I have come across an apparently simple exercise on Orlicz norms (exercise 2.2.3 of Van der Vaart's "Weak convergence and Empirical processes") which I cannot fully understand.
One has to show that, for any constant $C\geq 1$, random variable $X$, and convex, increasing function $\psi$, it holds that
$$\Vert X \Vert _{C\psi} \leq C \Vert X \Vert _{\psi}$$
It is suggested to use the fact that, being $\psi$ convex, we have
$$\mathbb{E}[\psi(\vert X \vert/C\Vert X \Vert_{\psi})] \leq 1/C \quad (\star)$$
Using this hint, solving the exercise is straightforward. In fact
$$ \Vert X \Vert_{C\psi} := \inf \Bigg\lbrace K > 0 \, : \, \mathbb{E} \bigg[C\psi\bigg(\frac{\vert X \vert}{K} \bigg) \bigg] \leq 1 \Bigg\rbrace = \inf \Bigg\lbrace K > 0 \, : \, \mathbb{E} \bigg[\psi\bigg(\frac{\vert X \vert}{K} \bigg) \bigg] \leq \frac{1}{C} \Bigg\rbrace$$
and so, thanks to $(\star)$, it follows that $\Vert X \Vert_{C\psi}\leq C\Vert X \Vert_{\psi}$.
My question is: how to prove $(\star)$? It does not seem obvious to me (maybe because I am quite rusty on convex functions...)
Clearly we have that $\mathbb{E}[\psi(\vert X \vert/\Vert X \Vert_{\psi})] \leq 1$, by definition of Orlicz norm. If for instance it were possible to show that $\psi(ax)\leq a\psi(x), \forall x, \forall a \in (0,1]$, then $(\star )$ would follow. In the special case $\psi(x)=x^p$, $p\geq1$, or more in general if $\psi$ is $p$-homogeneous, then the previous inequality holds.
Unfortunately I cannot figure out how to extend this to general $\psi$ with the required properties (if it is true). Any suggestions? Many thanks!
P.S.: In the definition of Orlicz norm it is also assumed that $\psi(0)=0$.