# Fourier coefficients of the logarithm of a given function

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 $$f$$: $$\mu_0 = \min \left\{\theta > 0, \exists x \in \mathbb{R}, f(x + i \theta) = 0 \right\}.$$

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 $$|c_n|=O(e^{-\lambda|n|})$$ 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 Cauchy-Hadamard 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 non-symmetric 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: $$\log f(z)=\sum_{j}k_j\log(z-z_j)+h(z),$$ 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.
• 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. Commented Dec 8, 2023 at 18:03