1

Let $A=\mathbb C[x] $ prove there is no norm on A in which it makes a C* algebra.

i think this is true because the spec(a) is infinity for any $a\in A$ ? but im not sure how to prove it.

I did try verifying the axioms for a norm but im not sure what a* in this case

Martin Argerami
  • 205,756
Faust
  • 5,669
  • First sentence doesn't make sense. Looks unfinished – DanielWainfleet Sep 20 '18 at 02:04
  • The idea is no norm exists that makes A into a C* algebra – Faust Sep 20 '18 at 02:14
  • I don't know of any norm on $\Bbb C[x]$ maybe that's why I upvoted your question; but can you give some definitions/links to how norms on $C[x]$ are supposed to work? – Robert Lewis Sep 20 '18 at 02:21
  • Supposed to prove that there does not exist a norm on the set A so that it is a C* algebra. So no I can't cause there isn't one... – Faust Sep 20 '18 at 02:22
  • It should break one of the rules of being a norm but im not sure which – Faust Sep 20 '18 at 02:25
  • Suppose we just wanted a plain old *algebra norm* on $\Bbb C[x]$; or even just a really plain old *Banach space* norm. How would that look? I've never see such a thing . . . – Robert Lewis Sep 20 '18 at 03:01
  • By the way, where did you get this question? – Robert Lewis Sep 20 '18 at 03:02
  • OK, here's an issue: what happens when we try to make $\Bbb C[x]$ complete in some norm? Will it still contain only polynomials? – Robert Lewis Sep 20 '18 at 03:23
  • 2
    @RobertLewis. An infinite-dimensional Banach space cannot have a countable Hamel (vector-space) basis. – DanielWainfleet Sep 20 '18 at 04:09
  • @RobertLewis you can consider B the field of quotients of A but this is also not a C* algebra because spec(a) is empty so r(a) is undefined, so no you cannot canonically complete it. – Faust Sep 20 '18 at 05:57
  • @Faust: how do you define $\text{spec}(A)$ and $r(A)$? – Robert Lewis Sep 20 '18 at 05:59
  • Should be clear in the answer now how they are defined for $a\in A$ you cant define them for the whole space i dont think. @RobertLewis – Faust Sep 20 '18 at 06:03
  • 2
    @RobertLewis: There are tons of algebra norms on $\mathbb{C}[x]$. Just fix some infinite bounded subset $X\subset\mathbb{C}$ and take the sup norm for polynomials as functions on $X$, for instance. Of course, these norms are not complete. – Eric Wofsey Sep 20 '18 at 06:04
  • 1
    @EricWofsey where were you for the last 3 hours that ive been trying to figure this out ;P – Faust Sep 20 '18 at 06:04

1 Answers1

1

First notice that for any $a\in A $ the $spec(a)=\{\lambda \in \mathbb{C}| \lambda 1 - a \space\text{is not invertable}\} $ and $r(a)=\sup\{|\lambda| s.t \lambda \in spec(a) \}$

Theorem: if a $*$-algebra posses a norm in which it is a C*algebra, then it possesses only one such norm.

Theorem: If a is a self adjiont element of a unital C*algebra A then $||a||=r(a)$ but $r(a) = \infty $ for everything which is clearly not finite.

Hence there is no norm on A that makes it into a C* algebra.

Faust
  • 5,669