Show that there exist $k\in\{1,2,\cdots,n\}$ such that $\frac{1}{n}\sum_{i=1}^{n}\left(\{kx_{i}\}-\frac{1}{2}\right)^2>\frac{1}{12}-\frac{1}{6n}$

The following question seems very interesting, perhaps from a question of expectation.

Show that: for any real numbers $x_{1},x_{2},\cdots,x_{n}$, there exist $k\in\{1,2,\cdots,n\}$ such that $$\dfrac{1}{n}\sum_{i=1}^{n}\left(\{kx_{i}\}-\dfrac{1}{2}\right)^2>\dfrac{1}{12}-\dfrac{1}{6n},$$ where $\{x\}=x-\lfloor x\rfloor$. (this problem Creat by Xi Xi )

Solution 1:

Not a complete answer but two hopefully interesting geometric interpretations.

The problem is non-trivial only for $n\geq 3$.
Let $X=(x_1,\ldots,x_n)$. $X,2X,\ldots nX$ are collinear on a line through the origin in $\mathbb{R}^n$, and we have to prove that at least one of these points is sufficiently far from $E=\left(\frac{1}{2},\ldots,\frac{1}{2}\right)+\mathbb{Z}^n$, precisely at a distance $\geq\sqrt{\frac{n}{12}-\frac{1}{6}}$. On the other hand, if we assume that $X,2X,\ldots,nX$ all lie in $F$, a $\sqrt{\frac{n}{12}-\frac{1}{6}}$ neighbourhood of $E$, then $F\cap\frac{F}{2}\cap\ldots\cap\frac{F}{n}$ has to be non-empty. The distance between $\left(\frac{1}{2},\ldots\frac{1}{2}\right)$ and $\left(\frac{1}{4},\ldots\frac{1}{4}\right)$ is $\frac{\sqrt{n}}{4}$, hence if $F_\rho$ is a $\rho$-neighbourhood of $E$ and $\rho<\frac{2}{3}\cdot\frac{\sqrt{n}}{4}=\frac{\sqrt{n}}{6}$ then $F_\rho\cap\frac{F_\rho}{2}=\emptyset$. By imposing that $\frac{F_\rho}{a}\cap\frac{F_\rho}{b}\neq\emptyset$ for any $a,b\in[1,n]$ we may improve such bound till reaching the wanted conclusion.

As an alternative, we may cover $(0,1)^n$ with $n-1$ sets $A_1,A_2,\ldots,A_{n-1}$ with the same diameter $d$. By the pigeonhole principle at least two points among $X,2X,\ldots,nX\pmod{1}$ belong to the same $A_j$, hence by linearity there is some point among $X,2X,\ldots,nX$ with a distance $< d$ from $\mathbb{Z}^n$, hence with a distance $\geq\frac{\sqrt{n}}{2}-d$ from $E$.

It might be useful to consider that $$ \left(\{x\}-\frac{1}{2}\right)^2 =\frac{1}{12}+\sum_{m\geq 1}\frac{\cos(2\pi m x)}{\pi^2 m^2}$$ By letting $x_0=0$ and $z_i=e^{2\pi i x_i}$ the question can be rephrased as $$\text{For some }k\in[1,n],\qquad p_k=\sum_{i=0}^{n}\text{Re}\,\text{Li}_2(z_i^k)>0. $$ Let us assume that the opposite inequality $p_k\leq 0$ holds for any $k\in[1,n]$ and let us consider $$ f(u) = \sum_{m\geq 1}-\frac{p_m}{m}u^m=\text{Re}\sum_{m\geq 1}-\frac{u^m}{m}\sum_{i=0}^{n}\sum_{h\geq 1}\frac{z_i^{mh}}{h^2}=\sum_{i=0}^{n}\text{Re}\sum_{h\geq 1}\frac{-\log(1-uz_i^h)}{h^2}. $$ By the assumption all the coefficients of $f(u)$ up to $u^n$ are non-negative.
By formally exponentiating both sides $$\exp\left(\frac{6}{\pi^2}f(u)\right) = \prod_{i=0}^{n}\prod_{h\geq 1}\frac{1}{|1-uz_i^h|^{\frac{6}{\pi^2 h^2}}}=\frac{1}{|1-u|}\prod_{i=1}^{n}\prod_{h\geq 1}\frac{1}{|1-uz_i^h|^{\frac{6}{\pi^2 h^2}}} $$ where the modulus of the LHS is $\gg 1+o(|u|^n)$ as $u\to 0$. This gives $$ \prod_{i=1}^{n}\prod_{h\geq 1}\frac{1}{|1-uz_i^h|^{\frac{6}{\pi^2 h^2}}} \gg |1-u|\left(1+o(|u|^n)\right) $$ leading to a contradiction for $u\to 0^+$ or $u\to 0^-$.