1
$\begingroup$

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!!

$\endgroup$
  • $\begingroup$ Hint: use the cauchy schwarz inequality $\endgroup$ – user392395 Sep 11 '17 at 20:13
  • $\begingroup$ How can i apply Cauchy Schwarz inequality here? $\endgroup$ – Justin Stevenson Sep 11 '17 at 20:39
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Sep 12 '17 at 6:37
0
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Justin Stevenson Sep 11 '17 at 22:15
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Sep 12 '17 at 6:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.