In a nutshell: Does the kernel of a bounded operator change "nicely" with the operator?
The details:
Let $(X,\| \|)$ be an infinite-dimensional real normed space. Let $A_t $ be a continuous family of bounded linear maps $X \to X$, that is we have a continuous map
$ (-\delta,\delta) \to \text{Hom}(X,X)$, given by $ t \to A_t$. (I consider $\text{Hom}(X,X)$ with the operator norm).
Suppose that $\dim(\ker A_t)=r$ for some finite $r$, for every $t$. (All the kernels are finite dimensional of the same dimension).
Let $S$ be the unit sphere of $(X,\| \|)$. 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 $(X,\| \|)$.
Let $\epsilon >0$. Does there exist $\delta_0>0$ such that for every $\delta <\delta_0$, $ 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).
I ask here if $A_{t}$ being close to $A_0$ implies $\ker A_t$ is close to $\ker A_0$. If $X$ were finite-dimensional we could ask directly whether the map $t \to \ker A_t$ is continuous as a map into the corresponding Grassmannian, which I guess should be sort of equivalent to my formulation with the unit spheres.
I think that for the finite-dimensional case, such a result needs to be true by standard perturbation theory of matrices, but I am not sure.
Best Answer
I don't think that this is possible. Let us consider $X = \ell^2$. First, we consider $t > 0$ of the form $t = 1/n$, $n \in \mathbb N$. We define $a_0, a_{1/n} \in \ell^\infty$ via $$(a_0)_1 = 0, \quad (a_0)_i = 1/i, \text{ for other $i$}$$ and $$(a_{1/n})_1 = 1/n, \quad (a_{1/n})_n = 0, \quad (a_{1/n})_i = 1/i \text{ for other $i$}.$$ Now, let $A_0, A_{1/n}$ be the multiplication operators on $X$ associated with $a_0$ and $a_{1/n}$. It can be checked that $A_{1/n} \to A_0$ and the kernels are one-dimensional. Yet, the kernels do not converge as $n \to \infty$.
It remains to fill the gaps between $A_{1/n}$ and $A_{1/(n+1)}$. For $t \in (1/n, 1/(n+1))$, we set $$(A_t x)_1 = t \, x_1,\quad (A_t x)_i = x_i / i \text{ for $i \not\in\{1,n,n+1\}$}.$$ In the two-dimensional subspace of coordinates $n$ and $n+1$, we need to go continuously from $$\begin{pmatrix} 0 & 0 \\ 0 & 1/(n+1)\end{pmatrix}$$ to $$\begin{pmatrix} 1/n & 0 \\ 0 & 0\end{pmatrix}$$ and we have to preserve the one-dimensional kernel. This is certainly possible. If we do it correctly, $t \mapsto A_t$ should become continuous from $[0,1]$ to $Hom(X,X)$.