\theta \in[0,2 \pi)$. For any complex number $z$, the inner product $\langle z u+v, z u+v\rangle \geqslant 0$ by (CI1). Now
\[
\langle z u+v, z u+v\rangle=|z|^2\|u\|^2+z\langle u, v\rangle+\bar{z}\langle v, u\rangle+\|v\|^2=|z|^2\|u\|^2+z\langle u, v\rangle+\overline{z\langle u, v\rangle}+\|v\|^2
\]
\[
=|z|^2\|u\|^2+2 \operatorname{Re}(z\langle u, v\rangle)+\|v\|^2 \text {. }
\]
Let $x$ be any real number $x$ and set $z=x \mathrm{e}^{-i \theta}$. Then $|z|^2=x^2$ and $\operatorname{Re}(z\langle u, v\rangle)=x|\langle u, v\rangle|$, since $\langle u, v\rangle=|\langle u, v\rangle| \mathrm{e}^{i \theta}$. The upshot is that
\[
x^2\|u\|^2+2 x|\langle u, v\rangle|+\|v\|^2 \geqslant 0 \text { for all } x \in \mathbb{R},
\]
and Cauchy's inequality follows just as in the case of a real inner product space.