Liouville's theorem (complex analysis)
Encyclopedia
In complex analysis
, Liouville's theorem, named after Joseph Liouville
, states that every bounded
entire function
must be constant. That is, every holomorphic function f for which there exists a positive number M such that |f(z)| ≤ M for all z in C is constant.
The theorem is considerably improved by Picard's little theorem
, which says that every entire function whose image omits at least two complex numbers must be constant.
about 0
where (by Cauchy's integral formula
)
and Cr is the circle about 0 of radius r > 0. We can estimate directly
where in the second inequality we have invoked the assumption that |f(z)| ≤ M for all z and the fact that |z|=r on the circle Cr. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives ak = 0 for all k ≥ 1. Thus f(z) = a0 and this proves the theorem.
where I is the value of the remaining integral. This shows that f is bounded and entire, so it must be constant, by Liouville's theorem. Integrating then shows that f is affine
and then, by referring back to the original inequality, we have that the constant term is zero.
f cannot be C. Suppose it was. Then, if a and b are two periods of f such that a⁄b is not real, consider the parallelogram
P whose vertices
are 0, a, b and a + b. Then the image of f is equal to f(P). Since f is continuous and P is compact
, f(P) is also compact and, therefore, it is bounded. So, f is constant.
The fact that the domain of a non-constant elliptic function f can not be C is what Liouville actually proved, in 1847, using the theory of elliptic functions. In fact, it was Cauchy
who proved Liouville's theorem.
in C. This might seem to be a much stronger result than Liouville's theorem, but it is actually an easy corollary. If the image of f is not dense, then there is a complex number w and a real number r > 0 such that the open disk centered at w with radius r has no element of the image of f. Define g(z) = 1/(f(z) − w). Then g is a bounded entire function, since
So, g is constant, and therefore f is constant.
Similarly, if an entire function has a pole at ∞, i.e. blows up like zn in some neighborhood of ∞, then f is a polynomial. This extended version of Liouville's theorem can be more precisely stated: if |f(z)| ≤ M.|zn| for |z| sufficiently large, then f is a polynomial of degree at most n. This can be proved as follows. Again take the Taylor series representation of f,
The argument used during the proof shows that
So, if k > n,
Therefore, ak = 0.
Complex analysis
Complex analysis, traditionally known as the theory of functions of a complex variable, is the branch of mathematical analysis that investigates functions of complex numbers. It is useful in many branches of mathematics, including number theory and applied mathematics; as well as in physics,...
, Liouville's theorem, named after Joseph Liouville
Joseph Liouville
- Life and work :Liouville graduated from the École Polytechnique in 1827. After some years as an assistant at various institutions including the Ecole Centrale Paris, he was appointed as professor at the École Polytechnique in 1838...
, states that every bounded
Bounded function
In mathematics, a function f defined on some set X with real or complex values is called bounded, if the set of its values is bounded. In other words, there exists a real number M...
entire function
Entire function
In complex analysis, an entire function, also called an integral function, is a complex-valued function that is holomorphic over the whole complex plane...
must be constant. That is, every holomorphic function f for which there exists a positive number M such that |f(z)| ≤ M for all z in C is constant.
The theorem is considerably improved by Picard's little theorem
Picard theorem
In complex analysis, the term Picard theorem refers to either of two distinct yet related theorems, both of which pertain to the range of an analytic function.-Little Picard:...
, which says that every entire function whose image omits at least two complex numbers must be constant.
Proof
The theorem follows from the fact that holomorphic functions are analytic. Since f is entire, it can be represented by its Taylor seriesTaylor series
In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point....
about 0
where (by Cauchy's integral formula
Cauchy's integral formula
In mathematics, Cauchy's integral formula, named after Augustin-Louis Cauchy, is a central statement in complex analysis. It expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk, and it provides integral formulas for all...
)
and Cr is the circle about 0 of radius r > 0. We can estimate directly
where in the second inequality we have invoked the assumption that |f(z)| ≤ M for all z and the fact that |z|=r on the circle Cr. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives ak = 0 for all k ≥ 1. Thus f(z) = a0 and this proves the theorem.
Fundamental theorem of algebra
There is a short proof of the fundamental theorem of algebra based upon Liouville's theorem.No entire function dominates another entire function
A consequence of the theorem is that "genuinely different" entire functions cannot dominate each other, i.e. if f and g are entire, and |f| ≤ |g| everywhere, then f = α·g for some complex number α. To show this, consider the function h = f/g. It is enough to prove that h can be extended to an entire function, in which case the result follows by Liouville's theorem. The holomorphy of h is clear except at points in g−1(0). But since h is bounded, any singularities must be removable. Thus h can be extended to an entire bounded function which by Liouville's theorem implies it is constant.If f is less than or equal to a scalar times its input, then it is linear
Suppose that f is entire and |f(z)| is less than or equal to M|z|, for M a positive real number. We can apply Cauchy's integral formula; we have thatwhere I is the value of the remaining integral. This shows that f is bounded and entire, so it must be constant, by Liouville's theorem. Integrating then shows that f is affine
Affine transformation
In geometry, an affine transformation or affine map or an affinity is a transformation which preserves straight lines. It is the most general class of transformations with this property...
and then, by referring back to the original inequality, we have that the constant term is zero.
Non-constant elliptic functions cannot be defined on C
The theorem can also be used to deduce that the domain of a non-constant elliptic functionElliptic function
In complex analysis, an elliptic function is a function defined on the complex plane that is periodic in two directions and at the same time is meromorphic...
f cannot be C. Suppose it was. Then, if a and b are two periods of f such that a⁄b is not real, consider the parallelogram
Parallelogram
In Euclidean geometry, a parallelogram is a convex quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure...
P whose vertices
Vertex (geometry)
In geometry, a vertex is a special kind of point that describes the corners or intersections of geometric shapes.-Of an angle:...
are 0, a, b and a + b. Then the image of f is equal to f(P). Since f is continuous and P is compact
Compact space
In mathematics, specifically general topology and metric topology, a compact space is an abstract mathematical space whose topology has the compactness property, which has many important implications not valid in general spaces...
, f(P) is also compact and, therefore, it is bounded. So, f is constant.
The fact that the domain of a non-constant elliptic function f can not be C is what Liouville actually proved, in 1847, using the theory of elliptic functions. In fact, it was Cauchy
Augustin Louis Cauchy
Baron Augustin-Louis Cauchy was a French mathematician who was an early pioneer of analysis. He started the project of formulating and proving the theorems of infinitesimal calculus in a rigorous manner, rejecting the heuristic principle of the generality of algebra exploited by earlier authors...
who proved Liouville's theorem.
Entire functions have dense images
If f is a non-constant entire function, then its image is denseDense set
In topology and related areas of mathematics, a subset A of a topological space X is called dense if any point x in X belongs to A or is a limit point of A...
in C. This might seem to be a much stronger result than Liouville's theorem, but it is actually an easy corollary. If the image of f is not dense, then there is a complex number w and a real number r > 0 such that the open disk centered at w with radius r has no element of the image of f. Define g(z) = 1/(f(z) − w). Then g is a bounded entire function, since
So, g is constant, and therefore f is constant.
Remarks
Let C ∪ {∞} be the one point compactification of the complex plane C. In place of holomorphic functions defined on regions in C, one can consider regions in C ∪ {∞}. Viewed this way, the only possible singularity for entire functions, defined on C ⊂ C ∪ {∞}, is the point ∞. If an entire function f is bounded in a neighborhood of ∞, then ∞ is a removable singularity of f, i.e. f cannot blow up or behave erratically at ∞. In light of the power series expansion, it is not surprising that Liouville's theorem holds.Similarly, if an entire function has a pole at ∞, i.e. blows up like zn in some neighborhood of ∞, then f is a polynomial. This extended version of Liouville's theorem can be more precisely stated: if |f(z)| ≤ M.|zn| for |z| sufficiently large, then f is a polynomial of degree at most n. This can be proved as follows. Again take the Taylor series representation of f,
The argument used during the proof shows that
So, if k > n,
Therefore, ak = 0.