This is a follow-up of [this question](https://math.stackexchange.com/questions/2889556/continuity-of-the-kernel-of-bounded-operators-under-perturbation).



> 
> In a nutshell: Does the kernel of a bounded operator change "nicely" with the operator?
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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](https://en.wikipedia.org/wiki/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?
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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](https://math.stackexchange.com/a/2889584/104576), 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.



Is the kernel of a Fredholm operator stable under perturbation?
The kernel of a Fredholm operator is not continuous with respect to small norm perturbations: For $t\geq 0$, consider the operator $S\_t:X\times Y \to X\times Y$ defined by $S(x,y)=(tx,y)$ where $X,Y$ are Banach spaces and $X$ is finite dimensional. 


However, when the operator $T:X\to Y$ has closed range, you have semicontinuity in some sense. See Proposition 3.1 
[here.](https://www.researchgate.net/publication/243033124_The_gap_between_subspaces_and_perturbation_of_non-semi-Fredholm_operators) And from this result (and finite dimension for the kernels) you can get a positive answer to your question.


When the operator $T:X\to Y$ has non-closed range and it is injective, you can find a compact perturbation $K:X\to Y$ with arbitrarily small norm so that the kernel of $T+K$ is infinite dimensional. So there is no semicontinuity. 


