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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.

Thanks!

t.b.
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Hui Yu
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    Very closely related: http://math.stackexchange.com/questions/2924/the-gelfand-transformation-on-ell1-mathbb-z-is-not-isometric-do-you-have. Also related: http://math.stackexchange.com/questions/3020/does-the-gelfand-transformation-on-ell1-mathbb-z-possess-a-continuous-inve – Jonas Meyer Dec 30 '11 at 18:11

1 Answers1

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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.

t.b.
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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 06:48
  • But $x^\ast = x$... (and that was a typo which I fixed immediately after posting) – t.b. Dec 30 '11 at 06:51
  • Oh! I see. But I think $x^{}x(1)=x_{1}x_{0}+x_{0}x_{1}=2$ and $x^{}x(2)=1$ and the equation holds for this example. – Hui Yu Dec 30 '11 at 06:56
  • You're right, I miscalculated. The example should be fixed now. – t.b. Dec 30 '11 at 07:17
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    @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 –  Dec 30 '11 at 07:45
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    @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
  • @t.b. Good point (could also use Grothendieck's inequality). I think the result I had in mind was that $\ell^1({\mathbb Z})$ is not even isomorphic as a Banach algebra to any closed subalgebra of $B(H)$ –  Dec 30 '11 at 17:11
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    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 07 '12 at 06:43
  • @t.b. is there any way to construct such a counter example rather than cramming it ?? – Devendra Singh Rana Jan 18 '18 at 11:03
  • It has been a while ago, but do you mind telling me how you computed the norm $||x^{*} \star x||_{1} = 5$ – Chengdu Nov 16 '21 at 16:43