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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.


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Very closely related:…. Also related:… – Jonas Meyer Dec 30 '11 at 18:11
up vote 19 down vote accepted

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.

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Hi! This is not what I mean. The condition is $\|x^{*}*x\|=\|x\|^2$, not $\|x*x\|=\|x\|^2$. – Hui Yu Dec 30 '11 at 6:48
But $x^\ast = x$... (and that was a typo which I fixed immediately after posting) – t.b. Dec 30 '11 at 6:51
@t.b. As you may know, one can prove something stronger: there is no Banach algebra isomorphism from $\ell^1({\mathbb Z})$ onto any $C(K)$ (we forget the involution and allow isomorphism rather than isometry). The only proofs I'm aware of require substantial machinery though – user16299 Dec 30 '11 at 7:45
@Yemon: Sorry I misread you when I first posted my request for a reference. Since $\ell^1$ is a Schur space and since $C(K)$ is never weakly sequentially complete if $K$ is infinite they aren't even isomorphic as Banach spaces. – t.b. Dec 30 '11 at 16:15
A way to see that $\ell^1(\mathbb Z)$, with the convolution multiplication, is not a C*-algebra with respect to any Banach algebra norm & involution, is to note that the Gelfand transform depends only on the algebra, and it is not onto (because there are continuous functions on the circle with non-absolutely convergent Fourier series). (The Gelfand map also has dense image, so if $\ell^1$ were given the norm making it isometric, then it would not be complete. This is related to Rasmus's questions linked in a comment on the question above.) – Jonas Meyer Jan 7 '12 at 6:43

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