I am hoping someone can clarify the application of the Schwarz inequality I have found in the following two situations.
$\textbf{1)}$ Consider the function in $L_2[0,1]$ $$\int_0^1\left(\int_0^txds\right)^2dt$$ By the Schwarz inequality, we have that $$\int_0^1\left(\int_0^txds\right)^2dt\le\int_0^1\left(\underbrace{\int_0^t 1^2ds}_{=t}\underbrace{\int_0^tx^2ds}_{=||x||^2}\right)dt=\left[\frac{1}{2}t^2||x||^2\right]_0^1=\frac{1}{2}||x||^2$$
Fine, I think I can follow. I'm no too familiar enough to know if this is correct but I can at least follow the logic.
$\textbf{2)}$ Now consider the function in $L_2[0,1]$ $$\int_0^1\left|\int_0^tx-yds\int_0^tx+yds\right|dt$$ Again by Schwarz inequality, we have that $$\int_0^1\left|\int_0^tx-yds\int_0^tx+yds\right|dt\le\int_0^1\sqrt{\int_0^1 1^2ds\int_0^1|x-y|^2ds}\sqrt{\int_0^1 1^2\int_0^1|x+y|^2ds}dt=||x-y||\cdot||x+y||$$
I guess specifically, why in this case did the integrals change from $\int_0^t$ to $\int_0^1$, thus just becoming $1$'s and effectively cancelling vs $t$'s which we needed to then integrate again? Thank you.
I answer the question 2 (I do not see any specific question in 1). The point here is that $$\begin{split} \left| \int_0^t (x(s) - y(s)) \, ds \right| \le \int_0^t |x(s) - y(s)| \, ds & \le \int_0^t |x(s) - y(s)| \, ds + \underbrace{\int_t^1 |x(s) - y(s)| \, ds}_{\ge 0} \\ & = \int_0^1 |x(s) - y(s)| \, ds \end{split}$$and then you apply Cauchy-Schwarz.