Closed subspace of the space of Holomorphic functions 
Let $H(\Omega)$ be the space of holomorphic functions on the simply connected open set $\Omega$ with $0\in\Omega$. $H(\Omega)$ is endowed with the topology of uniform convergence on compact subsets of $\Omega$.
Let $\phi\in H(\Omega)$ not identically $0$. Define $T:\phi H(\Omega)\to H(\Omega)$ as $T(f)(z)=\frac{1}{z}\int_0^z\frac{f(j)}{\phi(j)}dj$
 1) Show that $\phi H(\Omega)$ is a closed linear subspace of $H(\Omega)$.
 2) Show that $T$ is continuous

 A: I tried to answer part 1):
Let $f_n$ be a sequence of holomorphic functions such that $\phi f_n\to f$ uniformly on compact sets and let $A=\{a\in \Omega: \phi(a)=0\}$ be the zero set of $\phi$, which is discrete by the identity theorem. Clearly the function $f/\phi\in H(\Omega\setminus A)$. It suffices to show that this function extends to a holomorphic function $h$ on $\Omega$, since this would imply $f=\phi \cdot h$ and $f\in \phi H(\Omega)$. We fix a point $a\in A$ and choose $r>0$, such that $\bar{D}_r(a)\subset \Omega$ and $\phi$ does not vanish on $\bar{D}_r(a)$.
If $n$ is large enough, we have for all $z\in \partial D_r(a)$
\begin{equation}|\phi(z) f_n(z)|\leq |f(z)|+1\ \end{equation}
and since $\phi$ is nonvanishing on $\partial D_r(a)$ this implies that $|f_n|$ is uniformly bounded on $\partial D_r(a)$ by a constant $C$. Using the maximum principle we can conclude that the sequence $f_n$ is uniformly bounded on $\bar{D}_r(a)$ (by the same constant $C$).
Furthermore there exists an integer $k$ and a holomorphic function $g$ such that $g$ is nonvanishing on $\bar{D}_r(a)$ and $\phi(z)=(z-a)^kg(z)$ on a neigbourhood of the disk. If $z\in D_r(a)\setminus\{a\}$, we get
\begin{equation}
|\frac{f(z)}{(z-a)^k}|=\lim_{n\to \infty}|\frac{(z-a)^kg(z)f_n(z)}{(z-a)^k}|=\lim_{n\to\infty}|g(z)f_n(z)|\leq \sup_{z\in \bar{D}_r(a)}|g(z)| \cdot C<\infty
\end{equation}
Therefore the function $f(z)/(z-a)^k$ is bounded on a neighbourhood of $a$ and by Riemann's theorem on removable singularities there exists a holomorphic function $f_1\in H(D_r(a))$ such that $f(z)=(z-a)^kf_1(z)$. In particular $f$ has a zero of order at least $k$ at $a$, which implies that the singularity $a$ of $f/\phi$ is removable. Since $a\in A$ was arbitrary, this implies the claim.
