minkowski distance at the infinity

Question

Given that the Minkowski distance between points $X = (x_1,\dots,x_n)$ and $Y = (y_1,\dots ,y_n)$ is $$ d(X, Y) = \left(\sum_{i=1}^n|x_i−y_i|^p\right)^{1/p}, $$ I want to show that $$ \lim_{p\to\infty}d(X, Y) = \max_{i=1, \dots, n} |x_i-y_i|. $$
How can I prove that as $p$ approaches infinity, the Minkowski distance approaches maximum difference of coordinates?

Answer 1

Fix $X,\ Y$ And assume that $$ |x_1-y_1|\geq |x_i-y_i|\ \ast$$ Then \begin{align*} |x_1-y_1|&\leq d(X,Y) \\&=|x_1-y_1| \bigg(1+ \sum_{i=2}^n \frac{|x_i-y_i|^p}{|x_1-y_1|^p} \bigg)^\frac{1}{p} \\&\leq |x_1-y_1| n^\frac{1}{p} \end{align*}

Here second equality is followed from

$$\sum_i |x_i-y_i|^p=|x_1-y_1|^p \bigg\{ 1+ \sum_{i=2}^n |x_i-y_i|^p/|x_1-y_1|^p\bigg\} $$

And last inequality is followed from $\ast$

And

$$\lim_{p\rightarrow \infty}\ n^\frac{1}{p}=1$$

so that we complete the proof