I was reading Paul Garrett's note about Hartogs’ Theorem . In the first step to prove separate analyticity implies joint.
We assume $f$ is a separate analytic function on a polydisk $U\subset \Bbb{C}^n$, and $f$ is continuous on $\bar{U}$ , the proof goes as follows, however I can't see where use the continuity of $f$?
Since separately analytic we can apply Cauchy integral formula on each variable iteratively.
$$f(z)=\frac{1}{(2 \pi i)^n} \int_{C_1} \ldots \int_{C_n} \frac{f(\zeta)}{\left(\zeta_1-z_1\right) \ldots\left(\zeta_n-z_n\right)} d \zeta_1 \ldots d \zeta_n$$
Then we have the expansion :$$\frac{1}{\zeta_j-z_j}=\sum_{n \geq 0} \frac{z_j^n}{\zeta_j^{n+1}}$$
when $|z_j|<|\zeta_j|$ and the series converge uniformly and absolutely on a compact subset of this disk.
Therefore we can interchange the sum and integral and the result is a jointly analytic function.
I can't figure out why continuous of $f$ is needed in the first set, we only need boundness of $f$ in the above proof?
It is an assumption of Cauchy's theorem for $n=1$, and I presume also for $n\in\Bbb N$ arbitrary, that the function be continuous on the boundary as well as analytic on the interior of the contour (or open set),