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Topic: Singularities of holomorphic functions. (Read 797 times) |
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Icarus
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Singularities of holomorphic functions.
« on: Dec 26th, 2006, 6:38am » |
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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:
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.
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| « Last Edit: Dec 30th, 2006, 9:14am by Icarus » |
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Icarus
wu::riddles Moderator
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Re: Singularities of holomorphic functions.
« Reply #1 on: Jan 3rd, 2007, 7:31pm » |
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A second hint: Find a sequence in C - E which has E as its set of limit points.
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IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
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Re: Singularities of holomorphic functions.
« Reply #2 on: Jan 9th, 2007, 7:00pm » |
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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) =  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 <= 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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IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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