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This should be a basic, or even stupid, question, but I am really confused, and I cannot find any webpage that addresses my question.

From wikipedia (https://en.wikipedia.org/wiki/Resolvent_formalism), an operator $A$ has compact resolvent iff $(A - zI)^{-1}$ is compact for some $z$.

My confusion is that, compact operators cannot be invertible if the domain is infinite dimensional, but clearly $(A - zI)^{-1}$ is invertible by definition. Then this definition would not make sense!

Alex
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  • The definition of the resolvent (or spectrum) of operator $A$ involves the invertibility not of $A$ itself, but of the new operator, $(A - zI)$, where $z$ is a scalar. Thus, we are asking about the invertibility of operators in the family $A-zI$, parameterized by the scalar $z$.

    The resolvent of $A$ consists, by definition, of all the $z$ for which the operator $(A-zI)$ is invertible.

     – avs Aug 17 '16 at 20:39
  • I understand that when $z \in \rho(A)$, then $A - zI$ is invertible. My confusion is, how can $(A - zI)^{-1}$ then be compact? – Alex Aug 17 '16 at 20:46
  • I would recommend checking out Halmos's Introduction to Hilbert space. – avs Aug 17 '16 at 20:51
  • If $X$ is a Banach space and $L \in \mathcal{B}(X)$, then the standard terminology is that $L$ is invertible iff there exists $M\in\mathcal{B}(X)$ such that $LM=ML=I$. $(A-zI)^{-1}$ is not invertible if $A-zI \notin\mathcal{B}(X)$, which must be the case if $(A-zI)^{-1}$ is compact and $X$ is infinite-dimensional. – Disintegrating By Parts Aug 17 '16 at 21:31
  • But if $(A - zI)^{-1}$ is compact, then it is bounded. Since $(A - zI)^{-1}$ is bijective, open mapping theorem implies that its inverse is bounded, and hence $(A - zI)^{-1}$ is invertible. – Alex Aug 17 '16 at 21:36
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    @Alex : If $(A-zI)^{-1}\in\mathcal{B}(X)$ is compact, then $(A-zI)^{-1}$ cannot be surjective, unless $X$ is finite-dimensional. So you're typically dealing with closed, densely-defined $A$ that is not bounded. – Disintegrating By Parts Aug 17 '16 at 22:08
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    I see, since $A$ is not defined everywhere, $dom A$ is not necessarily complete, and open mapping theorems and its relatives do not apply. Thx a lot. – Alex Aug 20 '16 at 18:00

1 Answers1

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Indeed, when the resolvent is compact and the space is not finite dimensional, the operator must be unbounded. An example: $x=(x_1,x_2,...)\in \ell^2({\Bbb N})$ $$ A x= (x_1,2x_2,3x_3,...), \ \ \ (\lambda-A)^{-1} x = \left( \frac{1}{\lambda- k} x_k \right)_{k\geq 1}$$

which is readily verified to be compact for all $\lambda\notin {\Bbb N}$. Another example (more interesting) is the Laplacian on a bounded domain $\Omega\subset \Bbb{R}^d$.

H. H. Rugh
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  • Thank you for your example, but I am still confused. $(\lambda - A)^{-1}$ is compact, and hence bounded. Being invertible, it is also surjective, then open mapping theorem should imply that the inverse of $(\lambda - A)^{-1}$, which is $\lambda - A$, is bounded. But clearly $\lambda - A$ is not bounded. What is wrong with the above argument? – Alex Aug 17 '16 at 20:58
  • Injectivity does not imply surjectivity (only true in finite dimensions) so the open mapping theorem does not apply. – H. H. Rugh Aug 17 '16 at 21:00
  • Wait, the codomain of A is not $\ell_2$? – Alex Aug 17 '16 at 21:09
  • The range of $A$ is all but the domain of definition is a strict subset. For example $(1,\frac12,\frac13,...)$ is not in the domain. – H. H. Rugh Aug 17 '16 at 21:19
  • I thought the setup should be that $A: X \to X$? – Alex Aug 17 '16 at 21:30
  • I see, but no, the use of compact resolvent is really when you look at an unbounded operator $A$ defined on a certain domain $D(A)$ and having dense range (typically onto). Even better when $A$ is self-adjoint on a Hilbert space. The idea is that the unbounded operator may be difficult to analyze but when it has a compact resolvent things get much easier. – H. H. Rugh Aug 17 '16 at 21:52
  • I see, in that case $D(A)$ is not necessarily, closed, and hence not necessarily a Hilbert space, and hence open mapping theorem and its relatives do not apply. Thanks a lot. – Alex Aug 20 '16 at 17:59