I'm currently learning some stuff about Hardy-Spaces and there is this one part right at the beginning of a proof that i can't get my head around. Let $f\in\mathcal{H}^p\setminus\{0\}$, while $\mathcal{H}^p$ is the Hardy-Space of class $p$. Which especially means that our function is holomorphic on the unit disc $\mathcal{D}$. Using the identity principle we can assume that there exists any desired small $r\in (0,1)$ such that $0\not\in f(\partial\mathcal{D}_r(0)) $. Now i don't understand why we can do that. I know that it is not possible for the function to be constant $0$ on an arc but why can't it vanish on a point.
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If $f$ is not identically zero and analytic on some open $U$ and $C\subset U$ is compact, then $f$ can only have a finite number of zeros in $C$. So, there are lots or $r$s that will allow you to avoid a zero. – copper.hat Mar 28 '21 at 18:40
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Either $f(0)\ne0$ or $f(0)=0$. If $f(0)\ne0$, then by continuity, $f$ has no zeros on some disk centered at $0$. And, if $f(0)=0$, then, since the zeros of analytic functions with connected domain are isolated zeros (unless the function is the null function), there is some closed disk centered at $0$ such that $f$ has no zeros there, except at $0$ itself.
José Carlos Santos
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