Logarithm on $A(\overline{\mathbb{D}})$

Question

Let $\mathbb{D}$ be the unit disk and let $A(\mathbb{\overline{D}})$ be the space of functions which are continuous on $\overline{\mathbb{D}}$ and holomorphic on $\mathbb{D}$.
I can prove by means of the holomorphic functional calculus for Banach algebras that any $f \in A(\overline{\mathbb{D}})$ which has no zeros admits a logarithm. However, this seems like cracking a nut with a sledgehammer, so the question is:

Does anyone know a more elementary proof of this fact?

I tried to use the fact that the set of $f \in A(\overline{\mathbb{D}})$ that are analytically continuable beyond $\overline{\mathbb{D}}$ is dense in $A(\overline{\mathbb{D}})$ but without success.

Answer 1

$\newcommand\DC{\overline{\Bbb D}}$

Sledgehammer or not, are you sure your proof by the functional calculus is correct? I don't see how it can work; seems to me that to make any sort of functional-calculus argument work we need to know there exists $g$ at least continuous on $f(\DC)$ with $e^{g(z)}=z$, and that's simply not true (for exammple, $f(\DC)$ could be an annullus centered at the origin).

Since $\overline{\Bbb D}$ is simply connected there exists a continuous function $L:\overline{\Bbb D}\to\Bbb C$ with $$e^L=f.$$It's not hard to show that it follows that $L$ is holomorphic in $\Bbb D$.