$l^p$ spaces are increasing in $p$

Question

Problem

Let $\mathbb{K}$ be either $\mathbb{R}$ or $\mathbb{C}$. Let $x\in\mathbb{K}^\mathbb{N}$ be any sequence in $\mathbb{K}$. Let the $p$-norm be defined by $$\lVert x\rVert_p:=\left(\sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^p\right)^\frac{1}{p}$$ for $1\leq p\lneq\infty$ and $$\lVert x\rVert_\infty:=\lim_{p\to\infty}\lVert x\rVert_p=\sup_{n\in\mathbb{N}}\left\lvert x_n\right\rvert$$ for $p=\infty$. The $l^p$-space is defined by $$l^p:=l^p(\mathbb{K}):=\left\{x\in\mathbb{K}^\mathbb{N}\ \mid\ \lVert x\rVert_p\lneq\infty\right\}$$

Let $1\leq p \lneq q\leq\infty$. Prove $l^p\subsetneq l^q$.

My Attempt

• $1\leq p\lneq q\lneq \infty$:

\begin{align*} &x\in l^p\ \Leftrightarrow\ \lVert x\rVert_p\lneq\infty\\ \Leftrightarrow\ &\left(\sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^p\right)^\frac{1}{p}\lneq\infty\\ \Leftrightarrow\ &\sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^p\lneq\infty\\ \Leftrightarrow\ &\lim_{n\to\infty}\left\lvert x_n\right\rvert^p=0\ \Rightarrow\ \lim_{n\to\infty}\left\lvert x_n\right\rvert=0\\ \Rightarrow\ &\exists N\in\mathbb{N}:\forall n\geq N: \left\lvert x_n\right\rvert\leq 1&\left(p\lneq q,\ 0\leq\left\lvert x_n\right\rvert\leq1\right)\\ \Rightarrow\ &\exists N\in\mathbb{N}:\forall n\geq N: \left\lvert x_n\right\rvert^q\leq\left\lvert x_n\right\rvert^p &\left(\sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^p\lneq\infty\right)\\ \Rightarrow\ &\sum_{n\geq N} \left\lvert x_n\right\rvert^q\leq\sum_{n\geq N}\left\lvert x_n\right\rvert^p\lneq\infty &\left(\sum_{n<N}\left\lvert x_n\right\rvert^q \lneq\infty\right)\\ \Rightarrow\ &\sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^q\lneq\infty\\ \Leftrightarrow\ &\left(\sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^q\right)^\frac{1}{q}\lneq\infty\\ \Leftrightarrow\ &\lVert x\rVert_p\lneq\infty\ \Leftrightarrow\ x\in l^q \end{align*}

• $1\leq p\lneq q=\infty$: \begin{align*} &x\in l^p\ \Leftrightarrow\ \lVert x\rVert_p\lneq\infty\ \Leftrightarrow\ \left(\sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^p\right)^\frac{1}{p}\lneq\infty\\ &\Leftrightarrow\ \sum_{n\in\mathbb{N}}\left\lvert x_n\right\rvert^p\lneq\infty\ \Leftrightarrow\ \lim_{n\to\infty}\left\lvert x_n\right\rvert^p=0\ \Rightarrow\ \lim_{n\to\infty}\left\lvert x_n\right\rvert=0\\ &\Rightarrow\ \sup_{n\in\mathbb{N}}\left\lvert x_n\right\rvert\lneq\infty \Leftrightarrow x\in l^\infty. \end{align*} If $x=(1,1,1,\dots)$, then $\sup_{n\in\mathbb{N}}\left\lvert x_n\right\rvert=1\lneq\infty$, but $\lVert x\rVert_p=\infty$. We have $x\in l^\infty$ while $x\notin l^p$.

Questions

1. Is everything up to this point correct?
2. What is an easy example for $x\in l^q$ with $x\notin l^p$ for the case $1\leq p\lneq q\lneq \infty$?

Answer 1

Your answer for the first question seems to be correct for me. For the second question, if $1\le p < q \le \infty$, let $x = \left(\frac {1}{n^{\frac12\left(\frac{1}{p} + \frac{1}{q}\right)}}\right)_{n\in\mathbb N}$, $\left\|x\right\|_p = \left(\sum\limits_{n\in \mathbb N} \frac{1}{n^{\frac12\left(1+\frac{p}{q}\right)}}\right)^{\frac1p} = \infty$ and $\left\|x\right\|_q = \left(\sum\limits_{n\in \mathbb N} \frac{1}{n^{\frac12\left(1+\frac{q}{p}\right)}}\right)^{\frac1q} < \infty$

Answer 2

Your proof is OK but a little too complicated. It turns out that $\|x\|_q \le \|x\|_p$ whenever $0 < p \le q \le \infty$, which implies $\ell_p \subset \ell_q$ by virtue of the fact that $\|x\|_p < \infty \implies \|x\|_q < \infty$.

For any index $i$ you have $$|x_i| = \left( |x_i|^p \right)^{1/p} \le \left( \sum_n |x_n|^p \right)^{1/p} = \|x\|_p$$ and by taking the supremum over all $i$ you obtain $\|x\|_\infty \le \|x\|_p$. If $q < \infty$ then $$\|x\|_q^q = \sum_n |x_n|^q = \sum_n |x_n|^{q-p} |x_n|^p \le \|x\|_\infty^{q-p} \sum_n |x_n|^p = \|x\|_\infty^{q-p} \|x\|_p^p.$$ In light of the first result you obtain $\|x\|_q^q \le \|x\|_p^q$ and the result follows by taking the $q$ root on both sides.

Answer 3

My favourite proof of $\|x\|_p\le \|x\|_q$ for $0<q<p\le \infty$ is the following. By homogeneity it suffices to consider the case $\|x\|_q=1.$ Then $|x_n|\le 1$ and $$\|x\|_p^p =\sum |x_n|^p\le \sum |x_n|^q=1$$ Hence $\|x\|_p\le 1=\|x\|_q.$

My favourite easy example for $\displaystyle x\in \left (\bigcap_{q>p}\ell^q\right )\setminus \ell^p$ is $$x_n= n^{-1/p}$$