Solutions to homework # 4.1. The functionsd3andd4are not metrics since they do not satisfy axiom (a); for example,d3(1,-1) = 0, andd4(2,1) = 0.The remaining functions all satisfy (a) and (b), so itremains to see whether they satisfy the triangle inequality (c).To see thatd1fails thetriangle inequality, takex= 1,y= 0,z= 1/2. Thusd1is not a metric. The functiond2satisfies the trianngle inequality, since the condition|x-y| ≤|x-z|+|z-y|is equivalent to|x-y| ≤ |x-z|+|z-y|+ 2|x-z| · |z-y|and the latter is satisfied due to the triangle inequality for the absolute value function|·|anddue to the fact that the term 2|x-z| · |z-y|is nonnegative. So,d2is a metric. Finally,d5also satisfies the triangle inequality and is therefore a metric. Indeed, the triangle inequalityford5is equivalent (after some algebraic manipulations) to the inequality|x-y| ≤ |x-z|+|z-y|+ 2|x-z| · |z-y|+|x-y| · |x-z| · |z-y|,which holds because of the usual triangle inequality for| · |.
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