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 bounded linear maps $X \to X$ with closed images, that is we have a continuous map $ (-\delta,\delta) \to (\text{Hom}(X,X),\|\cdot\|_{op})$, given by $ t \to A_t$, such that $\text{Image}(A_t)$ is closed in $X$.

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 assumption of closed image, 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.

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.

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 bounded linear maps $X \to X$ with closed images, that is we have a continuous map $ (-\delta,\delta) \to (\text{Hom}(X,X),\|\cdot\|_{op})$, given by $ t \to A_t$, such that $\text{Image}(A_t)$ is closed in $X$.

Suppose that all the kernels $\ker A_t$ are finite-dimensional and have the same 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 assumption of closed image, 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.

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 bounded linear maps $X \to X$ with closed images, that is we have a continuous map $ (-\delta,\delta) \to (\text{Hom}(X,X),\|\cdot\|_{op})$, given by $ t \to A_t$, such that $\text{Image}(A_t)$ is closed in $X$.

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 assumption of closed image, 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.