TITLE: Luxemburg norm as argument of Young's function: $\Phi\left(\lVert f \rVert_{L^{\Phi}}\right)$
QUESTION [8 upvotes]: Let $\Phi$ be a Youngs's function, i.e.
$$ \Phi(t) = \int_0^t \varphi(s) \,\mathrm d s$$
for some $\varphi$ satifying

$\varphi:[0,\infty)\to[0,\infty]$ is increasing
$\varphi$ is lower semi continuous
$\varphi(0) = 0$
$\varphi$ 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!

Notes:

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.

REPLY [5 votes]: 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 $t$.