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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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    My function is analytic on the closure. So it's clear it has a minimum. It even takes its minimum at the boundary, right?2011-10-23

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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.