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Consider a measure space $(\Omega,\mu)$, where $\mu(\Omega)<\infty$.

It is classically known that if $p<q$, then $L^q(\Omega,\mu)\not\subset L^p(\Omega,\mu)$ (see For $1\leq p<q\leq\infty$ what function is in $L^p$ but not in $L^q$). The form of the argument on the link is that given $q>p\geq 1$, we construct a function $f \in L^p(\Omega,\mu)$ such that $f \not\in L^q(\Omega,\mu)$.

Is it possible to take one such function $f \in L^p(\Omega,\mu)$ such that $f \not \in L^{p+\epsilon}(\Omega,\mu)$ for every $\epsilon>0$?

gpr1
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  • With $\Omega = (0, 1)$, taking $f(x) = x^{\alpha}$ for appropriate $\alpha$ should work. – Aryaman Maithani Nov 04 '21 at 17:26
  • The thing is: if $\alpha=1$, then $f \not\in L^1(]0,1[)$. Therefore, we have to take $0<\alpha<1$ in order to assure that $f \in L^1(]0,1[)$. On the other hand: if $0<\alpha<1$, then there exists $\epsilon>0$ such that $f^{1+\epsilon} \in L^1(]0,1[)$. – gpr1 Nov 04 '21 at 17:31
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    For simplicity let $p=1$. Consider $f_n(x)=x^{-1+1/n} 1_{(0;1/n^2)}(x)$. Define $$g_n=f_n/(\Vert f_n\Vert_1).$$ Now define $$f(x)=\sum_{n\geq 1} 2^{-n} g_n(x+\sum_{j=1}^{n-1} 1/n^2).$$ As $\Vert g_n\Vert_1=1$, we get $f\in L^1$. On the other hand, given $\varepsilon>0$, there exists $n$ such that $g_n$ is not in $L^{1+\varepsilon}$. Thus, $f$ is not in $L^{1+\varepsilon}.$ – Severin Schraven Nov 04 '21 at 19:56
  • Excelent. Thank you, @SeverinSchraven. – gpr1 Nov 04 '21 at 20:01
  • In my example, we can pick $\Omega=(0;\pi^2/6).$ – Severin Schraven Nov 04 '21 at 20:01
  • I am not sure, we can say anything about general measure spaces. It seems hard to construct explicit examples. – Severin Schraven Nov 04 '21 at 20:02

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A necessary and sufficient condition to find an $f\in\mathbb L^p$ but not in $\mathbb L^{p+\varepsilon}$ is that the involved measure space contains sets of arbitrarily small measure. See here for the details.

Davide Giraudo
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