Differentiable Path of Operators and their Inverses Let $\mathcal{H}$ be a separable Hilbert space. Consider a differentiable map $\mathbb{R} \rightarrow \mathcal{B}(\mathcal{H}), t \mapsto A(t)$, where $\mathcal{B}(\mathcal{H})$ is the space of bounded linear operators from $\mathcal{H}$ into itself with respect to the operator norm (by differentiable I mean differentiable in the Frechet-sense). Furthermore, assume, for each $t$, that $A(t)$ has the following properties: 
(1) for each $t$ there is a orthogonal decomposition of $\mathcal{H} = \mathcal{H}_{t}^{1} \oplus \mathcal{H}_{t}^{2}$, where $\mathcal{H}_{t}^{1}$ is a finite dimensional subspace of $\mathcal{H}$; (2) $A(t)|_{\mathcal{H}_{t}^{1}} \equiv 0$, $A(t)$ are selfadjoint and $\dim{(\mathcal{H}_{t}^{1})} = n > 1$ (here $n$ is independent of $t$, hence for all $t$, $\mathcal{H}_{t}^{1}$ have the same dimension greater than $1$); (3) $A(t)|_{\mathcal{H}_{t}^{2}}:\mathcal{H}_{t}^{2} \rightarrow \mathcal{H}_{t}^{2}$ is an isomorphism; (4) The $(A(t)|_{\mathcal{H}_{t}^{2}})^{-1} : \mathcal{H}_{t}^{2} \rightarrow \mathcal{H}_{t}^{2}$ are bounded operators; (5) $(A(t)|_{\mathcal{H}_{t}^{2}})^{-1}A(t) : \mathcal{H} \rightarrow \mathcal{H}_{t}^{2}$ is an orthogonal projection.
Extend now the oprators $(A(t)|_{\mathcal{H}_{t}^{2}})^{-1}$ to the whole $\mathcal{H}$ by: $A(t)^{-1}x = 0$ if $x \in \mathcal{H}_{t}^{1}$ and $A(t)^{-1}|_{\mathcal{H}_{t}^{2}} = (A(t)|_{\mathcal{H}_{t}^{2}})^{-1}$.
Is it true that then the map $\mathbb{R} \rightarrow \mathcal{B}(\mathcal{H}), t \mapsto A(t)^{-1}$ is Frechet-differentiable? I'm sure I've mentioned too many assumptions. I know that if the operators $A(t)$ are invertible (on the whole space) then the map $t \mapsto A(t)^{-1}$ is Frechet-differentiable. But what about in this case?
Clark
 A: Yes, the map you describe is differentiable (in the Fréchet sense or the difference-quotient sense; it doesn't make any difference when the domain is ${\mathbb R}$). Here is a proof using the Dunford-Taylor integral to construct the operator you call $A(t)^{-1}$:


*

*Let $J\subset {\mathbb R}$ be any compact interval. By the hypotheses, whenever $\epsilon\ne0$ and $|\epsilon|$ is sufficiently small, $A(t)+\epsilon I$ has a bounded inverse for all $t\in J$.
Hence $0$ is the only
point in the spectrum of $A(t)$ inside a circle $\Gamma_\epsilon$ of some radius $\epsilon>0$, and is an eigenvalue of finite type.

*$A(t)$ is uniformly bounded on $J$, so its spectrum is contained 
inside a circle $\Gamma_R$ of sufficiently large radius $R$.

*Let $\Gamma$ be the boundary of the annulus $\epsilon<|z|<R$, tracing $\Gamma_R$ in the positive (counterclockwise) direction and tracing $\Gamma_\epsilon$ in the negative direction. The operator
$$
{\mathcal P}(t) = \frac1{2\pi i} \int_\Gamma (zI-A(t))^{-1}\,dz
$$
is the spectral projection of ${\mathcal H}$ onto ${\mathcal H}^2_t={\rm range\,}A(t)$. (${\mathcal P}(t)$ commutes with $A(t)$).
Moreover,
we have 
$$
A(t) = \frac1{2\pi i} \int_\Gamma (zI-A(t))^{-1}z\,dz
$$
and the operator
$$
B(t) = \frac1{2\pi i} \int_\Gamma (zI-A(t))^{-1}z^{-1}\,dz
$$
has the property that $B(t)A(t)=A(t)B(t)={\mathcal P}(t)$. That is, $B(t)$ is the operator you call $A(t)^{-1}$.  One can easily justify differentiation under the integral sign and infer that $B(t)$ is differentiable.
