数学-柯西不等式：三代人的传承
Given any $\alpha \in \mathbb{R}$
$$ \int_a^b \left|f(x)-\alpha g(x)\right|^2dx\geq 0 $$Break it Down and become
$$ \alpha^2\int_a^b g^2(x)dx - 2\alpha \int_a^b f(x)g(x)dx+\int_a^b f^2(x)dx\geq 0 $$to make it apply for any $\alpha$
$$ \Delta = 4\left(\int_a^b f(x)g(x)dx\right)^2 - 4\int_a^bf^2(x)dx\int_a^b g^2(x)dx \leq 0 $$which is
$$ \left(\int_a^b f(x)g(x)dx\right)^2 \leq \int_a^bf^2(x)dx\int_a^b g^2(x)dx $$遇到一道题目,$f(a)=0$
$$ \int_a^b f^2(x)dx\leq \frac{(b-a)^2}{2}\int_a^b[f'(x)]^2dx $$题眼：
$$ f^2(x) = \left(\int_a^x f'(x)dx\right)^2\leq \int_a^x 1^2 dx\int_a^x[f'(x)]^2dx $$