# For $p \in (0,\infty)$, is it possible to have a function $f \in L_q$ for $q\neq p$ and $f \notin L_p$?

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.

• Your example is not correct. Also, this question is a duplicate. Can you locate it on MSE? Nov 12, 2020 at 23:43
• @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 $q\neq p$.
– user811819
Nov 12, 2020 at 23:45

No, that's not possible: Suppose $$f\ge 0,$$ $$q_1 and $$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.$$