Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!).

share|improve this question
    
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: math.stackexchange.com/questions/468658/… –  Mike F Aug 16 '13 at 17:34

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.