Question about sum of l'p spaces

Question

Is this true or false: $$\bigcup_{1 \leq p < q} \ell^p = \ell^q?$$ I've tried to give an counter example, but I did not get any one.

Answer 1

Probably not what you are looking for but one COULD use some functional analysis to answer the question: If $\ell^q = \bigcup_{p<q} \ell^p =\bigcup_{n\in\mathbb N} \ell^{q-1/n}$ Baire's theorem implies that some $\ell^{q-1/n}$ would be of second category in $\ell^q$ and then the open mapping theorem for the inclusion $\ell^{q-1/n} \hookrightarrow \ell^q$ would imply the surjectivity of this inclusion. So, all you need to know is that $\ell^p \neq \ell^q$ for $p<q$.

Answer 2

For $q=\infty$, anything you can think of likely works as a counterexample.

Otherwise, use the fact that $\sum \frac{1}{n (\log n)^a} < \infty$ for $a>1$ to build counterexamples.

Answer 3

Consider $$x_n = \frac{1}{(n+2)^{1/q}(\ln (n+2))^{1/q} \ln (\ln (n+2))}, \quad n =1,2, \ldots$$ Then $\{x_n\}_1^{\infty} \in \ell^q$, but $\{x_n\}_1^{\infty} \notin \ell^p$ for $p<q$.