Proposition 29.5.3.label Let $A$ be a unital Banach algebra and $x \in A$, then

  1. (1)

    $\exp(x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$.

  2. (2)

    $\exp(x) \in G_{0}(A)$ with $\exp(x)^{-1}= \exp(-x)$.

  3. (3)

    For any $y \in A$ commuting with $x$, $\exp(x + y) = \exp(x)\exp(y)$.

  4. (4)

    Let $\ell: \complex \setminus (-\infty, 0] \to \complex$ be the principal logarithm. If $\sigma_{A}(x) \subset \bracs{\lambda \in \complex| \text{Im}(\lambda) \in (-\pi, \pi)}$, then $\ell(\exp(x)) = x$.

Proof. (1): By the power series representation of $\exp(x)$ and continuity of the holomorphic functional calculus.

(2): By the homomorphism property of the functional calculus, $\exp(x)^{-1}= \exp(-x)$. Thus $\exp(x)$ is connected to $1$ by the path $t \mapsto \exp(tx)$.

(3): Since the exponential is an entire function, the following series converges absolutely, and is eligible for arbitrary manipulations:

\begin{align*}\exp(x)\exp(y)&= \sum_{n = 0}^{\infty} \sum_{k = 0}^{\infty} \frac{x^{n}y^{k}}{n!k!}= \sum_{n = 0}^{\infty} \sum_{k = 0}^{n} \frac{x^{k}y^{n-k}}{k!(n-k)!}\\&= \sum_{n = 0}^{\infty} \frac{1}{n!}\sum_{k = 0}^{n}{n \choose k}x^{k}y^{n-k}\\&= \sum_{n = 0}^{\infty} \frac{(x + y)^{n}}{n!}= \exp(x + y)\end{align*}

(4): By the Spectral Mapping Theorem.$\square$

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