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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.

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  • $\begingroup$ What do you mean by $Hom(X,X)$? I would think that this are bijective bounded linear maps, but then the question does not make sense. $\endgroup$ – gerw Aug 21 '18 at 6:55
  • $\begingroup$ I meant for the space of bounded linear maps. (not necessarily bijective, injective or surjective). Perhaps $B(X)$ is a more standard name. (Usually when people mean for invertible maps they write $Iso$ instead of $Hom$. $\endgroup$ – Asaf Shachar Aug 21 '18 at 6:59
  • $\begingroup$ What example do you have in mind of a family $A_t$ all whose kernels are finite dimensional and of the same dimension? $\endgroup$ – uniquesolution Aug 21 '18 at 7:13
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    $\begingroup$ Tosio Kato's Perturbation Theory for Linear Operators? $\endgroup$ – user539887 Aug 21 '18 at 7:18
  • $\begingroup$ @uniquesolution Here is the example which inspired this question : The Laplacian on the space of differential $k$-forms on a Riemannian manifold, as a function of the metric. (That is, I think of the family $g_t \to \Delta_{g_t}$ where $g_t$ is a family of Riemannian metrics). The kernel $\ker \Delta_{g_t}$ is the space of $g_t$-harmonic forms, which is known to be finite dimensional, and metric independent. (It's a topological invariant of the manifold). Other (finite-dimensional) examples are kernels of bundle morphisms of constant rank, $\endgroup$ – Asaf Shachar Aug 21 '18 at 7:40
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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 $Hom(X,X)$.

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  • $\begingroup$ Thanks, That is a very nice counter-example. $\endgroup$ – Asaf Shachar Aug 21 '18 at 8:00
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    $\begingroup$ In my opinion, the essential feature of the counterexample is that the range of $A_0$ is not closed. I am not sure what happens if we add the requirement that $A_0$ has a closed range... $\endgroup$ – gerw Aug 21 '18 at 8:02
  • $\begingroup$ I would guess that under your constant-kernel-dimension assumption, the closedness of the range of $A_0$ should imply the closedness of the ranges of $A_t$ for small $t$. But I do not have any idea how to prove this. $\endgroup$ – gerw Aug 21 '18 at 8:09
  • $\begingroup$ To help future readers: The image of $A_0$ is not closed since it contains $(0,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},0,0,0,0,\dots)$ for every $n$, but not their limit, which is "supposed to be" the image of the element $(1,1,1,\dots)$. (This element is not in $\ell^2$). $\endgroup$ – Asaf Shachar Aug 21 '18 at 12:47
  • $\begingroup$ By the way, you were right about the crucial part played by the non-closed range here. When the range is closed, everything works smoothly, as can be seen here: math.stackexchange.com/a/2892319/104576 (In the answer there it is proved that the kernels converge in a strong sense). Thanks again for your insights. $\endgroup$ – Asaf Shachar Aug 28 '18 at 7:21

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