When do Orlicz norms tend to the uniform norm? It is well known that the $p$-norms tend to the $\infty$-norm, in that if $\lVert f \rVert_q < \infty$ for some $q \ge 1$ then $\lVert f \rVert_p \to \lVert f \rVert_\infty$ as $p \to \infty$.  Does this extend in some way to general Orlicz norms?  I have only found this article on the subject, and I am hoping for something more general, for example conditions on a sequence of Young functions $\Phi_n$ so that $\lVert f \rVert_{\Phi_n} \to \lVert f \rVert_\infty$.
 A: $\newcommand{\ep}{\varepsilon}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}$A natural generalization of the fact that the $p$-norm converges to the $\infty$-norm as $p\to\infty$ is as follows.
Let $\mu$ be a probability measure on a measurable space $(S,\Si)$. For each natural $n$, let $\Phi_n$ be a Young function normalized by the condition that $\Phi_n(1)=1$. Suppose that the sequence $(\Phi_n)$ satisfies the condition
\begin{equation*}
    0<x<y<\infty\implies\Phi_n(y/x)\to\infty  \tag{0}\label{0}
\end{equation*}
(as $n\to\infty$).
Take now any measurable function $f\colon S\to\R$. Then we have

Claim:
\begin{equation*}
    \|f\|_{\Phi_n}\to\|f\|_\infty.  \tag{1}\label{1}
\end{equation*}

Indeed, recall that
\begin{equation*}
    \|f\|_{\Phi_n}=\inf E_n\in[0,\infty],  
\end{equation*}
where
\begin{equation*}
    E_n:=\{k\in(0,\infty)\colon\int_S\Phi_n(|f|/k)\,d\mu\le1\}. 
\end{equation*}
Note that, since the Young function $\Phi_n$ is increasing, we have $E_n=[\|f\|_{\Phi_n},\infty)$ or $E_n=(\|f\|_{\Phi_n},\infty)$.
Take any $k\in(0,\|f\|_\infty)$ (if such a number $k$ exists). Take any $l\in(k,\|f\|_\infty)$. Then $\ep:=\mu([|f|>l])>0$, where $[|f|>l]:=\{s\in S\colon|f(s)|>l\}$. So,
\begin{equation*}
    \int_S\Phi_n(|f|/k)\,d\mu\ge\int_{[|f|>l]}\Phi_n(|f|/k)\,d\mu\ge\ep\Phi_n(l/k)\to\infty, 
\end{equation*}
by \eqref{0}. So, for all large enough $n$ we have $k\notin E_n$ and hence $k\le\|f\|_{\Phi_n}$, for any $k\in(0,\|f\|_\infty)$. So,
\begin{equation*}
    \liminf_{n\to\infty}\|f\|_{\Phi_n}\ge \|f\|_\infty. \tag{2}\label{2}
\end{equation*}
So, if $\|f\|_\infty=\infty$, there is nothing more to prove.
If now $\|f\|_\infty<\infty$, take any real $k>\|f\|_\infty$. Then $|f|/k\le1$ $\mu$-almost everywhere. Hence, in view of the condition $\Phi_n(1)=1$, we have $\Phi_n(|f|/k)\le1$ $\mu$-almost everywhere, so that $\int_S\Phi_n(|f|/k)\,d\mu\le1$, which means that $k\in E_n$ and therefore $\|f\|_{\Phi_n}\le k$, for every real $k>\|f\|_\infty$. So, $\|f\|_{\Phi_n}\le \|f\|_\infty$, for each $n$.
Thus, \eqref{1} folows from \eqref{2}.
