# 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.

• You mean open neighbourhood of $\gamma$? Feb 8, 2012 at 14:44
• @squark: Yes i mean open neighbourhood of $\gamma$. Feb 8, 2012 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 Feb 8, 2012 at 16:12
• contd:the compact subset is in the plane . Feb 8, 2012 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 Feb 8, 2012 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. Mar 19, 2012 at 11:52