# Isometry of two metric spaces

How to prove that $(\Bbb R^2, d_1)$ and $(\Bbb R^2, d_\infty)$ are isometric?

My approach

Let $f:(\Bbb R^2, d_1)\to(\Bbb R^2, d_\infty)$ be a function defined by $f(x, y) = (x+y,x-y)$. I can easily prove that $f$ is bijective, but I can not prove the isometry of $f$.

• Please write the question properly. What is $d_1$ and $d_{\infty}$? It will be better if you write those in your question, rather than asking people to guess. – Krish Aug 29 '17 at 8:56
• Is it $d_1(x,y)=|x_1-y_1|+|x_2-y_2|$ ? – Marios Gretsas Aug 29 '17 at 9:01
• @Marios yes absolutely correct – Anil Kumar Pandey Aug 29 '17 at 9:03
• The standard notation is $d_n(x, y) = \left(|x_1-y_1|^n + |x_2-y_2|^n\right)^{1/n}$, so that $d_1$ is the taxicab metric (as suggested by Marios), $d_2$ is the Euclidean metric, and $d_\infty = \lim_{n\to \infty}d_n$ (pointwise limit) is the max-metric (sup-metric for infinite dimensions, like polynomial spaces). – Arthur Aug 29 '17 at 9:03
• @JeanMarie: I was not aware of the terminology. Thanks Arthur – Krish Aug 29 '17 at 9:07

Since both metrics are induced by norms and your function is linear, it is enough to check that $f$ preserves norms, i.e. we must check that $$\max \{|x+y|,|x-y|\}=|x|+|y|$$ for all $x,y\in \Bbb R$.

We can assume WLOG that $x$ and $y$ and non-negative, since changing their sign would not change the absolute values in the RHS and would only swap the two terms in the LHS.

Then the RHS would be equal to $\max\{x+y,|x-y|\}$; but since $$|x-y|\leq |x|+|y|=x+y,$$ both sides must then be equal to $x+y$.

• In order to prove that $\max\{|x+y|,|x-y|\}=|x|+|y|$, I would square both sides. The square of the LHS is $\max\{x^2+y^2+2xy,x^2+y^2-2xy\}$ and the square of the RHS is $x^2+y^2+2|xy|$. It is clear that they are equal. – José Carlos Santos Aug 29 '17 at 9:18

The map $f$ maps the square $Q_1$ with vertices $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$ onto the square $Q_\infty\subset({\mathbb R}^2,d_\infty)$ with vertices $(1,1)$, $(1,-1)$, $(-1,-1)$, $(-1,1)$.

Since $Q_1$ is the closed unit ball in $({\mathbb R}^2,d_1)$, and $Q_\infty$ is the closed unit ball in $({\mathbb R}^2,d_\infty)$, it follows from general principles about norms and linear maps that $f$ is an isometry.

Here it is important to keep indices and dimensions straight. Remember that both the $d$'s take in two pairs of numbers as arguments, while $f$ only takes in one pair. Both a basic pair (vector / point) and a pair of pairs is tempting to write as $(x, y)$, so this is a breeding ground for confusion. I will try to keep it straight in my answer, using indexed variables for numbers and unindexed variables for points.

Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$. Then $d_1(x, y) = |x_1-y_1| + |x_2-y_2|$, while $$d_\infty(f(x), f(y)) = \max(|f(x)_1 - f(y)_1|, |f(x)_2 - f(y)_2|)\\ = \max(|(x_1 + x_2) - (y_1+y_2)|, |(x_1-x_2) -(y_1-y_2)|)\\ = \max(|(x_1 -y_1) + (x_2-y_2)|, |(x_1-y_1) - (x_2-y_2)|)$$ which is now of the form $\max(|a+b|, |a-b|)$, which is equal to $|a|+|b|$ (you can, for instance, argue by cases: do $a$ and $b$ have the same sign or not?), and we're done.

Let $(x,y), (u,v) \in \mathbb{R}^2$.

$$D =d_\infty(f(x,y),f(u,v))$$ $$=d_\infty((x+y,x-y),(u+v,u-v))$$ $$= \|x+y-u-v, x-y-u+v||_{\infty}$$ As, $$\|(A,B)\|_{\infty}= \max(A,B) =\frac{|A+B|+|A-B|}{2}$$

we can write $$D = \frac{1}{2}(|(x+y-u-v)+(x-y-u+v)|+|(x+y-u-v)-(x-y-u+v)|)$$

$$= \frac{1}{2}(|2x-2u|+|2y-2v|)$$ $$= |x-u|+|y-v|$$ $$=d_1((x,y),(u,v))$$