Metric Spaces: Why $L_infty$ selects the maximum value

I have a basic question about the metric spaces. There are several metric spaces like $L_1$, $L_2$ to $L_\infty$. The $L_p$ metric is defined by the following equation:

$$d_p(x,y)=\left(\sum_{i=1}^{n}|x_i-y_i|^p \right)^{1/p}$$

Now, when $p = \infty$, then the equation of the matrix is following:

$$d_{\infty}(x,y)=\max_{i=1,2,\ldots, n}|x_i-y_i|$$

Now, I am taking the value of $x$ as $x = [1,2,3]^T$ (Transpose) and compute $L_p$ metric for, $p = 1,2$ and $\infty$.

My question is, why $L_\infty$ metric choose the maximum value.

Answer

In the general case, let $x = (x_1,\dots,x_n)$ be a finite-length vector (in a finite dimensional space). The finite sequence of absolute values $|x_i|$ does attain its maximum (because the sequence is finite), denoted $M = \max_i |x_i|$.

Let $m$ be the (exact) number of coordinates in $x = (x_1,\dots,x_n)$ whose absolute value is equal to $M$. Thus, $1\le m\le n$. Then, we can lower and upper bound the $\ell_p$ norm of $x$ as follows: $$(mM^p)^{\frac{1}{p}}\le\ell_p(x)\le (nM^p)^{\frac{1}{p}}\,.$$

Both $m^\frac{1}{p}$ and $n^\frac{1}{p}$ tend to $1$ as $p\to\infty$, thus $\ell_p(x) \to M$, by the squeeze theorem.

[EDIT] To provide more concrete substance, let us see what happen with your example: $x=[1,2,3]^T$ (the transposition does not change the result): $$\|x\|_p = d_p(x,0) = (1^p+2^p+3^p)^{1/p} = 3\times\left(\left(\frac{1}{3}\right)^p+\left(\frac{2}{3}\right)^p+1\right)^{1/p}\,.$$

Both $\left(\frac{1}{3}\right)^p$ and $\left(\frac{2}{3}\right)^p$ tend to $0$ as $p \to \infty$. Thus, $\|x\|_p \to 3$.