- #1

mathmari

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MHB

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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)