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 $.


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 $r < 1$?

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    $\begingroup$ Looks like the optimal exponent is exactly max$(p,q)$. To see this, consider two strictly positive sequences: $a_n, b_n >0$. Then trivially $$ a*b(n) = \sum_{k\in \mathbb Z} a_{n-k} b_k \ge a_n b_0 ;$$ and similarly $a*b(n) \ge b_n a_0$. Then $a*b(n)$ cannot go to zero faster than the slower between $a$ and $b$. @Davide: if you want, you should post an answer. $\endgroup$ – Angelo Lucia Apr 9 '13 at 10:42

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 $$a*b(n)\geqslant \max\{a(n)b(0),a(0)b(n)\}.$$

  • $\begingroup$ Just one one: should it be "Fubini", not "Funini"? $\endgroup$ – Angelo Lucia Apr 9 '13 at 13:36

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