Continuous section inside a family of rank-varying operators

Question

Good morning everybody,

my question is as follows: let $K$ be a compact set and assume $F:K\to L(\mathcal H,\mathbb R^{m+1})$ be a continuous map from the compact set $K$ to the space of linear operators from a fixed infinite-dimensional Hilbert space $\mathcal H$ into $\mathbb R^{m+1}$, $m\geq 0$.

Moreover we assume that the rank of $F(k),\;k\in K$ to be either $m$ or $m+1$.

Assume now $\overline k$ to be a fixed element in $K$, and $\overline v=v(\overline k)$ to be an element of $\mathcal H$ such that $$F(\overline k)[\overline v]=0.$$

My question is:

is it possible to construct, in a neighborhood $k\in U\subset K$ a continuous extension $v=v(k),\; k\in U$ of $\overline v$ such that $F(k)[v]=0,\;\forall k\in U$?

Of course if $F(\overline k)$ is of rank $m+1$ this is guaranteed by the implicit function theorem, also if $F(\overline k)$ is of rank $m$ and there exists a neighborhood $U$ of $k$ such that $F(k)$ is of rank $m$ for any $k\in U$ then this is also possible, since we are basically working on a subbundle and I know how to proceed.

I am interested in the intermediate situations; is there any reasonable assumption for this to work? What is known in the literature? I googled a bit but without success.

To finish my discussion I think that the hypothesis of $\mathcal H$ to be infinite-dimensional is fundamental, as for example if we consider $K=[0,1]$, $F:[0,1]\to L(\mathbb R^2,\mathbb R)$ and $$F(k)=k(\cos(1/k),\sin(1/k)),\;k\neq 0,$$ $$F(0)=(0,0)$$ and $v(0)=(1,0)$, then the extension of $v$ is impossible, since basically $F(k)$ comes infinitely close to any point as $k\to 0$, but again I think this patology is partly due to the finite dimension of the kernel of our operators.

Any comment is welcomed, as well as any reference to known results.

Regards,

Guido Giuliani

Edit: A reasonable condition for my question to be true seems to be give in the comment below this post, namely it seems necessary for the operator norms of the $F(k)$ to be bounded away from zero. Can we say something more at least under this hypothesis?

Answer 1

Christian Remling pointed out that a counterexample to the original form of the question is given by $F: [0,1] \to L(H, \mathbb{R}^1)$, $F(t) = t\langle \cdot, e_1\rangle$, $v(0) = e_1$. Any continuous extension to a path $v(t)$, $t \in [0,1]$, will have $\langle v(t), e_1\rangle \neq 0$ for sufficiently small $t$, and therefore $F(t)(v(t)) \neq 0$ for small nonzero $t$.

The question then was modified to require that $\|F(t)\|$ be bounded away from zero. In that case the answer switches to yes for $m = 0$ since then the rank of $F(t)$ must be constantly 1. However, it is no for $m > 0$ by a simple modification of Christian's example. Namely, for $t \in [0,1]$ define $F(t): H \to \mathbb{R}^{m+1}$ by $F(t)(w) = (t\langle w, e_1\rangle, \langle w, e_2\rangle, \ldots, \langle w, e_{m+1}\rangle)$ and set $v(0) = e_1$. Clearly the norm is bounded away from zero but any continuous extension to a path $v(t)$ will have $\langle v(t), e_1\rangle \neq 0$ for sufficiently small $t$, and therefore the first coordinate of $F(t)(v(t))$ will be nonzero for small nonzero $t$.