Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Removable singularity
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
== Riemann's theorem == [[Bernhard Riemann|Riemann's]] theorem on removable singularities is as follows: {{math theorem| Let <math>D \subset \mathbb C</math> be an open subset of the complex plane, <math>a \in D</math> a point of <math>D</math> and <math>f</math> a holomorphic function defined on the set <math>D \setminus \{a\}</math>. The following are equivalent: # <math>f</math> is holomorphically extendable over <math>a</math>. # <math>f</math> is continuously extendable over <math>a</math>. # There exists a [[neighborhood (topology)|neighborhood]] of <math>a</math> on which <math>f</math> is [[bounded function|bounded]]. # <math>\lim_{z\to a}(z - a) f(z) = 0</math>.}} The implications 1 β 2 β 3 β 4 are trivial. To prove 4 β 1, we first recall that the holomorphy of a function at <math>a</math> is equivalent to it being analytic at <math>a</math> ([[Proof that holomorphic functions are analytic|proof]]), i.e. having a power series representation. Define :<math> h(z) = \begin{cases} (z - a)^2 f(z) & z \ne a ,\\ 0 & z = a . \end{cases} </math> Clearly, ''h'' is holomorphic on <math> D \setminus \{a\}</math>, and there exists :<math>h'(a)=\lim_{z\to a}\frac{(z - a)^2f(z)-0}{z-a}=\lim_{z\to a}(z - a) f(z)=0</math> by 4, hence ''h'' is holomorphic on ''D'' and has a [[Taylor series]] about ''a'': :<math>h(z) = c_0 + c_1(z-a) + c_2 (z - a)^2 + c_3 (z - a)^3 + \cdots \, .</math> We have ''c''<sub>0</sub> = ''h''(''a'') = 0 and ''c''<sub>1</sub> = ''h{{'}}''(''a'') = 0; therefore :<math>h(z) = c_2 (z - a)^2 + c_3 (z - a)^3 + \cdots \, .</math> Hence, where <math>z \ne a</math>, we have: :<math>f(z) = \frac{h(z)}{(z - a)^2} = c_2 + c_3 (z - a) + \cdots \, .</math> However, :<math>g(z) = c_2 + c_3 (z - a) + \cdots \, .</math> is holomorphic on ''D'', thus an extension of <math> f </math>.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)