# Show that $\ell^1(\mathbb{Z})$ is not a $C^*$-algebra.

I am currenctly working with some training exercises in operator algebra. I have to show that $$\ell^1(\mathbb{Z})$$ is not a $$C^\ast$$-algebra for the following involution defined as $$(x^\ast)_n:=\overline{x_{-n}}$$.

What I have in mind is that we can use the following $$x_n:=\left\{\begin{matrix} 1, & \text{ if } n=0\;\;\;\;\\ -2, & \text{ if } n=1,2\\ 0, & \text{ otherwise } \end{matrix}\right.$$ And now my goal is to show that it is not a $$C^\ast$$-algebra. Anyone who can help me to continue?

• What do you want to do with such $\{x_n\}$ ? Any plan ? Nov 15, 2021 at 15:12
• What is the definition of a $C^\star$ algebra for you ? Nov 15, 2021 at 15:14
• What I try to do is to make a contradiction with the identity but I am not sure how to compute the norms. I.e. $||x^\ast \ast x||_1$ and $||x||_1^2$. What I want is to get different numbers. Nov 15, 2021 at 15:16
• ok so, $\ell^1$ is equipped with $\vert \vert\cdot \vert\vert_1$. You may have to specify this in your question as it may have been equipped with a $C^\star$ norm. Nov 15, 2021 at 15:18
• So what are those numbers ? Nov 15, 2021 at 15:20

I assume the product on $$\ell^1(\mathbb Z)$$ is the pointwise product.
If it makes a $$C^*$$-algebra, then $$\|xx^*\|=\sum_n |x_n\overline {x_{-n}}| = \|x\|^2$$ for all $$x$$.
E.g. let $$(x_n) = \chi_{\{0, 1\}}$$ be the characteristic function of $$\{0, 1\}$$, then $$\sum_n |x_n \bar x_{-n}|=1$$ but $$\|x\|^2= 2^2=4$$.
• Interesting. We define $\ell^1(\mathbb{Z}):=\{(x_n)_{n\in\mathbb{Z}}:x_n\in\mathbb{C},\sum_{k\in\mathbb{Z}}|x_n|<+\infty\}$. A quick question which might be simple but how do you get 2 when computing the norm? Is it just the number of elements $\{0,1\}$ or am I wrong? Nov 15, 2021 at 16:49
• It's just $|1| + |1|$ where the first $1$ is the coefficient for $0$ and the second is for $1$. Nov 15, 2021 at 16:52
• BTW, the convolution can also make a Banach algebra out of $\ell^1(\mathbb Z)$, which is still not a $C^*$-algebra for similar reasons. (I saw you used $x^**x$ in the comment, but I'm not sure whether you mean convolution by $*$ in the middle. Nov 15, 2021 at 16:53