Extending a continuous function on circle to holomorphic function on unit disk There are some obvious functions like $$f(z)=\frac{1}{z}$$ which are continuous on the unit circle $\{ z \in \mathbb{C} \ | \ |z|=1 \}$ and have no holomorphic extension to the unit disk $\mathbb{D} = \{ z \in \mathbb{C} \ | \ |z|<1 \}$.
However, if we impose a stringent condition like
$$\lim_{z \rightarrow 0} z^nf(z) = 0$$
for all nonnegative integers $n$, can such a holomorphic extension exist?
What if we have, instead of the above, the slightly less onerous condition
$$\int_{|z|=1} z^nf(z) \text{d}z = 0$$
for all nonnegative integers $n$?
 A: If $g(z)$ is a continuous complex-valued function defined on the unit circle $|z|=1$, and $g(z)$ satisfies the condition
$$\int_{|z|=1} z^n g(z) \text{d}z = 0$$
for all nonnegative integers $n$, then there is a function $f(z)$ that is holomorphic on unit disk $|z|<1$, continuous up to its boundary, and agrees with $g(z)$ on the boundary. 
Proof:
For each integer $m$, $g(z)/z^{m+1}$ is well-defined and continuous on the compact set $|z|=1$. Define
$$c_m = \frac{1}{2\pi\text{i}} \int_{|z|=1} \frac{g(z)}{z^{m+1}} \text{d}z.$$
Then using the standard parameterization of the unit circle $z(t) = \operatorname{exp}(\text{i}t)$, $0 \leq t \leq 2\pi$, it can be easily seen that
$$c_m = \frac{1}{2\pi} \int_0^{2\pi} g(\operatorname{e}^{\text{i}t})\operatorname{e}^{-\text{i}mt} \text{d}t,$$
and so the infinite series 
$$\sum_{m = -\infty}^{+ \infty} c_m \operatorname{e}^{\text{i}mt}$$
is convergent for $0 \leq t \leq 2\pi$ and is the complex Fourier series representation of $g(\operatorname{e}^{\text{i}t})$ on the unit circle. Since the condition $\int_{|z|=1} z^n g(z) \text{d}z = 0$ for all nonnegative integers $n$ implies that $c_m = 0$ for all $m  \leq -1$,
$$g(z) = \sum_{m = 0}^{+ \infty} c_m \operatorname{e}^{\text{i}mt}.$$
The Ratio Test applied to $\sum_{m = 0}^{+ \infty} c_m \operatorname{e}^{\text{i}mt}$ gives us the information that
$$\lim_{m \to \infty} \left| \frac{c_{m+1}}{c_m} \right| \leq 1,$$
and so the same Ratio Test yields that $\sum_{m = 0}^{+ \infty} c_m z^m$ is absolutely convergent for $|z|<1$. Therefore, the function
$$f(z) = \sum_{m = 0}^{+ \infty} c_m z^m$$
is well-defined on the closed disk $|z|\leq 1$ and satisfies the requirements.
