Let be continous on and analytic on prove that is analytic on .
My idea was to transform this problem into a simpler one. If is analytic on a region minus a line segment and continous there, then it must be analytic everywhere on the region. Thus, perhaps a conformal map will transform the circle into a line segment.
Do you agree then that this completes the proof? Because is analytic on unit disk because it is continous the line segment (using the theorem). And so is analytic on upper-half plane.
Note: My professor said that any function can be analytically continued along an analytic curve. Meaning if is analytic on a region minus an analytic curve, but continous there. Then in fact is analytic on the curve too.
This is Mine 91th Post!!!