proof of Riemann mapping theorem
This proof relies on the existence of a solution to the Dirichlet problem, and hence is applicable to any domain for which it can be proven that the Dirichlet problem has a solution. For simplicity, I will assume that the region is bounded;
Since this proof uses real techniques to prove a complex result, a few simple conventions will make it easier to read. The letter and will always denote real quantities and will always equal . Functions of a complex variable will be written as functions of two real variables without further warning; thus, and will denote the same entity.
Consider the following boundary value problem:
when lies on the boundary of . Note that, since lies in the interior of , bounded on the boundary of . Hence, by the existence theorem, there exists a unique function satisfying this boundary value problem. Since the boundary values of are bounded, will be bounded on the interior of as well. By the regularity theorem for the Laplace equation, will be differentiable (in fact, analytic) on the interior of .
vanishes. Since is simply connected, Poincare’s lemma implies that there must exist a function such that this vector field is the gradient of . Since is only determind up to an additive constant, we may impose the condition . Written out explicitly, the condition that the gradient of equals the vector field looks like
Define , and as
Because and satisfies Laplace’s equation, and will also satisfy Laplace’s equation in . Note that, whilst is single-valued, is multiply-valued with branch point at . Upon circling once, the value of increases by . On the other hand, is single valued since the exponentiation operation cancels out the multiple-valuedness of . Because and satisfy the Cauchy-Riemann equations and the exponential function is analytic, is analytic.
By construction, when lies on the boundary of . It will now be shown that whenever . Since as and is finite, it follows that there exists , such that when Consider the region . For a point to lie on the boundary of , either or must lie on the boundary of . Either way, , so the maximum principle implies that whenever . We already saw that when , so whenever .
As a consequence, when because
Also, note that and that is real and positive because
since was chosen so that .
We will now show that is finite. The set
Choose such that whenever . Let denote the level set
We shall now show that is smooth and is homeomorphic to a circle. As the level set of a continuous function on a compact set, is compact. Let be an point of . By assumption, . Hence, by the inverse function theorem, there exists a neighborhood of on which is invertible. Furthermore, is an anlytic function. Since if and only if , it follows that is the image of an arc the circle under . Hence, is diffeomorphic to a line segment.
Since this is true of every point , is a compact one-dimensional manifold. A compact, one-dimensional manifold must be either a circle or a finite union of circles. Suppose that is a union of more than one circle. By the Jordan curve theorem, each of these circles divides the compex plane into an interior and an exterior reigion. Hence, given two of the circles which would comprise , one of these circles would have to lie inside the other and there would have to be an open set which has these two circles as boundary. However, it is assumed that is constant on both circles and assumes the same value on both. By the maximum principle, this would imply that is constant in the region between the circles which would, in turn, imply that is constant on , which is impossible. Hence, consists of a single circle.
Next, note that must lie inside . If did not lie in , it would follow that when lies outside of . But this is not possible because the boundary of lies on the outsde of as well and was chosen to satisfy the boundary value when lies on the boundary of .
Since lies on the interior of the circle , the winding number of about is 1. Hence, upon traversing once, the phase of will increase by .
Since is analytic, is not only homeomorphic to a circle, it is also a smooth curve. Hence it makes sense to speak of the tangent to and the normal to . In terms of the normal and tangential derivatives, the Cauchy-Riemann equations my be written as
Since when lies on and when lies in the interior of , it follows that and . By the Cauchy-Riemann eqations, this means that . It is not possible that because, if that were so, then all the derivatives of and would vanish, which would imply that , contrary to hypothesis. Hence is a monotonically decreasing function on . We already saw that decreases by upon traversing . These two facts together imply that is a bijection from to the unit circle. Hence is a bijection from to the circle of radius .
To finish the proof, we need to deal with the points such that . We shall use the argument principle to show that these points do not exist! As before, choose such that whenever . Then is a smooth close curve and we may apply the argument principle to along . Differentiating the definition,
Since the argument of a product is the sum of the arguments of the factors, we may consider and separately. The argument principle states that, because has only one simple zero (located at ) inside , the argument of will increase by upon traversing . To compute the argument of , we may make use of the fact that the derivative of an analytic function is the same no matter what direction one chooses to compute the derivative. To compute the derivative of at a point on , choose the normal direction. From what we had seen earlier, it follows that will point along the inward normal to . Since the inward normal rotates by by upon traversing the curve, the argument of changes by upon traversing . Hence, the argument of stays the same after traversing . Since is analytic inside , the argument principle states that can have no zeros inside of .
|Title||proof of Riemann mapping theorem|
|Date of creation||2013-03-22 14:35:17|
|Last modified on||2013-03-22 14:35:17|
|Last modified by||rspuzio (6075)|