$\|\cdot \|_p$ is not equivalent to $\|\cdot \|_q$ in $l^p$

Question

To clarify things,

let $\textbf{x}=(x_n) \in l^p$, then:

If $p\geq 1$, $$\| \textbf{x} \|_p=\Bigg(\sum_{n=1}^{\infty} |x_n|^{p}\Bigg)^{1/p}. $$ If $p=\infty$, $$\| \textbf{x} \|_\infty=\sup_{n\geq1} |x_n|. $$

First, note that for $q>p$, $\|\cdot \|_q$ is a well-defined norm on $l^p$.

Now, to show that $\|\cdot \|_p$ is not equivalent to $\|\cdot \|_q$ in $l^p$, I need to come up with counter-examples.


That is, we need to find a sequence of vector $\textbf{v}_n \in l^p$ such that $$\frac{\|\textbf{v}_n \|_p}{\|\textbf{v}_n \|_q} \to \infty \\\ \text{or}\ \\\ \frac{\|\textbf{v}_n \|_q}{\|\textbf{v}_n \|_p} \to \infty,$$




for the following cases:

$\textbf{Case 1}$: $p=1$ and $q=\infty.$

$\textbf{Case 2}$: $1 < p<q<\infty.$

$\textbf{Case 3}$: $1 < p<\infty$ and $q=\infty.$

I'm stuck with it for quite some time now. I have try to play with it using geometric series but have not succeed yet. Any suggestions?

Edit: Later on I found the answers to my own question so, I decide to put the answer in the Answer below.

Answer 1

These are my answers to the above cases.

$\textbf{Case 1}$: $p=1$ and $q=\infty.$

Let $\textbf{x}_n=(1,1,1,...,1,0,0,...) \in l^1.$ ($1$'s $n$-th times).

Hence, $$\frac{\|\textbf{x}_n \|_1}{\|\textbf{x}_n \|_\infty}=\frac{n}{1} \to \infty.$$

$\textbf{Case 2}$: $1 < p<q<\infty.$

Let $\textbf{x}_n=(1,1,1,...,1,0,0,...) \in l^p.$ ($1$'s $n$-th times).

Hence, $$\frac{\|\textbf{x}_n \|_p}{\|\textbf{x}_n \|_q}=\frac{n^{1/p}}{n^{1/q}}=n^{(p-q)/(pq)} \to \infty.$$

$\textbf{Case 3}$: $1 < p<\infty$ and $q=\infty.$

Let $\textbf{x}_n=(1,1,1,...,1,0,0,...) \in l^p.$ ($1$'s $n$-th times).

Hence, $$\frac{\|\textbf{x}_n \|_p}{\|\textbf{x}_n \|_\infty}=\frac{n^{1/p}}{1} \to \infty.$$

Answer 2

Let $1<p<q<\infty$ be fixed. Since $\frac pq < 1$ we can fix $\varepsilon>0$ small such that $$ (1+\varepsilon)\frac pq \leq 1. \tag{1} $$ Consider the sequence $X = \{ n^{-\frac{1+\varepsilon}{q}} \}_{n=1}^\infty$. Then $X \in l^q$, but $X \notin l^p$ since the series $\sum\limits_{n=1}^\infty \frac{1}{n^{(1+\varepsilon) \frac{p}{q}} }$ diverges in view of the condition $(1)$. Thus $l^p$ and $l^q$ norms are not equivalent.