# Clarification of showing that something is an involution e.g. *-algebra

Let $$\ell^1(\mathbb{Z}):=\{(x_n)_{n\in\mathbb{Z}}:x_n\in\mathbb{C},\sum_{k\in\mathbb{Z}}|x_n|<+\infty\}$$ . Given by $$(x^\ast)_n:=\overline{x_{-n}}$$, we want to show that for $$x \in \ell^{1}(\mathbb{Z})$$, $$\ell^{1}(\mathbb{Z})$$ is a $$*-algebra$$

claim:

1. $$(x^\ast)^\ast=x$$, $$\forall x\in \ell^1(\mathbb{Z})$$.

$$\begin{equation*} (x^\ast)^\ast=(\bar{x})^\ast=\bar{\bar{x}}=x. \end{equation*}$$ 2. Let $$x,y\in \ell^1(\mathbb{Z})$$ and $$a,b\in\mathbb{C}$$. Claim: $$\begin{equation*} (ax+by)^\ast=\bar{a}x^\ast+\bar{b}y^\ast=\bar{a}\bar{x}+\bar{b}\bar{y}=\overline{ax}+\overline{by}=(ax)^\ast+(by)^\ast=(ax+by)^\ast. \end{equation*}$$ 3. $$x,y\in\ell^1(\mathbb{Z})$$ claim: $$\begin{equation*} (xy)^\ast=y^\ast x^\ast \end{equation*}$$ But $$(xy)^\ast=(\overline{xy})=(yx)$$ and then what???

Furthermore, I must show that $$*$$ is an isometry, which can easily be done if I show that \begin{align*} ||x||^{2} \leq ||x^{*}x||, \end{align*} but I am not quite sure if that inequality holds $$\forall x \in \ell^{1}(\mathbb{Z})$$

This inequality $$\|x\|^2\le\|x^*x\|$$ is not true, there are very easy counter-examples; also why even bother to show this when being an isometry simply means $$\|x^*\|=\|x\|$$ which is much simpler to show?

A counter example to the inequality you are trying to show: take $$x=(x_n)$$ with $$x_n=0$$ if $$n\ne1,2$$ and $$x_1=1, x_2=-1$$. Then $$x^*=(y_n)$$ where $$y_n=0$$ for all $$n\ne-1,-2$$ where $$y_{-1}=1, y_{-2}=-1$$. We have $$\|x\|^2=(1+1)^2=4$$. On the other hand, $$\sum_{m\in\mathbb{Z}}y_mx_{n-m}=x_{n+1}-x_{n+2}=\begin{cases}1, n=-1\\ 0,n=0\\-1, n=1\\ 0, \text{else}\end{cases}$$ so $$\|x^*x\|=2<4$$.

However, $$-^*$$ is indeed an isometry: let $$x=(x_n)\in\ell^1(\mathbb{Z})$$. Then $$\|x^*\|=\|(\bar{x_{-n}})_{n\in\mathbb{Z}}\|=\sum_{n\in\mathbb{Z}}|\bar{x_{-n}}|=\sum_n|x_n|=\|x\|.$$

Edit: A proof of the computation $$(xy)^*=y^*x^*$$. Let $$x=(x_n)_n$$ and $$y=(y_n)_n$$. We have $$xy=\big(\sum_{m\in\mathbb{Z}}x_my_{n-m}\big)_n$$, so $$(xy)^*=\big(\sum_{m\in\mathbb{Z}}\bar{x}_m\overline{y}_{-n-m}\big)_n$$ On the other hand, $$y^*=(\bar{y}_{-n})_n, x^*=(\bar{x}_{-n})_n$$, so $$y^*x^*=\bigg(\sum_{l\in\mathbb{Z}}\bar{y}_{-l}\bar{x}_{-n+l}\bigg)_n=\bigg(\sum_{\in\mathbb{Z}}\bar{x}_{-n+l}\bar{y}_{-l}\bigg)_n$$ so one should show that $$\sum_{l}\bar{x}_{-n+l}\bar{y}_{-l}=\sum_m\bar{x}_m\bar{y}_{-n-m}$$ which is clear after doing the change of variable in the left sum $$k:=-n+l$$: as $$l$$ ranges all over $$\mathbb{Z}$$, so does $$k$$ and thus $$\sum_l\bar{x}_{-n+l}\bar{y}_{-l}=\sum_k\bar{x}_k\bar{y}_{-n-k}$$ as we wanted.

• @Chengdu Yes, they are on the right way. You've almost completed the proof, all that's left is to show that $(xy)^*=y^*x^*$. You seem to be confused about this, but if you right down explicitly each step, you will see that this is straight forward: write what $xy$ is, then take the adjoint. On another line, write $y^*$, $x^*$ and write what their product is. You will see that the two things are equal Nov 16, 2021 at 10:41
• I'll add the computation in my answer. Your notation is problematic and that's the reason that you get stuck, this is a computation that is definitely easy enough for you! Nov 16, 2021 at 11:30
• @Chengdu see it Nov 16, 2021 at 11:52
• @chengdu what are you referring to? I cant understand what you mean Nov 16, 2021 at 15:03
• @chengdu yes, the other notation is not even specified Nov 16, 2021 at 15:42