The set $B=${$f \in C(D): f(z) = \sum^{\infty}_{n=0}a_n z^n$ , with $(a_n) \in l^1(\mathbb{N})$} is "spectrally invariant" Let $D=${$z \in \mathbb{C}: |z| \leq 1$}
$C(D)=${$f:D\rightarrow \mathbb{C}: f$ is continuous}
Im trying to prove that the set $B=${$f \in C(D): f(z) = \sum^{\infty}_{n=0}a_n z^n$ , with $(a_n) \in l^1(\mathbb{N})$} is "spectrally invariant", that is, if $f \in B$ and $f(z) \neq 0  \ \  \forall z \in D $, then $1/f \in B$.
I know that $1/f$ is analytical in $D$, but I dont know how to use this to estimate the coeficients of the power series of $1/f$...
Any hints?
 A: Since the OP says he or she is studying Banach algebras, just a few hints.
(Readers unfamilar with Banach algebras should read the relevant chapter in Rudin Real and Complex Analysis. You'll be glad you did! It's lovely stuff, and the present exercise shows that it's very powerful. I mean how the hell does one get information about the Fourier coefficients of $1/f$ "directly"? That 100 pages of Wiener mentioned in a comment above is now just a bit of elegant algebra. The only thing in the chapter that seems somewhat "hard" to me is the proof of the Spectral Radius Formula; you can skip that for our purposes here.)
First show that $B$ is a Banach algebra with the norm $$\|f\|=\sum_n|a_n|\quad(f=\sum_na_nz^n).$$The least trivial part is showing that $\|fg\|\le\|f\|\,\|g\|$, which is not hard.
Now suppose $\phi$ is a complex homomorphism of $B$. (In this context that entails that $\phi$ is non-trivial: $\phi(1)=1$.) Let $\alpha = \phi(z)$. Note that $$|\alpha|\le\|z\|=1.$$
Since $\| f-s_n\|\to0$ (where $s_n$ is the $n$-th partial sum), and $\phi$ is bounded, it follows that $$\phi(f)=f(\alpha)\quad(f\in B).$$
And now you're done. If $f\in B$ and $f(z)\ne0$ for every $z$ in the closed disk then $\phi(f)\ne0$ for every complex homomorphism $\phi$, so the result that I've always felt should be the "Fundamental Theorem" of Banach algebra theory shows that there exists $g\in B$ with $fg=1$.
