First of all sorry for my bad English, I'm an Italian student, hope to let you understand!
I'm having a little troubles with the proof of the Riesz-Schauder theorem for Compact Operators.
Some infos in advance:
Let $B(\mathbb{H})$ the Algebra of bounded operators in the Hilbert space $\mathbb{H}$.
The resolvent of an operator $A$ is the subset of $\mathbb{C}$, $\rho(A)=\{{\lambda \in \mathbb{C}: (A-\lambda I)^{-1} \in B(\mathbb{H})}\}$.
The spectrum of an operator $A$ is the subset of $\mathbb{C}$, $\sigma(A)=\mathbb{C}\setminus \rho(A)=\{{\lambda \in \mathbb{C}: (A-\lambda I)^{-1} \notin B(\mathbb{H})}\}$.
The point spectrum of an operator $A$ is the subset of $\mathbb{C}$, $\sigma_p (A)$={$\lambda \in \mathbb{C}: Ay=\lambda y$ has a solution $y \ne 0$}, in other words $\sigma_p (A)$ is the set of eigenvalues of $A$ and holds that $\sigma_p (A) \subseteq \sigma(A)$.
For the proof of the Riesz-Schauder Theorem I will use the following:
Analytic Fredholm Theorem. Let $D$ an open connected subset of $\mathbb{C}$. Let $f: D$ $\rightarrow$ $B(\mathbb{H})$ be an analytic operator-valued function such that $f(z)$ is compact for all $z \in D$.
Then, either:
(a) $(I-f(z))^{-1}$ exists for no $z \in D$
or
(b) $(I-f(z))^{-1}$ exists for all $z \in D\setminus S$, where $S$ is a discrete subset of $D$ (i.e. a set with has no limit points in $D$).
(Note that $I$ is the identity operator of $B(\mathbb{H})$.)
Finally, the main theorem
Riesz-Schauder Theorem. The spectrum $\sigma(A)$ of a compact operator $A$ is a discrete set having no limit points except (perhaps) $0$. $\sigma(A)$ is finite or countable and in this case $0$ is the only limit point. Further, any nonzero $\lambda \in \sigma(A)$ is an eigenvalue of finite multiplicity.
For the proof (as suggested here http://www.macs.hw.ac.uk/~hg94/pdst11/pdst11_AFTcompact.pdf ) , I will proceed like this:
Let $f: \mathbb{C}$ $\rightarrow$ $B(\mathbb{H})$ such that $f(\lambda)=\lambda A$, where $A$ is a fixed compact operator.
It is easy to see that $f$ is analitical for all $\lambda \in \mathbb{C}$. Because $K(\mathbb{H})$={A $\in B(\mathbb{H})$: $A$ is compact operator} is a $\mathbb{C}$-subspace of $B(\mathbb{H})$ then $f(\lambda)$ is compact for all $\lambda \in \mathbb{C}$. So we can use the Analytical Fredholm Theorem.
Then either
(a) $(I-f(\lambda))^{-1}$ exists for no $z \in \mathbb{C}$
or
(b) $(I-f(\lambda))^{-1}$ exists for all $z \in \mathbb{C}\setminus S$, where $S$ is a discrete subset of $\mathbb{C}$ with no limit points.
But, if $\lambda=0$, then $(I-f(\lambda))^{-1}=(I-0A)^{-1}=I$ exists, so option (b) is true.
Furthermore, from the Analytic Fredholm Theorem proof we know that $S=${$\lambda \in \mathbb{C}$: $\lambda A y=y$ has a solution $y \ne 0$}, and it is easy to see that $0 \notin S$.
We can observe that if $1/\lambda \notin S$, then
$$(A-\lambda I)^{-1}=-1/\lambda (I-1/\lambda A)^{-1}$$
and so $\lambda \in \rho(A)$.
But holds even the viceversa so, $1/\lambda \notin S \Leftrightarrow \lambda \in \rho(A)$.
Then we have $1/\lambda \in S \Leftrightarrow \lambda \notin \rho(A)$, thus
$$1/\lambda \in S \Leftrightarrow \lambda \in \sigma(A).$$
Here are my doubts. How can I prove that $\sigma(A)$={$0$}$\cup \sigma_p (A)$?
Maybe it is possible to prove that $\sigma(A)=S\cup \{0\}$? In this way, it is easy to see that the only limit point is zero.
Instead, I've done something like this:
If $\lambda \in \sigma(A)$ and $\lambda \ne 0$ then $1/\lambda \in S$. Thus $1/\lambda A y =y$ has a solution $y \ne 0$, and $\lambda \in \sigma_p (A)$. So, because for a compact operator it holds that $0 \in \sigma(A)$, I can state that
$$\sigma(A)=\sigma_p (A) \cup \{0\}.$$
But in this way I have some troubles attempting to prove that zero is the only limit point.
Certainly I can state that either $\sigma_p (A)$ is finite, or, on the contrary, because $\sigma_p (A) \subseteq \sigma(A)$ and $\sigma(A)$ is a limited set, then for the Bolzano – Weirstrass theorem $\sigma_p (A)$ has a limit point. But $\sigma(A)$ is norm-closed, so if $\mu$ is a limit point for $\sigma_p (A)$ then $\mu \in \sigma(A)$. But how can I state that is $\mu = 0$??
Thank you in advance for your time and sorry again for my English!
Best Answer
http://individual.utoronto.ca/jordanbell/notes/SVD.pdf
check out the end of page 4; the equality you're trying to prove only holds for the infinite dimension case.