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

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