Taking the limit $$\lim_{p\rightarrow \infty} \left( \frac{\|f\|_\infty}{\|f\|_p}\right)^p$$

First I think the expression after taking the limit will depend on the function $f$.

In my attempt, because it is in the form $``1^\infty"$, I tried to use L'Hopital's rule. And we can calculate the limit (assuming the integrals are defined and finite, I just want to see what the limit might look like). \begin{align*} \lim_{p\rightarrow \infty} \left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)^{p} &= \lim_{p\rightarrow \infty} \exp\left( p\log\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)\right)\\ &=\lim_{p\rightarrow \infty} \exp\left( \frac{\log\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)}{\frac{1}{p}}\right)\\ &=\lim_{p\rightarrow \infty} \exp\left( \frac{\frac{d}{dp} \left[-\log\left(\frac{\|\nabla u \|_p}{\|\nabla u\|_\infty}\right)\right]}{\frac{-1}{p^2}}\right)\\ &=\lim_{p\rightarrow \infty} \exp\left( \frac{\left(\frac{\|\nabla u \|_\infty}{\|\nabla u\|_p}\right)\frac{\frac{d}{dp}\left[\|\nabla u\|_p \right]}{\|\nabla u \|_\infty}\}}{\frac{1}{p^2}}\right)\\ \end{align*} where \begin{align*} \frac{d}{dp}\left[\|\nabla u\|_p \right] &= \frac{d}{dp}\left[\left(\int_{\mathbb R^N} |\nabla u |^p dx \right)^{1/p} \right] \\ &=\frac{d}{dp}\left[\exp\left(\frac{1}{p} \log\left(\int_{\mathbb R^N} |\nabla u |^p dx \right)\right)\right] \\ &=\|\nabla u \|_p \left\{\frac{-1}{p^2}\log\left(\int_{\mathbb R^N} |\nabla u |^p dx \right) + \frac{1}{p} \frac{1}{\int_{\mathbb R^N} |\nabla u |^p dx }\int_{\mathbb R^N} |\nabla u|^p \log(|\nabla u |) dx \right\} \end{align*}

Putting it back into the limit we get $$\lim_{p\rightarrow \infty} \exp \left(-\log\left(\int_{\mathbb R^N} |\nabla u |^p dx \right) + p \frac{1}{\int_{\mathbb R^N} |\nabla u |^p dx }\int_{\mathbb R^N} |\nabla u|^p \log(|\nabla u |) dx \right)$$ which simplifies to $$\lim_{p\rightarrow \infty} \frac{\exp \left(p\frac{\int_{\mathbb R^N} |\nabla u|^p \log(|\nabla u |) dx}{\int_{\mathbb R^N} |\nabla u |^p dx } \right)}{\int_{\mathbb R^N} |\nabla u |^p dx }$$

And I am stuck. Is this a correct approach?

Thank you very much!

  • 3
    $\begingroup$ I think your original instinct is right. Consider $L^p[0,1]$. Then the limit for $f(x) = x$ is $\infty$; the limit for $f(x) = 1$ is $1$; and the limit for $f(x) = 1 + x$ is zero. So given that, it seems unlikely to me that there will be a simplier expression for that limit. $\endgroup$
    – Simon S
    Aug 12, 2015 at 13:10
  • $\begingroup$ Try using $\lim_{n\to\infty}(n+\frac1n)^n=e$ $\endgroup$
    – Rowan
    Aug 12, 2015 at 13:20
  • 1
    $\begingroup$ @user229922 : You mean $(1+1/n)^n$? ;) $\endgroup$ Aug 12, 2015 at 13:23
  • $\begingroup$ @PatrickDaSilva Yes. So sorry $\endgroup$
    – Rowan
    Aug 12, 2015 at 13:34
  • $\begingroup$ @SimonS For the case $f(x) = x+1,$ the limit is $\infty.$ $\endgroup$
    – zhw.
    Aug 12, 2015 at 19:04

1 Answer 1


Suppose $f\in L^\infty([0,1])$ and $\|f\|_\infty> 0$. Let $E = \{x:|f(x)|=\|f\|_\infty\}.$ Then

$$\lim_{p\to \infty} \left( \frac{\|f\|_\infty}{\|f\|_p}\right)^p = \frac{1}{m(E)}.$$

(If $m(E)=0,$ the conclusion is that the limit is $\infty.$)

Proof: Let $M= \|f\|_\infty$. Then the expression equals

$$\frac{M^p}{M^p\cdot m(E) + M^p\int_{[0,1]\setminus E}|f/M|^p}.$$

Cancel the $M^p$ terms and then apply the dominated convergence theorem to see the integral in the denominator $\to 0.$ That gives the result.

  • $\begingroup$ Nice answer and a nice result! $\endgroup$
    – Simon S
    Aug 12, 2015 at 19:41
  • $\begingroup$ thank you for the answer, but I feel like this result depends on the fact that the domain is $[0,1]$ therefore we always have $\|f\|_p \leq \|f\|_\infty$. $\endgroup$
    – Xiao
    Aug 13, 2015 at 7:18
  • $\begingroup$ Well it would work the same on any set of finite measure. In fact, it should work on any measurable set $A$ with one small difference. Let $E$ be as above. If $m(E) = \infty,$ then $\|f\|_p = \infty, 0<p<\infty,$ and all the quotients are $0.$ If $m(E) <\infty,$ then the proof should work as before and the answer should be $1/m(E).$ $\endgroup$
    – zhw.
    Aug 13, 2015 at 8:51
  • $\begingroup$ I know this is an older posted, but do you think this is related to something that I asked here: math.stackexchange.com/questions/3733478/… $\endgroup$
    – Boby
    Jun 25, 2020 at 0:49

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