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
