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I would like to solve the following exercise but there is something I am not clear about:

Let $A$ be the Banach algebra of $C^1([0,1])$ endowed with the norm $\|f\|=\|f\|_\infty + \|f'\|_\infty$. Show that the Gelfand representation is not surjective.

I believe the Gelfand representation is the map $f \mapsto \widehat{f}$ where $\widehat{f}$ is the map $\Omega (A) \to \mathbb C$ , $\tau \mapsto \tau (f)$.

Is this so?

If it is then I am not clear how it is possible that not every map $\Omega (A) \to \mathbb C$ is an evaluation map (this is, as I understand, what the exercise is saying)

Or can one use a non zero constant function $\Omega (A) \to \mathbb C$ as a counterexample? For some reason I think maps $\Omega (A) \to \mathbb C$ in this exercise are understood to be linear.

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  • $\begingroup$ No, the maps $\Omega(A)\to \mathbb{C}$ are only supposed continuous. The Gelfand representation is a map $A \mapsto C(\Omega(A),\mathbb{C})$, where $\Omega(A)$ carries the subspace topology of the weak$^\ast$ topology. If $A$ is a commutative $C^\ast$-algebra, the Gelfand representation is an isometric isomorphism. $\endgroup$ – Daniel Fischer Jun 28 '14 at 10:49
  • $\begingroup$ To elaborate on what Daniel said, you want to prove that the Gelfand representation is not an isometry, since this is a commutative $C^*$ algebra. $\endgroup$ – Ukhrir Jun 28 '14 at 10:52
  • $\begingroup$ @DanielFischer So then it should work for a counterexample to pick any constant map $\Omega (A) \to \mathbb C$ and show that no $f$ can map to it? $\endgroup$ – Student Jun 28 '14 at 11:16
  • $\begingroup$ No, constant won't do, since the unit maps to the constant function $1$. Find a model for $\Omega(A)$ [since $A$ is generated by $p(x) = x$, it suffices to know $\tau(p)$ for every $\tau\in\Omega(A)$ to know $\tau$]. Then it's pretty obvious what the Gelfand map is, and that the image is a proper subalgebra of $C(\Omega(A))$. $\endgroup$ – Daniel Fischer Jun 28 '14 at 11:28
  • $\begingroup$ @DanielFischer I had previously shown that $\Omega (A) \cong [0,1]$ but I wasn't really sure what to do then. Now it seems to me that $[0,1]^\ast$ is the set of all continuous linear (and therefore differentiable) functions $[0,1]\to \mathbb C$. So any non linear function should do for a counterexample? $\endgroup$ – Student Jun 28 '14 at 12:23

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