Bound of the norm of $f^{-1}$ in Wiener algebra

Question

Setup. Wiener algebra $W$ is a set of all functions $f(\zeta) = \sum_{n\in \mathbb{Z}}c_n\zeta^n$ on the unit circle with $\|f\|_W = \sum_{n\in \mathbb{Z}}|c_n| < \infty$. The famous Wiener $1/f$ theorem states that $f$ is invertible in $W$ iff $\displaystyle\inf_{|\zeta| = 1} |f(\zeta)| > 0$.

Question. Can the norm of $f^{-1}$ be bounded from above by the norm of $f$ and minimal value of $|f|$?

Thoughts. My guess that the answer is no, because the proofs of the theorem that I know do not provide such bound. However I can't come up with the examples or a general argument for that.

Shouldn't it be $\inf_{|\zeta|=1} |f(\zeta)| > 0$? – Gerd Mar 24 '23 at 12:18 — Gerd, Mar 24 '23 at 12:18
Oh, thank you @Gerd, of course. Fixed – Pavel Gubkin Mar 24 '23 at 13:09 — Pavel Gubkin, Mar 24 '23 at 13:09

score 1 · Accepted Answer · answered Mar 27 '23 at 13:44

This question is studied in the paper Estimates for resolvents in Beurling–Sobolev algebras by El-Fallah, Nikolskii and Zarrabi. Namely, if $\|f\|_W = 1$ and $\inf_{\mathbb{D}} |f(\zeta)| = \delta > 1/2$ then one can write $$ \|f^{-1}\|_W = \left\|\frac{1}{f(0)}\left(1 - \frac{f(0) - f}{f(0)}\right)^{-1}\right\|_W \le \frac{1}{\delta}\sum_{k \ge 0}\left\|\frac{f(0) - f}{f(0)}\right\|^k_W \\ \le \frac{1}{\delta}\sum_{k \ge 0}\left(\frac{\left\|f\right\|_W - |f(0)|}{\delta}\right)^k \le \frac{1}{\delta}\frac{1}{1 - \frac{1 - \delta}{\delta}} = \frac{1}{2\delta - 1}. $$ This bound is also precise. However the construction of the example is more involved and is not explicit.

Bound of the norm of $f^{-1}$ in Wiener algebra

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