Is the kernel of a Fredholm operator stable under perturbation?

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.

• Probably you want to exclude the kernel $\{0\}$ from the setup? – Hannes Aug 22 '18 at 7:10
• Yes, of course. I will mention this explicitly. – Asaf Shachar Aug 22 '18 at 7:12
• @Hannes All the kernels are assumed to have the same dimension, so if that dimension is $0$ the kernels are all $\{0\}$ and the answer is trivially yes. – Robert Israel Aug 22 '18 at 7:12
• Yes, sure, I just wanted to point out that the further setup doesn't fit to that case before someone else does it ;-) – Hannes Aug 22 '18 at 7:16
• Just a small remark: If a bounded operator has finite dimensional kernel and closed range, then it is a so-called 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. – Jochen Glueck Aug 22 '18 at 7:35

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. 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.
• You can modify $S_t$ $(t\neq 0$) so that all their kernels have the same positive dimensión. – M.González Aug 22 '18 at 10:05
• $S_t:X_1\times X_2\times Y\to X_1\times X_2\times Y$ given by $S_t(x_1,x_2,y)=(0,tx_2,y)$, $X_1, X_2$ finite dimensional. – M.González Aug 22 '18 at 10:11
• I am not sure I follow: In both your examples, $\dim(\ker S_0) \neq \dim(\ker S_t)$ for $t>0$. (while I assumed the dimension is independent of $t$). Anyway, I think that there is a chance that proposition 3.1 would suffice for my purposes (though I need to think some more about this). Thanks again for your help. – Asaf Shachar Aug 22 '18 at 10:17
• I did not carefully read your question. If you assume $dim (\ker S_0)=dim (\ker S_t)<\infty$, closed range for all the operators, and $\|S_t-S_0\|\to 0$, then the distance from $\ker S_t$ to $\ker S_0$ tends to $0$. This is a consequence of the result I mentioned plus finite dimensión. – M.González Aug 22 '18 at 10:52