Let $f$ be a $1$periodic real function that I know is bounded away from zero: $$ f(x) = \sum_{n = \infty}^\infty c_n e^{2\pi i n x} $$ Let me also assume that $f$ is analytic with Fourier coefficients satisfying $$ c_n \leq C e^{\lambda n}. $$ I would like to have good estimates on the Fourier coefficients $d_n$ of $\log(f(x))$: $$ \log(f(x)) = \sum_{n = \infty}^\infty d_n e^{2\pi i n x}. $$ In particular, can we say that asymptotically $$ d_n \leq C' e^{\mu n}, $$ where $C'$ is an explicit constant and $\mu$ is the minimum between $\lambda$ and the smallest positive imaginary part $\mu_0$ of a zero of $f$: $$ \mu_0 = \min \left\{\theta > 0, \exists x \in \mathbb{R}, f(x + i \theta) = 0 \right\}. $$
1 Answer
What you stated is correct. The crude statement is $d_n\leq C(\mu)e^{\mu n}$, for every $\mu<\mu_0$, where $$\mu_0=\min\{\lambda,\theta_0\},\; \theta_0=\min\{\theta: f(x+i\theta)=0\}.$$ Proof. The assumption that $c_n=O(e^{\lambdan})$ implies that your function has an analytic continuation to the strip $\{ z:{\mathrm{Im}}\, z<\lambda\}$. If $x+i\theta_0$ is the zero of $f$ with the smallest absolute value of imaginary part, then $\log f$ is analytic in the strip $\{ z:{\mathrm{Im}}\, z<\theta_0\}$. You can write $\log f$ in the form of Laurent series $$\log f=\sum_{\infty}^\infty d_n\zeta^n,\quad \zeta=e^{iz}.\quad\quad\quad\quad (1)$$ Then CauchyHadamard formula for the radii of convergence of a Laurent series gives the required estimate for $d_n$.
One can improve this estimate by considering the zeros of $f$ with the smallest positive imaginary part and with the largest negative imaginary part (which are responsible for the largest nonsymmetric strip in which $\log f$ is analytic.
ADDED: If you are interested in the estimate for $\mu=\mu_0$. It is actually true that $$d_n=o(e^{\mu_0 n}),$$ since we know that function (1) has only finitely many logarithmic singularities on the circles of convergence. In other words: $$\log f(z)=\sum_{j}k_j\log(zz_j)+h(z),$$ where the sum is finite, $h$ has strictly greater radius of convergence, and $z_j$ belong to the circles of convergence. So for the sum of the logs we have an explicit expansion, and $h$ constibutes an exponentially small term.

$\begingroup$ Thank you for your answer. Indeed, I wrote the least positive imaginary part because the function $f$ satisfies $f(\overline{z}) = \overline{f(z)}$ because $f$ is holomorphic and real on the real line. My interest would be in having an upper bound for the constant $C'$ and also understand what happens for the decay $\mu = \mu_0$ as you might expect a branch cut for $\log(f)$ at the "first" zero. $\endgroup$ Commented Dec 8, 2023 at 18:03