Vtyjkyuk

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 $B(mathbbH)$ the Algebra of bounded operators in the Hilbert space $mathbbH$.

The resolvent of an operator $A$ is the subset of $mathbbC$, $rho(A)=lambda in mathbbC: (A-lambda I)^-1 in B(mathbbH)$.

The spectrum of an operator $A$ is the subset of $mathbbC$, $sigma(A)=mathbbCsetminus rho(A)=lambda in mathbbC: (A-lambda I)^-1 notin B(mathbbH)$.

The point spectrum of an operator $A$ is the subset of $mathbbC$, $sigma_p (A)$=$lambda in mathbbC: Ay=lambda y$ 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 $mathbbC$. Let $f: D$ $rightarrow$ $B(mathbbH)$ 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 $z in Dsetminus S$, 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 $B(mathbbH)$.)

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 $f: mathbbC$ $rightarrow$ $B(mathbbH)$ such that $f(lambda)=lambda A$, where $A$ is a fixed compact operator.
It is easy to see that $f$ is analitical for all $lambda in mathbbC$. Because $K(mathbbH)$=A $in B(mathbbH)$: $A$ is compact operator is a $mathbbC$-subspace of $B(mathbbH)$ then $f(lambda)$ is compact for all $lambda in mathbbC$. So we can use the Analytical Fredholm Theorem.
Then either
(a) $(I-f(lambda))^-1$ exists for no $z in mathbbC$
or
(b) $(I-f(lambda))^-1$ exists for all $z in mathbbCsetminus S$, where $S$ is a discrete subset of $mathbbC$ 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=$$lambda in mathbbC$: $lambda A y=y$ has a solution $y ne 0$, and it is easy to see that $0 notin S$.
We can observe that if $1/lambda notin S$, then
$$(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 $sigma(A)=Scup 0$? 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!

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.