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Question: I want to show now that if $r>s>0,f \in F_s (\Omega), $ and $u \in F_r (\Omega)$, then for any $i$,
$$f \frac{\partial u}{\partial z_i} \in F_s(\Omega).$$
My work so far: This will follow if I can show that $\partial u / \partial z_i \in F_s(\Omega).$ Since $u \in F_r (\Omega)$, $u$ must be analytic in $\Omega_r$ and so $\partial u / \partial z_i$ must also be analytic in $\Omega_r$. Then at the very least, $\partial u / \partial z_i$ must be in $E_s(\Omega)$. But I have no idea how to show that $\partial u / \partial z_i$ is in $F_s(\Omega)$, or equivalently, that it is in the closure of $\mathscr{H}(\mathbb{C}^n)$ in $E_s(\Omega)$.
The Cauchy integral provides an estimate of the derivatives $\frac{\partial f}{\partial z_k}(w)$ of a holomorphic function depending on a bound on the absolute value of $f(z)$ on the (distinguished) boundary of a polydisk $P$ for $w$ in any smaller polydisk $P' \subset\mspace{-2mu}\subset P$. ($A \subset\mspace{-2mu}\subset B$ means that $A$ is relatively compact in $B$, the closure of $A$ is compact and contained in $B$.)
As a consequence of that, if $f_n$ are holomorphic functions converging to $g$ uniformly on an open set $U$, all partial derivatives (of all orders) of the $f_n$ converge to the corresponding partial derivative of $g$ locally uniformly on $U$.
Since for $0 < s < r$ you have $\Omega_s \subset\mspace{-2mu}\subset \Omega_r$, if $f_n \to u$ uniformly on $\Omega_r$, then $\frac{\partial f_n}{\partial z_i} \to \frac{\partial u}{\partial z_i}$ uniformly on $\Omega_t$ for all $t < r$.
Since partial derivatives of entire functions are entire, you have
$$u \in F_r(\Omega) \Rightarrow \frac{\partial u}{\partial z_i} \in F_s(\Omega),\quad 0 < s < r.$$