TITLE: Frechet derivative of square root on positive elements in some $C^*$-algebra
QUESTION [14 upvotes]: Let $A$ - is some unital $C^*$ algebra, and $P$ is set of all strictly positive elements in $A$. We can define map $\sqrt{?} : P \to A$ which takes positive element and returns its (unique) strictly positive square root. How to evaluate its Frechet derivative?

REPLY [3 votes]: Let $P$ be the cone of strictly positive elements and let $V$ the the real linear combinations of these. $V$ is a real Banach space and $P$ is an open subset of $V$.
Consider the maps $s:P\to P, a\mapsto a^2$ and $r:P\to P,a\mapsto a^{1/2}$, you have that $s\circ r=\mathrm{id}\lvert_P=r\circ s$. It follows with the chain rule that
$$d(s\circ r)(a)=ds(r(a))\cdot dr(a) = \mathbb1\lvert_V=dr(s(a))\cdot ds(a)=d(r\circ s)(a)$$
So $dr(a)$ is the same as $ds(r(a))^{-1}$. Specifically $ds(r(a))=L(r(a))+R(r(a))$ where $L$ and $R$ are the left and right multiplication operations on $A$.
In the case of a commutative algebra the inverse becomes $\frac1{2\sqrt{a}}$, but I cannot find a nice expression in the general case, but I think this approach can be fruitful.