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Question

Math 6420 • PDE

Spring 2017 • Question 5

Let \(T:H\to H\) be a compact, self-adjoint linear operator defined on a separable Hilbert space \(H\). Let \((v_n)\) be a complete orthonormal sequence of eigenvectors of \(T\) and let \((\lambda _n)\) be the corresponding eigenvalues.

  1. Let \(\lambda \in \rho (T)\). Show that the resolvent of \(T\) at \(\lambda \), acting on some \(x\in H\) can be written as \[ R_\lambda x=\sum _{n=1}^{\infty }\frac {1}{\lambda _n-\lambda }\langle x,v_n\rangle v_n. \]
  2. Use part (a) to show that if \(\lambda ,\mu \in \rho (T)\) then the first resolvent identity holds: \[ R_\lambda -R_\mu =(\lambda -\mu )R_\lambda R_\mu . \]

Raw LaTeX

Let \(T:H\to H\) be a compact, self-adjoint linear operator defined on a separable Hilbert
space \(H\). Let \((v_n)\) be a complete orthonormal sequence of eigenvectors of \(T\) and let
\((\lambda_n)\) be the corresponding eigenvalues.
\begin{enumerate}
\item[(a)] Let \(\lambda\in\rho(T)\). Show that the resolvent of \(T\) at \(\lambda\), acting
on some \(x\in H\) can be written as
\[
  R_\lambda x=\sum_{n=1}^{\infty}\frac{1}{\lambda_n-\lambda}\langle x,v_n\rangle v_n.
\]
\item[(b)] Use part (a) to show that if \(\lambda,\mu\in\rho(T)\) then the first resolvent
identity holds:
\[
  R_\lambda-R_\mu=(\lambda-\mu)R_\lambda R_\mu.
\]
\end{enumerate}