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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.

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    $\begingroup$ 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). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 20, 2019 at 20:49
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    $\begingroup$ 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 $R'$? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 20, 2019 at 21:10
  • $\begingroup$ 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. $\endgroup$
    – Sylvain JULIEN
    Commented Nov 20, 2019 at 21:31
  • $\begingroup$ 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. $\endgroup$
    – Sylvain JULIEN
    Commented Nov 20, 2019 at 21:37
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    $\begingroup$ 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. $\endgroup$
    – Lasse Rempe
    Commented Nov 20, 2019 at 21:40

1 Answer 1

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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 $\delta<K/4$ you obtain that $f$ has no zeros in the disk of radius $K/4$. Of course this can be improved, depending on your needs.

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