Openness of invertibility in Fréchet spaces for families parameterized by compact spaces

Question

Consider the following setup. Let $K$ be a compact topological space, $X$ a Fréchet space and $T:K \times X \to X$ a continuous family of linear maps (i.e. $T$ is a continuous map and $T_k \equiv T(k, \cdot): X \to X$ is a linear for all all $k \in K$).

I would like to have a result similar to the following statement:

For every $k_0$ with $T_{k_0} = \mathrm{id}_X$, there exist an open neighborhood $U$ of $k_0$ in $K$ such that $T_k: X \to X$ is a topological isomorphism for every $k \in U$ and the map $T^{-1}: U \times X \to X$ defined by $(k, x) \mapsto T_k^{-1}(x)$ is continuous.

Since the space of invertible operators is not open in the space of all linear operators on a Fréchet space, such a result may be a bit much to ask but I was hoping that compactness of $K$ may help in this situation. I'm grateful for every pointers to results that go into this direction.

Answer 1

The statement is indeed overoptimistic.

As a counterexample, consider the continuous family $T: [0,1] \times C^\infty([0, 1]) \to C^\infty([0, 1])$ of linear differential operators defined by \begin{equation} T_r (g) (x) = g - r x g' \end{equation} for $r \in [0,1]$ and $ g \in C^\infty([0, 1]) $. Clearly, $ T_0 $ is the identity operator on $ C^\infty([0, 1]) $. On the other hand, for every $ n \in \mathbb{N} $, the operator $ T_{\frac{1}{n}} $ annihilates the function $ g_n(x) = x^n $. Hence, $ T_r $ fails to be injective for arbitrarily small $ r $.