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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}) $

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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.

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    $\begingroup$ @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 $\endgroup$ Nov 16, 2021 at 10:41
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    $\begingroup$ 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! $\endgroup$ Nov 16, 2021 at 11:30
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    $\begingroup$ @Chengdu see it $\endgroup$ Nov 16, 2021 at 11:52
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    $\begingroup$ @chengdu what are you referring to? I cant understand what you mean $\endgroup$ Nov 16, 2021 at 15:03
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    $\begingroup$ @chengdu yes, the other notation is not even specified $\endgroup$ Nov 16, 2021 at 15:42

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