Suppose that are analytic on with on . Prove that for all implies
Define new function h(z)
h is continuous on the closure of D
Since on , then by the Maximum Modulus Principle, is not zero in .
Hence, since and are analytic on and on , then is analytic on which implies is continuous on
h is analytic on D
This follows from above
Case 1: h(z) non-constant on D
If is not constant on , then by Maximum Modulus Principle, achieves its maximum value on the boundary of .
But since on (by the hypothesis), then
In particular , or equivalently
Case 2: h(z) constant on D
Suppose that is constant. Then
Then from hypothesis we have for all ,
Hence, by maximum modulus principle, for all
Since , we also have