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.

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    $\begingroup$ 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. $\endgroup$ – JavaMan Aug 2 '11 at 18:42
  • $\begingroup$ Oh right. That's a fair point. Thanks! In fact, that might be the easier case to deal with, now that you mention it. $\endgroup$ – user1736 Aug 2 '11 at 18:44

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 $\mathbb{R}^n$.

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    $\begingroup$ 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. $\endgroup$ – t.b. Aug 2 '11 at 19:21
  • $\begingroup$ @Didier: Thanks for editing. $\endgroup$ – t.b. Aug 2 '11 at 19:21
  • $\begingroup$ 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? $\endgroup$ – user1736 Aug 2 '11 at 19:29
  • $\begingroup$ @user1736: Yes, fixed. Thanks. Here's a link in return to illustrate my previous comment. $\endgroup$ – 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 $p=1$.

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  • $\begingroup$ I'm not following, is $\frac1{x^{1/p}\log^2(x^{1/p})}$ the inverse of $\left(\frac1{x\log^2x}\right)^p$ ? $\endgroup$ – Weltschmerz Aug 13 '14 at 23:16
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    $\begingroup$ Is this one really divergent for p<1? $\endgroup$ – C-Star-W-Star Aug 24 '14 at 16:58
  • $\begingroup$ How the p>1 case is working, I am not getting it anyone? $\endgroup$ – Devendra Singh Rana Aug 16 '19 at 9:08

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

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    $\begingroup$ Thanks for the great example. I accepted Theo's because it provided more generality, but yours works well too! $\endgroup$ – user1736 Aug 2 '11 at 19:34
  • $\begingroup$ Comparison test either on $(0,1)$ or $(1,\infty)$. $\endgroup$ – Robert Israel Mar 20 '19 at 1:07

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 $p=7\pm\varepsilon$.

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