Consider $U\subset \mathbb C^{n}$, an open domain and denote by $Hol(U)$ the space of holomorphic complex valued functions defined on $U$.
Prove that the topologies induced by $C^{k}$ uniform convergence on compacts are all the same.
One inclusion is trivial and for the other inclusion I think it is sufficient to show that $T_1 = T_0$, because you can do a recursive argument with the partial derivatives. where $T_k$ represents the respective topology. But i do not know how to prove it.
Note Since every holomorphic functions is analytic, does this result implies that the partial derivatives are continuous?
Suppose that the sequence $(f_n)_{n\geq0}$ of holomorphic functions on $U$ converges uniformly on compact sets of $U$ to $f$, let $k\geq1$ and let us show that the sequence $(f_n^{(k)})_{n\geq0}$ of the $k$th derivatives converges uniformly on compact sets of $U$ to $f^{(k)}$.
Let $p\in U$, let $B$ be a closed disc centered at $p$ and contained in $U$ and let $\gamma$ be the boundary of a closed disc centered at $p$ which contains $B$ in its interior and is itself contained in $U$. Let $\rho$ be the absolute value of difference of the radii of $B$ and $\gamma$: we then have $|t-z|\geq\rho$ for all $t\in\gamma$ and all $z\in B$.
According to Cauchy's integral formulas, for all functions $g$ holomorphic on $U$ we have that $$g^{(k)}(z) = \frac{k!}{2\pi i}\oint_\gamma\frac{g(t)}{(t-z)^{k+1}}\,\mathrm dt$$ for all $z\in B$. In particular, we have for such $z$ that $$f^{(k)}(z)-f^{(k)}(z)=\frac{k!}{2\pi i}\oint_\gamma\frac{f_n(t)-f(t)}{(t-z)^{k+1}}\,\mathrm dt,$$ so that $$ |f^{(k)}(z)-f^{(k)}(z)| \leq\frac{k!\operatorname{length(\gamma)}}{2\pi}\sup_{t\in\gamma}\left|\frac{f_n(t)-f(t)}{(t-z)^{k+1}}\right| \leq\frac{k!\operatorname{length(\gamma)}}{2\pi\rho^{k+1}}\sup_{t\in\gamma}|f_n(t)-f(t)| $$ for all $\in B$. As $f_n$ converges to $f$ uniformly on the compact set $\gamma$, we see that $f_n^{(k)}$ converges uniformly to $f^{(k)}$ on $B$.
It follows that we have uniform convergence of all derivatives on all closed balls contained in $U$ and, therefore, on all compact subsets of $U$.