Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an exercise from a book called "theory of complex functions" I want to solve:

Let $c\in G$ and $B\subset G$ a disc around c. When is the $\mathbb{C}$-algebra homomorphism $\mathcal{O}(G) \rightarrow \mathcal{O} (B)$ defined by $f\mapsto f_{|B}$ one-to-one? When is it onto?

I am not sure what this exercise asks. The $\mathbb{C}$-algebra maps a function to a function with its ensemble of definition restricted to the map of its disc? Or a domain is mapped to a disc in the domain?

If the second is the case, then I would say that it is never one-to-one, because the ensemble of definition has a higher cardinality than the ensemble of the map, but always onto.

Merci for any explaination.

share|improve this question
    
What do we know about $G$? –  Davide Giraudo Dec 6 '11 at 19:48
    
I assume $\cal O(B)$ is the set of holomorphic functions on $B$. The question(s) you are supposed to answer are: i) if $f, g$ are holomorphic on $G$ and $f=g$ on $B$, does it follow that $f=g$ on $G$? -- this is the one-to-one part -- and ii) if $f$ is a holomorphic function on $B$, is it true that there is a holomorphic $F$ defined on $G$ such that the restriction of $F$ to $B$ coincides with $f$ -- this is the onto part. –  user20266 Dec 6 '11 at 19:49
    
Merci Thomas Andrews, i) no, because the radius of convergence of f and g can be different, so that from $f=g$ on $B$ it doesn't follow that $f=g$ on G. ii) Yes, because the radius of convergence of f and F can be different, but the one of f is always smaller, so they will still both converge in the same disc. ? –  VVV Dec 6 '11 at 20:11
    
i) careful, the key words you want to look up are 'analytical continuation' and 'connectedness'. ii) assume the disc has radius 1 and $G$ is the disc of radius 2. Try to find a holomorphic function on the disc of radius 1 which cannot be extended beyond that radius. –  user20266 Dec 6 '11 at 20:18
    
i) yes, because f-g is an analytic function which vanishes in the open, connected domain B , so it must vanish on its entire domain. ii) $log(z)$ is holomorphic, but it can't be extended beyond the radius 1, since 0 doesn't work with it. ? Is this what you meant? Merci –  VVV Dec 6 '11 at 20:44
show 1 more comment

1 Answer 1

  • It is one-to-one if and only if $G$ is connected. If $G$ is connected, you can apply the identity theorem. If $G$ is not connected, then two functions can be equal on the component containing $B$ while being different on a component not containing $B$.
  • It is onto if and only if the component of $G$ containing $B$ is equal to $B$. If the latter condition holds, then each $f\in \mathcal O(B)$ extends to a holomorphic function on $G$ by setting the values to $0$ (for example) on $G\setminus B$. If the condition does not hold, then there exists a $b\in G\cap\partial B$, and $z\mapsto\frac{1}{z-b}$ is an element of $\mathcal O(B)$ that has no extension in $\mathcal O(G)$.
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.