Extension of a continuous function on the boundary of a disc Suppose $f$ is continuous on the boundary of a disc $D \subset \mathbb{C}$. Suppose further that $$\int_{\partial D} f(\zeta) \zeta^k d\zeta =0$$ for all $k  \in \mathbb{Z}_{\geq 0}$. Is it possible to extend $f$ to be a holomorphic function within $D$? 
I'm almost certain that if so, we would require the use of the Cauchy integral formula. I've been attempting to mess around with the difference $$\Bigg \| f(z) - \int_{\Gamma} \frac{f(\zeta)}{\zeta - z}d\zeta \Bigg \|,$$ but have yet to obtain anything useful. 
 A: *

*Assume that $f$ is Hölder continuous  on the boundary. Then for $|z| = 1$ : $f(z) = \sum_{k=-\infty}^\infty c_k z^k$ where $c_k = \frac{1}{2i\pi} \int_{|z|=1} f(z) z^{-k-1}dz$ are the Fourier coefficients. 
$\int_{|z| = 1} f(z)z^{-k-1} dz = 0$ for $k < 0$ means that $c_k = 0$ for $k < 0$, i.e. $$ f(z) = \sum_{k=0}^\infty c_k z^k \qquad (|z|=1)$$
and by the properties of power series, this series converges and is analytic on $|z| < 1$ and is continuous on $|z| \le 1$.

*If we only know that $f$ is continuous, then its Fourier series doesn't have to converge, but the same analysis shows that $g(z) = \sum_{k=-\infty}^\infty c_k z^k = \sum_{k=0}^\infty c_k z^k $ is analytic on $|z| < 1$ and this is again the analytic extension we want, its limit as $|z| \to 1$ again coincides with $f$.
A: No, not necessarily. Take for example $f(z) = \frac{1}{z}$ and $D$ as the unit disc. Then the integral vanishes for every $k \ge 1$, but $f$ can't be extended to a holomorphic function (since the integral doesn't vanish for $k=0$).
