Which condition give us equality in Cauchy Schwartz?

Question

I know that using Cauchy-Schwarz $$\left(\int_x^y u(t)dt\right)^2\leq (y-x)\int_x^y u^2(t)dt,$$ but I know that I can multiply $u$ by a certain function to have an equality like $$\left(\int_{x}^y u(t)dt\right)=\left(\int_x^yu(t)\frac{\varphi(t)}{\varphi(t)}dt\right)^2=\int_x^yu^2(t)\varphi^2(t)dt\int_x^y \frac{1}{\varphi^2(t)}dt,$$ but what can be $\varphi$ ? And what condition on $u$ to have such a $\varphi$ ? (I heard about decreasing of $u$...)

Answer 1

The Cauchy Schwarz inequality $\left \langle u,v\right\rangle \leq \|u\|\|v\|$ is an equality iff $v=\lambda u$ for some $\lambda\in \mathbb{R}$. In your case, you need to find a $\varphi$ such that $u\varphi = \frac{\lambda}{\varphi}$, i.e. $\varphi=\sqrt{\frac{\lambda}{u}}$. In order to have such a function:

1) $u$ must not change sign, and $u\neq 0$ a.e.;

2) $u\varphi$ and $\frac{1}{\varphi}$ must be in $L^2$, i.e. $$\int_x^y u^2\frac{\lambda}{u}dt<\infty , \int_{x}^{y}\frac{u}{\lambda}dt<\infty \Leftrightarrow u\in L^1(x,y)$$