Let $D \subset \mathbb{C}^n$ be a bounded convex domain. Let $F_j : D \times D \rightarrow D$ be a sequence of holomorphic functions such that $F_j(q,q) = q$ for all $j$. Then $\{F_j\}_j$ is a compact normal family.
I don't understand why this conclusion is true. I understand that a uniformly bounded set of holomorphic functions is normal, but why is it compact? Please help.
I don't think the conclusion is true. For instance, let $\pi_1\colon D\times D\to D$ and $\pi_2\colon D\times D\to D$ be the two projection maps, and let $F_j\colon D\times D\to D$ be $$F_j = \frac{1}{j}\pi_1 + \frac{j-1}{j}\pi_2$$ for all $j\geq 1$. Note, $F_j$ has image in $D$ for each $j$ precisely because $D$ is convex.
The limit of this sequence $F_j$ is $\pi_2$, which does not belong to the family $\{F_j\}_{j\geq 1}$, so $\{F_j\}_{j\geq 1}$ is not closed (hence not compact).