# If $h$ is self-adjoint then $e^{ih}$ is unitary

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 $$e^{ih}e^{-ih}=e^{ih-ih}=I.$$
• $\bar f=f^{-1}$ and $e^{ih}$ are slightly ambiguous, with $f(h) = \exp(ih)$, in general $f(h)^{-1} = f(-h)$ and when $h = h^*$ then $f(h)^* = \exp((ih)^*) = f(-h)$ – reuns Oct 10 '18 at 19:02