Union of $l^p$ spaces for $p\in[1,q)$

Asked Nov 08 '23 at 23:38

Active Nov 09 '23 at 11:44

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Inclusion of $l^p$ space for sequences has been discussed here: for $1≤p<q≤\infty$, $l^p\subset l^q$.

then I have 2 questions in mind:

Given $1<q<\infty$. Is it true that $\bigcup_{1 \leq p <q }l^p=l^q$?

and

Given $1<p<\infty$. Is it true that $\bigcap_{p <q\le\infty }l^q=l^p$?

Inclusion of $L^p$ spaces for functions on $[0,1]$ has been discussed here: For $1≤p<q≤∞$, $L^q⊂L^p$.

similar questions:

Given $1<q<\infty$. Is it true that $\bigcap_{1 \leq p <q }L^p=L^q$?

and

Given $1<p<\infty$. Is it true that $\bigcup_{p <q\le\infty }L^q=L^p$?

edited Nov 09 '23 at 11:44

asked Nov 08 '23 at 23:38

hbghlyj

What is $l^p$? As written, without more details, it's not true that $l^p \subset l^q$ if $p<q$ – Andrew Nov 08 '23 at 23:45
1

The inequality you speak of for $L^p$ is only true for finite measures and it's false, for instance, for $L^p(\Bbb R)$. Be as it may, for your questions it's Thatcher's answer: "No, no, no(, no)." – Sassatelli Giulio Nov 09 '23 at 00:07
@SassatelliGiulio Yes, I have added "and finite measures" – hbghlyj Nov 09 '23 at 00:52
For the second, consider $(n^{-1/p})_n$ – Brian Moehring Nov 09 '23 at 05:07
1

For the others, I believe https://math.stackexchange.com/q/1243910/694754 should suffice. – Brian Moehring Nov 09 '23 at 07:45
@SassatelliGiulio I changed $X$ to $[0,1]$. Thanks for the information – hbghlyj Nov 09 '23 at 11:43

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