Convolution of $l^2$ functions is not an $l^2$ function. Consider $$l^2(\mathbb{Z})= \left\{f: \mathbb{Z} \to \mathbb{C}: \ \sum_{n \in \mathbb{Z}} |f(n)|^2 < \infty\right\}$$
and define the convolution of two functions $f,g: \mathbb{Z} \to \mathbb{C}$ by
$$(f*g)(x):= \sum_{n \in \mathbb{Z}} f(n) g(x-n), \quad x \in \mathbb{Z}$$
I'm looking for two functions $f,g \in l^ 2(\mathbb{Z})$ with $f*g \notin l^2(\mathbb{Z})$ but couldn't find any examples. Is there some intuition that can guide me to a correct example? I tried with harmonic series but the calculations always get messy.
Thanks in advance.
 A: Consider $f$ supported on $\mathbb{Z}_{>0}$ with $f(n) = n^\alpha$ for $n > 0$, where $\alpha$ is a fixed constant which we will choose appropriately. Note that if $\alpha < 0$, the term $f(k)f(n-k) = (k(n-k))^\alpha$ is minimized at $k = n/2$, hence the bound
$$(f * f)(n) = \sum_{k=1}^{n-1} f(k) f(n-k) \geq (n-1) (n/2)^{2\alpha} \geq (n/2)^{2\alpha + 1}$$
holds for $n \geq 2$. (In fact you can prove the stronger result that if $f$ and $g$ are supported on $\mathbb{Z}_{>0}$ with $f(n) \asymp n^\alpha$ and $g(n) \asymp n^\beta$, then $(f * g)(n) \asymp n^{\alpha + \beta + 1}$ whenever $\alpha, \beta > -1$).
But $f$ is in $\ell^2$ so long as $\alpha < -\frac{1}{2}$, and by the above we see that $f * f$ is not in $\ell^2$ when $2\alpha + 1 \geq -\frac{1}{2}$, i.e. $\alpha \geq -\frac{3}{4}$. Then setting $\alpha = -\frac{2}{3}$, $f$ is in $\ell^2$, but $f * f$ is not.
A: If that was true, then applying Fourier transform you would see that the pointwise product of $L^2$ functions on the interval $[0,1]$ is also $L^2$.
