# Fredholm Alternative and Riesz Schauder Theorem

Theorem (Analytic Fredholm Alternative) Let ${D}$ be a connected open subset of ${{\mathbb C}}$ and ${\mathcal{H}}$ be a separable Hilbert space. Suppose that ${f : D \longrightarrow \mathcal{L} (\mathcal{H})}$ is an analytic operator-valued function such that ${f(z)}$ is compact ${\forall z \in D}$. Then either

1. ${(I - f(z)) ^{-1}}$ does not exist ${\forall z \in D}$.
2. ${(I - f(z)) ^{-1}}$ exists ${\forall z \in D\setminus S}$, where ${S}$ is discrete.

Proof: We will prove it in 6 steps. First we will prove the result holds in the neighborhood of any ${z_0 \in D}$, then by a connectedness argument to get the result on all ${D}$.

Step 1:

Let ${z_0 \in D}$, there exists ${r > 0}$ such that if ${|z - z_0| < r}$, then ${\| f(z) - f(z_0)\| \le \frac{1}{2}}$. Since ${f(z_0)}$ is compact, there exists a finite rank operator ${F}$ such that ${\| F - f(z_0)\| \le \frac{1}{2}}$. This implies ${\| F - f(z)\| \le 1}$ for all ${z \in D_r:= \{ z \in D : |z-z_0| < r\}}$. So ${(I - f(z) +F )^{-1}}$ exists and is bounded and analytic on ${D_r}$.

Step 2 (Relation between ${I - f(z)}$ and ${I - f(z) +F}$):

Let ${g(z) = F(I - f(z) + F)^{-1}}$, then one has

$\displaystyle I - f(z) = (I - g(z)) (I - f(z) +F).$

1. ${I - f(z)}$ is invertible iff ${I - g(z)}$ is invertible.
2. ${f(z) \psi = \psi}$ has non-zero solution iff ${g(z) \varphi = \varphi}$ has non-zero solution.

Step 3 (Analysis of ${g(z)}$):

Since ${F}$ is finite rank, there exists orthonormal vectors ${\psi_1, \cdots, \psi_N}$ such that ${F(\varphi) = \sum_{i=1}^N \alpha_i (\varphi) \psi_i}$ where ${\alpha_i \in H^*}$. By Riesz representation, there exists vectors ${\phi_1, \cdots \phi_N}$ such that ${\alpha_i( \varphi) = \langle \varphi, \phi_i \rangle}$. Therefore ${F(\varphi) = \sum_{i=1}^N \langle \varphi, \phi_i \rangle \psi_i}$. And we have

$\displaystyle g(z) v = \sum_{i = 1}^N \langle (I - f(z) + F)^{-1} v ,\phi_i \rangle \psi_i = \sum_{i = 1}^N \langle v , [(I - f(z) + F)^{-1}]^* \phi_i \rangle \psi_i.$

Denoting ${\varphi_i(z) = [(I - f(z) + F)^{-1}]^* \phi_i}$, we get that ${g(z) = \sum_{i=1}^N \langle \cdot , \varphi_i(z) \rangle \psi_i}$.

Step 4:

If ${g(z) \varphi = \varphi}$, then ${\varphi = \sum_{n=1}^N \beta_n \psi_n}$ and ${\sum_{m=1}^N \langle \psi_m , \varphi_n(z) \rangle \beta_m = \beta_n}$ for ${m = 1, 2, \cdots , N}$, and some ${\vec{\beta} = (\beta_1, \cdots, \beta_N)}$. This is equivalent to say

$\displaystyle \sum_{m=1} ( \delta_{nm} - \langle \psi_m , \varphi_n(z) \rangle ) \beta_m = 0.$

Since ${ \langle \psi_m , \varphi_n(z) \rangle}$ is analytic on ${D_r}$, ${d(z) := \det ( \delta_{nm} - \langle \psi_m , \varphi_n(z) \rangle )}$ is also analytic. This implies the set ${S = \{z \in D_r: d(z) = 0\}}$ is either all of ${D_r}$ or a discrete set in ${D_r}$. If ${S = D_r}$, then the first alternative in the theorem holds.

Step 5:

Suppose ${d(z) \neq 0}$, we will show that ${I- g(z)}$ is invertible. Indeed, ${I - g(z)}$ is injective by step 4. By Inverse Mapping Theorem it remains to show it is also onto. Let ${\psi \in \mathcal{H}}$, we need to solve the equation ${(I- g(z)) \varphi = \psi}$ for some ${\varphi \in \mathcal{H}}$. Let ${\varphi = \psi + p}$, then ${p}$ has to solve

$\displaystyle (I - g(z)) (\psi + p) = \psi$

$\displaystyle \Leftrightarrow \quad p = g(z) (\psi + p).$

But ran ${g(z) \subset span \{ \psi_1 ,\cdots \psi_N\}}$. So ${p = \sum_{n=1}^N \gamma_n \psi_n}$ for some ${(\gamma_1 , \cdots , \gamma_N) \in {\mathbb C}^N}$. Substitute back and use the formula for ${g(z)}$, we get ${\gamma_n = \langle \psi, \varphi_n (z) \rangle + \sum_{m =1}^N \gamma_m \langle \psi_m , \varphi_n (z) \rangle}$. Therefore

$\displaystyle \sum_{m=1}^N (\delta_{mn} - \langle \psi_m , \varphi_n (z) \rangle ) \gamma_m = \langle \psi , \varphi_n (z) \rangle.$

Since ${d(z) \neq 0}$, this inhomogeneous system has a solution. This implies ${I - g(z)}$ is invertible if ${d(z) \neq 0}$.

Step 6:

We have proved for every ${z \in D}$, there is a neighborhood where the Analytic Fredholm Alternative holds. Define the sets ${O_1 := \{ z \in D:}$ there exists a neighborhood around ${z}$ where the first alternative holds\}, ${O_2 := \{ z \in D:}$ there exists a neighborhood around ${z}$ where the second alternative holds\}. Therefore ${O_1 \cap O_2 = \emptyset}$ and ${D = O_1 \cup O_2}$. By the definition of ${O_1}$ and ${O_2}$, we know both sets are open. Since ${D}$ is connected, we know either ${D = O_1}$ or ${D = O_2}$. This concludes the proof. $\Box$

Corollary 1 (Fredholm Alternative) If ${A}$ is compact, ${\lambda \in {\mathbb C}}$ nonzero, then either ${(\lambda I - A)^{-1}}$ exists or ${A \psi = \lambda \psi}$ has a nonzero solution.

Proof: Apply the theorem to ${f(z) = zA}$ with ${D = {\mathbb C}}$ at ${z = \frac{1}{\lambda}.}$ $\Box$

Corollary 2 (Riesz Schauder Theorem) Let ${A}$ be a compact operator on a separable Hilbert space ${\mathcal{H}}$, then the spectrum ${\sigma(A)}$ is a discrete having no limit points except perhaps at ${\lambda = 0}$. Furthermore, any nonzero ${\lambda \in \sigma(A)}$ is an eigenvalue with finite multiplicity.

Proof: Let ${f(z) = zA}$, define the set ${S:= \{ z \in {\mathbb C} : zA \psi = \psi \text{ has a nonzero solution} \}}$. Since ${ 0 \not\in S}$, by the Analytic Fredholm Alternative, ${S}$ is discrete. If ${\frac{1}{\lambda} \not\in S}$, then ${(I -\frac{1}{\lambda}A)^{-1}}$ exists so ${(\lambda - A)^{-1} = \lambda^{-1} (I -\frac{1}{\lambda}A)^{-1}}$ exists. Hence ${\lambda \not\in \sigma(A)}$. Therefore ${\sigma(A) \subset \{0\} \cup \{ \lambda : \frac{1}{\lambda} \in S\}}$.

The fact that nonzero eigenvalues have finite multiplicity follows from compactness. Otherwise we get infinite orthonormal set of eigenvectors with eigenvalue ${\lambda \neq 0}$, and it have no convergence subsequence. $\Box$