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Let $f:\mathbf{H}\to \mathbf{C}$ be a holomorphic function on the complex upper-half plane and let $U$ be a bounded open subset in $\mathbf{H}$ contained in $$\{\tau \in \mathbf{H}: \mathrm{Im}(\tau) > 1\}.$$ Suppose that $f$ does not vanish on the closure of this open subset in $\mathbf{H}$.

Is the absolute value of $f$ bounded from below by a positive constant on $U$?

I'm thinking this should follow from applying the maximum-principle to $1/f$.

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is f cont on the closure of H? – Digital Gal Oct 23 '11 at 17:55
Yes, that's correct. In fact, this is true for any holomorphic function $U \to \mathbb{C}$ on any open set $U \subset \mathbb{C}$. – Ted Oct 23 '11 at 18:15
If $f$ extends continuously to the closure of $H$, then this is trivial, and analyticity is not required (a continuous function on a compact set takes a minimal value). If $f$ does not extend continuously to the closure of $H$, then the statement is false. – Lasse Rempe-Gillen Oct 23 '11 at 18:57
My function is analytic on the closure. So it's clear it has a minimum. It even takes its minimum at the boundary, right? – Bana Oct 23 '11 at 19:22
up vote 1 down vote accepted

Since $U$ is contained in $\{z \in \mathbf{H}: \mathrm{Im}(z) > 1\}$, the closure of $U$ in $\mathbf{H}$ is equal to the closure of $U$ in $\mathbb{C}$, and so is compact. Thus $|f|$ takes a minimum value on the closure of $U$, which by hypothesis is not $0$ and so is strictly positive. All we need to assume about $f$ is that $|f|$ is continuous; the complex analysis in the question is irrelevant.

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