proof of identity theorem of holomorphic functions

Since z0D, there exists an ϵ0>0 the closed disk of radius ϵ about z0 is contained in D. Furthermore, both f1 and f2 are analyticPlanetmathPlanetmath inside this disk. Since z0 is a limit pointMathworldPlanetmathPlanetmath, there must exist a sequence xkk=1 of distinct points of s which converges to z0. We may further assume that |xk-z0|<ϵ0 for every k.

By the theorem on the radius of convergenceMathworldPlanetmath of a complex function, the Taylor seriesMathworldPlanetmath of f1 and f2 about z0 have radii of convergence greater than or equal to ϵ0. Hence, if we can show that the Taylor series of the two functions at z0 ar equal, we will have shown that f1(z)=g1(z) whenever z<ϵ.

The n-th coefficient of the Taylor series of a function is constructed from the n-th derivativePlanetmathPlanetmath of the function. The n-th derivative may be expressed as a limit of n-th divided differences


Suppose we choose the points at which to compute the divided differences as points of the sequence xi. Then we have


Since f1(xi)=f2(xi), it follows that f1(n)(z0)=f2(n)(z0) for all n and hence f1(z)=f2(z) when |z-z-0|<ϵ0.

If D happens to be a circle centred about z0, we are done. If not, let z1 be any point of D such that |z1-z0|ϵ. Since every connected open subset of the plane is arcwise connected, there exists an arc C with endpoints z1 and z0.

Define the function M:D as follows


Because D is open, it follows that 0<M(z)1 for all zD.

We will now show that M is continuousMathworldPlanetmathPlanetmath. Let w1 and w2 be any two distinct points of D. If M(w1)>|w1-w2|, then a disk of radius M(w1)-|w1-w2| about w2 will be contained in the disk of radius M(w1) about w1. Hence, by the definition of M, it will follow that M(w2)M(w1)-|w1-w2|. Therefore, for any two points w1 and w2, it is the case that |M(w1)-M(w2)||w1-w2|, which implies that M is continuous.

Since M is continuous and the arc C is compactPlanetmathPlanetmath, M attains a minimum value m on C. Let μ>0 be chosen smaller strictly less than both m/2 and ϵ0. Consider the set of all open disks of radius μ centred about ponts of C. By the way μ was selected, each of these disks lies inside D. Since C is compact a finite subset of these disks will serve to cover D. In other words, there exsits a finite setMathworldPlanetmath of points y1,y2,yn such that, if zC, then |z-yj|<μ for some j{1,2,,n}. We may assume that the yj’s are ordered so that, as one traverses C from z0 to z1, one encounters yj before one encounters yj+1. This imples that |yj-yj+1|<μ. Without loss of generality, we may assume that y1=z0 and yn=z1.

We shall now show that f1(z)=f2(z) when |z-yj|<μ for all j by inductionMathworldPlanetmath. From our definitions it follows that f1(z)=f2(z) when |z-y1|<μ. Next, we shall now show that if f1(z)=f2(z) when |z-yj|m/2, then f1(z)=f2(z) when |z-yj+i|m/2. Since |yj-yj+1|<μ, there exists a point wC and a constant ϵ>0 such that |w-z|<ϵ implies |z-yj|μ and |z-yj+i|μ. By the induction hypothesis, f1(z)=f2(z) when |z-yj|<μ. Consider a disk of radius m about w. By the definition of m, this disk lies inside D and, by what we have already shown, f1(z)=f2(z) when |z-w|m. Since |w-yj+1|<μ<m/2, it follows from the triangle inequalityMathworldMathworldPlanetmathPlanetmath that f1(z)=f2(z) when |z-yj+1|<μ.

In particular, the propositionPlanetmathPlanetmath just proven implies that f1(z1)=f1(z1) since z1=yn. This means that we have shown that f1(z)=f2(z) for all zD.

Title proof of identity theorem of holomorphic functions
Canonical name ProofOfIdentityTheoremOfHolomorphicFunctions
Date of creation 2013-03-22 14:40:41
Last modified on 2013-03-22 14:40:41
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 12
Author rspuzio (6075)
Entry type Proof
Classification msc 30A99