TITLE: Orlicz Norm and A result on expectation
QUESTION [6 upvotes]: I am reading paper which is mainly about Dobrushin's contraction coefficient and its generalization. In page 27, the following is defined: 
Consider arbitrary, non-negative, convex function $\psi:\mathbb{R}\mapsto \mathbb{R}$ with $\psi(0)<1$. Then for an integrable random variable $X$ 
$$||X||_{\psi}:=\inf \{c>0:~\mathbb{E}[\psi(\frac{X}{c})]\leq 1\}.$$
Then letting $\psi^*$ be the convex conjugate of $\psi$, we can write the following for arbitrary random variables $X$ and $Y$ by Young's inequality: 
$$XY\leq \psi(X)+\psi^*(Y)$$ and hence, 
$$\mathbb{E}[XY]\leq 2||X||_{\psi}||Y||_{\psi^*}.$$
First of all, in the typical definition of Orlicz norm, it is assumed that $\psi(0)=0$ and also $\psi$ is non-decreasing. Why here those properties are removed? 
Secondly, the last inequality is not clear to me. Can anybody give me a hint on how to show that?
Thanks

REPLY [2 votes]: It seems that the author applies the result with $\psi_1(x):=e^x /2$ and $\psi_2(x):=e^{x^2}/2$ in order to exploit the identity $\lVert X^2\rVert_{\psi_1}=\lVert X\rVert_{\psi_2}^2$. It turns out that with these definition, $\lVert X\rVert_{\psi_j}$ is equal to the norm associated to the true Orlicz functions $\phi_1(x):=e^x-1$ and $\phi_2(x)=e^{x^2}-1$. 
As suggested by Christian Remling, we have for fixed $c,c'\gt 0$ the inequality 
$$\frac Xc\frac Y{c'} \leqslant \psi\left(\frac Xc\right)+\psi^*\left(\frac Y{ c'} \right),$$
hence if $\psi\left(\frac Xc\right)\leqslant 1$ and $\psi\left(\frac Y{c'} \right)\leqslant 1$, the inequality 
$$\mathbb E[XY] \leqslant 2cc'$$
takes place. We conclude by taking the infimum over those $c$ and $c'$.