Let $\omega\in L^1\_{\text{loc}}(\mathbb R^N$) be given. We assume that $\omega\geq 1$, l.s.c, and satisfies, for a constant $C>0$,
$$
\frac{1}{|B(x,r)|}\int\_{B(x,r)}\omega(y)dy\leq C\omega(x)
$$
for any $x\in\mathbb R^N$ and $r>0$.


We define the weighted $L^1\_\omega$ space by, for function $u$,
$$
\int\_{\mathbb R^N}|u(x)|\omega(x)dx<\infty
$$


It is well known that $L^1(\mathbb R^N)\ast L^1(\mathbb R^N)\subset L^1(\mathbb R^N)$, i.e., the convolution for two $L^1$ function is still $L^1$.


Now my question, for what condition do I need for $\omega$ that 
$$
L^1\_\omega(\mathbb R^N)\ast L^1(\mathbb R^N)\subset L^1\_\omega(\mathbb R^N)?
$$


I feel I may need some growth condition on $\omega$ but I am not sure. I also search online. There are many papers dealing with weighted $L^p$ space but they are just interested in the case $p>1$ and $\omega=|x|^\alpha$ for some $\alpha>0$. But I am really just interested in the case $p=1$...


Thank you!




---


PS: Of course making $\omega$ be bounded above will work but it makes this theorem not so useful. I wish to keep $\omega$ has the ability to blow up to infinity still.



The convolution between weighted $L^1$ space and normal $L^1$ space
This will never work in the kind of situation you outline. As soon as $\omega$ gets large somewhere, you're doomed. For instance, let's assume that we can find disjoint balls $B\_n$ of radius $1$, such that $\int\_{B\_n} \omega\ge n^2$. Then take $u$ as the characteristic function of the ball of radius $3$ about the origin, and $v=\sum n^{-2}\chi\_{-B\_n}$. Then $u\in L\_{\omega}^1$, $v\in L^1$, but
$$
\| v\*u\|\_{L\_{\omega}^1} = \int dx\, v(x) \int dt\, \omega(t)u(t-x) \ge \sum n^{-2}n^2 = \infty ,
$$
so $u\*v\notin L\_{\omega}^1$.


