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Section 8.1 Spectrum and resolvent of an operator

The most fundamental concept in spectral theory is the resolvent of an operator. Since it has to do with inverses, we start with some general consideration.

Let \(B\) be a Banach space and let \((A,D(A))\) be an operator on \(B\text{,}\) not necessarily continuous nor closed. Assume that \(A\) is bijective, namely \(\ker A=\{0\}\) and \(R(A)=B\text{.}\) Is \(A^{-1}\) necessarily continuous? The next example shows that this is not the case.

Take \(X=\ell^2\) and let \(e_n\) be the standar orthonormal basis. Extend \(e_n\) to a Hamel basis \(B=\{e_n,f_\alpha\}\text{,}\) namely every vector can be written as the finite linear combination of elements in \(B\text{.}\) Define \(T\) as \(Te_n=e_n/n\) for the \(e_n\) and as \(Tf_\alpha=f_\alpha\) for every other element of the basis and extend \(T\) to \(X\) by linearity. Clearly \(T\) is surjective by construction. On the other side, \(T^{-1}\) is not bounded because
\begin{equation*} \frac{\|T^{-1}e_n\|}{\|e_n\|} = n\to\infty\text{ as }n\to\infty. \end{equation*}
Notice that, on the contrary, the operator \(S\) defined on \(e_n\) and extended to \(X\) via series as
\begin{equation*} S(x) = \sum_{i=1}^\infty \frac{x_n}{n}e_n \end{equation*}
is not surjective. For instance, the element \(x_n=1/n\in\ell^2\) is not in the image of \(S\) since \((1,1,\dots)\not\in\ell^2\text{.}\)

On the other side, if \(A\) is a closed surjective operator, its inverse is necessarily continuous. Indeed, the inverse of a closed operator is closed (their graphs are one the specular image of the other) and closed operators defined on the whole space are necessarily continuous.

This is a first sign of the fact that closed operators are the optimal type of operator for our goals. In fact, from now on, all operators we deal with, unless otherwise specified, will be assumed to be closed.

Definition 8.1.2. Resolvant and spectrum.
We call resolvant of \(A\) the set
\begin{equation*} \rho(A) = \{\lambda\in\bC:\lambda Id - A\text{ is invertable}\}. \end{equation*}
We call spectrum of \(A\) the set
\begin{equation*} \sigma(A) = \bC\setminus \rho(A). \end{equation*}
We sort the spctrum into three components:
\begin{gather*} \sigma_p(A) = \{\lambda\in\bC:\lambda Id - A\text{ is not injective}\}\\ \sigma_c(A) = \{\lambda\in\bC:\lambda Id - A\text{ is injective with dense image}\}\\ \sigma_r(A) = \{\lambda\in\bC:\lambda Id - A\}\text{ is injective with non-dense range}\} \end{gather*}
If \(A\) is inside the unit ball, \(I-A\) is invertble.

Assume that \(\|A\|<1\text{.}\) Then
\begin{equation*} T = Id + A + A^2 + \dots \end{equation*}
is bounded and is the inverse of \(Id - A\text{.}\) Indeed,
\begin{equation*} \|T\| \leq \|Id\| + \|A\| + \|A^2\| + \dots \leq 1 + \|A\| + \|A\|^2 + \dots = \frac{1}{1-\|A\|} \end{equation*}
and
\begin{equation*} T(Id-A) = (Id + A + A^2 + \dots)(Id-A) = Id + A + A^2 + \dots - A - A^2 - \dots = Id. \end{equation*}
Assume that \(\lambda\in\rho(A)\text{,}\) so that \(\lambda Id-A\) has a bounded inverse. Then
\begin{equation*} \lambda' Id-A = (\lambda'-\lambda) Id + \lambda Id -A = - (\lambda Id -A)(Id - (\lambda'-\lambda)(\lambda Id -A)^{-1}). \end{equation*}
By the result above, if
\begin{equation*} |\lambda'-\lambda|\|(\lambda Id -A)^{-1}\|<1, \end{equation*}
namely if
\begin{equation*} |\lambda'-\lambda|<\frac{1}{\|(\lambda Id -A)^{-1}\|}, \end{equation*}
the series converges and does converge to the bounded operator \((\lambda' Id-A)^{-1}\text{.}\)
By the argument above,
\begin{equation*} \lambda Id-A = \lambda(Id -\frac{A}{\lambda}) \end{equation*}
has a bounded inverse if
\begin{equation*} \frac{\|A\|}{|\lambda|}=\left\|\frac{A}{\lambda}\right\|<1. \end{equation*}
This shows that bounded operators are not good candidates to represent physical quantities that can assume arbitrarily high values.
Consider the multiplication operator \(M\) on \(L^2([0,1])\) given by
\begin{equation*} (Mf)(x) = xf(x). \end{equation*}
The reader can verify that
\begin{equation*} \|M\| = 1. \end{equation*}
We claim that \(M\) has no eigenvalues. Indeed, for any \(\lambda\text{,}\) the condition
\begin{equation*} xf(x) = \lambda f(x) \end{equation*}
has, modulo a zero-measure set, the unique solution
\begin{equation*} f(x) = 0. \end{equation*}
On the other side,
\begin{equation*} \sigma(A)\supset[0,1]. \end{equation*}
Indeed, the operator \(\lambda Id-A\) is not surjective for any \(\lambda\in[0,1]\text{.}\) If it were, then we could solve in \(L^2([0,1])\) the equation
\begin{equation*} \lambda f - xf = g, \end{equation*}
that has the unique solution
\begin{equation*} f(x) = \frac{g(x)}{\lambda -x}. \end{equation*}
But \(1\in L^2([0,1])\) while \(\frac{1}{\lambda -x}\not\in L^2([0,1])\) for \(\lambda\in[0,1]\text{.}\) One can prove that the range of \(M\) is dense, so this spectrum is purely continuous.