# Showing these two metrics are equivalent

Let $(X,D)$ and $(X,d)$ both be metric spaces on the set $X \subset \mathbb{R}^2$ with, $$D(\vec{x},\vec{y})=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

$$d(\vec{x},\vec{y})=|x_2-x_1| + |y_2-y_1|$$

Show that the metrics $D$ and $d$ are equivalent to each other.

The definition of equivalence being used the following,

Two metrics $d_1$ and $d_2$ of a set $X$ are equivalent if and only if for any $p \in X$ the following holds for some $\delta_1,\delta_2,\epsilon >0$ $$N_{d_1}(p, \delta_1) \subset N_{d_2}(p, \epsilon)$$ $$N_{d_2}(p,\delta_2) \subset N_{d_1}(p,\epsilon)$$ My attempt: Looking at $N_D(0,1)$ and $N_d(0,1)$ helped me gain some intuition as to what I need to do as $N_D(0,1)$ represents an open unit circle, and $N_d(0,1)$ represents a diamond completely enclosed within it. Thus showing that $N_d(x,\delta) \subset N_D(x,\epsilon)$ isn't to bad.
What I really need some help on is showing the converse is true. Through some geometry I think that setting $\delta=\sqrt{2}\,\epsilon$ will work, i'm just having a tough time actually showing this to be the case. Any help or suggestions are appreciated, Thanks!!

• Hint: use the cauchy schwarz inequality – user392395 Sep 11 '17 at 20:13
• How can i apply Cauchy Schwarz inequality here? – Justin Stevenson Sep 11 '17 at 20:39
• As you formulate it you only need to have the ball inclusions for some $\delta_1, \delta_2, \varepsilon$. This is not correct. See my answer for a more accurate condition in the same spirit. – Henno Brandsma Sep 12 '17 at 6:37

The exact condition for equivalence you want is:

$$\forall p \in X: \forall \epsilon>0: \left(\exists \delta_1>0: N_{d_1}(p, \delta_1) \subseteq N_{d_2}(p,\varepsilon)\right) \land \left(\exists \delta_2 > 0 : N_{d_2}(p, \delta_2) \subseteq N_{d_1}(p,\varepsilon)\right)$$

In this case we can prove a global inequality that implies it.

For all $x,y \in \mathbb{R}^2$: $D(x,y)^2 = (x_1 - y_1)^2 + (x_2 - y_2)^2 \le 2\max(|x_1- y_1| ,|x_2- y_2|)^2$, so that $D(x,y) \le \sqrt{2}\max(|x_1 - y_1|,|x_2 -y_2|)\le \sqrt{2}(|x_1 - y_1| + |x_2 - y_2|) =\sqrt{2}d(x,y)$. So

$$D(x,y) \le \sqrt{2}d(x,y)$$

On the other hand $d(x,y) = |x_1 - y_1| + |x_2 - y_2| \le 2\max(|x_1 - y_1|,|x_2 - y_2|)$ and as clearly $\max(|x_1 - y_1|,|x_2 - y_2|)^2 \le D(x,y)^2$, taking roots and combining we get

$$d(x,y) \le 2D(x,y)$$

So for $d_1 = D, d_2 = d$ in the condition:
for $\varepsilon > 0$ we can take $\delta_1 = \frac{\varepsilon}{\sqrt{2}}$ and $\delta_2 = \frac{\varepsilon}{2}$ for all $p$. (check the two ball inclusions)

• Maybe i'm not understanding correctly but using $\delta_1=\frac{\epsilon}{\sqrt{2}}$ and letting $\epsilon=1$ which would be the unit circle and the line $y=\frac{1}{\sqrt{2}}-x$ which does have a section of the circle not covered by the diamond. – Justin Stevenson Sep 11 '17 at 22:15
• What I claim is that the $D$-ball with radius $\frac{\varepsilon}{\sqrt{2}}$ is contained in the $d$-ball of radius $\varepsilon$ (notice the rôles of $d_1 = D$, $d_2 =d$), and also that the $d$-ball of radius $\frac{\varepsilon}{2}$ is contained in the $D$-ball of radius $\varepsilon$. Both are correct (draw the pictures for $\varepsilon=1$) @JustinStevenson. – Henno Brandsma Sep 12 '17 at 6:34