> [!definition] > > Let $\alg$ be a unital complex [[Banach Algebra|Banach algebra]] and $x \in \alg$, then the **spectrum** of $x$ is > $ > \spec{x} = \bracs{\lambda \in \complex: \lambda e - x \text{ not invertible}} > $ > [!theorem] > > Let $A: \cx \to \cx$ be a bounded linear operator, then its spectrum can be decomposed into > 1. Point spectrum: $\bracs{\lambda \in \sigma(A): (A - \lambda I) \text{ not injective}}$. > 2. Continuous spectrum: $\bracs{\lambda \in \sigma(A): (A - \lambda I) \text{ injective}, \ol{A(\cx)} = \cx}$. > 3. Residual spectrum: $\bracs{\lambda \in \sigma(A): (A - \lambda I) \text{ injective}, \ol{A(\cx)} \ne \cx}$. > [!theorem] > > Let $x \in \alg$, then $\spec{x}$ is a closed subset of $\bracs{\lambda: \abs{\lambda} \le \norm{x}}$. > > *Proof*. Since $\lambda e - x$ is invertible for all $\lambda$ with $\abs{\lambda} > \norm{x}$, $\spec{x}$ is a subset of the disc. On the other hand, since the set of invertible elements is open, $\spec{x}$ is closed. > [!theorem] > > Let $x \in \alg$, then $\spec{x}$ is non-empty. > > *Proof*. Suppose that $\spec{x}$ is empty, then $R(\lambda)$ would be holomorphic on $\complex$ with it being bounded. By Liouville's Theorem applied to $\phi \circ R(\lambda)$ for all $\phi \in \alg^*$, $R(\lambda)$ is identically zero. > [!theorem] > > Let $\ch$ be a [[Hilbert Space|Hilbert space]] and $T \in L(\ch, \ch)$ be a [[Bounded Linear Operator|bounded linear operator]]. If $\lambda \in \sigma_r(T)$, then $\lambda \in \sigma_p(T^*)$. > > *Proof*. If $\lambda \in \sigma_r(T)$, then $\ol{(T - \lambda I)(\ch)} \subsetneq \ch$. Let $x \in \ol{(T - \lambda I)(\ch)}^\perp$. Thus for any $y \in \ch$, > $ > \angles{x, (T - \lambda I)y} = \angles{(T^* - \lambda I)x, y} = 0 > $ > Taking $y = (T^* - \lambda I)x$ yields that $\lambda \in \sigma_p(T^*)$. > [!theorem] > > Let $\alg$ be a unital Banach algebra and $\cb \subset \alg$ be a closed subalgebra containing $e$. If $x \in \cb$ and $\sigma_\cb(x)$ is nowhere dense in $\complex$, then $\sigma_\alg(x) = \sigma_\cb(x)$. > > *Proof*. Let $\lambda \in \sigma_\cb(x)$, then there exists a sequence $\lambda_n \in \complex \setminus \sigma_b(x)$ converging to $\lambda$. Therefore $\norm{(\lambda_n e - x)^{-1}} \to \infty$, so $\lambda e - x$ is not invertible in $\alg$. > [!theorem] > > Let $\alg$ be a unital $C^*$-algebra and $\cb \subset \alg$ be a closed subalgebra containing $e$. > 1. If $x \in \cb$ and $x$ is invertible in $\alg$, then $x^{-1} \in \cb$. > 2. If $x \in \cb$, then $\sigma_\alg(x) = \sigma_\cb(x)$. > > *Proof*. Let $x \in \cb$, $y = x^*x$, and $\cc \subset \alg$ be the closed subalgebra generated by $y$ and $e$, with $\cc \subset \cb \subset \alg$. If $x$ is invertible in $\alg$, then so is $y$, and $0 \not\in \sigma_\alg(y)$. Since $\mathcal C$ is a commutative $C^*$ algebra, $\sigma_\cc(y) = \sigma_\alg(y)$. Thus $0 \not\in \sigma_c(y)$ and $y$ is invertible in $\cc$. Then we can express $x^{-1} = y^{-1}x^* \in \cb$, so $x$ is invertible in $\cb$. > [!theorem] > > Let $\alg$ be a unital $C^*$-algebra. > 1. If $x \in \alg$ with $xx^* = x^*x$ (normal), then $\rho(x) = \norm{x}$. > 2. If $\cb$ is a Banach $*$-algebra and $\phi: \cb \to \alg$ is a $*$-homomorphism, then $\norm{\phi} \le 1$. > > *Proof*. Let $\cc$ be the closed subalgebra generated by $x$, $x^*$, and $e$, then $\cc$ is a commutative $C^*$-algebra. By the [[Gelfand-Naimark Theorem]], $\rho(x) = \norm{\Gamma_\cc x}_u = \norm{x}$. > > If $\phi$ is bounded, then for any $y \in \cb$, > $ > \norm{\phi(y^*y)^n} \le C\norm{(y^*y)^n} \le \norm{y^*y}^n \le \norm{y}^{2n} > $ > On the other hand, if $x = \phi(y^*y)$, then > $ > \norm{\phi(y)}^2 = \norm{\phi(y^*y)} = \limv{n}\norm{\phi(y^*y)^n}^{1/n} \le \limv{n}\norm{\phi}^{1/n}\norm{y^2} = \norm{y^2} > $ > so $\norm{\phi(y)} \le \norm{y}$.