Why can entire function be written as exponential, and why is it bounded in this way?

Question

Let $A$ be a commutative complex Banach algebra with unit element $e$. Suppose now that $f(x) \in \sigma(x)$ for every $x \in A$ where $\sigma(x)$ denotes the spectrum of $x$.

Now, let $x\in A$ and define $$\phi(\lambda) = f[\exp(\lambda x)]$$ where $\lambda \in \mathbb C$. Clearly $\phi(\lambda)$ is entire. Since $f(x)\in \sigma(x)$ for every $x \in A$ we know that $\phi(\lambda) \neq0$.

The paper then goes on to say the following:

1. Since $\phi(\lambda)\neq0$ it can be written in the form $\phi(\lambda)=\exp[\psi(\lambda)]$ for some entire complex function $\psi(\lambda)$.
2. We have the following estimation: $|\phi(\lambda)|\le\|f\|\exp(|\lambda|\|x\|)$

Why is (1) and (2) true? I cannot seem to see why.

Answer 1

If $\phi(\lambda) = 0$ for some $\lambda$, then $f(\exp(\lambda x)) = 0$ for some $\lambda$, which by the assumption on $f$ says that $0$ is in the spectrum of $\exp(\lambda x)$. However the spectral mapping theorem says that the spectrum of $\exp(\lambda x)$ would be $\exp(\lambda \sigma(x))$, since $\exp$ is entire, which does not include zero.

Since $\phi$ is entire and never zero, you can write it as an exponential of another function. The obvious choice would be to have $\phi(\lambda) = \exp(\log(\phi(\lambda)))$. However there is no guarantee that $\log\phi$ is entire. However the idea of using $\log$ is a good one. The way to go about it is to define $\psi$ to be such that

$$\psi'(\lambda) = \frac{\phi'(\lambda)}{\phi(\lambda)}.$$

Note that this is nothing more than the derivative of $\log\phi$ in disguise. This ratio is extremely useful and important in complex analysis. Since $\phi$ is nonzero, the above ratio defines an entire function and so an antiderivative of $\psi'$ exists everywhere and is entire.

As for the last part, recall that $|f(y)| \le \|f\|\|y\|$ and make use of the series for $\exp$.

Answer 2

My guess: $$ f(e^{\lambda x})=f\left(\sum_{n=0}^{\infty}\frac{\lambda^{n}}{n!}x^n\right) = \sum_{n=0}^{\infty}\frac{\lambda^n f(x^n)}{n!} = \sum_{n=0}^{\infty}\frac{\lambda^n f(x)^n}{n!}=e^{\lambda f(x)}. $$