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Compact Operators have been the major topic in our Operator Theory course for the past few weeks.

All the theorems which tell us whether a operator is compact or not are clear to me, but I still don't know why the compactness of an operator is bothering us in the first place?

We haven't been given an introduction why the compactness of an operator should bother us. We also haven't been given a reason why we are talking about the compactness of an adjoint and dual operator, too.

So my simple question is: Why is compactness of an operator important? Why are compact operators important? And also, what's the use and motivation of adjoint and dual operators (and whether they are compact, or not)?

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There is a big difference between finite-dimensional and infinite-dimensional spaces. We know for example that the unit ball in a normed vector space is compact if and only if the vector space is finite-dimensional.

But we are most familiar with finite-dimensional spaces. We are happy if our operator at hand has at least finite-dimensional range (we call them finite-dimensional operators), because we have a good feeling for these. Now, compact operators are in a sense closest to finite-dimensional operators as you can get. For example, in a Hilbert space every compact operator is the limit of a sequence of finite-dimensional operators.

You also get some feeling for this when you consider the spectrum of an operator. Let $A$ be a bounded operator from a Banach space $X$ into itself. Its spectrum is defined as the set of all complex numbers $z$ for which $A - zI$ is not bijective. In finite dimensions this is just the set of eigenvalues, right? The spectrum of a general bounded operator is always compact. However, it can be rather weird. A simple example is the multiplication operator $T : L^2(0,1)\to L^2(0,1)$ mapping a function $f$ to $xf(x)$. You can easily compute its spectrum. It is the whole interval $[0,1]$, but it has no eigenvalues at all. However, when $A$ is a compact operator all points in its spectrum (except $z=0$) are eigenvalues and you have have Jordan blocks as in the finite-dimensional situation. The only difference is that there might be infinitely many eigenvalues (which then accumulate to $z=0$) and that $z=0$ might be not an eigenvalue, although it is always a spectral point. There are also weird compact operators like the Volterra operator whose only spectral point is $z=0$, which is not an eigenvalue.

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  • $\begingroup$ I don't understand your example with the spectrum. Our definition was $\sigma (T) = \left\{ \lambda \in \mathbb{C} \: : \: (T - \lambda \cdot I)^{-1} \notin \mathcal{B}(H) \text{ for }T \in \mathcal{B}(H) \right\}$, and not something that it's not bijective, so I don't see the analogy here. Back to compact operators: So this means that we like compact operators because they are the closest ones to being finite-rank operators? What does "closest ones" mean? And we like finite-rank operators, because our intuition is more familiar with finite-rank space than infinite-rank space? $\endgroup$
    – anon
    Oct 21, 2021 at 17:11
  • $\begingroup$ The limit of a sequence of finite-rank operators is always compact. That's what's meant by "closest". At least in a Hilbert space the space of compact operators is the closure of the space of finite-rank operators. Your last question: yes, of course. Concerning the spectrum: Your definition is bad because there is an object in the definition, that sometimes doesn't exist: the inverse of $T - \lambda I$. $\endgroup$
    – amsmath
    Oct 21, 2021 at 18:01
  • $\begingroup$ What it wants to say is that $\lambda\in\sigma(T)$ if $T - \lambda I$ is either not injective or, if it is, then the range of $T - \lambda$ is not $H$. That is the same as to say that $T - \lambda$ is not bijective. $\endgroup$
    – amsmath
    Oct 21, 2021 at 18:01
  • $\begingroup$ Can we rescue our definition for $\sigma (T)$ somehow? Because the theorems, properties and lemmas which came next were all related to our given definition of $\sigma (T)$. Also, what does "spectrum" mean here? Is it kind of like with eigenvalues in matrices? Or how should I understand what a spectrum is (on an easy example)? $\endgroup$
    – anon
    Oct 21, 2021 at 18:17
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    $\begingroup$ As I already wrote, in the finite-dimensional case the spectrum is the set of eigenvalues. You can simply define the spectrum as the set of all $\lambda$ for which $(T - \lambda I)^{-1}$ does not exist as a bounded operator. The rest of your questions are not understandable for me. $\endgroup$
    – amsmath
    Oct 21, 2021 at 18:20

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