Since it is a Reinhardt domain it can be represented by plotting it on the plane as follows.
Figure 1: Hartogs triangle
It is obvious then where the name comes from. To see that this is a domain of holomorphy, then given a boundary point we wish to exhibit a holomorphic function on the whole Hartogs triangle which does not extend beyond that point. First note that on the top boundary is anything and for some , so will not extend beyond . Now for the diagonal boundary this is where , that is for some , so will do not extend beyond .
One of the many properties of this domain is that if is the Hartogs triangle, then it is a domain of holomorphy, but if we take a sufficently small neighbourhood of (the closure of ), then any function holomorphic on is holomorphic on the polydisc (just fill in everything below the triangle in Figure 1). So if does not include all of then it is not a domain of holomorphy. This is because a Reinhardt domain that contains zero (the point ) is a domain of holomorphy if and only if it is a logarithmically convex set and any neighbourhood of does contain zero while itself does not.
- 1 Steven G. Krantz. , AMS Chelsea Publishing, Providence, Rhode Island, 1992.
|Date of creation||2013-03-22 14:31:08|
|Last modified on||2013-03-22 14:31:08|
|Last modified by||jirka (4157)|