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Section 8.5 Fredholm Alternatives

Definition 8.5.1.
A linear operator \(A:X\to Y\) has finite rank, if its range is finite dimensional.
Consider a sequence \(\{x_n\}_n\) in \(H\) with \(\|x_n \|=1\) such that
\begin{equation*} \|(A-\lambda I) x_n\|\to 0, \qquad \mbox{for}\quad n\to \infty. \end{equation*}
Since the range of \(A\) is finite, there exists a convergent subsequence in the range of \(A\text{:}\)
\begin{equation*} \|A x_{n_j}-y\| \to 0, \qquad \mbox{for}\quad n\to \infty. \end{equation*}
Then
\begin{equation*} x_{n_j} = \frac{1}{\lambda} \Bigl(\underbrace{(\lambda I - A) x_{n_j}}_{\to 0 } + \underbrace{A x_{n_j}}_{\to y} \Bigr) \to \frac{1}{\lambda} y. \end{equation*}
For the norm of \(x_{n_j}\) we have
\begin{equation*} 1=\|x_{n_j}\| \to 1 = \frac{1}{\lambda}\|y\|, \qquad \mbox{ and} \qquad \|y\|\neq 0. \end{equation*}
Furthermore,
\begin{equation*} A x_{n_j} \to y \qquad \mbox{and} \qquad A x_{n_j} \to \frac{1}{\lambda} A y, \end{equation*}
which implies that \(Ay = \lambda y\) and \(\lambda\in \sigma_p(A)\text{.}\)
If \(\lambda\not\in \sigma_p(A)\) then, by the previous Proposition [cross-reference to target(s) "p:finiterank" missing or not unique], there exists a \(c>0\) such that
\begin{equation*} \|(A-\lambda I) x\| \geq c\|x\| \qquad \mbox{for all} \quad x\in H. \end{equation*}
Consider \(y\in \overline{\mbox{Range}{(A-\lambda I)}}\text{.}\) Then there exists a sequence \(\{x_n\}_n\) with
\begin{equation*} (A-\lambda I) x_n \to y, \qquad\mbox{as}\quad n\to\infty. \end{equation*}
Then
\begin{equation*} \|x_n - x_m\| \leq c^{-1} \|(A-\lambda I) x_n - (A-\lambda I) x_m\| \end{equation*}
and \(\{x_n\}_n\) is a Cauchy sequence with a limit in \(H\text{:}\) \(x_n \to x\text{.}\) Consequently, \((A-\lambda I) x = y\) and \(R_\lambda(A)y\) exists. Hence \(\mbox{Range}(A-\lambda I) = \mbox{domain}(R_\lambda(A))\) is closed. Since \(\lambda\not\in \sigma_p(A)\) it is also not in \(\sigma_p(A^*)\) and
\begin{equation*} \mbox{Range}(A-\lambda I) = (\mbox{ker}(A-\lambda I)^*)^\perp = H. \end{equation*}
Now let
\begin{equation*} T y := \{\mbox{ unique element } x \mbox{ such that } \quad (A-\lambda I)x =y\}. \end{equation*}
Then \((A-\lambda I) Ty = y \) and
\begin{equation*} c \|Ty\| \leq \|(A-\lambda I) Ty\| = \|y\| , \end{equation*}
which implies that \(T\) is bounded. In fact \(T=(A-\lambda I)^{-1}\) so \(\lambda\in \rho(A). \) If \(\lambda\in \sigma_p(A)\) then the eigenspace must be finite dimensional, since \(A\) has finite rank.
We consider \(\{\psi_i\}_{i=1\dots n} \) a finite basis of the finite dimensional space \(M\subset H\text{.}\) Then \(F\) has a spectral representation
\begin{equation*} F(\lambda) \phi = \sum_{i=1}^n (\gamma_i(\lambda), \phi) \psi_i, \end{equation*}
and the maps \(\lambda\mapsto \gamma_i(\lambda)\) are analytic in \(G\text{.}\) We define
\begin{equation*} \Lambda_{ij}(\lambda ) := (\gamma_i(\lambda), \psi_j). \end{equation*}
The inverse \((I- F(\lambda))^{-1}\) does not exist when \(F(\lambda)\phi = \phi\) has a nontrivial solution. We write this equation in components for \(\phi = \sum_{i=1}^n a_i \psi_i\) as
\begin{equation*} \phi = \sum_{i=1}^n a_i \psi_i = F(\lambda) \phi = \sum_{i=1}^n \left(\gamma_i(\lambda), \sum_{j=1}^n a_j \psi_j\right) \psi_i. \end{equation*}
Using orthogonality of the eigenfunctions, we find
\begin{equation*} a_i = \sum_{j=1}^n (\gamma_i(\lambda), \psi_i) a_j = \sum_{j=1}^n \Lambda_{ij} a_j, \end{equation*}
which we write in matrix notation as
\begin{equation*} (I-\Lambda(\lambda))a = 0 ,\qquad a=(a_1,\dots, a_n), \quad \Lambda= (\Lambda_{ij}). \end{equation*}
For a non-trivial solution we need
\begin{equation*} d(\lambda)=\det(I-\Lambda(\lambda)) =0. \end{equation*}
By the assumption \(d(\lambda)\) is analytic in \(\lambda\in G\text{,}\) hence \(d(\lambda)\) is either identically zero in \(G\) (case 1.), or the zeroes form a discrete set.

Finally, since \(F(\lambda)\) has finite rank, we can only have finitely many solutions of \(F(\lambda)\phi = \phi\text{.}\)
[Proof-Sketch] Use a spectral representation of \(B(\lambda)\) and approximate \(B(\lambda)\) by considering finite sums. Use the previous result and let \(n\) go to infinity.
[Spectrum of an integral operator] Compute the spectrum of
\begin{equation*} u \mapsto Ku = \int_0^1 e^{-\gamma(x-y)} u(y) dy \end{equation*}
in
\begin{equation*} D(K) = (L^2(0,1))_+ = \{ u\in L^2(0,1): u\geq 0\}. \end{equation*}
Since the kernel of the integral operator
\begin{equation*} k(x,y) = e^{-\gamma(x-y)} \in L^2((0,1)\times (0,1)), \end{equation*}
the operator \(K\) is a compact Hilbert-Schmidt operator. Hence it has a discrete spectrum of eigenvalues with a possible limit point at \(0\text{.}\)

To solve \(Ku = \lambda u\text{,}\) that is
\begin{equation*} \int_0^1 e^{-\gamma(x-y)} u(y) dy = \lambda u(x), \end{equation*}
we multiply by the kernel \(e^{-\gamma(z-x)}\) and integrate:
\begin{align*} \int_0^1 e^{-\gamma(z-x)}\int_0^1 e^{-\gamma(x-y)} u(y) \, dy\, dx \amp=\amp \lambda \int_0^1 e^{-\gamma(z-x)} u(x) dx\\ \amp=\amp \lambda^2 u(z). \end{align*}
On the other hand we compute directly
\begin{align*} \int_0^1 e^{-\gamma(z-x)}\int_0^1 e^{-\gamma(x-y)} u(y) \, dy\, dx \amp=\amp \int_0^1 \int_0^1 e^{-\gamma z+\gamma y} u(y) dy\, dx\\ \amp=\amp \int_0^1 e^{-\gamma(z-y)} u(y) dy\\ \amp=\amp \lambda u(z). \end{align*}
Hence for \(u\neq 0\) we have \(\lambda^2 = \lambda\text{,}\) which implies \(\lambda =0\) or \(\lambda =1\text{.}\)

If \(\lambda =0\text{,}\) then we have
\begin{equation*} \int_0^1 e^{-\gamma(x-y)} u(y) dy =0 \end{equation*}
and since \(u\geq 0\) this implies that \(u=0\) a.e. in \([0,1]\text{.}\) Since we work in \(L^2\) this means \(u=0\text{,}\) and \(u\) is not an eigenfunction. Hence \(\lambda=0\) is not an eigenvalue.

If \(\lambda = 1\text{,}\) then we need to solve
\begin{equation*} \int_0^1e^{-\gamma(x-y)} u(y) dy = u(x). \end{equation*}
We try \(u(x) = e^{-\gamma x}\text{:}\)
\begin{equation*} \int_0^1e^{-\gamma(x-y)} u(y) dy = e^{-\gamma x} \int_0^1 e^{\gamma y} e^{-\gamma y} dy = e^{-\gamma x} = u(x). \end{equation*}
Hence \(u(x) = e^{-\gamma x}\) is an eigenfunction of \(K\) with eigenvalue \(1\text{.}\) The spectrum is
\begin{equation*} \sigma(K) = \{1\}. \end{equation*}
Moreover, since \(k(x,y)>0\) we can apply the Krein-Rutman Theorem \cite{KreinRutman} (not covered here), and find that \(\lambda=1\) has multiplicity one with a unique positive eigenfucntion \(u(x) = e^{-\gamma x}\text{.}\)

Let \(G:=\CC\backslash\{0\}\) and \(B(\lambda) = \frac{1}{\lambda} A\text{.}\) We apply the previous Compact Fredholm Alternative. Note that \(\frac{1}{\lambda}\to \infty\) is possible, i.e., \(\lambda\to 0\) is a possible limit point.
All eigenvalues are real, since \(A\) is self-adjoint. Based on the Spectral Fredholm Alternative the eigenvalues form a disctrete set, each with finite multiplicity. Thus we enumerate them according to their multiplicity \(\{\lambda_i\}_i\) with eigenvectors \(\{\psi_i\}_i\text{,}\) which are orthogonal. We can always normalize them to norm \(1\text{.}\)

Now we set
\begin{equation*} M:= \overline{\mbox{span}\{\psi_i, i=1, \dots, \infty\}} \end{equation*}
and we show that \(\mbox{Range}(A)\subset M\text{.}\)

First, since \(A\) is compact, we have \(AM=M\text{.}\) For \(\phi\in M^\perp\) we have that \((A\phi, \psi) = (\phi, A\psi) = 0\) for all \(\psi\in M\text{.}\) Hence \(A\phi\in M^\perp\) as well. This implies that both \(M\) and \(M^\perp\) are invariant under \(A\text{.}\) We define
\begin{equation*} \hat A := A\Big|_{M^\perp} \end{equation*}
and we will show that \(\hat A=0\text{.}\) Indeed, \(\hat A\) is self adjoint and compact (same as \(A\)), and each spectral value is an eigenvalue. But each eigenvector belongs already to \(M\text{.}\) Hence \(\sigma_p(\hat A) =\emptyset\text{.}\) Hence \(\hat A=0\text{.}\) It follows that \(\{\psi_i\}_i\) is a basis of \(\mbox{Range}(A)\) and we write
\begin{equation*} A(u) =\sum_{i=1}^\infty (\psi_i, Au) \psi_i = \sum_{i=1}^\infty (A\psi_i, u) \psi_i = \sum_{i=1}^\infty \lambda_i(\psi_i, u) \psi_i. \end{equation*}