# A corollary to Stone-Weierstrass theorem

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
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
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
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