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Suppose $p \leq q$. Does there exist a space $X$ such that $L^p(X) \subset L^q(X)$ or $L^q(X) \subset L^p(X)$? What about a space $X$ such that $L^p(X) - L^q(X)$ nonempty, $L^q(X) - L^p(X)$ nonempty?

EDIT: Here is what I have so far:

(a) For $L^p(X) \subsetneq L^q(X)$, take $X = [0,1]$ which holds since $[0,1]$ has finite measure.

(b) For $L^q(X) \backslash L^p(X)$, take $x \in (1, \infty)$ and the function $1/x$. This function is in $L^2(1,\infty)$ but not $L^1(1, \infty)$.

(c) $L^p(X)\backslash L^q(X)$ - I still have nothing.

(d) $L^q(X) \subsetneq L^p(X)$ I still have nothing. I want to consider a set $X$ which does not contain a subset of arbitrarily small measure?

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    Many of the answers you seek can be found here. Here are a few points. If $X$ is a single point, then these function spaces are all equal to $\mathbb{R}$. If $X$ has finite measure, then $L^q(X) \subset L^p(X)$. If $X=[1,\infty)$, then $f(x)=x^{-1/p}$ is in $L^q(X)$ but not $L^p(X)$. – angryavian Feb 21 '17 at 04:01
  • You want $p<q,$ right? – zhw. Feb 21 '17 at 07:24
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    For all cases, you can use $X = \mathbb R$ equipped with the Lebesgue measure. For $p \ne q$ you neither have $L^p(X) \subset L^q(X)$ nor $L^q(X) \subset L^p(X)$. – gerw Feb 21 '17 at 07:55
  • @angryavian What about $L^p \subset L^q$? And for $L^q \subset L^p$, simply take $X = [0,1]$? And lastly, what about show that $L^p \backslash L^q$ can be nonempty? –  Feb 23 '17 at 00:19
  • @zhw. I have made an update. –  Feb 26 '17 at 04:16
  • @angryavian I have made an update. –  Feb 26 '17 at 04:16
  • @gerw I have made an update. –  Feb 26 '17 at 04:16

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