Exponential map in $C^{*}$-algebra and unitary invariance Let $A$ be a unital $C^{*}$-algebra. 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 $A$. I want to show that $ba-ab\in X$ whenever $a\in X$ and $b\in A$.
A previous part of the problem was to show that the map $f:\mathbb{C}\rightarrow A,f(\lambda)=e^{i\lambda b}ae^{-i\lambda b}$, where $a,b\in A$, is differentiable and $f'(0)=i(ba-ab)$. I've done this part but I can't get the connection between this and the problem above. The expression I got for the derivative is $f'(\lambda)=ie^{i\lambda b}bae^{-i\lambda b}-ie^{i\lambda b}abe^{-i\lambda b}$. The main problem is that I don't see a unitary that allows me to use the assumption on $X$.
 A: Good question, +1!  Interesting . . .
The way I see it, the main difficulty here is that, for general $b \in A$ and $\lambda \in \Bbb C$, $e^{i \lambda b}$ is not a unitary element of $A$; thus we have no assurance that $e^{i \lambda b} a (e^{i \lambda b})^* = e^{i \lambda b} a e^{-i \bar \lambda b^*} \in X$.  Indeed, we can't even conclude that $e^{i \lambda b} e^{-i \bar \lambda b^*} = e^{i \lambda b -i \bar \lambda b^*}$ without for example the assurance that $bb^* - b^*b = [b, b^*] = 0$, i. e., that $b$ is a normal element of $A$.  If, however, we restrict $b$ to being Hermitian, i. e.,   assume that $b^* = b$, and $\lambda$ to being real, $\lambda \in \Bbb R$, then progress can be made, as follows:
For $b \in A$ such that $b^* = b$ and $\lambda \in \Bbb R$, it is evident that $e^{i \lambda b}$ is a unitary in $A$, since in this case $(e^{i \lambda b})^* = e^{-i \lambda b^*} = e^{-i \lambda b}$, whence $e^{i \lambda b} (e^{i \lambda b})^* =1$.  Thus we have, by hypothesis, the entire path $f(\lambda) = e^{i \lambda b} a e^{-i \lambda b} \in X$ for all $\lambda \in \Bbb R$.  Furthermore, $f'(\lambda) \in X$ for all $\lambda \in \Bbb R$ as well.  This follows since for $\lambda \ne \mu$ distinct elements of $\Bbb R$, then
$(f(\lambda) -f(\mu)) / (\lambda - \mu) \in X \tag{1}$
by the linearity of $X$; we also have that, if it exists,
$f'(\lambda) = \lim_{\mu \to \lambda}((f(\lambda) - f(\mu)) / (\lambda - \mu)) \in X, \tag{2}$
since $X$ is closed.  The remarks of our OP learner affirm that this limit does exist and that additionally
$f'(\lambda)=ie^{i\lambda b}bae^{-i\lambda b}-ie^{i\lambda b}abe^{-i\lambda b}; \tag{3}$
more immediately, both assertions follow from the Leibniz rule for the derivative of a product and the definition of the exponential function on $A$.  In any event, by (2) and (3) we have, upon taking $\lambda = 0$,
$i[b, a] = iba-iab = f'(0) \in X, \tag{4}$
which in turn shows that
$[b, a] \in X, \tag{5}$
again by the linearity of $X$.  Thus we have established that for all $a \in X$ and $b = b^* \in A$, $[b, a] \in X$; the case of $b$ self-adjoint or Hermitian is thus resolved in the affirmative.
Next, we consider the case of $b$ skew-adjoint or skew-Hermitian, that is, $b^* = -b$.  We observe that, for such $b$, $ib$ is in fact self adjoint, for
$(ib)^* = -ib^* = -i(-b) = ib, \tag{6}$
and similarly, for $b = b^*$, $ib$ is skew:
$(ib)^* = -ib^* = -ib. \tag{7}$
We exploit (6) for skew-adjoint $b$:  for all $\lambda \in \Bbb R$, we have
$f(\lambda) = e^{i \lambda (ib)} a e^{-i \lambda (ib)} \in X, \tag{8}$
and
$f'(\lambda)=ie^{i\lambda (ib)}(ib)ae^{-i\lambda (ib)}-ie^{i\lambda (ib)}a(ib)e^{-i\lambda (ib)} \in X; \tag{9}$
again taking $\lambda = 0$ we see that
$-[b, a] = i[ib, a] \in X, \tag{10}$
whence
$[b, a] \in X, \tag{11}$
again by the linearity of the subspace $X$.  We now have $[b, a] \in X$ for $a \in X$ and
all $b^* = b \in A$ and $b^* = -b \in A$.
Finally, for arbitrary $b \in A$, we set
$b_+ = (1 / 2)(b + b^*), \tag{12}$
and
$b_- = (1 / 2)(b - b^*); \tag{13}$
it is manifestly evident that
$b_+^* = b_+, \; b_-^* = -b_-, \tag{14}$
and
$b = b_+ + b_-. \tag{15}$
Also,
$[b, a] = ba - ab = (b_+ + b_-)a - a(b_+ + b_-) = [b_+, a] + [b_-, a] \in X, \tag{16}$
since $X$ is closed under addition.  We have thus proved that $[b, a] \in X$ for all $a \in X$, $b \in A$, as per request.  QED.
Hope this helps.  Yuletide Joy to One an All,
and as always, especially on these cold, dark days,
Fiat Lux!!!
