Difference between revisions of "Liouville's Theorem (complex analysis)"
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m (does not add up, one R gets canceled there, in other proofs and in parametric form of the contour it also ends up with M/R and not M/R^2) |
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counterclockwise circle of radius <math>R</math> centered at <math>z_0</math>. Then | counterclockwise circle of radius <math>R</math> centered at <math>z_0</math>. Then | ||
<cmath> \lvert f'(z_0) \rvert = \biggl\lvert \frac{1}{2\pi i} | <cmath> \lvert f'(z_0) \rvert = \biggl\lvert \frac{1}{2\pi i} | ||
| − | \int_C | + | \int_C \frac{f(z)}{(z-z_0)^2}dz \biggr\rvert \le \frac{M}{R} .</cmath> |
Since <math>f</math> is holomorphic on the entire complex plane, <math>R</math> can | Since <math>f</math> is holomorphic on the entire complex plane, <math>R</math> can | ||
be arbitrarily large. It follows that <math>f'(z) = 0</math>, for every | be arbitrarily large. It follows that <math>f'(z) = 0</math>, for every | ||
Latest revision as of 19:30, 16 January 2024
In complex analysis, Liouville's Theorem states that a bounded holomorphic function on the entire complex plane must be constant. It is named after Joseph Liouville. Picard's Little Theorem is a stronger result.
Contents
Statement
Let 
be a holomorphic function.
Suppose there exists some real number 
such that
for all 
. Then 
is a
constant function.
Proof
We use Cauchy's Integral Formula.
Pick some 
; let 
denote the simple
counterclockwise circle of radius 
centered at 
. Then
![\[\lvert f'(z_0) \rvert = \biggl\lvert \frac{1}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^2}dz \biggr\rvert \le \frac{M}{R} .\]](http://latex.artofproblemsolving.com/f/4/e/f4edd238953e3bece2f6f49800c9e2dd5c982418.png)
Since is holomorphic on the entire complex plane, can
be arbitrarily large. It follows that 
, for every
point . Now for any two complex numbers 
and 
,
![\[f(B) - f(A) = \int_{A}^B f'(z)dz = 0 ,\]](http://latex.artofproblemsolving.com/3/e/0/3e0a144bac3b1f27d9d0bd2fb2aa5cfc17756d60.png)
so is constant, as desired. 
Extensions
It follows from Liouville's theorem if is a non-constant entire
function, then the image of is dense in
; that is, for every , there exists
some 
that is arbitrarily close to .
Proof
Suppose on the other hand that there is some not in the
image of , and that there is a positive real 
such
that 
has no point within of . Then
the function
![\[g(z) = \frac{1}{f(z) - z_0}\]](http://latex.artofproblemsolving.com/e/7/6/e766977739d0c8f795d38ab403ae09d73ffbe928.png)
is holomorphic on the entire complex plane, and it is bounded by
. It is therefore constant. Therefore is constant.
Picard's Little Theorem offers the stronger result that if
avoids two points in the plane, then it is constant. It is
possible for an entire function to avoid a single point, as
avoids 0.
