Graduate 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.