This is a follow-up of this question.

In a nutshell: Does the kernel of a bounded operator change "nicely" with the operator?

Let $(X,\| \cdot \|)$ be an infinite-dimensional normed vector space.

Let $A_t $ be a continuous family of Fredholm operators of index $0$ on $X$- that is we have a continuous map $ (-\delta,\delta) \to (\text{Hom}(X,X),\|\cdot\|_{op})$, given by $ t \to A_t$, such that each $A_t$ is Fredholm of index $0$.

Suppose that all the kernels $\ker A_t$ are **finite**-dimensional and have the same **positive** dimension.

Let $S$ be the **unit sphere** of $(X,\|\cdot \|)$. Define $S_t=\ker A_t \cap S$. Set
$$ d(\ker A_t,\ker A_0):=d_H(S_t,S_0),$$
where $d_H$ is the Hausdorff distance of $S_t,S_0$ inside $(X,\|\cdot \|)$.

Let $\epsilon >0$. Does there exist $\delta>0$ such that for every $t <\delta$, $ d(\ker A_t,\ker A_0)<\epsilon$ holds?

Stating it explicitly, I ask whether for every $v_t \in S_t$ there exist $v_0 \in S_0$ such that $||v_t-v_0||<\epsilon$. (and vice versa, since the Hausdorff distance is "symmetric". However, I am also interested in the "one-way closedness" described above).

*Comment:* In this answer, there is a counter-example when $X=\ell^2$, and $A_0$ has a non-closed image. In the construction there, $\ker(A_{\frac{1}{n}})=\text{span}\{e_n\}$ so the kernels "run away".
I hope that under the additional "Fredholm" assumption, there would be a chance for a positive answer.

I looked at Kato's book "Perturbation Theory for Linear Operators" but didn't find anything which seemed relevant.

upper semi-Fredholm operator. If for some reason, say in a concrete application, you know that $0$ is in the topological boundary of the spectrum, then it follows that the operator is even a Fredholm operator with Fredholm index $0$; maybe this is helpful to prove the desired stability result in this special case. $\endgroup$ – Jochen Glueck Aug 22 '18 at 7:351more comments