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jkramar@gmail.com
science forum beginner
Joined: 24 May 2006
Posts: 1
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 Posted: Wed May 24, 2006 1:59 am Post subject:
how strong is liouville's theorem?
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My question is about bounded holomorphic functions f on domains D in
the complex plane. Liouville's Theorem says that if D=C then f must be
constant. It follows easily that if the complement of D is
topologically discrete then f must also be constant. My question out of
curiosity is: under what conditions on D does the implication hold? For
instance if the complement of D contains a continuous curve
(nonconstant continuous image of R) then the Riemann mapping theorem
tells us that D can be holomorphically mapped into a disc, even
injectively. What if D is the complement of a nondiscrete set which
doesn't contain any continuous curves, such as the Cantor set? |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Wed May 24, 2006 6:05 am Post subject:
Re: how strong is liouville's theorem?
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In article
<1148435989.301700.176750@38g2000cwa.googlegroups.com>,
"jkramar@gmail.com" <jkramar@gmail.com> wrote:
| Quote: | My question is about bounded holomorphic functions f on domains D in
the complex plane. Liouville's Theorem says that if D=C then f must be
constant. It follows easily that if the complement of D is
topologically discrete then f must also be constant. My question out of
curiosity is: under what conditions on D does the implication hold? For
instance if the complement of D contains a continuous curve
(nonconstant continuous image of R) then the Riemann mapping theorem
tells us that D can be holomorphically mapped into a disc, even
injectively. What if D is the complement of a nondiscrete set which
doesn't contain any continuous curves, such as the Cantor set?
|
Suppose K it the standard Cantor set of measure 0 on the real
axis, and f is bounded and holomorphic on C \ K. Let's try to
emulate Morera's theorem: The contour integral of f over the
boundary of every triangle in C is well defined (because m_1(K) =
0), and by looking at subtriangles and taking limits using
Cauchy's theorem, you can show that these integrals are all 0. So
it makes sense to define F(z) = int_[0,z] f(w) dw for all z in C,
and we will have F'(z) = f(z) for all z in C \ K by the standard
argument. But F is clearly continuous on C, so now we can use the
actual Morera to see that F is entire. But F' = f on C \ K, and f
is bounded, so F' is entire and bounded on all of C, hence is
constant by Liouville. Therefore f is constant. |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Wed May 24, 2006 2:04 pm Post subject:
Re: how strong is liouville's theorem?
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On 23 May 2006 18:59:49 -0700, "jkramar@gmail.com" <jkramar@gmail.com>
wrote:
| Quote: | My question is about bounded holomorphic functions f on domains D in
the complex plane. Liouville's Theorem says that if D=C then f must be
constant. It follows easily that if the complement of D is
topologically discrete then f must also be constant. My question out of
curiosity is: under what conditions on D does the implication hold? For
instance if the complement of D contains a continuous curve
(nonconstant continuous image of R) then the Riemann mapping theorem
tells us that D can be holomorphically mapped into a disc, even
injectively. What if D is the complement of a nondiscrete set which
doesn't contain any continuous curves, such as the Cantor set?
|
Say K is a compact set and D is the complement of K (K has to
be closed so D is open so we know what "holomorphic in D" means).
By definition, K has "zero analytic capacity" if and only if
every bounded holomorphic function in D is constant. In
some sense that answers your question, although it doesn't
really say anything because we don't know which sets have
zero analytic capacity (let's abbreviate that to "ac(K)=0".)
I believe that someone has fairly recently proved a highly
non-trivial theorem that characterizes ac(K)=0 in terms
of metric, "geometric measure theory" conditions. There
are simpler results that are not quite definitive:
Say h_1(K)=0 if for every eps > 0 K is contained in a
countable union of disks such that the sum of the
radii is less then eps. (Ie K has zero one-dimensional
"Hausdorff measure"). It's not hard to show that
h_1(K)=0 implies ac(K)=0; this follows from the
Cauchy integral formula:
f(z) = int_C f(w)/(w-z),
where C consists of a large circle minus the union
of those small circles in the definition of h_1(K)=0.
Let epsilon -> 0 and it follows that f is the integral
over the large circle, but the integral over the large
circle extends to be holorphic across points of K.
So f extends to a bounded entire function, which is
constant.
I _think_ that if K is a subset of the real line then
the converse holds, ac(K)=0 implies h_1(K) = 0. Not
sure about that. The converse definitely does not
hold in general: Vitushkin gave a complicated example
of a K with ac(K)=0 but h_1(K) > 0. Garnett gave a
much simpler example:
K is going to be a sort of two-dimensional Cantor
set. Start with a square. Divide it into 16 equal
sub-squares, and keep the 4 subsquares which lie
in the cornners of the original square. Repeat
on each subsquare, etc, and take the intersection.
************************
David C. Ullrich |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Thu May 25, 2006 3:00 pm Post subject:
Re: how strong is liouville's theorem?
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On Wed, 24 May 2006 09:04:16 -0500, David C. Ullrich
<ullrich@math.okstate.edu> wrote:
| Quote: | On 23 May 2006 18:59:49 -0700, "jkramar@gmail.com" <jkramar@gmail.com
wrote:
My question is about bounded holomorphic functions f on domains D in
the complex plane. Liouville's Theorem says that if D=C then f must be
constant. It follows easily that if the complement of D is
topologically discrete then f must also be constant. My question out of
curiosity is: under what conditions on D does the implication hold? For
instance if the complement of D contains a continuous curve
(nonconstant continuous image of R) then the Riemann mapping theorem
tells us that D can be holomorphically mapped into a disc, even
injectively. What if D is the complement of a nondiscrete set which
doesn't contain any continuous curves, such as the Cantor set?
Say K is a compact set and D is the complement of K (K has to
be closed so D is open so we know what "holomorphic in D" means).
By definition, K has "zero analytic capacity" if and only if
every bounded holomorphic function in D is constant. In
some sense that answers your question, although it doesn't
really say anything because we don't know which sets have
zero analytic capacity (let's abbreviate that to "ac(K)=0".)
I believe that someone has fairly recently proved a highly
non-trivial theorem that characterizes ac(K)=0 in terms
of metric, "geometric measure theory" conditions. There
are simpler results that are not quite definitive:
Say h_1(K)=0 if for every eps > 0 K is contained in a
countable union of disks such that the sum of the
radii is less then eps. (Ie K has zero one-dimensional
"Hausdorff measure"). It's not hard to show that
h_1(K)=0 implies ac(K)=0; this follows from the
Cauchy integral formula:
f(z) = int_C f(w)/(w-z),
where C consists of a large circle minus the union
of those small circles in the definition of h_1(K)=0.
Let epsilon -> 0 and it follows that f is the integral
over the large circle, but the integral over the large
circle extends to be holorphic across points of K.
So f extends to a bounded entire function, which is
constant.
I _think_ that if K is a subset of the real line then
the converse holds, ac(K)=0 implies h_1(K) = 0.
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Yes, that's so. There's an overview of results
about analytic capacity, which is to say results
concering exactly the question you asked, at
http://www.math.u-psud.fr/~biblio/pub/1998/abs/ppo1998_90.html
At the top of p.3 it says that this result, in fact
the corresponding result for sets which lie on
sufficiently smooth curves, is "not too hard". Then
on p.5 there's a hint towards the proof, with a
reference for the details.
************************
David C. Ullrich |
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