Proving the limit of lipschitz, complex-analytic function is analytic Suppose that a sequence of analytic functions is locally Lipschitz in an open connected set $D$ of the complex plane. (That is, for all $z\in D$, there is a constant $L$ such that $|f_n(z_1)-f_n(z_2)|\leq L|z_1-z_2|$ if $z_1,z_2$ are in a neighborhood of $z$.)
I want to show that $f(z)$ must also be analytic, and moreover $f_n'(z)\to f'(z)$, where $f(z)$ is the pointwise limit of $f_n$.
I think I can show that $f$ is continuous using a $\epsilon/3$ type argument (ie split up $|f_n(z)-f(z)|$ using the triangle inequality), but I'm not sure how this shows analyticity of the  function $f$. I would really like to say that we in fact have uniform convergence, but I'm not sure if the condition given is strong enough to prove that.
Any help is appreciated!
 A: Recall Montel's theorem. It says if $U\subseteq \mathbb{C}$ is a domain and $(f_n)_n$ is a family of complex-analytic functions $f_n : U \rightarrow \mathbb{C}$ that is locally uniformly bounded (i.e for any compact $D\subseteq U$ there exists a constant $C_D$ such that $\sup_{n} \Vert f_n \Vert_{L^\infty(D)}<C_D)$ we have that there exists a subsequence $(f_{n_k})_k$ that converges uniformly on compact subsets of $U$.
Now we show that $(f_n)_n$ and $(f_n')_n$ are both locally uniformly bounded if $(f_n)_n$ is locally equi-lipschitz (this means locally we can find an upper bound for the lipschitz constant of all functions in our family).
It is clear that $\vert f_n'\vert$ on $D$ is bounded by the Lipschitz constant of $f_n$ on $D$. Thus, locally equi-lipschitz implies that $(f_n')_n$ is locally uniformly bounded. On the other hand, if we fix some compact $D\subseteq U$, and some point $z_0\in D$, then we get
$$ \vert f_n(z) \vert \leq \vert f_n(z_0) \vert + \vert f_n(z)-f_n(z_0) \vert 
\leq \sup_{n} \vert f_n(z_0) \vert + \text{diam}(D) \Vert f_n'\Vert_{L^\infty(D)}.$$
As $(f_n(z_0))_n$ converges and $(f_n')_n$ is locally uniformly bounded, we get that $(f_n)_n$ is locally uniformly bounded too.
In particular, we can now use Montel's theorem and pick a subsequence $(f_{n_k})_k$ and some functions $h,g$ such that $(f_{n_k})_k$ converges locally uniformly to $h$ and $(f_{n_k}')_k$ converges locally uniformly to $g$. However, as $(f_n)_n$ converges pointwise to $f$, we get that $h=f$. Also, as $f$ is locally the uniform limit of complex-analytic functions, we get that $f$ is complex-analytic.
As $(f_{n_k})_k$ converges locally uniformly to $f$ and $(f_{n_k}')_k$ converges locally uniformly to $g$ one can use the standard argument to show that $g=f'$. Namely, fix $z_0, z\in U$ and path $\gamma$ that connects $z_0, z$. As the path is compact, we have that $(f_{n_k})_k$,$(f_{n_k}')_k$ converges uniformly on $\gamma$ and hence, we obtain
$$ f(z) = \lim_{k\rightarrow \infty} f_{n_k}(z) = \lim_{k\rightarrow \infty} \left( f_{n_k}(z_0)+ \int_\gamma f_{n_k}'(\tau) d\tau \right) = f(z_0) + \int_\gamma g(\tau) d\tau. $$
Taking a derivative in $z$ yields $f'(z) = g(z)$ (by the fundamental theorem of calculus).
Finally, we can now start with any subsequence $(f_{n_m})_m$ and find a subsequence $(f_{n_{m_\ell}})_\ell$ such that $(f_{n_{m_\ell}})_\ell$ converges locally uniformly to $f$ and $(f_{n_{m_\ell}}')_\ell$ converges locally uniformly to $f'$. This means, that if you fix with some $z\in U$, then there exists for every subsequence $(f_{n_m}'(z))_m$ a subsequence $(f_{n_{m_\ell}}')_\ell$ that converges to $f'(z)$. However, this implies that $(f'_n(z))_n$ converges to $f'(z)$, which is what we wanted to show.
