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. $\endgroup$ – Jochen Wengenroth Jun 13 '19 at 12:36
  • 1
    $\begingroup$ @JochenWengenroth: Thanks for being honest with me! $\endgroup$ – Hirsch Jun 13 '19 at 12:38

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 $t$.

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  • 1
    $\begingroup$ Nice counterexample! $\endgroup$ – Dirk Jun 14 '19 at 11:52
  • $\begingroup$ Indeed it is - thank you, @harfe! $\endgroup$ – Hirsch Jun 16 '19 at 7:44

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