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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.

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    It seems as we just emphasize this difference for the operator, but not for the Banach algebra, why?2012-07-29

2 Answers 2

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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.

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    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)?"2012-07-25
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    Well... this explanation shows that the two cases *are* different, so it is natural that they are distinguished!2012-07-25
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    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.2012-07-25
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    Oh, my English is also poor. In fact, I want to know why do we emphasize the "almost surjective" from the "nonsurjective".2012-07-25
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    @Strongart, I think that another reason of such emphasizing is given in AlbertH's answer2012-07-25
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    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.2012-07-26
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    It is not true that $T$ is invertibel iff it is bounded from below and have a dense range. The reason that for invertability we need bijectivity.2012-07-26
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    That is from the book "Banach Algebra Techniques in Operator Theory" by Ronald G. Douglas P76.proposition 4.8. Maybe it is because the difference about the definitions.2012-07-27
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    @Strongart I don't have this book, but if you want answer with definition of invertability from this book please write it here2012-07-27
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    I read the prove again and because of the below bounded condition, we can get the range is closed first,so Ran(T) is the whole space. But I also see some book treat the invertability as one-to-one, a Chinese popular functional analysis book by Gongqing Zhang is an example.2012-07-28
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    @Norbert: Yes it is true. Boundedness from below, i.e. existence of $C>0$ with $\|Tx\|\geq C\|x\|$ for all x, is equivalent to injectivity+closed image. So if you add 'dense image' then you get 'bijective', in both directions.2012-07-29
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    @wildildildlife, closedness of image follows from boundedness below if space $X$ is Banach. Otherwise, you can consider $$T:(C^1([0,1]),\Vert\cdot\Vert_\infty)\to (C([0,1]),\Vert\cdot\Vert_\infty):x\mapsto x$$2012-07-30
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    Obviously X is assumed to be Banach, otherwise we can't even apply the open mapping theorem and this whole discussion would be ridiculous.2012-07-30
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    Yes, Banach space is the stage. If X is not Banach space, we have another way to break the inverse of λI-T,(λI-T)^(-1) is exist but unbounded, I do not know what is the name for such spectrum.2012-07-30
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    In Kreyszig's functional analysis, the continuous spectrum is defined as $\sigma_c(T) = \{\lambda \in \mathbb C: R_\lambda(T) $ exists, $R_\lambda(T)$ is defined on a set which is dense in $X$, but $R_\lambda(T)$ is not bounded $\}$. Is this definition the same as yours? i.e. if it is "almost surjective" but not injective, does it imply that $R_\lambda(T)$ not bounded?2015-05-03
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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.

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    Yes, it is also true for the norm operator.2012-07-30