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[SOLVED] Strong equivalence of metrices

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Hey!! :eek:

Consider the following metrices in $\mathbb{R}^2$. For $x,y\in \mathbb{R}^2$ let \begin{align*}&d_1(x,y)=|x_1-y_1|+|x_2-y_2| \\ &d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\ &d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}\end{align*}

Draw the unit ball $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\}$ in each metric.

Show that the metrices $d_1, d_2, d_{\infty}$ are strongly equivalent.


Could you give me a hint for the first part?

As for the second part:

Without loss of generality, we assume that $\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|$.

\begin{align*}d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|\leq |x_1-y_1|+|x_2-y_2|=d_1(x,y)\end{align*}

\begin{align*}d_1(x,y)&=|x_1-y_1|+|x_2-y_2|\leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\}=2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_1(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_1(x,y)$ are strongly equivalent? (Wondering)

\begin{align*}d_{\infty}(x,y)&=\max \{|x_1-y_1|,|x_2-y_2|\} =|x_2-y_2|=\sqrt{|x_2-y_2|^2}=\sqrt{(x_2-y_2)^2}\\ & \leq \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=d_2(x,y)\end{align*}

\begin{align*}d_2(x,y)&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}\leq \sqrt{(x_1-y_1)^2}+\sqrt{(x_2-y_2)^2}\\ & =|x_1-y_1|+|x_2-y_2| \leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\} = 2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_2(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_2(x,y)$ are strongly equivalent? (Wondering)


Do we have to show that also for $d_1$ and $d_2$ ? Or does this follow from the above? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Hey!! :eek:

Consider the following metrices in $\mathbb{R}^2$. For $x,y\in \mathbb{R}^2$ let \begin{align*}&d_1(x,y)=|x_1-y_1|+|x_2-y_2| \\ &d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\ &d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}\end{align*}

Draw the unit ball $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\}$ in each metric.

Show that the metrices $d_1, d_2, d_{\infty}$ are strongly equivalent.


Could you give me a hint for the first part?
We have for instance $d_1(0,y) = |0-y_1|+|0-y_2| = |y_1|+|y_2|$ don't we? (Wondering)

So $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\} = \{y\in X\mid |y_1|+|y_2|<1\} = \{(x,y)\in \mathbb R^2\mid |x|+|y|<1\}$.

We can draw it like:



As for the second part:

Without loss of generality, we assume that $\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|$.

\begin{align*}d_{\infty}(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\}=|x_2-y_2|\leq |x_1-y_1|+|x_2-y_2|=d_1(x,y)\end{align*}

\begin{align*}d_1(x,y)&=|x_1-y_1|+|x_2-y_2|\leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\}=2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_1(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_1(x,y)$ are strongly equivalent? (Wondering)

\begin{align*}d_{\infty}(x,y)&=\max \{|x_1-y_1|,|x_2-y_2|\} =|x_2-y_2|=\sqrt{|x_2-y_2|^2}=\sqrt{(x_2-y_2)^2}\\ & \leq \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=d_2(x,y)\end{align*}

\begin{align*}d_2(x,y)&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}\leq \sqrt{(x_1-y_1)^2}+\sqrt{(x_2-y_2)^2}\\ & =|x_1-y_1|+|x_2-y_2| \leq \max \{|x_1-y_1|,|x_2-y_2|\}+\max \{|x_1-y_1|,|x_2-y_2|\}\\ & =2\max \{|x_1-y_1|,|x_2-y_2|\} = 2d_{\infty}(x,y)\end{align*}

So we get \begin{equation*}d_{\infty}(x,y)\leq d_2(x,y)\leq 2d_{\infty}(x,y)\end{equation*} Does this mean that $d_{\infty}(x,y)$ and $d_2(x,y)$ are strongly equivalent?
Yep. (Nod)

Do we have to show that also for $d_1$ and $d_2$ ? Or does this follow from the above?
Strong equivalence of metrics is an equivalence relationship, which implies indeed that $d_1$ and $d_2$ are strongly equivalent by the transitivity property. (Nod)

Alternatively we can show it using the inequalities you found above:
$$\frac 12 d_1(x,y) \le d_\infty(x,y) \le d_2(x,y) \le 2 d_\infty(x,y) \le 2 d_1(x,y)$$
So:
$$\frac 12 d_1(x,y) \le d_2(x,y) \le 2 d_1(x,y)$$
Therefore $d_1$ and $d_2$ are strongly equivalent. (Nerd)

We can also illustrate it graphically:

(Cool)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
We have for instance $d_1(0,y) = |0-y_1|+|0-y_2| = |y_1|+|y_2|$ don't we? (Wondering)

So $B_i(0,1)=\{y\in X\mid d_i(0,y)<1\} = \{y\in X\mid |y_1|+|y_2|<1\} = \{(x,y)\in \mathbb R^2\mid |x|+|y|<1\}$.
How do we draw the last one?

We have $d_{\infty}(0,y) = \max\{|0-y_1|, |0-y_2|\} = \max\{|y_1|, |y_2|\}$. So we get $B_{\infty}(0,1)=\{y\in X\mid d_{\infty}(0,y)<1\} = \{y\in X\mid \max\{|y_1|, |y_2|\}<1\} = \{(x,y)\in \mathbb R^2\mid \max\{|y_1|, |y_2|\}<1\}$.

But which is the graph of that? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
How do we draw the last one?

We have $d_{\infty}(0,y) = \max\{|0-y_1|, |0-y_2|\} = \max\{|y_1|, |y_2|\}$. So we get $B_{\infty}(0,1)=\{y\in X\mid d_{\infty}(0,y)<1\} = \{y\in X\mid \max\{|y_1|, |y_2|\}<1\} = \{(x,y)\in \mathbb R^2\mid \max\{|y_1|, |y_2|\}<1\}$.

But which is the graph of that?

(Emo)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Ahh yes (Tmi)

Thank you so much!!! (Blush)