Let ‎$‎A$ ‎be a‎ ‎unital ‎$C^*‎$‎-algebra.‎ Let $A$ ‎be a‎ ‎unital ‎‎$C^*$-algebra.‎
‎
(a)‎ If ‎$‎‎a,b\in A$, ‎show ‎that ‎the ‎map ‎‎‎$‎‎f:\mathbb{C}\to A$ defined by $f(\lambda)=e^{-i\lambda b}ae^{i\lambda b}$ ‎is ‎differentible ‎and ‎that ‎‎$‎‎f^\prime (0)=i(ba-ab)$‎‎
(b) Let $X$ be a‎ ‎closed ‎vector ‎subspace ‎of $A$‎ which ‎is ‎unitarily ‎invariant ‎in ‎the ‎sense ‎that ‎‎$‎‎uXu^*‎\subseteq X‎$ ‎for ‎all ‎unitaries ‎‎$‎‎u$ ‎of ‎‎$‎‎A$. ‎Show ‎that ‎‎$‎‎ba-ab\in X$‎ if ‎$‎‎a\in X$ and ‎$‎‎b\in A$‎.‎
(c) Deduce that the closed linear span $X$ ‎of ‎the ‎projections ‎in $A$ has ‎the ‎property ‎that ‎‎$‎‎a\in X$ ‎and ‎‎$‎‎b\in A$ ‎implies ‎that ‎‎$‎‎ba-ab\in X$.‎
 A: Hint (a):
Note you have a product rule for differentiable maps $\mathbb C\to A$, $\frac d{dt}(f(t)\cdot g(t))=\frac{df(t)}{dt}g(t)+f(t)\frac{dg(t)}{dt}$. This product rule can be derived in the same way as the product rule for maps $\mathbb R\to\mathbb R$). The first part is then reduced to finding out what $\frac{d\exp({i\,t b})}{dt}$ is.
To find that you can use the chain rule:
$$\frac{d}{dt}\exp({i\,tb})=(D\exp)(ibt)\cdot (ib)$$
here $D\exp$ is the derivative of the exponential function. What is this derivative? Here you need to look back into your analysis course to remember how to derive power series in Banach-algebras inside of their radius of convergence. Doing this will give you $D\exp=\exp$. If you plug everything in you will see the result of (a).
Hint (b):
If $b\in A$ then $e^{i\lambda b}$ is a unitary for $\lambda\in\mathbb R$. It follows from $a\in X$ that $\gamma(\lambda):=e^{i\lambda b}ae^{-i\lambda b}$ lies in $X$ for all $\lambda\in\mathbb R$. This is a path that lies entirely in the closed vector subspace $X$. Then the differential of the map $\gamma:\mathbb R\to X$ is always element of $X$, but this differential evaluated at $0$ is $i(ba-ab)$.
Hint (c):
The span of projections is unitarily invariant, since $uPu^*$ is again a projection for $P$ a projection and $u$ unitary. If you verify that the closure of this is also unitarily invariant you can use (b) to get the result.
