# The convolution between weighted $L^1$ space and normal $L^1$ space

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.

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 $u*v\notin L_{\omega}^1$.
• I understand. However, if I assume that, both $u$ and $v$ are compact supported, $u\in L^1_\omega(\mathcal R^N)$ and $v\in L^1(\mathbb R^N)$, then do I have $u\ast v \in L^1_\omega(\mathbb R^N)$? or do I need to assume $\omega$ is locally bounded? – JumpJump Aug 12 '15 at 15:22
Let $N=1$ and $\omega(x)=1+|x|^{-m}$ with $0<m<1$ (it satisfies your conditions, with $C=2^m/(1-m)$, if I'm not mistaken -- I've checked it for intervals in $\mathbb R^+$ only...).
Let $u$ vanish outside $[1,2]$ with $u(x)=(x-1)^{-\alpha}$ for $x$ in that interval, and $v(x)=u(-x)$. Clearly, both $u$ and $v$ are in $L^1_\omega (\subset L^1)$ if $0<\alpha<1$. But, for $0<x<1$,$$u*v(x)=\int_x^{1} y^{-\alpha}(y-x)^{-\alpha}\ dy=x^{1-2\alpha}\int_1^{1/x} t^{-\alpha}(1-t)^{-\alpha}\ dt$$and $x^{-m-2\alpha+1}$ is not integrable near 0 if $\alpha\ge 1-\frac{m}{2}$, so $u*v\notin L^1_\omega$.
• Thank you for your answer! I actually found a paper says that if and only if $\omega$ is locally bounded, the convolution will be finite. – JumpJump Aug 17 '15 at 2:02