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Let $L$ be a bounded linear operator acting on a complex Banach space $B$. If there exists a nonzero continuous linear functional $\ell \colon B \to \mathbb{C}$ such that $\ell(Lx)=\ell(x)$ for all $x \in B$, then $1$ is by definition an eigenvalue of the adjoint operator $L^* \colon B^* \to B^*$, and a well-known consequence of this is that $1$ belongs to the spectrum of $L$ (though it is not necessarily an eigenvalue of $L$). My question is: if we have only that $\ell \colon D \to \mathbb{C}$ is a nonzero closed linear functional defined on a dense linear space $D \subset B$, and $\ell(Lx)=\ell(x)$ for all $x \in D$, does it still follow that $1$ belongs to the spectrum of $L$?

I would guess that this implication does not hold in general, but I am having difficulty thinking up a counterexample (not an unusual situation for me in functional analysis!).

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Be cautious: $Lx$ might not be in $D$, what is $\ell (Lx)$ in this case? – Florian Sep 19 '11 at 15:51
Oops: yes, L preserves D (otherwise the question is nonsense). – Ian Morris Sep 19 '11 at 17:09
Hi, it might interest you to know that there are no closed, densely-defined linear functionals that aren't actually bounded. This just came up here:… – Mike F Aug 16 '13 at 17:34
up vote 1 down vote accepted

Assuming $L$ maps $D$ to $D$, it is actually straightforward to show that it's true: if $1$ would not be in the spectrum, $(L-1)^{-1}$ exists and maps $D$ to $D$. Choose $y\in D$ with $\ell(y)\ne 0$ and set $x:=(L-1)^{-1}(y)$ to it, then $\ell(Lx)\ne \ell(x)$. No density of $D$ or closedness of $\ell$ is needed here.

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Excellent, thank you! This is perhaps a story with a moral: I should remember what the definition of the spectrum actually is, rather than just treating it as some object... – Ian Morris Sep 19 '11 at 17:11
Actually, how do we ensure that $(L-I)^{-1}$ maps $D$ to $D$? – Ian Morris Sep 20 '11 at 14:00

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