Suppose $U_1,...,U_n$ is a simple sampling from a uniform distribution $U(0,1)$ and $G_n(u)$ is an empirical distribution function. Prove that
$$ \begin{gathered} n \int_0^1\left(G_n(s)-s\right)^2 d s=\frac{1}{12 n}+\sum_{k=1}^n\left(U_{(k)}-\frac{2 k-1}{2 n}\right)^2, \\ \int_0^1\left(G_n(s)-s\right)^2 d s \leq \frac{1}{3} \end{gathered} $$
Look, I don't even know where to start. This is a recent task, but I don't remember any integral-related theorems or properties, I am not very comfortable with the Law of Large Numbers (even if applicable here).
I am, of course, not looking for a complete solution, but any hints at which direction to even look.
The first equality has nothing to do with Law of Large Numbers, just some algebraic manipulations.
\begin{align} &\quad n \int_0^1\left(G_n(s)-s\right)^2 ds\\ &=n \int_0^1\left(s^2+G_n(s)^2-2sG_n(s)\right) ds\\ &=\frac13n+n\left[\int_0^{U_{(1)}}0^2ds+\int_{U_{(1)}}^{U_{(2)}}(1/n)^2ds+\int_{U_{(2)}}^{U_{(3)}}(2/n)^2ds+\dots+\int_{U_{(n)}}^{1}(n/n)^2ds\right] \\ &\quad-2\left[\int_0^{U_{(1)}}0sds+\int_{U_{(1)}}^{U_{(2)}}1sds+\int_{U_{(2)}}^{U_{(3)}}2sds+\dots+\int_{U_{(n)}}^{1}nsds\right]\\ &=\frac13n+\frac1n\left[1^2\cdot (U_{(2)}-U_{(1)})+2^2\cdot (U_{(3)}-U_{(2)})+3^2\cdot (U_{(4)}-U_{(3)})+\dots+n^2\cdot (1-U_{(n)})\right]\\ &\quad-\left[1\cdot(U_{(2)}^2-U_{(1)}^2)+2\cdot(U_{(3)}^2-U_{(2)}^2)+\dots+n\cdot(1^2-U_{(n)}^2)\right]\\ &=\frac13n-\frac1n\left[1\cdot U_{(1)}+3\cdot U_{(2)}+\dots+(2k-1)\cdot U_{(k)}+\dots+(2n-1)\cdot U_{(n)}-n^2\right]\\&\quad+\left[U_{(1)}^2+U_{(2)}^2+U_{(3)}^2+\dots+U_{(n)}^2-n\right]\\ &=\frac13n+(U_{(1)}-\frac1{2n})^2+(U_{(2)}-\frac3{2n})^2+\dots+(U_{(n)}-\frac{2n-1}{2n})^2\\&\quad-(\frac1{2n})^2-(\frac3{2n})^2-\dots-(\frac{2n-1}{2n})^2\\ &=\sum_{k=1}^n\left(U_{(k)}-\frac{2 k-1}{2 n}\right)^2+\left[\frac13n-\frac{1^2+3^2+\dots+(2n-1)^2}{4n^2}\right]\\ &=\sum_{k=1}^n\left(U_{(k)}-\frac{2 k-1}{2 n}\right)^2+\left[\frac13n-\frac{n(2n-1)(2n+1)}{12n^2}\right]\\ &=\frac{1}{12 n}+\sum_{k=1}^n\left(U_{(k)}-\frac{2 k-1}{2 n}\right)^2 \end{align} As to the second inequality, I currently have no idea.