Show that this parameterized family of operators is right-differentiable in its parameter Let $H$ be a $\mathbb R$-Hilbert space, $\kappa^{(i)}$ be a linear self-adjoint contraction (i.e. a bounded operator with operator norm at most $1$) on $H$ and $\lambda\in(0,1)$. It's easy to see that $$Q_\alpha:=(1-\alpha)\kappa^{(1)}+\alpha\kappa^{(2)}$$ is again a contraction on $H$ and hence the spectrum $\sigma(Q_\alpha)$ of $Q_\alpha$ is contained in $[-1,1]$. Thus, $$A_\lambda(\alpha):=(1-\lambda Q_\alpha)^{-1}(1+\lambda Q_\alpha)$$ is well-defined for all $\alpha\in[0,1]$.

I guess it's trivial, but for the moment I can't see why $$B_\lambda(\alpha):=\lim_{t\to0+}\frac{A_\lambda(\alpha+t)-A_\lambda(\alpha)}t\tag1$$ exists for all $\alpha\in[0,1)$. In particular, with respect to which operator topology do we need to understand $(1)$? The strong or even the uniform operator topology?

 A: Let $GL(H)$ denote the invertible elements of $B(H)$. Most of the work is understanding why the map $\iota: GL(H)\to GL(H)$, $x\mapsto x^{-1}$ is smooth. These statements are classical statements about invertible elements of Banach algebras, the relevant ingredient here is the von Neumann series:

Suppose $\|a\|<1$, then
  $$(1-a)\sum_{n=0}^\infty a^n = \sum_{n=0}^\infty a^n - \sum_{n=0}^\infty a^{n+1}=1$$
  and $(1-a)$ is invertible. We will use this to show that $GL(H)$ is open and to calculate the differential of $\iota$.

To see that $GL(H)$ is open consider an $x\in GL(H)$ then look at the ball $B_{\|x^{-1}\|^{-1}}(x)$. If $y$ is in that ball then
$$\|1-x^{-1}y\|=\|x^{-1}(x-y)\|<\|x^{-1}\|\,\|x^{-1}\|^{-1}=1$$
hence the von Neumann series $\sum_{n=0}^\infty (1-x^{-1}y)^{n}$ is the inverse of $1-(1-x^{-1}y)=x^{-1}y$. As $x$ and $x^{-1}y$ are invertible, so to is $y$. 
The next step is to see that the map $\iota:GL(H)\to GL(H)$, $x\mapsto x^{-1}$ is differentiable. First note that (for $h$ small enough):
$$(x+h)^{-1}-x^{-1}= (x(1+x^{-1}h))^{-1}-x^{-1} = ((1+x^{-1}h)^{-1}-1)x^{-1},$$
in particular we may evaluate $1+x^{-1}h = \sum_{n=0}^\infty (-1)^n (x^{-1}h)^n$ for $h$ small enough. Plugging that into the equation we just got retrieves:
$$\frac{\iota(x+h)-\iota(x)}{\|h\|} = - \frac{x^{-1}hx^{-1}}{\|h\|}+ o(h),$$
from which you may calculate that the differential of $\iota$ at $x$ exists and is equal to the map $h\mapsto -x^{-1}hx^{-1}$.
Finally we apply this to our problem. $\|Q_\alpha\|≤1$ for $\alpha\in[0,1]$ and the norm varies continuously with $\alpha$, hence there is some $\epsilon$ so that $\|Q_\alpha\|<\frac1\lambda$ for all $\alpha\in[-\epsilon,1]$ and a fixed $\lambda\in(0,1)$. Thus $1-\lambda Q_\alpha$ is invertible for all $\alpha\in [-\epsilon,1]$ and since $Q_\alpha$ is actually smooth in $\alpha$, the map $(1-\lambda Q_\alpha)^{-1}$ is also smooth in $\alpha$ (here $\alpha \in (-\epsilon,1)$).
Then the product $(1-\lambda Q_\alpha)^{-1}(1+\lambda Q_\alpha)$ is smooth on $(-\epsilon, 1)$, in particular smooth at $0$. One may evaluate the differential:
$$d_\alpha = (1-\lambda Q_\alpha)^{-1} \lambda(\kappa^{(2)}-\kappa^{(1)})(1-\lambda Q_\alpha)^{-1}(1+\lambda Q_\alpha) +(1-\lambda Q_\alpha)^{-1} \lambda (\kappa^{(2)}-\kappa^{(1)})$$
(One remark that might still need to be made, is that if $f, g :U\to B(H)$ are differentiable, then $f\cdot g$ is differentiable with differential $d_x(f\cdot g)[h]= d_xf[h]\cdot g(x) + f(x)\cdot d_xg[h]$, this calculation is the same as the usual product rule.)
