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 1


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.

  • $\begingroup$ 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

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