# Holomorphic extensions of a non-vanishing real-analytic function

Let f(z) be a holomorphic function defined on an open neighborhood $$R$$ of the interval $$I=[0,1]\subset \mathbb{R}$$. Assume $$f$$ does not vanish on $$I$$. Then $$g(x) = |f(x)|$$ is a real-analytic function on $$I$$, and thus extends to a holomorphic function $$g$$ on some neighborhood of $$R'\subset R$$ of $$I$$.

Now, $$g$$ does not necessarily extend to all of $$R$$. Take, for instance, $$f(z) = z + i$$, which is of course entire. Then $$g(x) = |f(x)| = \sqrt{x^2+1}$$ cannot be extended to any domain including $$-i$$, as then we hit the branch point of the square-root function at $$0$$.

May we actually construct a region $$R'$$ where $$g$$ is guaranteed to be holomorphic, given $$R$$ and, say, a lower bound on $$(\min_{x\in I} |f(x))/(\max_{x\in I} |f(x)|)$$? If not, what other conditions could be helpful?

Added remark: what if we have a lower bound on $$\frac{\min_{z\in R} |f(z)|}{\max_{z\in R} |f(z)|}$$?

Note: the question came out of a conversation with F. Johannson. The added remark is also his.

• The upper bound on the oscillation of $|f|$ over $I$ is clearly not sufficient: $f(z) = \exp(i k^2 z) - e^{-k}$ is entire, $(\max_{[0,1]} |f|) / (\min_{[0,1]} |f|) \leqslant (1 + e^{-k}) / (1 - e^{-k})$, but the holomorphic extension of $|f(x)| = ((\cos(k^2 x) - e^{-k})^2 + (\sin(k^2 x))^2)^{1/2}$ breaks at $z = \pm 1/k$ (the holomorphic extension of $|f(x)|^2$ has a simple zero there). Nov 20 '19 at 20:49
• One more random comment: since $|f(x)| = (f(x) \overline{f(x)})^{1/2}$, the question really asks for region where $g(z) := f(z) \overline{f(\bar z)}$ is zero-free (or at least it only has zeroes of even order). It is hence sufficient to answer the following question: what assumptions on a holomorphic function $g$ in $R$ which is real-valued on $[0,1]$ imply that $g$ is zero-free in $R'$? Nov 20 '19 at 21:10
• Maybe a way to answer Mateusz' question would be to require that the absolute value of the curvature of $g$ is fast decreasing provided the slope of $g$ is close to $0$ around the boundary of $I$ in a suitable sense. Nov 20 '19 at 21:31
• So one can imagine to make $g$ undergo some kind of "curvature diffusion" process with the constraint of remaining holomorphic during such a process, maybe a holomorphic analogue of the Ricci flow. Nov 20 '19 at 21:37
• I feel that it will be difficult to give a simple condition, involving only the values on the interval. For example, after all we can approximate any function on the interval arbitrarily closely by a polynomial that has a zero as close to the interval as we want. Nov 20 '19 at 21:40

To say something about the size of $$R'$$, you need a bound for the distance of the nearest zero to the real line. Such a bound can be obtained if your conditions on your function define a normal family in $$R$$. No reasonable condition involving only restriction of $$f$$ on $$I$$ will do this. However one can give bounds in terms of $$K:=\min_I|f|/\max_R|f|$$. Namely, there exists $$\delta>0$$ which depends on $$K$$ and $$R$$, such that $$f$$ has no zero in a $$\delta$$-neighborhood of $$I$$. To give an explicit estimate of this $$\delta$$ one needs to know the shape of $$R$$.
To obtain an explicit estimate, you may argue as follows. Let $$f$$ be a function in the unit disk. $$|f(z)|\leq 1$$, and $$|f(0)|\geq K$$. Then by Cauchy, $$|f'(z)|\leq 4, |z|\leq 1/2,$$ so $$|f(z)-f(0)|\leq 4|z|,\quad |z|\leq 1/2,$$ so $$|f(z)|\geq K-4\delta\quad |z|\leq\delta<1/2.$$ taking $$\delta you obtain that $$f$$ has no zeros in the disk of radius $$K/4$$. Of course this can be improved, depending on your needs.