Distance inequality

Question

Consider $H=\{ (x,y) \in {\bf R}^2\mid x$ or $y$ is an integer $\}$ If $d$ is canonical distance in ${\bf R}^2$, show that if $d(x):=d(x,H)$, (1) $$ d(x) - d(y) \leq d(x,y) $$ if $x,\ y$ are in same square determined by $H$, for instance we can assume that $x,\ y\in [0,1]^2$, and (2) $$ \sqrt{ d(x)^2+ d(y)^2 }\leq d(x,y) $$ if $x,\ y$ are in different square

Proof : (1) Divide a small square $[0,1]^2$ into four triangles by using two diagonals.

(1.1) Then if $x,\ y$ are in a triangle then from $x,\ y$ we draw orthogonal lines $xh_1,\ yh_2$ to the base on the triangle Hence we have a polygon $xh_1h_2y$

That is, $|d(x)-d(y)|^2 + d(h_1,h_2)^2 = d(x,y)^2 $ by Pythagoras theorem So we solved this case

(1.2) If $x,\ y$ are in opposite triangles, then define $\epsilon,\ \delta$ s.t. $$ d(x)+\epsilon= d(y)+\delta =\frac{1}{2} $$ Here $\epsilon,\ \delta \geq 0$

Consider a parallel lines passing through $x,\ y$ s.t. these lines meet orthogonally to $H$ and the distance of two lines is $\epsilon +\delta $. That is, $$ |d(x)-d(y)| = |\epsilon -\delta | \leq \epsilon +\delta \leq d(x,y)$$

(Since $x,\ y$ are in two parallel lines, $d(x,y)$ is larger than the distance of two lines)

Here equality holds iff $x,\ y$ give a distance of two parallel lines

(1.3) If $x,\ y$ are in different triangles and if they are adjacent, I have no idea

Thank you in advance

Answer 1

Q.1. applies to any non-empty $H\subset S$ where $(S,d)$ is any metric space, provided that we define $d(x)=\inf \{d(x,h):h\in H\}.$

By contradiction, suppose $d(x)-d(y)>d(x,y).$ Then let $d(x)=k+d(y)+d(x,y)$ where $k>0.$

There exists $h\in H$ such that $d(y)< d(h,y)+k/2.$ Then, since $d(h,x)\geq d(x),$ we have $$d(x)=k+d(y)+d(x,y)=$$ $$=k+[\;d(y)-d(h,y)\;]+[\;d(h,y)+d(y,x)\;]\geq $$ $$\geq k+\;[d(y)-d(h,y)\;]+d(h,x)\geq$$ $$\geq k+[\;d(y)-d(h,y)\;]+d(x)>$$ $$>k-k/2+d(x).$$ So we have $d(x)>k-k/2+d(x)$, which implies $k<0,$ a contradiction.

Answer 2

Let $g(t) = \min (t,1-t) = \frac 12 -|t-\frac 12|.$

Observe that from triagnale inequality,

$$(*)\quad \quad |t_1 - t_2| \ge |t_1 - \frac 12| - |t_2 - \frac 12| = g(t_1) -g(t_2).$$

For a point ${\bf x}=(x_1,x_2)\in {\mathbb R}^2$. Then

$$ d({\bf x}) = \min (g(x_1-\lfloor x_1\rfloor),g(x_2-\lfloor x_2\rfloor)).$$

Suppose that ${\bf x}$ and ${\bf y}$ are in the same box, and assume WLOG that $d({\bf y})=g(y_1-\lfloor y_1\rfloor)$. Then

$$ d ({\bf x}) -d ({\bf y})\le g(x_1-\lfloor x_1\rfloor)-g(y_1-\lfloor y_1\rfloor)\le |(x_1-\lfloor x_1\rfloor)-(y_1-\lfloor y_1\rfloor)|= |x_1-y_1|\le d ({\bf x},{\bf y}),$$

where $(*)$ was used for the second inequality, and the last equality uses the fact that $\lfloor x_1\rfloor =\lfloor y_1 \rfloor$.

Now suppoose that ${\bf x}$ and ${\bf y}$ are in disjoint boxes. Without loss of generality, assume that $\lfloor x_1\rfloor \ne \lfloor y_1\rfloor$. Then

$$g(x_1 - \lfloor x_1\rfloor)+ g(y_1-\lfloor y_1\rfloor) \le |x_1 - y_1|,$$

because we need to pass the gridline to get from $x_1$ to $y_1$, and the distance then larger than the sum of the distances from $x_1$ to the gridline in the direction of $y_1$ and from $y_1$ to the gridline in the direction of $x_1$, which are larger or equal to $ g(x_1 - \lfloor x_1\rfloor)$ and $g(y_1-\lfloor y_1\rfloor)$, respectively. Thus,

$$\begin{align*} d^2({\bf x})+d^2({\bf y})&\le (d({\bf x})+d({\bf y}))^2\\ & \le ( g(x_1 - \lfloor x_1\rfloor)+ g(y_1-\lfloor y_1\rfloor))^2\\ & \le |x_1-y_1|^2\\ & \le d({\bf x},{\bf y}). \end{align*}$$