Theorem (Analytic Fredholm Alternative)Let be a connected open subset of and be a separable Hilbert space. Suppose that is an analytic operator-valued function such that is compact . Then either

- does not exist .
- exists , where is discrete.

*Proof:* We will prove it in 6 steps. First we will prove the result holds in the neighborhood of any , then by a connectedness argument to get the result on all .

Step 1:

Let , there exists such that if , then . Since is compact, there exists a finite rank operator such that . This implies for all . So exists and is bounded and analytic on .

Step 2 (Relation between and ):

Let , then one has

- is invertible iff is invertible.
- has non-zero solution iff has non-zero solution.

Step 3 (Analysis of ):

Since is finite rank, there exists orthonormal vectors such that where . By Riesz representation, there exists vectors such that . Therefore . And we have

Denoting , we get that .

Step 4:

If , then and for , and some . This is equivalent to say

Since is analytic on , is also analytic. This implies the set is either all of or a discrete set in . If , then the first alternative in the theorem holds.

Step 5:

Suppose , we will show that is invertible. Indeed, is injective by step 4. By Inverse Mapping Theorem it remains to show it is also onto. Let , we need to solve the equation for some . Let , then has to solve

But ran . So for some . Substitute back and use the formula for , we get . Therefore

Since , this inhomogeneous system has a solution. This implies is invertible if .

Step 6:

We have proved for every , there is a neighborhood where the Analytic Fredholm Alternative holds. Define the sets there exists a neighborhood around where the first alternative holds\}, there exists a neighborhood around where the second alternative holds\}. Therefore and . By the definition of and , we know both sets are open. Since is connected, we know either or . This concludes the proof.

Corollary 1 (Fredholm Alternative)If is compact, nonzero, then either exists or has a nonzero solution.

*Proof:* Apply the theorem to with at

Corollary 2 (Riesz Schauder Theorem)Let be a compact operator on a separable Hilbert space , then the spectrum is a discrete having no limit points except perhaps at . Furthermore, any nonzero is an eigenvalue with finite multiplicity.

*Proof:* Let , define the set . Since , by the Analytic Fredholm Alternative, is discrete. If , then exists so exists. Hence . Therefore .

The fact that nonzero eigenvalues have finite multiplicity follows from compactness. Otherwise we get infinite orthonormal set of eigenvectors with eigenvalue , and it have no convergence subsequence.