关于Cauchy–Schwarz inequality的疑问

hbghlyj

内积空间的定义中有Positive-definiteness: if $x$ is not zero, then $\left<x,x\right>>0$.
如果减弱为if $x$ is not zero, then $\left<x,x\right>\ge0$. 这样还成立Cauchy–Schwarz inequality吗?
我看了一下Cauchy–Schwarz inequality用到positive而主要部分似乎没有用到definite, 只是取等条件需要从$\left<x,x\right>=0$.
问题来源于: Cauchy–Schwarz inequality对于平方可积函数空间的形式$\displaystyle\left|\int_{0}^{1} f(x) g(x) d x\right|^{2} \leq \int_{0}^{1} f(x)^{2} d x \cdot \int_{0}^{1} g(x)^{2} d x$
$C[0,1]$是内积空间, 因为可以证明$\int_0^1f(x)^2dx=0$ if and only if $f(x)≡0$.

证明

otherwise if $f(x_0)\ne0$, by continuity, $∃δ$ such that $|x-x_0|<δ⇒|f(x)-f(x_0)|<\frac12f(x_0)⇒f(x)>\frac12f(x_0)$. Thus $0=\int_0^1f(x)^2dx\ge\int_{\max(0,x_0-\delta)}^{\min(1,x_0+\delta)}f(x)^2dx\ge\frac14f(x_0)^2·δ$ 积分上下限至少有一个没有从[0,1]出去,所以留在里面的长度$≥δ$

对于平方可积函数空间, $\int_0^1f(x)^2dx=0$ 不能推出 $f(x)≡0$. 这样还成立Cauchy–Schwarz inequality吗?

hbghlyj

The square integrable functions (in the sense mentioned in which a "function" actually means an equivalence class of functions that are equal almost everywhere) form an inner product space with inner product given by $$\langle f,g\rangle =\int _{A}{\overline {f(x)}}g(x)\,\mathrm {d} x$$
通常这个术语不是指某个特定函数，而是指几乎处处相等的一组函数。

看作是几乎处处相等的等价类就能说满足positive-definiteness, 从而构成内积空间了

hbghlyj

math.stackexchange.com/questions/354161/
只要是symmetric positive semidefinite bilinear form就成立Cauchy–Schwarz inequality，definiteness是不必要的，也就是说$F(w_i,w_i)$可以等于零，所以不能作分母。

First, recall the following form of the Cauchy-Schwarz inequality: let $V$ be a real vector space, and suppose $(\cdot, \cdot) : V \times V \to \mathbb{R}$ is a symmetric bilinear form which is positive semidefinite, that is, $(x,x) \ge 0$ for all $x$.  Then for any $x,y \in V$ we have $|(x,y)|^2 \le (x,x) (y,y)$.
I'd like to know what happens if we ...