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Can i get the answer to the following problem. I am having a proof, i feel there is something wrong here..Can you please point out!

Let $D\subset \mathbb C$ be a simply connected domain, and $\gamma: [0,1]\to D$, be a smooth embedding. Given a continuous one form $\phi$ along $\gamma$ and $\epsilon >0$, Does there exists a holomorphic function $h$ on some open neighborhood $U$ of $\gamma$, $U\subset D$ such that $|dh-\phi|<\epsilon$.

Suggested Proof:

Without loss of generality we can assume that $0\notin D$. We can write $\phi= \phi_1 d\zeta$, with $\phi_1$ a continuous function on $\gamma$. We can uniformly approximate $\phi_1$ by Laurent polynomials of the form $\psi_1(\zeta)= \sum_{-k}^k a_j\zeta^k$. As $0\notin D$, we have $\psi_1(\zeta)$ analytic on some possibly small simply connected subdomain of $D$ which we will denote by $D$ itself.

We know that if D is a simply connected domain and $\psi_1$ is analytic in D, then $\psi_1$ has an antiderivative at all points of D. Hence take $h(z)= \int \psi_1(\zeta)$ which will be our required holomorphic function.

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You mean open neighbourhood of $\gamma$? –  Squark Feb 8 '12 at 14:44
    
@squark: Yes i mean open neighbourhood of $\gamma$. –  zapkm Feb 8 '12 at 14:48
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There is a theorem of Hartogs Rosenthal which states that any continuous function on a compact subset of measure zero can be uniformly approximated by functions holomorphic in a neighbourhood of the compact set .The proof follows fairly immeditely from the Cauchy Pompeiu formula –  Mohan Ramachandran Feb 8 '12 at 16:12
    
contd:the compact subset is in the plane . –  Mohan Ramachandran Feb 8 '12 at 16:13
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It still has terms involving zeebar.The proof of hartogs rosenthal is pretty easy .First approximate uniformly on gamma by a smooth defined in a small neighbourhood of gamma ,then apply the Cauchy pompeiu formula to this smooth approximation.This is essentially the original proof .The original paper of Hartogs Rosenthal is in MathAnnalen vol 104 yr 1931 pages 606-610 –  Mohan Ramachandran Feb 8 '12 at 20:21
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1 Answer

up vote 4 down vote accepted

In your case we can find a holomorphic function on the plane that uniformly approximates the given continuous function .It is a consequence of the following .Suppose K is a compact measure zero subset of the plane whose complement in the plane is connected then every continuous function on K can be uniformly approximated by entire functions. Hartogs-Rosenthal says any continuous function on K can be uniformly approximated by functions holomorphic in a neighbourhood of K.By Runge's theorem, functions holomorphic in a neighbourhood of K can be uniformly approximated on K by entire functions. In your case since gamma is an arc its complement is connected .Since it is smooth it has measure zero .

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Thanks for the answer: But i didn't see this extended version of Hartogs-Rosenthal theorem: I thought that theorem guarantees for the approximation by RATIONAL function. It will be very helpful for me if you can provide some reference. Thanks a lot. –  zapkm Feb 8 '12 at 19:50
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Yes but the poles of the rational function are outside a neighbourhood of gamma . –  Mohan Ramachandran Feb 8 '12 at 21:19
    
May be i am making mistake.. Please have the following example: Take K any smooth curve which doesn't passes through $(0,0)∈\mathbb C$. Take $f(z)=\frac{1}{z}$, on K, f is continuous, and $\mathbb C−K$ is connected. But f can't be extended to a entire function.... So the theorem you mention seems to have some problem. –  zapkm Mar 19 '12 at 11:52
    
You are confusing approximation and extension. You can certainly approximate f on K uniformly by polynomials .In your example this is easy to do directly . –  Mohan Ramachandran Mar 19 '12 at 14:36
    
contd:the polynomials referred to are holomorphic. –  Mohan Ramachandran Mar 19 '12 at 14:38
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