Definition of the residual spectrum

[AxiomOfChoice]

If $A$ is a bounded operator on a Hilbert space $H$, isn't the following true of the residual spectrum $\sigma_r(A)$:

$\lambda \in \sigma_r(A)$ iff $(\forall \psi \in H, \psi \neq 0)((\lambda - A) \psi \neq 0)$ iff $\ker (\lambda - A) = \{0\}$ iff $\lambda - A$ is injective?

So isn't the condition that $\lambda$ isn't an eigenvalue equivalent to the condition $\lambda - A$ is injective?

 

[micromass]

If $A$ is a bounded operator on a Hilbert space $H$, isn't the following true of the residual spectrum $\sigma_r(A)$:

$\lambda \in \sigma_r(A)$ iff $(\forall \psi \in H, \psi \neq 0)((\lambda - A) \psi \neq 0)$ iff $\ker (\lambda - A) = \{0\}$ iff $\lambda - A$ is injective?

No, $\lambda$ is in the residual spectrum if $\lambda-A$ is injective AND does not have dense range. You're missing here that $\lambda-A$ can not have dense range.

So isn't the condition that $\lambda$ isn't an eigenvalue equivalent to the condition $\lambda - A$ is injective?

This is true.

It seems you're mistaking the residual spectrum for the part of the spectrum which are not eigenvalues. This is not true. The spectrum can be divided in 3 parts:

- The point spectrum: all the eigenvalues
- The continuous spectrum: These are the $\lambda$ such that $\lambda - A$ is injective and has dense range.
- The residual spectrum:These are the $\lambda$ such that $\lambda - A$ is injective and does not have dense range.

So the elements which are not eigenvalues are either contained in the continuous or the residual spectrum.