TITLE: Convolution in $\ell_p$ when $0<p<1$
QUESTION [6 upvotes]: Background
It is known that given real sequences $a = (a_n)_{n \in \mathbb Z} \in \ell_p$ and $b = (b_n)_{n \in \mathbb Z} \in \ell_q$, their convolution defined as
$$ a * b (n) = \sum_{k \in \mathbb Z} a_{n-k} b_k $$
is in $\ell_r$ if $1 \le p, q < \infty$ and $\frac 1 r = \frac 1 p + \frac 1 q -1 $.
Question
What happens when $0 < p, q < 1$? Obviously, since $a$ and $b$ are in $\ell_1$, their convolution $a * b$ is in $\ell_1$. Can we say better, i.e. $a*b$ is in $\ell_r$ for some $r < 1$?

REPLY [2 votes]: If $0< p<1$, then $(u+v)^p\leqslant u^p+v^p$ for any non-negative numbers $u$ and $v$. Consequently, working first on finite sums and then after taking the limit, we can see that
$$|a*b(n)|^r\leqslant \sum_{k\in\mathbb Z}|a(k)|^r\cdot |b(n−k)|^r$$
for any integer $n$ and any $r\geqslant \max\{p,q\}$. Then, by an application of Fubini-Tonelli's theorem, we get that $a*b\in\ell_{\max(p,q)}$. 
Angelo's remarks shows that this is optimal, because assuming that $a$ and $b$ are sequences of positive numbers, then 
$$a*b(n)\geqslant \max\{a(n)b(0),a(0)b(n)\}.$$