3
$\begingroup$

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 $L^1 \log L^1$ ?

$\endgroup$
1
  • 2
    $\begingroup$ What does it mean for $g$ to be $L^s\log L$? $\endgroup$ May 2, 2022 at 12:58

1 Answer 1

2
$\begingroup$

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 \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 $fg\notin L\ln^a L$.

$\endgroup$
1
  • $\begingroup$ Thank you very much ! $\endgroup$
    – Dorian
    May 13, 2022 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.