Liouville's Theorem

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That name denotes a theorem in complex analysis discovered by Joseph Liouville (1809 Mar 24 - 1882 Sep 08) that has a remarkable application in physics and its description of Reality. Simply put, Liouville's theorem tells us that if a bounded complex function, f(Z), is continuous and differentiable over the entire complex plane, it must be constant.

Consider a complex function f(Z) that is analytic on a region bounded by a circle C centered on the point Z0. The circle consists of the points z=Z0+Reiθ, so Cauchy's integral formula tells us that

(Eq'n 1)

for all points Z within the circle. But we can replace the denominator in the integrand of that equation by z-Z0 in accordance with

(Eq'n 2)

in which we have converted the unit fraction of the second factor on the second line into its equivalent infinite series, exploiting the convenient fact that (Z-Z0)/(z-Z0)<1. So now Equation 1 becomes

(Eq'n 3)

That equation gives us the equivalent of a Taylor series representation of a complex function,

(Eq'n 4)

The coefficients kn in that equation must represent constants, so comparing Equations 3 and 4 shows us that we must have as true to mathematics

(Eq'n 5)

In that equation I used our definition of z above to rewrite the integrand. Because R represents an arbitrary number, the only way in which the integral can yield a constant is for kn=0 for all n 0. For n=0 we have

(Eq'n 6)

In that case the fact that k0 must represent a constant and that we integrate and evaluate the function only over an angle necessitates that the function have no dependence upon the value of R. Thus we must have k0=f(Z0), which leads to f(Z)=f(Z0), which makes the function a constant everywhere in the region enclosed by the circle C.

Now let C expand to enclose the entire complex plane. Because the complex plane consists of an infinite set of points, such an enclosure is an impossibility: the circle must expand endlessly because the complex plane has no end. Thus we must express our result as a limit. We also need to specify that the function is bounded (i.e. having only finite values) to exclude functions that "blow up" somewhere "way out there". So we have

(Eq'n 7)

And that is Liouville's theorem.

We can also write the Taylor series of Equation 3 with the coefficients represented by derivatives;

(Eq'n 8)

We can represent f(Z) as the sum of two real-valued functions of the complex variable,

(Eq'n 9)

To differentiate that function with respect to Z=x+iy we use the complex differentiation operator

(Eq'n 10)

For our first differentiation we get

(Eq'n 11)

Because the coefficients in Equation 8 for all n>0 must equal zero, the derivatives in Equation 8 must all equal zero. We can meet that criterion in Equation 11 if we require that u(Z) and v(Z) have such a form that

(Eq'n 12)

and

(Eq'n 13)

We call those criteria the Cauchy-Riemann equations, which are only satisfied by functions that are analytic in the region in which we apply them. And since that first derivative in Equation 11 must be an inherent zero, rather than a calculated zero, all of the other derivatives zero out as well and we have, as we must,

(Eq'n 14)

And again we get Liouville's theorem.

That theorem transforms the complex plane into an almost magical realm. Consider the fact that on the real line we have many functions that are bounded (that is, have only finite values) and analytic (that is, have derivatives of all orders). Sin(ax) provides a good example of such a function. Complex functions, on the other hand, under the constraint of Liouville's theorem, are either not bounded (that is, they have infinite values at some points in the complex plane) or they are not analytic (differentiable) at some points in the complex plane, but they cannot be both. One of the consequences of that fact is the fundamental theorem of algebra.

That theorem states that any n-th order polynomial p(x) has n roots; that is,

(Eq'n 15)

In that equation the roots zi are, in general, complex numbers (for example, the roots of x2+1=0 are +i and -i). To prove and verify that theorem we define f(z)=1/p(z) and then assume that there exists no point in the complex plane for which p(z)=0. That means that f(z) has no infinities on the complex plane (it is bounded). The inverse of a polynomial is endlessly differentiable, so we have f(z) both bounded and analytic on the entire complex plane, which contradicts Liouville's theorem. That contradiction obliges us to dismiss at least one of our premisses and the only one available is our assumption that p(z) has no zeroes in the complex plane. Thus there must exist at least one point z1 at which p(z) goes to zero. We can then factor our polynomial, p(x)=(x-z1)q(x). We then apply the same analysis to q(x) and so on to generate the product form of p(x) in Equation 15. Thus we verify the fundamental theorem of algebra.

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