Is there such a function?

[esisk]

Hi,

Is there a function holomorphic on the open unit disk and continuoes on the closed disk such that f(z)= 1/z on the unit circle?


I will also like to know if somebody can help:
There are several approximation theorems out there, say Mergelyan, Runge, etc. Can somebody point at the salient features of these(i.e. when, what applies), or direct me to a source that is clear to a beginner. This sounds like spoon feeding, but I had to do it, bear with me. Thanks

 

[Hawkeye18]

If f is an analytic function in the open unit disc continuous in the closed disc, can you say what is
\int_C f(z) dz
where $C$ is the unit circle?

If the function f is analytic in a bigger disc, the answer follows immediately from the Cauchy Theorem. In the general case you can consider the functions f(rz), r<1 which are analytic in the disc of radius 1/r, and they take limit as r\to 1+ .

Next, what is the same integral for f(z) =1/z ? If you answer these 2 questions, the answer to your first question will be obvious.

Mergelyan theorem is a much harder result than Runge theorem. For a beginner, Runge theorem is enough, do not worry about Mergelyan yet.

Note that the Runge theorem with a pole at infinity gives you a "baby version" of the Mergelyan theorem.

 

[esisk]

Thank you for the response Hawkeye...
I am testing my understanding of your hints:

First we suppose that f were analythic in the interior of the circle ,say r=2. Then the integral around the unit circle would be zero (by Cauchy),
Whereas...,if f=1/z on the unit circle, then f=1/z on a set that has a limit point and therefore f=1/z on the interior of the circle with r=2. But the integral of 1/z around the unit circle is not zero.

I am still thinking about the "general case"
Thank you again

 

[mathwonk]

notice that 1/z reverses orientation of the circle, a problem for analytic functions on the closed disk.