MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As we know, continuous spectrum and residual spectrum are two cases in the spectrum of an operator, which only appear in infinite dimension.

If $T$ is a operator from Banach space $X$ to $X$, $aI-T$ is injective, and $R(aI-T)$ is not $X$. If $R(aI-T)$ is dense in $X$, then $a$ belongs to the continuous specturm, it not, $a$ belongs to the residual spectrum.

1) I want to know why do we care about whether $R(aI-T)$ is dense.

2) It seems as we just emphasize this difference for the operator, but not for the Banach algebra, why?

share|cite|improve this question
Crossposted from math.SE: Strongart, you need to explicitly mention (and provide a link) when you ask the same question on multiple fora. – Zev Chonoles Jul 29 '12 at 6:15
Thanks for your remind, I will do it next time. – Strongart Jul 31 '12 at 11:29
up vote 3 down vote accepted

I'm afraid that this is more or less a reformulation of what you asked, but: a bounded operator $T$ on a Banach space $X$ is invertible if and only if it is bounded below (i.e. there is some constant $C>0$ such that $||Tx|| \geq C||x||$ for all $x\in X$) and has dense range. Bounded below implies injective, and it also implies that the range is closed; the range is closed and dense, hence is everything.

So these two different types of spectrum distinguish different ways that the operator can fail to be invertible.

Why is this distinction not made for Banach algebras in general? Well, it doesn't make sense to ask this unless the Banach algebra is presented as a subalgebra of operators on some Banach space. I suppose that you could always consider the left-regular representation of the algebra on itself and then interpret those two conditions in that setting. If $A$ is a Banach algebra, let $a \mapsto L_a$ denote the left-regular representation $L_ab = ab$.

If $L_a$ does not have dense range then $\overline{aA}$ is not all of $A$, which condition I don't think can be reduced to anything simpler.

If $L_a$ is not bounded below, that means that there is a sequence $(x_n)$ in $A$ with $||x_n||=1$ such that $ax_n \to 0$, i.e. $a$ is what is called a (left?) topological zero divisor.

share|cite|improve this answer

For $\lambda$ in the continuous spectrum, while there may not be genuine eigenvectors, there are approximate eigenvectors.

That point is related to the fact that (as Bob Israel mentions) normal operators on Hilbert spaces, which do have a spectral theorem, do not have residual spectrum. It is only slightly frivolous to observe that an operator on Hilbert space with residual spectrum could not admit a reasonable spectral theorem.

(By "reasonable spectral theorem" we need not require decomposition into actual eigenvectors, but something much weaker, than can be described in various ways.)

share|cite|improve this answer

The most important thing about residual spectrum, I think, is that some classes of operators, e.g. normal operators on Hilbert space, don't have any.

share|cite|improve this answer

Your Answer


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.