Assume $X$ is a Banach space, $\Omega \subseteq X$ is an open set, $K\in {C}^{1}( \overline{\Omega}, X)$ is a nonlinear compact map, Is $\operatorname{Id}-K$ a proper map? I think maybe it has something to do with the spectrum of the compact operator. Could you give me some hints?
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1$\begingroup$ In the case $X = \Omega = \mathbb{R}$, the mapping $K = Id$ is compact, but $Id - K$ is not proper. $\endgroup$– Willie WongCommented Dec 19, 2023 at 3:35
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$\begingroup$ yeah, I agree with you. $\endgroup$– boundaryCommented Dec 19, 2023 at 3:47
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$\begingroup$ Is there any sufficient condition？ $\endgroup$– boundaryCommented Dec 19, 2023 at 3:48
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$\begingroup$ A sufficient condition is of course that $K(\Omega)$ is relatively compact. If you use the usual definition of compact map (maps bounded sets into relatively compact sets), it is thus sufficient that $\Omega$ is bounded, $\endgroup$– Martin VäthCommented Dec 19, 2023 at 19:25
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$\begingroup$ Another often-used sufficient condition for unbounded $\Omega$ is that $\lVert K(x)\rVert/\lVert x\rVert\to\infty$ as $\lVert x\rVert\to\infty$ (which implies that $(I-K)^{-1}(B)$ is bounded for bounded $B$). $\endgroup$– Martin VäthCommented Dec 19, 2023 at 19:29
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