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let $f$ be holomorphic and bounded on $\mathbb{C}\backslash K$ with $K = \{0\}\cup\{\frac11, \frac12, \cdots\}$ and further $f(2) = 1$. Determine such f.


What we tried

Since f is bounded, points in $K$ give essential or removable singularities. If every $k \in K$ had a removable singularity, then $f$ could be extended to a holomorph function on $\mathbb{C}$, and hence by Liouville's theorem $1$ everywhere.

Now we only have to look at functions with at least one essential singularity in some $k \in K$. Any hints?

Kees

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2 Answers 2

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I'm not sure what this talk of Picard, Casorati-Weierstrass is all about. Suppose $|f|\le M$ on $\mathbb {C}\setminus K.$ Each $1/n, n \in \mathbb {N},$ is an isolated singularity of $f.$ Since $f$ is bounded, $f$ has a removable singularity at each of these points. Thus $f$ extends to be holomorphic on $\mathbb {C}\setminus \{0\},$ with $|f| \le M$ on this larger domain. Now $0$ wasn't an isolated singularity of $f$ at the outset, but it is now, and the boundedness of $f$ tells us this is a removable singularity after all. Thus $f$ extends to be entire, with $|f|\le M$ everywhere. By Liouville $f$ is constant, and of course $f(2)=1$ implies $f \equiv 1.$

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As Kees suggests, now we only have to look at the case where there is at least one essential singularity from $K$, but by Great Picard's theorem, $f$ is unbounded in any neighborhood of an essential singularity. So any possible singularity is removable and $f$ must be constant and equal to 1.

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    $\begingroup$ Can use Casorati-Weierstrass aswell. It is much less deep than Picard. $\endgroup$
    – Lukas Betz
    May 11, 2015 at 11:16

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