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Let $d$ and $d^*$ be two metrics on a set $X$. If there exists two real numbers $k_1,k_2 >0$ such that $$ k_1d(x,y) \leq d^*(x,y) \leq k_2 d(x,y), $$ for $x,y \in X$, then the metrics $d$ and $d^*$ are equivalent.

Using the above definitions, the three metrics defined on $\mathbb{R}^2$, \begin{align*} d(x,y)= \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} \\ d^*(x,y)=\max \{|x_1-y_1|,|x_2-y_2|\} \\ d^{**}(x,y)=|x_1-y_1|+|x_2-y_2| \end{align*} are equivalent as $$ d^*(x,y) \leq d(x,y) \leq \sqrt{2} d^*(x,y) $$ and $$ d^*(x,y) \leq d^{**}(x,y) \sqrt{2} d^*(x,y). $$ My question is how these constants are found here? Suggest me a method to find the constants using the definition .

David Kraemer
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Kavita
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1 Answers1

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First of all, note that

$$\begin{align}d(x,y) \\&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\&\le \sqrt{2\max\{|x_1-y_1|^2,|x_2-y_2|^2\}}\\&=\sqrt2 \max\{|x_1-y_1|,|x_2-y_2|\}\\&=\sqrt2 d^*(x,y)\end{align}.$$

On the other hand,

$$\begin{align}d^*(x,y)\\&=\max\{|x_1-y_1|,|x_2-y_2|\}\\&=\sqrt{\max\{|x_1-y_1|^2,|x_2-y_2|^2\}}\\&\le \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}\\&=d(x,y).\end{align} $$

Next, $$d^*(x,y)=\max\{|x_1-y_1|,|x_2-y_2|\}\le |x_1-y_1|+|x_2-y_2|=2d^{**}(x,y).$$ And

$$d^{**}(x,y)=|x_1-y_1|+|x_2-y_2|\le 2 \max\{|x_1-y_1|,|x_2-y_2|\}=d^{*}(x,y).$$

Finally, using the inequalities above,

$$d^{**}(x,y)\le d^*(x,y)\le d(x,y)$$ and

$$d(x,y)\le \sqrt2 d^*(x,y)\le \sqrt2 d^{**}(x,y).$$

mfl
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