Background
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[It is known](https://mathoverflow.net/questions/107980/convolution-of-sequences) 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
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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 $r < 1$?



Convolution in $\ell_p$ when $0<p<1$
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 
$$a\*b(n)\geqslant \max\{a(n)b(0),a(0)b(n)\}.$$


