Question #217041

Give an example of two metric on R2 which are equivalent ,substantiate your claim


1
Expert's answer
2021-07-23T12:03:27-0400

Solution:

Consider two metric spaces on R2;

1.Euclidean metric de

de(x,y)=(x1y1)2+(x2y2)2d_e(\underline{x},\underline{y})=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}

2. The max-metric dm

dm(x,y)=d_m(\underline{x},\underline{y})=max{ |x1-y1|,|x2-y2|}

We have;

dm(x,y)d_m(\underline{x},\underline{y})=max{|x1-y1|,|x2-y2|}\leq (x1y1)2+(x2y2)2\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} =de(x,y)d_e(\underline{x},\underline{y})

Also;

de(x,y)=(x1y1)2+(x2y2)2=(x1y1)2+(x2y2)2d_e(\underline{x},\underline{y})=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}=\sqrt{(|x_1-y_1|)^2+(|x_2-y_2|)^2}

de(x,y)dm2(x,y)+dm2(x,y)d_e(\underline{x},\underline{y})\leq\sqrt{d_m^2(\underline{x},\underline{y})+d_m^2(\underline{x},\underline{y})} =2dm(x,y)\sqrt{2}d_m(\underline{x},\underline{y})

Hence,

dm(x,y)de(x,y)2dm(x,y)d_m(\underline{x},\underline{y})\leq d_e(\underline{x},\underline{y})\leq \sqrt{2}d_m(\underline{x},\underline{y})

for all x,yϵ\epsilon X.

Hence, metrics dm and de are equivalent.





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