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 $u$...)
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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. $$\int_x^y u^2\frac{\lambda}{u}dt<\infty , \int_{x}^{y}\frac{u}{\lambda}dt<\infty \Leftrightarrow u\in L^1(x,y)$$
Lorenzo Q
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Thanks you. Is it the same for Holder in general ? – MSE Dec 09 '17 at 11:54
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If $1<p,q<\infty$ (with $1/p+1/q=1$) it is true that $\int |uv|= ||u||_p ||v||_q$ iff $|u|^p=\lambda |v|^q$ a.e. for some $\lambda \in \mathbb{R}$. See this thread, https://math.stackexchange.com/questions/87636/on-the-equality-case-of-the-h%C3%B6lder-and-minkowski-inequalites – Lorenzo Q Dec 09 '17 at 17:32