stronger variant of "$l^p$ spaces are increasing in $p$"： $\bigcup_{1\leq p\lneq q}l^p\neq l^q$

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 $1\leq q\leq\infty$. Prove: $$\bigcup_{1\leq p\lneq q}l^p\subsetneq l^q$$

My Attempt

I can prove the following weaker statement: Let $1\leq p\lneq q\leq\infty$, then $l^p\subsetneq l^q$.

$$\forall x\in\mathbb{K}^\mathbb{N}: \lVert x\rVert_q\leq\lVert x\rVert_p\ \Rightarrow\ \left(\lVert x\rVert_p\lneq\infty \Rightarrow \lVert x\rVert_p\lneq\infty\right)\ \Rightarrow\ l^p\subseteq l^q$$

• $1\leq p\lneq q\lneq \infty$: Consider $x=\left(\frac{1}{n^{\frac{1}{p}}}\right)_{n\in\mathbb{N}}$, then $\lVert x\rVert_p=\infty$ and $\lVert x\rVert_q\lneq\infty$, because $\sum_{n\in\infty}\frac{1}{n^r}\lneq\infty \Leftrightarrow r>1$. Therefore $x\in l_q$ and $x\notin l_p$.
• $1\leq p\lneq q=\infty$: Consider $x=(1)_{n\in\mathbb{N}}$, then $\lVert x\rVert_\infty=1\lneq\infty$ and $\lVert x\rVert_p=\infty$. Therefore $x\in l_\infty$ and $x\notin l_p$.

Question

How can I prove $$\bigcup_{1\leq p\lneq q}l^p\neq l^q$$ for $1\leq p\lneq q\lneq \infty$?

Remark

A discussion of the weaker statement can be found here.

Answer 1

Hint: Play around with $$\sum_{n=2}^\infty\frac1{n^\alpha(\log n)^\beta},\tag{$*$}$$ where $\alpha,\beta>0$.

Indeed, $(*)$ is convergent if and only if ($\alpha>1$) or $(\alpha=1$ and $\beta>1$). Hence, the series $\sum_{n=2}^\infty \frac1{n(\log n)^2}$ is convergent and $\sum_{n=2}^\infty \left(\frac1{n(\log n)^2}\right)^a$ is divergent for all $a\in(0,1)$.

Therefore, for $q\in(1,\infty)$, consider $$x=\left(\frac1{n^{\frac1q}(\log n)^{\frac2q}}\right)_{n\geq 2}, $$ then $$\displaystyle\|x\|_{\ell^q}^q=\sum_{n=2}^\infty \frac1{n(\log n)^2}<\infty$$ and for any $p\in[1,q)$ we have $$\|x\|_{\ell^p}^p=\sum_{n=2}^\infty \left(\frac1{n(\log n)^2}\right)^\frac pq=\infty.$$