0

I was trying to think of some examples of $C^\ast$-algebras and I think $\ell^p$ with pointwise multiplication would be a good example. My reasoning is that if $a_n, b_n$ are in $\ell^p$ then eventually $|a_n b_n| \le |a_n|$ so this is closed with respect to multiplication.

Is this correct? The $\ast$-operation will naturally be complex conjugation. The equations $\|a^\ast\| = \|a\| $ and $\|a^\ast a\| = \|a\|^2$ seem to hold too. Am I missing anything?

1 Answers1

1

It is true that $\ell^p$ is closed with respect to multiplication, because convergent sequences are bounded. But I don't see how you get that the C$^*$-equality holds. How do you define $a^*$? If you take the most reasonable route, defining $a^*(n)=\overline{a(n)}$, then $\|a^*a\|=\|a\|^2$ would be $$ \left(\sum_n|a(n)|^{2p}\right)^{1/p}=\left(\sum_n|a(n)|^p\right)^{2/p}. $$ Why would you expect that equality to hold? For instance take $p=1$, $a=(1,1,0,\ldots)$. Then $$ \|a^*a\|=1+1=2, \ \ \|a\|^2=(1+1)^2=4. $$

Martin Argerami
  • 205,756
  • Thank you. I thought that in the sum I get the sum of squares didn't see that the equality only holds for a single complex number but obviously not for a sum. It is surprising thought that not even $\ell^2$ is a $C^\ast$-algebra. –  Oct 31 '14 at 22:43
  • If you think about it, it is not that surprising. The one that is a C$^$-algebra is $\ell^\infty$. Note that any abelian C$^$-algebra is $C_0(X)$ for some locally compact $X$, and any finite-dimensional one is a direct sum of full matrix algebras. – Martin Argerami Oct 31 '14 at 23:09