Why is $\int_{[0,1]} \frac{dw}{1-wz}$ is holomorphic in unit disc?

Solution 1:

You can interchange the integrals due to Fubini's theorem. Since the integrand is continuous on $[0,1]$ and the image of $\gamma$, the integral of its absolute value is finite.

For $w=0$, you just don't do (A). Keep it as $$\int_{[0,1]}\int_\gamma\frac{dz}{1}$$ As before the integral $\int_\gamma dz=0$ by Cauchy's theorem, or if you like by using that $\frac11$ has $z$ as anti-derivative. The step (A) wasn't really necessary. You can argue that the inner integral in (*) is zero by Cauchy's theorem, without taking the $w$ out.

In (2), if you are going to show that $f$ has a derivative, then why do anything else? You can prove continuity directly. The function $g(w,z)=\frac{1}{1-wz}$ is continuous for $w\in[0,1]$ and $z$ in the open unit disc, and therefore $g$ is uniformly continuous on $[0,1]\times K$ for $K$ a compact subset of the unit disc. Therefore, integrating with respect to $w$ in $[0,1]$ results in a continuous function on $K$.