Define the operator $\text{Ord} : \mathbb{R}^d \rightarrow \mathbb{R}^d$, that takes a vector $x = (x_1,\cdots,x_d)$ and returns the ordered vector $x^\ast = (x_{\sigma_x(1)},\cdots,x_{\sigma_x(d)})$ satisfying $x_{\sigma_x(i)} \le x_{\sigma_x(j)}$ if $i\le j$. Show that $\|\text{Ord}(x) - \text{Ord}(y)\|_\infty \le \|x-y\|_\infty$ for any $x,y \in \mathbb{R}^d$.
There's a hint to show and use that $x_{\sigma_x(j)} = \inf \{t\in \mathbb{R}:\#\{i \in [d]:x[i]\le t\}\ge j\}$. Which is a trivial enough result.
I have no idea how to use the hint so I tried to use $A_x$ and $A_y$ as the permutation matrices defined by $\sigma_x$ and $\sigma_y$, arriving at: $$ \|\text{Ord}(x)-\text{Ord}(y)\|_\infty = \|A_xx-A_yy\|_\infty \\ \le \|A_xx-A_xy\|_\infty + \|A_xy-A_yy\|_\infty \le \|x-y\|_\infty + 2\|y\|_\infty$$ Through simmetry, I can improve this to $\|\text{Ord}(x)-\text{Ord}(y)\|_\infty \le \|x-y\|+2\min\{\|x\|,\|y\|\}$ and this proves continuity at least. Of course this doesn't use any of the structure of the ordering function other than it being a distinct permutation on each vector.
I don't really understand what I'm supposed to show using the hint.
Managed to solve it after thinking for a bit. Assume without loss of generality that $\|\text{Ord}(x)-\text{Ord}(y)\|_\infty = x^*[j] - y^*[j]$. The $\#$ of elements bigger than or equal to $x^*[j]$ is at least $d-j+1$, the $\#$ of elements less than or equal to $y^*[j]$ is at least $j$, therefore there is a common $k$ such that $x[k] \ge x^*[j]$ and $y[k] \le y^*[j]$ by the pigeonhole principle, therefore: $$|\text{Ord}(x)-\text{Ord}(y)|_\infty = x^*[j]-y^*[j] \le x[k] - y[k] \le \|x-y\|_\infty $$