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.

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

  • $\begingroup$ It seems as we just emphasize this difference for the operator, but not for the Banach algebra, why? $\endgroup$
    – Strongart
    Jul 29, 2012 at 6:03

2 Answers 2


The point $\lambda\in\mathbb{C}$ belongs to the spectrum of operator $T$ if the operator $T_\lambda:=T-\lambda I$ is not invertible.

What can prevent $T_\lambda$ from being invertible? Recall that we are working in Banach space $X$ so invertibility is equivalent to bijectivity. Thus we need to study reasons why operator $T_\lambda$ can't be bijective. We can distinguish two cases:

  • the operator $T_\lambda$ is not injective

  • operator $T_\lambda$ is injective but not surjective

Now we discuss these cases.

1) The first one is the most common. In this case $\mathrm{Ker}(T_\lambda)$ is non-trivial so $T_\lambda$ is not invertible, and we say that $\lambda$ is in the point spectrum. If $X$ is finite dimensional this is the only possible case for operator not to be bijective. The reason is that an injective operator on a finite dimensional space is automatically surjective. But in case $X$ is infinite dimensional there are examples of injective but not surjective operators!

2) In the second case we have injective but not surjective operators. This means that the image of the operator $\mathrm{Im}(T)$ (which is a linear subspace) is not the whole space $X$. If $X$ is finite dimensional it is impossible for the operator $T_\lambda$ to be injective but not surjective, so this is not the case. If $X$ is infinite dimensional there two possibilities for the subspace $\mathrm{Im}(T_\lambda)$ not to be the whole $X$. Here we have two cases:

2.1) $\overline{\mathrm{Im}(T_\lambda)}=X$, speaking informally $T_\lambda$ is "almost surjective". In this case we say that $\lambda$ is in continuous spectrum.

2.2) $\overline{\mathrm{Im}(T_\lambda)}\neq X$, speaking informally $T_\lambda$ is "essentially non-surjective", even the closure of its image is a proper subspace of $X$! In this case we say that $\lambda$ is in the residual spectrum.

There are other classifications of points of the spectrum, but this one is the most common.

  • 5
    $\begingroup$ Yes, this is nice, but I don't quite understand how this answers the question which is "why do we distinguish cases 2.1) and 2.2)?" $\endgroup$
    – t.b.
    Jul 25, 2012 at 2:31
  • 1
    $\begingroup$ Well... this explanation shows that the two cases are different, so it is natural that they are distinguished! $\endgroup$ Jul 25, 2012 at 2:48
  • $\begingroup$ Theo, thanks for your edit. I see that my English is so poor... As for your question, I think we distinguish this cases because the measure "non-surjectiveness" of our operator. May this a bad explanation, this is just how I understand that. $\endgroup$
    – Norbert
    Jul 25, 2012 at 9:16
  • $\begingroup$ Oh, my English is also poor. In fact, I want to know why do we emphasize the "almost surjective" from the "nonsurjective". $\endgroup$
    – Strongart
    Jul 25, 2012 at 15:29
  • 1
    $\begingroup$ That is a good reason, but I do not think it is enough. Maybe the following claim is helpful: T is invertible iff T is bounded below and dense range. $\endgroup$
    – Strongart
    Jul 26, 2012 at 15:25

As pointed out by M. Reed, B. Simon in the section VI.3 of their "Methods of Modern Mathematical Physics: Functional Analysis":

The reason that we single out the residual spectrum is that it does not occur for a large class of operators, for example, for self-adjoint operators.

  • 1
    $\begingroup$ Yes, it is also true for the norm operator. $\endgroup$
    – Strongart
    Jul 30, 2012 at 15:30

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