It is well known that [the $p$-norms tend to the $\infty$-norm](https://math.stackexchange.com/q/242779/791850), 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](https://en.wikipedia.org/wiki/Orlicz_space)? I have only found [this article](https://arxiv.org/abs/2012.00707) on the subject, and I am hoping for something more general, for example conditions on a sequence of Young functions $\Phi\_n$ so that $\lVert f \rVert\_{\Phi\_n} \to \lVert f \rVert\_\infty$.



When do Orlicz norms tend to the uniform norm?
$\newcommand{\ep}{\varepsilon}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}$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](https://en.wikipedia.org/wiki/Orlicz_space#Formal_definition) 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}.


