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Let $L(t)$, $t\in\mathbb{R}$ be a family of bounded linear operators, where $L(t):X\to X$ for some Banach space $X$. Let $t_n\to t$, and $L(t_n)$ converge to some $L(t)$ in the operator norm. If $L(t_n)$ is non-invertible for all $n$, is it true that $L(t)$ must also be non-invertible? This is clearly true in finite dimensions; an $n\times n$ matrix is singular iff it has zero determinant, and the determinant is continuous, but it's not clear how one would extend this to infinite dimensions.

In my case, $X$ is the space of $2\pi$-periodic continuous functions.

Phil
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    If the limit was invertible then in a small neighborhood of $L(t)$ we would have $L(t_n)$ invertible so the answer must be no. – copper.hat Sep 10 '20 at 03:40
  • Of course, completely forgot that the space of invertible operators is open, so the result follows immediately. – Phil Sep 10 '20 at 03:41
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    For the sake of having an answer, there is a proof of the fact that the space of invertible bounded linear operators from a Banach space to itself is open in Corollary 4.13 here. – Phil Sep 10 '20 at 03:45

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