Spectra of Bounded Operators in Banach Space

Kreyszig Functional Analysis

We let $X$ denote an arbitrary complex Banach space throughout these notes. The following theorem makes spectral theory in such spaces relatively well-behaved:

Theorem 1: Let $T:X\to X$ be a closed linear operator and $\lambda$ be a regular value. The resolvent $R_{\lambda }(T)$ must be a bounded linear operator defined on the entire space $X$ (i.e. $R_{\lambda }(T)\in B(X,X)$).

Proof) As $\lambda \in \rho (T)$, $R_{\lambda }(T)$ must, by definition, be bounded and linear. Remember that $T$ is closed iff $\mathcal{G}=\{(x,Tx)\}$ is closed in $X\times X$. One can see that $\{(x,Tx)\}$ is closed iff $\{(x,Tx-\lambda x)\}$ is closed, and thus $T$ is closed iff $T_{\lambda }$ is closed (this is generally true for any arbitrary $\lambda \in \mathbb{C}$). Noting that $R_{\lambda }(T)$ is the inverse of $T_{\lambda }$, we can see $R_{\lambda }(T)$ thus must be closed. $R_{\lambda }(T)$ is thus a closed and bounded linear operator whose range is a Banach space, forcing its domain $\mathcal{D}(R_{\lambda }(T))$ to be closed. As, by definition, $\mathcal{D}(R_{\lambda }(T))$ is dense in $X$, we have $\mathcal{D}(R_{\lambda }(T))=\overline{\mathcal{D}(R_{\lambda }(T))}=X$. We therefore have that $R_{\lambda }(T)$ is a bounded linear operator defined on the entire space $X$.

Theorem 2 (Neumann Series): Let $T\in B(X,X)$ and $\Vert T\Vert <1$. Then, $(I-T{)}^{-1}$ exists as a bounded linear operator defined on the whole space $X$, and

$$(I-T{)}^{-1}=\underset{n\to \infty }{\lim }\sum _{j=0}^n{T}^j=I+T+T^2+\dots$$

The limit is taken in $B(X,X)$, which is also a Banach space.

Proof) Note that the series in the RHS is absolutely convergent, as $\Vert T^j\Vert \le \Vert T{\Vert }^j$ and $\sum \Vert T{\Vert }^j$ is a convergent geometric series. Thus, it is Cauchy and converges in $B(X,X)$ to some bounded linear operator $Q\in B(X,X)$. It suffices to show $Q=(I-T{)}^{-1}$, or equivalently, that $Q(I-T)=(I-T)Q=I$. This is a well-behaved problem as both $I-T$ and $Q$ are everywhere defined on $X$.

We will be cautious here as operator composition is not native to Banach space structure. We first show that for any uniformly convergent sequence of operators $f_n\in B(X,X)$ such that $f_n\to f$ and $g\in B(X,X)$, $g{f}_n$ is uniformly convergent and $g{f}_n\to gf$. This is because, by completeness, for any $ϵ>0$ there exists an $N$ such that for any $n,m>N$ $\Vert f_n-f_m\Vert <ϵ/\Vert g\Vert$. Operator composition satisfies $\Vert g{f}_n\Vert \le \Vert g\Vert \Vert f_n\Vert$; therefore, for any $n,m>N$, $\Vert g{f}_n-g{f}_m\Vert =\Vert g(f_n-f_m)\Vert <ϵ$. $g{f}_n$ is thus Cauchy and uniformly convergent in $B(X,X)$.

We prove that $(I-T)Q=I$ ($Q(I-T)=I$ can then be shown in the same way).

$$(I-T)Q=(I-T)\left(\underset{n\to \infty }{\lim }\sum _{j=0}^n{T}^j\right)=\underset{n\to \infty }{\lim }(I-T)\left(\sum _{j=0}^n{T}^j\right)=\underset{n\to \infty }{\lim }(I-T^{n+1})=I$$

The sequence $I-T^{n+1}$ uniformly converges to both $I$ and $(I-T)Q$ in $B(X,X)$, and by the uniqueness of limits this implies $(I-T)Q=I$.

Theorem 3: The resolvent set $\rho (T)$ of a bounded linear operator $T\in B(X,X)$ is open.

Proof) Take any ${\lambda }_0\in \rho (T)$. By Theorem 1, $R_{{\lambda }_0}(T)\in B(X,X)$. For any arbitrary $\lambda \in \mathbb{C}$, $T-\lambda I=(T-{\lambda }_{0}I)(I-(\lambda -{\lambda }_0)R_{{\lambda }_0}):=(T-{\lambda }_{0}I)V_{\lambda }$. If $\lambda$ satisfies $|\lambda -{\lambda }_0|<1/\Vert R_{{\lambda }_0}\Vert$, then by Theorem 2 $V_{\lambda }^{-1}$ exists and $V_{\lambda }^{-1}\in B(X,X)$. As $T-{\lambda }_{0}I$, by definition, has the inverse $R_{{\lambda }_0}\in B(X,X)$, we conclude that $T-\lambda I=(T-{\lambda }_{0}I)V_{\lambda }$ must have the inverse

$$(T-\lambda I{)}^{-1}=V_{\lambda }^{-1}{R}_{{\lambda }_0}(T)=\left(\underset{n\to \infty }{\lim }\sum _{j=0}^n{\left((\lambda -{\lambda }_0)R_{{\lambda }_0}(T)\right)}^j\right)R_{{\lambda }_0}(T)$$

By the logic of Theorem 2, this equals

$$=\underset{n\to \infty }{\lim }\left(\sum _{j=0}^n[(\lambda -{\lambda }_0)R_{{\lambda }_0}(T){]}^j{R}_{{\lambda }_0}(T)\right)=\underset{n\to \infty }{\lim }\left(\sum _{j=0}^n(\lambda -{\lambda }_0{)}^j{R}_{{\lambda }_0}(T{)}^{j+1}\right)=\sum _{j=0}^{\infty }(\lambda -{\lambda }_0{)}^j{R}_{{\lambda }_0}(T{)}^{j+1}$$

As this is the composition of two functions in $B(X,X)$, $(T-\lambda I{)}^{-1}\in B(X,X)$, which, a fortiori, implies $\lambda \in \rho (T)$ for all $\lambda \in \mathbb{C}$ satisfying $|\lambda -{\lambda }_0|<1/\Vert R_{{\lambda }_0}\Vert$. Thus, $\rho (T)$ is open in $\mathbb{C}$.

Theorem 4: The spectrum $\sigma (T)$ of a bounded linear operator $T\in B(X,X)$ lies in the disc given by $|\lambda |\le \Vert T\Vert$.

Proof) It suffices to show that for any $\lambda$ such that $|\lambda |>\Vert T\Vert$, $\lambda \in \rho (T)$. Note that

$$T_{\lambda }=(T-\lambda I)=-\lambda (I-\frac{1}{\lambda }T)$$

Thus, for any $\lambda$ such that $|\lambda |>\Vert T\Vert$, by Theorem 2, $T_{\lambda }$ has a well-defined inverse in $B(X,X)$ given by

$$R_{\lambda }(T)=-\frac{1}{\lambda }\sum _{j=0}^{\infty }{\left(\frac{1}{\lambda }T\right)}^j$$

A fortiori, $\lambda \in \rho (T)$.

Theorems 3 and 4 have two notable implications. First, as $\sigma (T)$ is bounded and closed in $\mathbb{C}$, it must be compact. Furthermore, as $\sigma (T)$ is bounded we can safely define the spectral radius

$$r_{\sigma }(T)=\underset{\lambda \in \sigma (T)}{\sup }|\lambda |$$

Theorem 5 (Spectral Mapping for Polynomials): Let $T\in B(X,X)$ and $p(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_0$, where $a_n\ne 0$. Then,

$$\sigma (p(T))=p(\sigma (T))$$

where $p(T)$ is the operator

$$p(T)=a_n{T}^n+a_{n-1}{T}^{n-1}+\dots +a_{0}I$$

Proof) We take it as a given the result that $\sigma (T)$ is nonempty for any $T\in B(X,X)$.

We first prove that $\sigma (p(T))\subset p(\sigma (T))$. Let $\eta$ be an arbitrary element of $\sigma (p(T))$. Note that, by the fundamental theorem of algebra, $p(x)-\eta =a_n(x-{\gamma }_1)(x-{\gamma }_2)\dots (x-{\gamma }_n)$ for complex numbers ${\gamma }_i\in \mathbb{C}$. This gives $p(T{)}_{\eta }=p(T)-\eta I=\prod (T-{\gamma }_{i}I)=\prod T_{{\gamma }_i}$. If ${\gamma }_i\in \rho (T)$ for every $i$, the product $p(T{)}_{\eta }$ must also have an inverse in $B(X,X)$ and $\eta \in \rho (p(T))$, giving a contradiction. WLOG, say ${\gamma }_1\in \sigma (T)$. Then, $\eta =p({\gamma }_1)\in p(\sigma (T))$.

We now show $p(\sigma (T))\subset \sigma (p(T))$. Take any arbitrary element $\eta \in \sigma (T)$ and let $\kappa =p(\eta )$. It suffices to show that $\kappa \in \sigma (p(T))$. We consider the cases where $\eta$ is in the point spectrum or residual spectrum of $T$. Note that the continuous spectrum of $T$ is empty as if $R_{\eta }(T)$ exists and is defined on a dense subset of $X$, by closure properties it must be defined on the entire set $X$ and thus bounded by the closed graph theorem.

If $\eta$ is in the point spectrum, there is some nonzero $x\in X$ such that $(T-\eta I)x=0$, or $Tx=\eta x$. This same $x$ must then satisfy $p(T)x=p(\eta )x$, and thus $p(\eta )$ is in the point spectrum of $p(T)$. That is, $p(\eta )\in \sigma (p(T))$. If $\eta$ is in the residual spectrum, the range of $T-\eta I$ is a proper subset of $X$. Note that $x-\eta$ divides $p(x)-p(\eta )$, and thus $p(T)-p(\eta )I=(T-\eta I)g(T)$ where $g(T)$ is some polynomial of $T$. Thus, the range of $p(T{)}_{p(\eta )}$ is a subset of the range of $T_{\eta }$, which is a proper subset of $X$. Thus, $p(\eta )\in \sigma (p(T))$.