Clarification of an existing answer pertaining to the proof that $||x||_\infty = max_{i=1\to n}{}|x_i| = \lim_{p\to\infty}||x||_p$

Question

In this answer to another question, a step is skipped and I do not understand the mechanics behind it. The sum is multiplied by the infinity norm of x, therefor we divide each element of the sum by the same amount to keep the result consistent. At this point, I would expect each term of the sum to look like:

$$\frac{|x_i|^p}{\|x\|_\infty}$$

But the answer jumps straight to:

$$(\frac{|x_i|}{\|x\|_\infty})^p$$

To my understanding this shouldn't be possible. Can anyone explain what's going on here? I feel that I am missing / forgetting something critical that I will need in order to pursue the rest of the material in the course I am taking.

Answer 1

Suppose WLOG $$\max\{|x_1|,...,|x_n|\}=|x_1|.$$

$$\left(\sum_{k=1}^n |x_i|^p\right)^{1/p}=|x_1|\left(1+\left(\frac{|x_2|}{|x_1|}\right)^p+...+\left(\frac{|x_n|}{|x_1|}\right)^p\right)^{1/p}.$$

Using the fact that $$\frac{|x_i|}{|x_1|}< 1$$ for all $i$, you have that $$\lim_{p\to \infty }\left(\frac{|x_i|}{|x_1|}\right)^p=0,$$ and thus, the fact that $$\lim_{p\to \infty }\left(1+\left(\frac{|x_2|}{|x_1|}\right)^p+...+\left(\frac{|x_n|}{|x_1|}\right)^p\right)^{1/p}=1,$$ is a common result (and also easy to prove).

Answer 2

Actually it jumps to \begin{align}\|x\|_p &= \left(\sum_{i=1}^n|x_i|^p\right)^{1/p}=\|x\|_\infty\frac{\bigl(\sum_{i=1}^n(|x_i|)^p\bigr)^{1/p}}{\|x\|_\infty}=\|x\|_\infty\left(\sum_{i=1}^n \frac{|x_i|^p}{\|x\|_\infty^p}\right)^{1/p}\\ &=\|x\|_\infty\left(\sum_{i=1}^n\left(\frac{|x_i|}{\|x\|_\infty}\right)^p\right)^{1/p}. \end{align}