Working with the Banach algebra $\ell^1(\mathbb{Z})$ and the involution $f^* (n)=\bar{f(-n)}$, I want to show this is not a $C^*$ algebra.
I know I need to find some function $f$ such that $||f^* f||_1 \neq ||f||_1^2$.
If I take $f=a_n e^{inx}$ for $a_n \in \ell^1(\mathbb{Z})$, $f^* =b_n e^{inx}$ for $b_n=\bar{a_{-n}}$.
I'm just left struggling to determine the values of $||f^* f||_1$ and $||f||_1$ if this function works to show that these two are not equal and hence it is not a $C^*$ algebra.
I think this give a proof. Consider $a = \delta_0 + i\delta_1 +i\delta_2,$ where $\delta_j(i) = \delta_{ij},$ $\delta_j \in \ell_1(\mathbb{Z}).$ Then $a^* = \delta_0 - i\delta_{-1} -i\delta_{-2}.$ Thus $$\|aa^*\| =\|-i\delta_{-2}+(1-i)\delta_{-1} + 3\delta_0 + (1+i)\delta_{1} + i\delta_2\| = 5+2\sqrt{2},$$ whereas $\|a\|=3.$
Question related to:
Why is $\ell^1(\mathbb{Z})$ not a $C^{*}$-algebra?
and Does the Gelfand transformation on $\ell^1(\mathbb Z)$ possess a continuous inverse on its image?
and The Gelfand transformation on $\ell^1(\mathbb Z)$ is not isometric. Do you have an example?
