That is, is it possible for a function to be in all $L_p$ spaces except for a single value of $p \in (0,\infty)$?

For $p=1$ $q \in (1,\infty)$, an example would be \begin{align*} f(x) = \begin{cases} \frac1x \text{ if }x\geq 1\\ 0 \text{ if }x< 1 \end{cases}. \end{align*}

This example wouldn't work for values of $p,q\in (0,1)$ though.

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    $\begingroup$ Your example is not correct. Also, this question is a duplicate. Can you locate it on MSE? $\endgroup$ Nov 12, 2020 at 23:43
  • $\begingroup$ @KaviRamaMurthy I couldn't locate a duplicate of this same question. The ones I found were in the other direction where $f\in L_p$ for one value of $p$ but $f\notin L_q$ for $q\neq p$. $\endgroup$
    – user811819
    Nov 12, 2020 at 23:45

1 Answer 1


No, that's not possible: Suppose $f\ge 0,$ $q_1<p<q_2,$ and $f$ is in both $L^{q_1}, L^{q_2}.$ Then

$$\int_{f\le 1} f^p \le \int_{f\le 1} f^{q_1} <\infty$$


$$\int_{f>1 } f^p \le \int_{f\le 1} f^{q_2} <\infty,$$

proving $f\in L^p.$


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