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Based on this question, I asked myself the following:

Let $\Omega=(0,1)$ and let $X=\overline{\text{span}\{x^{1/n-1}:n\in\mathbb{N}\}},$ where the closure is taken in $L^1(0,1)$. Does $X$ satisfy the conditions for the claim in the linked question?

My intuition is that it shouldn't, otherwise this would mean that there is a $q>1$ such that $X\subset L^q(0,1)$. But then if $1\leq n/(n-1)<q$ then $x^{-1+1/n}\in L^q(0,1)$ which I'm guessing is incorrect. But at first glance, it seems that $X$ satisfies the properties.

UserA
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1 Answers1

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You basically showed already that there does not exist a $q>1$ such that $X\subset L^q$ holds.

Using the statement from the linked question, this implies that the condition $$ \tag{*} X\subset \bigcup_{1<p\leq\infty} L^p(0,1) $$ is false (if it were true then $X\subset L^q$ would follow for some $q>1$).

You might be falsely thinking that (*) should be correct, because for all $n\in\Bbb N$ we have $x^{1/n-1}\in L^p$ for some $p>1$. However, this argument is not sufficient, because we do not have the same statement for elements in the closure.

For example, consider the element $$ z := \sum_{n\in\Bbb N} x^{1/n-1} 2^{-n}. $$ It can be shown that $z\in X\subset L^1(0,1)$ and that $z\not\in L^p(0,1)$ for any $p>1$.

UserA
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supinf
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  • So this means that it's hard to naively construct such a set $X$! – UserA Sep 22 '20 at 08:42
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    @UserA well, you could start with a set $X\subset L^p$ for some $p>1$, but constructing interesting sets $X$ would be hard. – supinf Sep 22 '20 at 08:44