If we have a product of functions $fg$ with $f\in L^r$ and $g\in L^s$ for some $s,r>1$ satisfying $1/r+1/s=1$, then we know that $fg\in L^1$.  

But if $g$ is a little bit more than $L^s$, say $L^s \log L$ , can we say that $fg$ is a little bit more than $L^1$ ? For instance $L^1 \log L^1$ ?



Hölder inequality between different Orlicz spaces
Yes, we can say so. Indeed, let us show that the conditions $f\in L^r$ and $g\in L^s\ln L$ imply $fg\in L\ln^t L$ for $t:=1/s$. Moreover, we shall show that the value $t=1/s$ here is optimal, as it cannot be replaced by any greater value. Of course, by $h\in L\ln^t L$ we mean $\int |h|\ln^t(|h|+1)<\infty$.


let $\psi\colon[0,\infty)\to[0,\infty)$ be any continuous strictly increasing function with $\psi(0)=0$. For real $x,y\ge0$, let
\begin{equation\*}
 \Psi(y):=\int\_0^y\psi(v)\,dv,\quad \Phi(x):=\int\_0^x\psi^{-1}(u)\,du ; 
\end{equation\*}
[then](https://en.wikipedia.org/wiki/Young%27s_inequality_for_products#Standard_version_for_increasing_functions)
\begin{equation\*}
 xy\le\Phi(x)+\Psi(y). \tag{1}\label{1}
\end{equation\*}


Let now
\begin{equation}
 \Psi(y):=y^s \ln(y+1) \tag{2}\label{2}
\end{equation}
for real $y\ge0$, so that $\psi(v)=\Psi'(v)\asymp v^{s-1}\ln(v+1)$, $\psi^{-1}(u)\asymp \dfrac{u^{r-1}}{\ln^{r-1}(v+1)}$,
\begin{equation}
 \Phi(x)\asymp \dfrac{x^r}{\ln^{r-1}(x+1)} \tag{3}\label{3}
\end{equation}
for real $v,u,x\ge0$. We write $A\ll B$ if $A\le CB$ for some real $C>0$ depending only on $r$, and we write $A\asymp B$ if $A\ll B\ll A$.


Without loss of generality, $f,g\ge0$. Let
\begin{equation\*}
 t:=1/s,
\end{equation\*}
so that $t\in(0,1)$. Then
\begin{equation\*}
 fg\ln^t(fg+1)\le [f\ln^t(f+1)]\,g+fg\ln^t(g+1). \tag{4}\label{4}
\end{equation\*}
By \eqref{1} with $\Phi$ and $\Psi$ as in \eqref{3} and \eqref{2},
\begin{equation\*}
 [f\ln^t(f+1)]\,g\ll f^r+g^s\ln(g+1),
\end{equation\*}
so that
\begin{equation\*}
 \int[f\ln^t(f+1)]\,g<\infty 
\end{equation\*}
assuming $f\in L^r$ and $g\in L^s\ln L$:
\begin{equation\*}
\int f^r<\infty,\quad \int g^s\ln(g+1) <\infty. \tag{5}\label{5}
\end{equation\*}
Also, conditions \eqref{5} imply $\int fg\ln^t(g+1)<\infty$, by the standard Hölder inequality. So, by \eqref{4}, $\int fg\ln^t(fg+1)<\infty$; that is, $fg\in L\ln^t L$ for $t=1/s$, as desired.




---


Note that the exponent $t=1/s$ cannot be improved -- that is, it cannot be replaced by any $a>1/s$. Indeed, let $g\ge0$ be such that $g\in L^s\ln L$ but $g\notin L^s\ln^b L$ for any $b>1$ -- that is, $\int g\ln(g+1)<\infty$ but $\int g\ln^b(g+1)=\infty$ for any $b>1$.


Let $f:=g^{s/r}\ln^{1/r}(g+1)$. Then for any real $a>1/s$ we have $a+1/r>1$ and
\begin{equation}
 fg\ln^a(fg+1)\asymp g^{s/r+1}\ln^{a+1/r}(g+1)=g^s\ln^{a+1/r}(g+1),
\end{equation}
so that $\int fg\ln^a(fg+1)\asymp\int g^s\ln^{a+1/r}(g+1)=\infty$ and $fg\notin L\ln^a L$.


