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Title: Singularities of holomorphic functions. Post by Icarus on Dec 26th, 2006, 6:38am Define a "fundamental singularity" of a holomorphic function f to be a point p such that: (i) p is in the closure of D(f), the domain of f. (ii) For any non-empty open set A in D(f), there is no holomorphic function g and integer n such that g(z) = (z-p)nf(z) on A and p is in D(g). (I.e., even locally, f is not extensible to include p in its domain). (The definition and terminology are my own.) Another way of explaining this is that a fundamental singularity is a place other than a pole where the global function (and not just the branch) fails to be holomorphic. This includes all essential singularities - but (depending on your definition of "essential singularity) it may also include other singularities. For example, if we define D(log) to be all non-negative & non-zero complex numbers, 0 is a fundamental singularity, but -1 is not. Show that a set E is the set of fundamental singularities for some analytic function if and only if E is closed and has no interior. [Edit: I've revamped this one, including changing the thread name, after a search of all references I could find showed that all of them defined "essential singularity" only for isolated singularities - though I did find a number of references to "isolated essential singularities", a redundant phrase if e.s. are isolated by definition.] I have discovered a very nice little proof for this, so I'm not letting it drop. As a small hint: [hide]Note that since a meromorphic function is just a holomorphic function with the poles considered to be included in the domain, the theorem is for meromorphic functions as well.[/hide] |
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Title: Re: Singularities of holomorphic functions. Post by Icarus on Jan 3rd, 2007, 7:31pm A second hint: [hide]Find a sequence in C - E which has E as its set of limit points.[/hide] |
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Title: Re: Singularities of holomorphic functions. Post by Icarus on Jan 9th, 2007, 7:00pm Since no one is biting, and I think this is a nice result, I'm going to give my proof: We start with the topological lemma I mentioned in the second hint: Lemma: If E is a closed subset of C without interior, then there exists a countable set A in C - E such that E is the set of limit points of A. Proof: For each n, divide the plane up into squares with sidelength 1/n. No such square is entirely within E, as E has no interior. For each square that intersects E, choose an arbitrary point in the square and not in E to put in A. Since for each n, the number of squares is countable, so is the number of points added to A. Therefore A can be expressed as the countable union of countable sets. If x is in E, then every neighborhood of x contains a square of size 1/n for some suitable large n, which in turn contains x. Since the square interects E, it also contains a member of A. Hence every point in E is a limit point of A. Conversely, if x is in C - E, then since E is closed, there is a minimum distance d from x to any point of E. Once 4/n < d, any square intersecting E is at least d/2 from x. Therefore only a finite number of points of A can be within d/2 of x. So x is not a limit point of A. Thus A is a countable subset of C-E having E as its set of limit points. Once the lemma is proved, the rest is almost trivial: Let {an} be an enumeration of A, and define f(z) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif 2-n/(z - an). If z is in C-E, then for all but a finite number of the terms, |z - an| > d for some d (dependent on z, but not n). If we leave the exceptions out of the sum, the absolute value of the rest is <= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif 2-n/d = 2/d. Hence the series is absolutely convergent everywhere in C-E, except for having poles. On the other hand, any point of E has an infinite number of poles of f in every neighborhood. Therefore f (that is, any branch of f) cannot be extended to include any point of E. |
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