$\lim_{p\to\infty}{\ell^p}$ and how this relates to the maximum metric.

Question

If the $\ell^p$ norm metric is defined as $$ \ell^p=\left[\sum_{n=1}^{m}{x_{n}^p}\right]^{1/p} $$ and the $\ell^\infty$ norm metric is defined as $$ \ell^\infty=\max(|x_1|,|x_2|,|x_3|...,|x_{m-1}|,|x_m|) $$ Does this imply that $$ \lim_{p\to \infty}\left[\sum_{n=1}^{m}{x_{n}^p}\right]^{1/p}=\max(|x_1|,|x_2|,|x_3|...,|x_{m-1}|,|x_m|)? $$

Answer 1

The correct formula is indeed $l_p = (\sum_{n=1}^{m} |x_n|^p)^{1/p}$.

Let $|x_k|$ be the largest component of the vector$(|x_1|, ..., |x_m|)^T$. It is $l_{\infty} = |x_k|$ as claimed in the question.

Proof:

$l_p = |x_k|(\sum_{n=1}^{m}(\frac{|x_n|}{|x_k|})^p)^{1/p} = |x_k|(1 + \sum_{n=1, n\neq k}^{m}(\frac{|x_n|}{|x_k|})^p)^{1/p}.$

$lim_{p \rightarrow \infty}(1 + \sum_{n=1, n\neq k}^{m}(\frac{|x_n|}{|x_k|})^p)^{1/p} = 1$, which can be shown by sandwiching the term between 1 and $\sqrt[p]{p}$ as follows:

Since $\frac{|x_n|}{|x_k|} \leq 1 \Rightarrow \sum_{n=1, n\neq k}^{m}(\frac{|x_n|}{|x_k|})^p$ does not increase in p.

$\Rightarrow \exists p \in N : \sum_{n=1, n\neq k}^{m}(\frac{|x_n|}{|x_k|}) < p $. $ \Rightarrow \sum_{n=1, n\neq k}^{m}(\frac{|x_n|}{|x_k|})^{2p} \leq \sum_{n=1, n\neq k}^{m}(\frac{|x_n|}{|x_k|}) < p < 2p$.

$t \geq 2p+1 \Rightarrow 1 \leq (1 + \sum_{n=1, n\neq k}^{m}(\frac{|x_n|}{|x_k|})^t)^{1/t} \leq \sqrt[t]{t}$. But $lim_{t \rightarrow \infty} \sqrt[t]{t} = 1$.

Answer 2

For each $i$ we have $|x_i| = (|x_i|^{p})^{1/p} \leq (\sum_{i=1}^{m} |x_i|^{p})^{1/p}$. So $\max \{|x_1|,|x_2|,...,|x_m|\} \leq (\sum_{i=1}^{m} |x_i|^{p})^{1/p}$. Let $j$ be an index such that $|x_j| =\max \{|x_1|,|x_2|,...,|x_m|\}$. Then $(\sum_{i=1}^{m} |x_i|^{p})^{1/p} \leq (m|x_j|^{p})^{1/p} =m^{1/p} |x_j| \to |x_j| = \max \{|x_1|,|x_2|,...,|x_m|\}$. Combining the two statement just proved we see that $(\sum_{i=1}^{m} |x_i|^{p})^{1/p} \to \max \{|x_1|,|x_2|,...,|x_m|\}$