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Given a linear operator $T:D(T)\to B_2$, with domain $D(T)\subseteq B_1$, and with $B_i$ Banach spaces. Let $\sigma(T)$ denote its spectrum. As is nicely explained (among other places) in

we define:

  • The approximate point spectrum $\sigma_{\rm ap}(T)$ as the set of $\lambda\in\mathbb C$ such that $T-\lambda I$ is not bounded below.
  • The continuous spectrum $\sigma_c(T)$ as the set of $\lambda\in\mathbb C$ such that $T-\lambda I$ is injective and its range is a proper dense subset of $B_2$.

At the same time, as mentioned e.g. here, by the open mapping theorem we know that if $T:B_1\to B_2$ is a bounded linear operator (still between Banach spaces), and $T$ is invertible, then its inverse $T^{-1}$ must also be bounded, and therefore $T$ must be bounded below.

This should mean that $\lambda\in\rho(T)$ iff $T-\lambda I$ is bounded below. Or in other words $\lambda\in \sigma(T)$ iff $T-\lambda I$ is not bounded below, i.e. $\lambda\in \sigma_{\rm ap}(T)$ is an approximate eigenvalue in the approximate point spectrum.

But the way these things are discussed makes me doubt my conclusion. For example the Wiki page says at some point that "It can be shown that the approximate point spectrum of a bounded multiplication operator equals its spectrum", and the qualifier "multiplication" here would imply my conclusion is wrong. I can also see that $\sigma_{\rm ap}(T)$ and $\sigma(T)$ might differ when $T$ is unbounded, but I couldn't find an explicit example of this.

So in summary, does $\sigma_{\rm ap}(T)=\sigma(T)$ when $T$ is bounded? And if it's true that $\sigma_{\rm ap}(T)\neq \sigma(T)$ only when $T$ is unbounded, what is an example where they differ?

glS
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  • $\lambda \in \rho (T)$ iff $T - \lambda \cdot I$ is bounded from below is false, because $T- \lambda \cdot I$ bounded from below is equivalent to $T- \lambda \cdot I $ is injective and that the inverse (which is only defined on the range of $T- \lambda \cdot I$) is bounded, but this does not imply that $T - \lambda \cdot I$ is surjective – jd27 Aug 22 '24 at 18:35
  • @jd27 mh, sorry, I think I'm confused on a few things. Bounded below$\iff$injective only holds for bounded operators, right? Otherwise I might have an injective operator with unbounded inverse which is thus not bounded below. For bounded ops, you're saying $T-\lambda I$ might be bounded below, thus injective, but not surjective and thus not in $\rho(T)$. I think I'm starting to see the problem in my logic then. So continuous and point spectrum are always in $\sigma_{\rm ap}$, but the residual spectrum might have elements not in the approximate point spectrum? What's an example of that? – glS Aug 23 '24 at 07:39
  • or in other words, while $\sigma_c(T),\sigma_p(T)\subseteq \sigma_{\rm ap}(T)$, residual eigenvalues $\lambda\in\sigma_r(T)$ might or might not be approximate eigenvalues? – glS Aug 23 '24 at 07:40

1 Answers1

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As pointed out by @jd27, the correct proposition regarding this is that a bounded linear operator between normed spaces, let's say $A:X\rightarrow Y$, is bounded below if and only if $A: X\rightarrow \mathcal{R}(A)$ is invertible and its inverse is continuous. This points to the fact that $R(\lambda, T)=T-\lambda I$ being bounded below might not be enough for $\lambda\in\rho(T)$, since $R(\lambda, T)$ might not be surjective.

For a counterexample, consider the (right) shift operator $S:\ell^2(\mathbb{C})\rightarrow\ell^2(\mathbb{C})$ defined by $$S(x_1,x_2,x_3,\dots)=(0,x_1,x_2,\dots).$$ Observe that $||Sx||=||x||$ and in particular $||Sx||\geq 1\cdot||x||$, so $S$ is bounded below. Now note that $S$ is of course not invertible, since for example $e_1=(1,0,0,\dots)$ doesn't have a pre-image under $S$ (here we are taking advantage that we know that if there is a problem, it has something to do with surjectivity).

Finally, we can easily transform this into a proper spectral theory example by noting that the latter implies that $S-0\cdot I$ is bounded below, so $0\notin \sigma_{ap}(S)$, however $0\in \sigma(S)=[0,1]$ (proving this last equality is a cool exercise, however if you just want to see a [kinda boring] proof of this see this post).

Esteban G.
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