proof of identity theorem of holomorphic functions
Since , there exists an the closed disk of radius about is contained in . Furthermore, both and are analytic inside this disk. Since is a limit point, there must exist a sequence of distinct points of which converges to . We may further assume that for every .
By the theorem on the radius of convergence of a complex function, the Taylor series of and about have radii of convergence greater than or equal to . Hence, if we can show that the Taylor series of the two functions at ar equal, we will have shown that whenever .
Suppose we choose the points at which to compute the divided differences as points of the sequence . Then we have
Since , it follows that for all and hence when .
If happens to be a circle centred about , we are done. If not, let be any point of such that . Since every connected open subset of the plane is arcwise connected, there exists an arc with endpoints and .
Define the function as follows
Because is open, it follows that for all .
We will now show that is continuous. Let and be any two distinct points of . If , then a disk of radius about will be contained in the disk of radius about . Hence, by the definition of , it will follow that . Therefore, for any two points and , it is the case that , which implies that is continuous.
Since is continuous and the arc is compact, attains a minimum value on . Let be chosen smaller strictly less than both and . Consider the set of all open disks of radius centred about ponts of . By the way was selected, each of these disks lies inside . Since is compact a finite subset of these disks will serve to cover . In other words, there exsits a finite set of points such that, if , then for some . We may assume that the ’s are ordered so that, as one traverses from to , one encounters before one encounters . This imples that . Without loss of generality, we may assume that and .
We shall now show that when for all by induction. From our definitions it follows that when . Next, we shall now show that if when , then when . Since , there exists a point and a constant such that implies and . By the induction hypothesis, when . Consider a disk of radius about . By the definition of , this disk lies inside and, by what we have already shown, when . Since , it follows from the triangle inequality that when .
In particular, the proposition just proven implies that since . This means that we have shown that for all .
|Title||proof of identity theorem of holomorphic functions|
|Date of creation||2013-03-22 14:40:41|
|Last modified on||2013-03-22 14:40:41|
|Last modified by||rspuzio (6075)|