Let $\Phi$ be a Youngs's function, i.e. $$ \Phi(t) = \int_0^t \varphi(s) \,\mathrm d s$$ for some $\varphi$ satifying

  1. $\varphi:[0,\infty)\to[0,\infty]$ is increasing
  2. $\varphi$ is lower semi continuous
  3. $\varphi(0) = 0$
  4. $\varphi$ is neither identically zero nor identically infinite

and define the Luxemburg norm of $f:\Omega\to\mathbb{R}$ as $$ \lVert f \rVert_{L^{\Phi}} := \inf \left\{\gamma\,\middle|\,\gamma>0,\,\int_{\Omega} \Phi\left(\frac {\lvert f(x)\rvert}{\gamma} \right)\,\mathrm{d}x\leq 1\right\}.$$

Question: What can we say about $\Phi\left(\lVert f \rVert_{L^{\Phi}}\right)$? In particular, I'd like to know, if $$\Phi\left(\lVert f \rVert_{L^{\Phi}}\right) \leq C \int_{\Omega}\Phi(\lvert f(x)\rvert) \,\mathrm d x$$ holds for some $C$ independent of $f$.

Any idea or hint for a reference is welcome!


  • The above inequality trivially holds for $\Phi(t) = t^p$, where $p>1$
  • Maybe it's appropriate to consider this question in the more general framework of Musielak-Orlicz spaces. However, e.g. in Lebesgue and Sobolev Spaces with Variable Exponents I was unable to find an appropriate result.
  • I have asked this question on Math.Stackexchange without luck, so I'm trying here.
  • $\begingroup$ Meanwhile he/she changed the name. I have deleted my comment. $\endgroup$ Jun 13, 2019 at 12:36
  • 1
    $\begingroup$ @JochenWengenroth: Thanks for being honest with me! $\endgroup$
    – CallMeStag
    Jun 13, 2019 at 12:38

1 Answer 1


The conjectured inequality does not hold.

For a counterexample, consider $\Phi(t)=\max(t^2,t^3)$ and $\Omega=(0,1)$. Let $f=a\chi_{(0,b)}$ for $a,b\in (0,1)$. It can be calculated that $\|f\|_{L^\Phi}= a b^{1/3}$. Then the inequality can be written as \begin{equation*} a^2 b^{2/3} \leq C a^2 b \end{equation*} which is not possible for a constant $C$ independent of $b$.

Intuitively, this is because the left-hand side is mostly determined by the values of $\Phi$ for large $t$, whereas the right-hand side is (for small $a,b$) independent of the values of $\Phi$ for large $t$.

  • 2
    $\begingroup$ Nice counterexample! $\endgroup$
    – Dirk
    Jun 14, 2019 at 11:52
  • $\begingroup$ Indeed it is - thank you, @harfe! $\endgroup$
    – CallMeStag
    Jun 16, 2019 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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