Deriving the Chebyshev Distance Using Limits

For a vector $𝑧 = (𝑧_{1}, \dots, 𝑧_{𝑛}) \in ℝ^{𝑛}$ , with $𝑛 \geq 1$ , the $ℓ_{𝑝}$ norm is

{‖ 𝑧 ‖}_{𝑝} = {(\sum_{𝑖 = 1}^{𝑛} {| 𝑧_{𝑖} |}^{𝑝})}^{\frac{1}{𝑝}}, 1 \leq 𝑝 < \infty .

We will show that its limit as $𝑝 \to \infty$ is the largest component magnitude.

Simple proof. Let $𝑀 = {max}_{1 \leq 𝑖 \leq 𝑛} | 𝑧_{𝑖} |$ . At least one term in the sum equals $𝑀^{𝑝}$ , and every term is at most $𝑀^{𝑝}$ . Hence

𝑀^{𝑝} \leq \sum_{𝑖 = 1}^{𝑛} {| 𝑧_{𝑖} |}^{𝑝} \leq 𝑛 𝑀^{𝑝} .

Taking the $𝑝$ th root gives

𝑀 \leq {‖ 𝑧 ‖}_{𝑝} \leq 𝑛^{\frac{1}{𝑝}} 𝑀 .

Since

lim_{𝑝 \to \infty} 𝑛^{\frac{1}{𝑝}} = lim_{𝑝 \to \infty} 𝑒^{\frac{log 𝑛}{𝑝}} = 1,

the squeeze theorem yields

lim_{𝑝 \to \infty} {‖ 𝑧 ‖}_{𝑝} = max_{1 \leq 𝑖 \leq 𝑛} | 𝑧_{𝑖} | = {‖ 𝑧 ‖}_{\infty} .

This argument also covers $𝑧 = 0$ , when both bounds are zero.

Chebyshev distance. Applying the result to $𝑧 = 𝑥 - 𝑦$ gives the $ℓ_{\infty}$ (Chebyshev) distance:

𝑑_{\infty} (𝑥, 𝑦) = lim_{𝑝 \to \infty} {(\sum_{𝑖 = 1}^{𝑛} {| 𝑥_{𝑖} - 𝑦_{𝑖} |}^{𝑝})}^{\frac{1}{𝑝}} = max_{1 \leq 𝑖 \leq 𝑛} | 𝑥_{𝑖} - 𝑦_{𝑖} | .

Note. The $ℓ_{\infty}$ norm selects a component with maximum magnitude and returns that magnitude, not its sign or index. For distance, it selects the largest absolute coordinate difference. Ties do not affect the value.

Deriving the Chebyshev Distance Using Limits

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