# Is it possible for a function to be in LpL^p for only one pp?

I’m wondering if it’s possible for a function to be an $L^p$ space for only one value of $p \in [1,\infty)$ (on either a bounded domain or an unbounded domain).

One can use interpolation to show that if a function is in two $L^p$ spaces, (e.g. $p_1$ and $p_2$,with $p_1 \leq p_2$ then it is in all $p_1\leq p \leq p_2$).

Moreover, if we’re on a bounded domain, we also have the relatively standard result that if $f \in L^{p_1}$ for some $p_1 \in [1,\infty)$, then it is in $L^p$ for every $p\leq p_1$ (which can be shown using Hölder’s inequality).

Thus, I think that the question can be reduced to unbounded domains if we consider the question for any $p>1$.

Intuitively, a function on an unbounded domain is inside an $L^p$ space if it decrease quickly enough toward infinity. This makes it seem like we might be able to multiply the function by a slightly larger exponent. At the same time, doing this might cause the function to blow up near zero. That’s not precise/rigorous at all though.

So I’m wondering if it is possible to either construct an example or prove that this can’t be true.

Robert’s and joriki’s examples are of course nice and explicit, but you can get examples on any subset of $\mathbb{R}^n$ with infinite measure. Here’s how:
Take a function $f$ that is in $L^p$ but not in $L^q$ for $q \gt p$ (on the unit ball $B$ around zero, say). Now take a sequence $x = (x_n)$ that is in $\ell^p$ but not in $\ell^q$ for $q \lt p$ (there are standard examples for both of these things). Now take disjoint balls $B_n$ of volume $1$ (disjoint from $B$) and consider $g = f + \sum x_n \cdot [B_n]$ where $[B_n]$ denotes the characteristic function of $B_n$. Obviously, $\|g\|_{q}^q = \|f\|_{q}^q + \|x\|_{q}^{q}$ is in $L^q$ if and only if $q = p$. If $q \lt p$ then $\|x\|_q = \infty$ and if $q \gt p$ then $\|f\|_{q}^q = \infty$.
I leave it to you to make that explicit and to modify it when your domain is not all of $\mathbb{R}^n$.