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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.

Community
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ophilbinbriscoe
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2 Answers2

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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).

Surb
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  • Hi, thanks for your answer. It illustrates the proof nicely, but unfortunately I understand the larger trend of the proof already, it's the small step that the your and the original answer gloss over of how you rip the exponent off of the numerator to have it enclose the entire fraction. This is what I do not understand. – ophilbinbriscoe Sep 17 '16 at 17:48
  • @ophilbinbriscoe: I'm not sure about your question. But for example, if $n=2$, $$(|x_1|^p+|x_2|^p)^{1/p}=\left(|x_1|^p\right)^{1/p}\left(1+\frac{|x_2|^p}{|x_1|^p}\right)^{1/p}=|x_1|\left(1+\left(\frac{|x_2|}{|x_1|}\right)^p\right)^{1/p}.$$ Does this answer to your question ? – Surb Sep 17 '16 at 17:51
  • @ophilbinbriscoe: Or differently: By the norm property, $$\frac{|x|_p}{|x|_∞}=\left|\frac{x}{|x|_∞}\right|_p.$$ – Lutz Lehmann Sep 17 '16 at 17:54
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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}

Bernard
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