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\begin{document}

\begin{center}
{\Large \em Theorems on Spectral Representation of Operators}\\
\vspace{.15in}
\end{center}

\vspace{.4in}

\large
Let $L:D(L)\subset H \rightarrow H$ be an operator on Hilbert
space $H$ with domain $D(L)$. Let $\sigma (L)$ be the spectrum of $L$.
The resolvent operator of $L$ is
\eq
R(\lambda;L) \equiv (L-\lambda I)^{-1} \quad , \lambda \in \complex
\endeq
where ever it is defined. The following theorems can be found in
Linear Operators Part I, by Dunford and Schwartz.

\vspace{.2in}
\noindent
{\bf Theorem:} {\em Let $L$ be a bounded operator defined on $H$
and $f(z)$ an analytic function defined on an open set $U \subset
\complex$ containing $\sigma (L)$. If $f(z) \ne 0$ on $\sigma (L)$
and the boundary curve $C$ of $U$ (counterclockwise sense, 
rectifiable Jordan) then the operator
\eq
f(L) = - \frac{1}{2\pi i} \int_C f(\lambda) R(\lambda;L) d\lambda
\endeq
is well defined on $H$ in the sense that}
\eq
f(L)u = - \frac{1}{2\pi i} 
\int_C f(\lambda) R(\lambda;L) \ u \  d\lambda \quad ,  u \in H
\endeq

\vspace{.2in}
\noindent
In particular note the identities if $f(z)=1$ or $f(z)=z$. Also, if
$L$ is bounded it can be shown $\sigma (L)$ is so that an appropriate
$U$ can always be found.

If $L$ is a closed operator (not necessarily bounded) 
with nonempty resolvent set (complement
of $\sigma (L)$ in $\complex$) there is a similar theorem:

\vspace{.2in}
\noindent
{\bf Theorem:} {\em Let $L$ be a closed operator defined on
 $D(L)\subset  H$
and $f(z)$ an analytic function defined on an open set $U \subset
\complex$ containing $\sigma (L)$. If $f(z) \ne 0$ on $\sigma (L)$
and the boundary curve $C$ of $U$ (counterclockwise sense, 
and possibly unbounded)  and $f$ is analytic at $\infty$ then the
 operator
\eq
f(L) = f(\infty) I - \frac{1}{2\pi i} \int_C f(\lambda) R(\lambda;L) d\lambda
\endeq
is well defined on $D(L)$.}

\vspace{0.1in}
\noindent
Here closed means that if $\{u_n\}\subset D(L)$, $u_n \rightarrow u$
and $Lu_n \rightarrow v$ then $u \in D(L)$ and $Lu=v$.
\end{document}