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Let $C^{n}[0, 1]$ be the space of all $n$-times continously differentiable functions endowed with the norm $$\|f\| = \sum_{k=0}^n \frac{\sup_{t \in [0, 1]}{|f^{(k)}(t)|}}{k!}$$

Let $*$ be the involution map that acts as follows: $f^{*}(t) = \overline{f(t)}$.

I would like to show that $A = (C^{n}[0, 1], \| \cdot \|, *)$ is not a C*-algebra.

So the first step is to show that there exists a sequence of self-adjoint (w.r.t. to the involution) functions of a unit norm, e.g. $\|f_n\| = 1$ so that $ f^2_n \rightarrow 0$ (if so, then the C*-identity would fail).

Apparently, this example is a sort of a classical one for the first course on functional analysis, but i bumped into troubles while trying to invent something suitable. What are the possible approaches/examples?

Update Looks as if $f_{n} = \frac{x^{n}}{n}$ works, since $f^{2}_{n} = \frac{x^{2n}}{n^{2}}$ converges to $0$ as $n \rightarrow + \infty$, but $||f_{n}|| = \frac{1}{n} + 1 \rightarrow 1$

hyperkahler
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    One slightly long winded proof is to show that the Gelfand transform is not an isomorphism, which it must be for commutative unital $C^{\ast}$-algebras. See https://math.stackexchange.com/questions/1855607/is-the-range-of-the-gelfand-transform-closed – Prahlad Vaidyanathan May 22 '17 at 12:59
  • @PrahladVaidyanathan Very nice and elegant approach, thanks! – hyperkahler May 23 '17 at 03:50

1 Answers1

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Let $n=1$ and $f(x)=x+1$. $$ \|f\|^2=(2+1)^2=9, $$ while $$ \|f^2\|=4+4=8 $$ So $\|f^*f\|=\|f^2\|\ne\|f\|^2$ and the C$^*$-identity fails.

Martin Argerami
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