Let $A$ be a unital C*-algebra. If $h \in A$ is self adjoint, then $e^{ih}$ is unitary.

This proof is in a lecture note, which I am not able to understand. They prove it in the following way:

For any $f \in \mathcal{C}(\mathrm{sp}(h))$, $f(h)^* =\bar{f}(h) =f^{-1}(h)$.

I do not understand where the equality follow from. Why is $f$ invertible, and what if $f$ has zero in its range?

Edit: By $e^{ih}$ I mean the element in $A$ we get by functional calculas, $C^*(1,h) \cong \mathcal{C}(sp(h))$. For $e^{ix} \in \mathcal{C}(sp(h))$ we get an element $e^{ih} \in C^*(1,h) \subseteq A$.


That's not true in general, but it is if you consider $f(t)=e^{it}$. For this particular $f$ you have $\bar f=f^{-1}$.

A proof involving less theory is to use that the adjoint operation is continuous: then $$ (e^{ih})^*=\left(\sum_{k=0}^\infty \frac{i^k h^k}{k!}\right)^* =\sum_{k=0}^\infty \frac{(-i)^k h^k}{k!}=e^{-ih}. $$ Then use that $e^{a+b}=e^ae^b$ when $ab=ba$ to get $$ e^{ih}e^{-ih}=e^{ih-ih}=I. $$

  • $\begingroup$ $\bar f=f^{-1}$ and $e^{ih}$ are slightly ambiguous, with $f(h) = \exp(ih)$, in general $f(h)^{-1} = f(-h)$ and when $h = h^*$ then $f(h)^* = \exp((ih)^*) = f(-h)$ $\endgroup$ – reuns Oct 10 '18 at 19:02
  • $\begingroup$ @Argerami In infinite sum can we distribute the * on each elemnt? $\endgroup$ – Arindam Oct 11 '18 at 15:21
  • $\begingroup$ The adjoint is continuous, so yes. $\endgroup$ – user42761 Oct 11 '18 at 15:36

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