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.

  • 5
    I haven't given this question much thought, (and while I don't think it is true), I would think that the most likely candidate for a counterexample would be $L^1$. In other words, we cannot necessarily rule out bounded domains. – JavaMan Aug 2 '11 at 18:42
  • Oh right. That's a fair point. Thanks! In fact, that might be the easier case to deal with, now that you mention it. – user1736 Aug 2 '11 at 18:44
up vote 57 down vote accepted

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 $\mathbb{R}^n$.

  • 3
    It is rather treacherous to think of $L^p$ functions of functions with any kind of decay property. In fact, you should try and construct an example of an $L^p$ function that is unbounded on every open set. – t.b. Aug 2 '11 at 19:21
  • @Didier: Thanks for editing. – t.b. Aug 2 '11 at 19:21
  • Hm, I think I see what you are getting at. Thanks for the detailed exposition! By the way, you meant to say not in $l^q$ for $q<p$ right? – user1736 Aug 2 '11 at 19:29
  • @user1736: Yes, fixed. Thanks. Here's a link in return to illustrate my previous comment. – t.b. Aug 2 '11 at 19:36

Since $1/x$ is the border case in both directions, the most promising candidate would be a modified version of $1/x$ that just converges but won't converge if you nudge it ever so slightly. We have

$$\int_2^\infty \frac1{x\log^2x}\mathrm dx=\left[-\frac1{\log x}\right]_2^\infty=\frac1{\log2}\;,$$


$$\int_2^\infty \left(\frac1{x\log^2x}\right)^p\mathrm dx$$

with $p<1$ diverges. If we stitch this function together with its inverse to get convergence at $0$, we get an integral

$$\int_2^\infty \frac1{x^{1/p}\log^2(x^{1/p})}\mathrm dx=\frac1{p^2}\int_2^\infty \frac1{x^{1/p}\log^2x}\mathrm dx\;,$$

which again diverges for $p>1$, so we only have convergence on both sides if $p=1$.

  • I'm not following, is $\frac1{x^{1/p}\log^2(x^{1/p})}$ the inverse of $\left(\frac1{x\log^2x}\right)^p$ ? – Weltschmerz Aug 13 '14 at 23:16
  • 2
    Is this one really divergent for p<1? – C-Star-Puppy Aug 24 '14 at 16:58

Try $$f(x) = \frac{1}{x^{1/p} (\ln(x)^2+1)} \qquad \text{on} \qquad (0, \infty)$$

  • 1
    Thanks for the great example. I accepted Theo's because it provided more generality, but yours works well too! – user1736 Aug 2 '11 at 19:34

Consider a real monomial: $x^\alpha$

Now poles give a restriction of the form $\alpha>\alpha_0$
while decay rates give a restriction of the form $\alpha<\alpha_0$.

This can be used to glue a desired example, e.g. if $$f(x):=x^{\alpha_0}\chi_{[-1,1]}+x^{\alpha_\infty}\chi_{(-\infty,-1)\cup(1,\infty)}$$ then the integrability is restricted to $$-\frac{1}{\alpha_0}<p<-\frac{1}{\alpha_\infty}$$

Unfortunately this technique only restricts up to a neighborhood like $p=7\pm\varepsilon$.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.