Let $D$ be the open unit disc and $C$ be the unit circle. Suppose $$f:C\to\mathbb{C}$$ is continuous. Show that $$g(w)=\dfrac{1}{2\pi i}\int_{C} \dfrac{f(z)}{z-w} \rm{d}z$$ is an analytic function of $w$ for $w\in D$.
Not sure how to approach this, the fact that $f$ is only defined on $C$ is posing a problem. The form that the RHS is in only reminds me of CIF... Could someone give me a hint?
(ps wasn't sure what to put as the title, feel free to edit)
I'll leave it to you to verify that $g$ is continuous. Assuming that, let $\gamma$ be an arbitrary simple closed curve inside $D$. Then \begin{align} \int_\gamma g(w)\,dw &= \int_\gamma \left( \frac{1}{2\pi i} \int_C \frac{f(z)}{z-w}\,dz \right)\,dw \\ &= \frac{1}{2\pi i} \int_C \left( \int_\gamma \frac{f(z)}{z-w}\,dw \right)\,dz\\ &= \frac{1}{2\pi i} \int_C 0\,dz = 0 \end{align} by Fubini's theorem and Cauchy's integral theorem (note that $\frac{f(z)}{z-w}$ is holomorphic in $w$; it's only singularity is at $w=z$, but $z \in C$ is outside $\gamma$).
Hence, Morera's theorem implies that $g$ is holomorphic on $D$.
