Why is $\ell^1(\mathbb{Z})$ not a $C^{*}$-algebra?

When $\ell^1(\mathbb Z)$ is equipped with the convolution as multiplication and $a^{*}_{n}=\bar{a}_{-n}$, I can prove it satisfies all conditions except $\|a^{*}a\|=\|a\|^2$, which I cannot prove nor find a counter example.

I wonder whether anyone can give a hint on this.

Thanks!

Take $x \in \ell^1(\mathbb{Z})$ to be $x(0) = 1$, $x(1) = x(2) = -1$, and $x(n) = 0$, otherwise.

Then compute $(x^\ast \ast x)(n) = \begin{cases} 3, & n = 0, \\ -1, & n = \pm 2, \\ 0, & \text{otherwise.}\end{cases}$

This gives $\|x^\ast \ast x\|_1 = 5$ while $\|x\|_{1}^2 = 9$.

Later: The above example shows that the $C^\ast$-identity isn't satisfied, which is of course enough to conclude. Let me point out that there is no way at all to turn $\ell^{1}(\mathbb{Z})$ into a commutative unital $C^\ast$-algebra: this is because $\ell^1(\mathbb{Z})$ would then have to be isomorphic to a space of the form $C(K)$ with $K$ compact (metrizable and infinite). However, this can't be because $\ell^1(\mathbb{Z})$ has the Schur property, which implies that it is weakly sequentially complete while it is not difficult to show that $C(K)$ isn't weakly sequentially complete as soon as $K$ is infinite. Other ways of seeing this are outlined in the comments below.